Define the function $\mathcal{I}:(0,1]\rightarrow\mathbb{R}$ via the definite integral
$$\mathcal{I}{\left(a\right)}:=\int_{0}^{a}\mathrm{d}x\,\frac{\sqrt{a^{2}-x^{2}}}{1-x^{2}}\operatorname{artanh}{\left(x\right)},$$
where the real inverse hyperbolic tangent is defined here by
$$\operatorname{artanh}{\left(z\right)}:=\int_{0}^{z}\mathrm{d}y\,\frac{1}{1-y^{2}}\stackrel{y\mapsto zt}{=}\int_{0}^{1}\mathrm{d}t\,\frac{z}{1-z^{2}t^{2}}=\frac12\ln{\left(\frac{1+z}{1-z}\right)};~~~\small{z\in(-1,1)}.$$
It can be shown that $\mathcal{I}{\left(1\right)}=2C$, where here $C$ denotes Catalan's constant:
$$\begin{align}
\mathcal{I}{\left(1\right)}
&=\int_{0}^{1}\mathrm{d}x\,\frac{\sqrt{1-x^{2}}}{1-x^{2}}\operatorname{artanh}{\left(x\right)}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{artanh}{\left(x\right)}}{\sqrt{1-x^{2}}}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(\frac{1+x}{1-x}\right)}}{2\sqrt{1-x^{2}}}\\
&=-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(\frac{1-x}{1+x}\right)}}{2\left(1+x\right)\sqrt{\frac{1-x}{1+x}}}\\
&=-\int_{1}^{0}\mathrm{d}y\,\frac{(-2)}{(1+y)^{2}}\cdot\frac{\ln{\left(y\right)}}{2\left(\frac{2}{1+y}\right)\sqrt{y}};~~~\small{\left[x=\frac{1-y}{1+y}\right]}\\
&=-\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(y\right)}}{2(1+y)\sqrt{y}}\\
&=-\int_{0}^{1}\mathrm{d}t\,\frac{2t\ln{\left(t^{2}\right)}}{2(1+t^{2})t};~~~\small{\left[\sqrt{y}=t\right]}\\
&=-2\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(t\right)}}{1+t^{2}}\\
&=2\int_{0}^{1}\mathrm{d}t\,\frac{\arctan{\left(t\right)}}{t};~~~\small{I.B.P.}\\
&=2C.\\
\end{align}$$
It remains to evaluate $\mathcal{I}{\left(a\right)}$ in the $a\in(0,1)$ case.
Suppose $a\in(0,1)$.
$$\begin{align}
\mathcal{I}{\left(a\right)}
&=\int_{0}^{a}\mathrm{d}x\,\frac{\sqrt{a^{2}-x^{2}}}{1-x^{2}}\operatorname{artanh}{\left(x\right)}\\
&=\int_{0}^{a}\mathrm{d}x\,\frac{\sqrt{a^{2}-x^{2}}}{1-x^{2}}\cdot\frac12\ln{\left(\frac{1+x}{1-x}\right)}\\
&=\int_{0}^{1}\mathrm{d}y\,\frac{a\sqrt{a^{2}-a^{2}y^{2}}}{1-a^{2}y^{2}}\cdot\frac12\ln{\left(\frac{1+ay}{1-ay}\right)};~~~\small{\left[x=ay\right]}\\
&=\int_{0}^{1}\mathrm{d}y\,\frac{a^{2}\left(1-y^{2}\right)}{1-a^{2}y^{2}}\cdot\frac{\ln{\left(\frac{1+ay}{1-ay}\right)}}{2\sqrt{1-y^{2}}}\\
&=\int_{0}^{1}\mathrm{d}y\,\left[1-\frac{1-a^{2}}{1-a^{2}y^{2}}\right]\frac{\ln{\left(\frac{1+ay}{1-ay}\right)}}{2\sqrt{1-y^{2}}}\\
&=\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(\frac{1+ay}{1-ay}\right)}}{\sqrt{1-y^{2}}}-\frac{1-a^{2}}{2}\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(\frac{1+ay}{1-ay}\right)}}{\left(1-a^{2}y^{2}\right)\sqrt{1-y^{2}}}\\
&=\mathcal{J}{\left(a\right)}-\mathcal{K}{\left(a\right)},\\
\end{align}$$
where in the last line above we've introduced the auxiliary functions
$$\mathcal{J}{\left(a\right)}:=\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(\frac{1+ay}{1-ay}\right)}}{\sqrt{1-y^{2}}};~~~\small{a\in(0,1)},$$
$$\mathcal{K}{\left(a\right)}:=\frac{1-a^{2}}{2}\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(\frac{1+ay}{1-ay}\right)}}{\left(1-a^{2}y^{2}\right)\sqrt{1-y^{2}}};~~~\small{a\in(0,1)}.$$
Suppose $a\in(0,1)$. Then $0<\frac{1-a}{1+a}<1$. Setting $p:=\sqrt{\frac{1-a}{1+a}}$, we have $0<p<1\land a=\frac{1-p^{2}}{1+p^{2}}$, and then
$$\begin{align}
\mathcal{J}{\left(a\right)}
&=\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(\frac{1+ay}{1-ay}\right)}}{\sqrt{1-y^{2}}}\\
&=\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(\frac{1+ay}{1-ay}\right)}}{\sqrt{\left(1-y\right)\left(1+y\right)}}\\
&=\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(\frac{1+ay}{1-ay}\right)}}{\left(1+y\right)\sqrt{\frac{1-y}{1+y}}}\\
&=\frac12\int_{1}^{0}\mathrm{d}t\,\frac{(-2)}{(1+t)^{2}}\cdot\frac{\ln{\left(\frac{1+a(\frac{1-t}{1+t})}{1-a(\frac{1-t}{1+t})}\right)}}{\left(\frac{2}{1+t}\right)\sqrt{t}};~~~\small{\left[y=\frac{1-t}{1+t}\right]}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(\frac{(1+t)+a(1-t)}{(1+t)-a(1-t)}\right)}}{2(1+t)\sqrt{t}}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(\frac{1+a+(1-a)t}{1-a+(1+a)t}\right)}}{2(1+t)\sqrt{t}}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(\frac{1+(\frac{1-a}{1+a})t}{(\frac{1-a}{1+a})+t}\right)}}{2(1+t)\sqrt{t}}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(\frac{1+p^{2}t}{p^{2}+t}\right)}}{2(1+t)\sqrt{t}}\\
&=\int_{0}^{1}\mathrm{d}u\,\frac{\ln{\left(\frac{1+p^{2}u^{2}}{p^{2}+u^{2}}\right)}}{1+u^{2}};~~~\small{\left[\sqrt{t}=u\right]}\\
&=\frac12\int_{0}^{1}\mathrm{d}u\,\frac{2\ln{\left(\frac{1+p^{2}u^{2}}{1+u^{2}}\right)}}{1+u^{2}}-\frac12\int_{0}^{1}\mathrm{d}u\,\frac{2\ln{\left(\frac{p^{2}+u^{2}}{1+u^{2}}\right)}}{1+u^{2}}\\
&=\frac12G{\left(1,p,1\right)}-\frac12G{\left(p,1,1\right)},\\
\end{align}$$
where in the last line above we've introduced another auxiliary function $G:\mathbb{R}_{>0}^{3}\rightarrow\mathbb{R}$, defined via the definite integral
$$G{\left(a,b,z\right)}:=2\int_{0}^{z}\mathrm{d}t\,\frac{\ln{\left(\frac{a^{2}+b^{2}t^{2}}{1+t^{2}}\right)}}{1+t^{2}}.$$
We similarly calculate for the integral $\mathcal{K}$
$$\begin{align}
\mathcal{K}{\left(a\right)}
&=\frac{1-a^{2}}{2}\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(\frac{1+ay}{1-ay}\right)}}{\left(1-a^{2}y^{2}\right)\sqrt{1-y^{2}}}\\
&=\frac{1-a^{2}}{2}\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(\frac{1+ay}{1-ay}\right)}}{\left(1-ay\right)\left(1+ay\right)\sqrt{\left(1-y\right)\left(1+y\right)}}\\
&=\frac{1-a^{2}}{2}\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(\frac{1+ay}{1-ay}\right)}}{\left(1-ay\right)\left(1+ay\right)\left(1+y\right)\sqrt{\frac{1-y}{1+y}}}\\
&=\frac{1-a^{2}}{2}\int_{1}^{0}\mathrm{d}t\,\frac{(-2)}{\left(1+t\right)^{2}}\cdot\frac{\ln{\left(\frac{1+a\left(\frac{1-t}{1+t}\right)}{1-a\left(\frac{1-t}{1+t}\right)}\right)}}{\left[1-a\left(\frac{1-t}{1+t}\right)\right]\left[1+a\left(\frac{1-t}{1+t}\right)\right]\left(\frac{2}{1+t}\right)\sqrt{t}};~~~\small{\left[y=\frac{1-t}{1+t}\right]}\\
&=\frac{1-a^{2}}{2}\int_{0}^{1}\mathrm{d}t\,\frac{\left(1+t\right)\ln{\left(\frac{(1+t)+a(1-t)}{(1+t)-a(1-t)}\right)}}{\left[\left(1+t\right)-a\left(1-t\right)\right]\left[\left(1+t\right)+a\left(1-t\right)\right]\sqrt{t}}\\
&=\frac{1-a^{2}}{2}\int_{0}^{1}\mathrm{d}t\,\frac{\left(1+t\right)\ln{\left(\frac{(1+a)+(1-a)t}{(1-a)+(1+a)t}\right)}}{\left[\left(1-a\right)+\left(1+a\right)t\right]\left[\left(1+a\right)+\left(1-a\right)t\right]\sqrt{t}}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{p^{2}\left(1+t\right)\ln{\left(\frac{1+p^{2}t}{p^{2}+t}\right)}}{2\left(p^{2}+t\right)\left(1+p^{2}t\right)\sqrt{t}}\\
&=\int_{0}^{1}\mathrm{d}u\,\frac{p^{2}\left(1+u^{2}\right)\ln{\left(\frac{1+p^{2}u^{2}}{p^{2}+u^{2}}\right)}}{\left(p^{2}+u^{2}\right)\left(1+p^{2}u^{2}\right)};~~~\small{\left[\sqrt{t}=u\right]}.\\
\end{align}$$
Then,
$$\begin{align}
\mathcal{K}{\left(a\right)}
&=\int_{0}^{1}\mathrm{d}u\,\frac{p^{2}\left(1+u^{2}\right)\ln{\left(\frac{1+p^{2}u^{2}}{p^{2}+u^{2}}\right)}}{\left(p^{2}+u^{2}\right)\left(1+p^{2}u^{2}\right)}\\
&=\int_{0}^{1}\mathrm{d}u\,\left[\frac{p^{2}}{\left(1+p^{2}\right)\left(1+p^{2}u^{2}\right)}+\frac{p^{2}}{\left(1+p^{2}\right)\left(p^{2}+u^{2}\right)}\right]\ln{\left(\frac{1+p^{2}u^{2}}{p^{2}+u^{2}}\right)}\\
&=\frac{p}{1+p^{2}}\int_{0}^{1}\mathrm{d}u\,\frac{p\ln{\left(\frac{1+p^{2}u^{2}}{p^{2}+u^{2}}\right)}}{1+p^{2}u^{2}}+\frac{p}{1+p^{2}}\int_{0}^{1}\mathrm{d}u\,\frac{p\ln{\left(\frac{1+p^{2}u^{2}}{p^{2}+u^{2}}\right)}}{p^{2}+u^{2}}\\
&=\frac{p}{1+p^{2}}\int_{0}^{p}\mathrm{d}v\,\frac{\ln{\left(\frac{p^{2}+p^{2}v^{2}}{p^{4}+v^{2}}\right)}}{1+v^{2}};~~~\small{\left[u=p^{-1}v\right]}\\
&~~~~~+\frac{p}{1+p^{2}}\int_{0}^{p^{-1}}\mathrm{d}v\,\frac{\ln{\left(\frac{1+p^{4}v^{2}}{p^{2}+p^{2}v^{2}}\right)}}{1+v^{2}};~~~\small{\left[u=pv\right]}\\
&=\frac{p}{1+p^{2}}\int_{0}^{p}\mathrm{d}v\,\frac{\ln{\left(p^{2}\right)}}{1+v^{2}}-\frac{p}{1+p^{2}}\int_{0}^{p}\mathrm{d}v\,\frac{\ln{\left(\frac{p^{4}+v^{2}}{1+v^{2}}\right)}}{1+v^{2}}\\
&~~~~~-\frac{p}{1+p^{2}}\int_{0}^{p^{-1}}\mathrm{d}v\,\frac{\ln{\left(p^{2}\right)}}{1+v^{2}}+\frac{p}{1+p^{2}}\int_{0}^{p^{-1}}\mathrm{d}v\,\frac{\ln{\left(\frac{1+p^{4}v^{2}}{1+v^{2}}\right)}}{1+v^{2}}\\
&=\frac{2p}{1+p^{2}}\ln{\left(p\right)}\arctan{\left(p\right)}-\frac{2p}{1+p^{2}}\ln{\left(p\right)}\arctan{\left(p^{-1}\right)}\\
&~~~~~-\frac{p}{2\left(1+p^{2}\right)}\int_{0}^{p}\mathrm{d}v\,\frac{2\ln{\left(\frac{p^{4}+v^{2}}{1+v^{2}}\right)}}{1+v^{2}}\\
&~~~~~+\frac{p}{2\left(1+p^{2}\right)}\int_{0}^{p^{-1}}\mathrm{d}v\,\frac{2\ln{\left(\frac{1+p^{4}v^{2}}{1+v^{2}}\right)}}{1+v^{2}}\\
&=-\frac{2p}{1+p^{2}}\ln{\left(p\right)}\left[\frac{\pi}{2}-2\arctan{\left(p\right)}\right]\\
&~~~~~-\frac{p}{2\left(1+p^{2}\right)}\left[G{\left(p^{2},1,p\right)}-G{\left(1,p^{2},p^{-1}\right)}\right]\\
&=-\sqrt{1-a^{2}}\ln{\left(\sqrt{\frac{1-a}{1+a}}\right)}\left[\frac{\pi}{2}-2\arctan{\left(\sqrt{\frac{1-a}{1+a}}\right)}\right]\\
&~~~~~-\frac{\sqrt{1-a^{2}}}{4}\left[G{\left(p^{2},1,p\right)}-G{\left(1,p^{2},p^{-1}\right)}\right]\\
&=\sqrt{1-a^{2}}\operatorname{artanh}{\left(a\right)}\arcsin{\left(a\right)}\\
&~~~~~-\frac{\sqrt{1-a^{2}}}{4}\left[G{\left(p^{2},1,p\right)}-G{\left(1,p^{2},p^{-1}\right)}\right].\\
\end{align}$$
Recall the definition of the Clausen function (of order $2$) for real argument via the integral representation
$$\operatorname{Cl}_{2}{\left(\theta\right)}:=-\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(\left|2\sin{\left(\frac{\varphi}{2}\right)}\right|\right)};~~~\small{\theta\in\mathbb{R}}.$$
For $0<|r|<1\land0<\theta<\pi$, it can be shown that
$$\begin{align}
\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(1-2r\cos{\left(\varphi\right)}+r^{2}\right)}
&=\operatorname{Cl}_{2}{\left(2\theta+2\omega\right)}-\operatorname{Cl}_{2}{\left(2\theta\right)}-\operatorname{Cl}_{2}{\left(2\omega\right)}-2\omega\ln{\left(|r|\right)},\\
\end{align}$$
where
$$\omega:=\arctan{\left(\frac{r\sin{\left(\theta\right)}}{1-r\cos{\left(\theta\right)}}\right)}.$$
Suppose $(a,b,z)\in\mathbb{R}_{>0}^{3}$, and set $c:=\frac{a}{b}\in\mathbb{R}_{>0}$. For the integral $G$, we find
$$\begin{align}
G{\left(a,b,z\right)}
&=2\int_{0}^{z}\mathrm{d}t\,\frac{\ln{\left(\frac{a^{2}+b^{2}t^{2}}{1+t^{2}}\right)}}{1+t^{2}}\\
&=2\int_{0}^{z}\mathrm{d}t\,\frac{\ln{\left(\frac{b^{2}c^{2}+b^{2}t^{2}}{1+t^{2}}\right)}}{1+t^{2}}\\
&=2\int_{0}^{z}\mathrm{d}t\,\frac{\ln{\left(b^{2}\right)}+\ln{\left(\frac{c^{2}+t^{2}}{1+t^{2}}\right)}}{1+t^{2}}\\
&=4\ln{\left(b\right)}\int_{0}^{z}\mathrm{d}t\,\frac{1}{1+t^{2}}+2\int_{0}^{z}\mathrm{d}t\,\frac{\ln{\left(\frac{c^{2}+t^{2}}{1+t^{2}}\right)}}{1+t^{2}}\\
&=4\ln{\left(b\right)}\arctan{\left(z\right)}+2\int_{0}^{z}\mathrm{d}t\,\frac{\ln{\left(\frac{c^{2}+t^{2}}{1+t^{2}}\right)}}{1+t^{2}}.\\
\end{align}$$
Setting $\gamma:=\arctan{\left(c\right)}\in\left(0,\frac{\pi}{2}\right)\land\theta:=2\arctan{\left(z\right)}\in\left(0,\pi\right)$,
$$\begin{align}
G{\left(a,b,z\right)}
&=4\ln{\left(b\right)}\arctan{\left(z\right)}+2\int_{0}^{z}\mathrm{d}t\,\frac{\ln{\left(\frac{c^{2}+t^{2}}{1+t^{2}}\right)}}{1+t^{2}}\\
&=4\ln{\left(b\right)}\arctan{\left(z\right)}+2\int_{0}^{\tan{\left(\frac{\theta}{2}\right)}}\mathrm{d}t\,\frac{\ln{\left(\frac{\tan^{2}{\left(\gamma\right)}+t^{2}}{1+t^{2}}\right)}}{1+t^{2}}\\
&=4\ln{\left(b\right)}\arctan{\left(z\right)}+2\int_{0}^{\frac{\theta}{2}}\mathrm{d}\tau\,\frac{\sec^{2}{\left(\tau\right)}\ln{\left(\frac{\tan^{2}{\left(\gamma\right)}+\tan^{2}{\left(\tau\right)}}{1+\tan^{2}{\left(\tau\right)}}\right)}}{1+\tan^{2}{\left(\tau\right)}};~~~\small{\left[t=\tan{\left(\tau\right)}\right]}\\
&=4\ln{\left(b\right)}\arctan{\left(z\right)}+2\int_{0}^{\frac{\theta}{2}}\mathrm{d}\tau\,\ln{\left(\frac{\tan^{2}{\left(\gamma\right)}+\tan^{2}{\left(\tau\right)}}{\sec^{2}{\left(\tau\right)}}\right)}\\
&=4\ln{\left(b\right)}\arctan{\left(z\right)}+2\int_{0}^{\frac{\theta}{2}}\mathrm{d}\tau\,\ln{\left(\frac{2\sin^{2}{\left(\gamma\right)}\cos^{2}{\left(\tau\right)}+2\cos^{2}{\left(\gamma\right)}\sin^{2}{\left(\tau\right)}}{2\cos^{2}{\left(\gamma\right)}}\right)}\\
&=4\ln{\left(b\right)}\arctan{\left(z\right)}+2\int_{0}^{\frac{\theta}{2}}\mathrm{d}\tau\,\ln{\left(\frac{1-\cos{\left(2\gamma\right)}\cos{\left(2\tau\right)}}{2\cos^{2}{\left(\gamma\right)}}\right)}\\
&=4\ln{\left(b\right)}\arctan{\left(z\right)}+\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(\frac{1-\cos{\left(2\gamma\right)}\cos{\left(\varphi\right)}}{2\cos^{2}{\left(\gamma\right)}}\right)};~~~\small{\left[\tau=\frac{\varphi}{2}\right]}\\
&=4\ln{\left(b\right)}\arctan{\left(z\right)}+\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(\frac{1}{2\cos^{2}{\left(\gamma\right)}}\right)}\\
&~~~~~+\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(1-\cos{\left(2\gamma\right)}\cos{\left(\varphi\right)}\right)}\\
&=4\ln{\left(b\right)}\arctan{\left(z\right)}+\theta\ln{\left(\frac{1}{2\cos^{2}{\left(\gamma\right)}}\right)}\\
&~~~~~+\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(1-\cos{\left(2\arctan{\left(c\right)}\right)}\cos{\left(\varphi\right)}\right)}\\
&=4\ln{\left(b\right)}\arctan{\left(z\right)}+2\arctan{\left(z\right)}\ln{\left(\frac{1+c^{2}}{2}\right)}\\
&~~~~~+\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(1-\frac{1-c^{2}}{1+c^{2}}\cos{\left(\varphi\right)}\right)}\\
&=2\ln{\left(\frac{a^{2}+b^{2}}{2}\right)}\arctan{\left(z\right)}+\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(1-\frac{1-c^{2}}{1+c^{2}}\cos{\left(\varphi\right)}\right)}.\\
\end{align}$$
Setting $r:=\frac{1-c}{1+c}$, we have $-1<r<1\land c=\frac{1-r}{1+r}\land\frac{1-c^{2}}{1+c^{2}}=\frac{2r}{1+r^{2}}$. We then have
$$\begin{align}
G{\left(a,b,z\right)}
&=2\ln{\left(\frac{a^{2}+b^{2}}{2}\right)}\arctan{\left(z\right)}+\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(1-\frac{1-c^{2}}{1+c^{2}}\cos{\left(\varphi\right)}\right)}\\
&=2\ln{\left(\frac{a^{2}+b^{2}}{2}\right)}\arctan{\left(z\right)}+\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(1-\frac{2r}{1+r^{2}}\cos{\left(\varphi\right)}\right)}\\
&=2\ln{\left(\frac{a^{2}+b^{2}}{2}\right)}\arctan{\left(z\right)}+\int_{0}^{\theta}\mathrm{d}\varphi\,\left[\ln{\left(1-2r\cos{\left(\varphi\right)}+r^{2}\right)}-\ln{\left(1+r^{2}\right)}\right]\\
&=2\ln{\left(\frac{a^{2}+b^{2}}{2}\right)}\arctan{\left(z\right)}-\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(1+r^{2}\right)}\\
&~~~~~+\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(1-2r\cos{\left(\varphi\right)}+r^{2}\right)}\\
&=2\ln{\left(\frac{a^{2}+b^{2}}{2}\right)}\arctan{\left(z\right)}-\theta\ln{\left(1+r^{2}\right)}\\
&~~~~~+\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(1-2r\cos{\left(\varphi\right)}+r^{2}\right)}\\
&=2\ln{\left(\frac{a^{2}+b^{2}}{2}\right)}\arctan{\left(z\right)}-2\ln{\left(1+r^{2}\right)}\arctan{\left(z\right)}\\
&~~~~~+\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(1-2r\cos{\left(\varphi\right)}+r^{2}\right)}\\
&=4\ln{\left(\frac{a+b}{2}\right)}\arctan{\left(z\right)}+\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(1-2r\cos{\left(\varphi\right)}+r^{2}\right)}.\\
\end{align}$$
Then, for $(a,b,z)\in\mathbb{R}_{>0}^{3}\land a\neq b$ with $c:=\frac{a}{b}\land r:=\frac{1-c}{1+c}\land\theta:=2\arctan{\left(z\right)}$,
$$\begin{align}
G{\left(a,b,z\right)}
&=2\int_{0}^{z}\mathrm{d}t\,\frac{\ln{\left(\frac{a^{2}+b^{2}t^{2}}{1+t^{2}}\right)}}{1+t^{2}}\\
&=4\ln{\left(\frac{a+b}{2}\right)}\arctan{\left(z\right)}+\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(1-2r\cos{\left(\varphi\right)}+r^{2}\right)}\\
&=4\ln{\left(\frac{a+b}{2}\right)}\arctan{\left(z\right)}+\operatorname{Cl}_{2}{\left(2\theta+2\omega\right)}-\operatorname{Cl}_{2}{\left(2\theta\right)}-\operatorname{Cl}_{2}{\left(2\omega\right)}-2\omega\ln{\left(|r|\right)},\\
\end{align}$$
where
$$\begin{align}
\omega
&=\arctan{\left(\frac{r\sin{\left(\theta\right)}}{1-r\cos{\left(\theta\right)}}\right)}\\
&=\arctan{\left[\frac{r\sin{\left(2\arctan{\left(z\right)}\right)}}{1-r\cos{\left(2\arctan{\left(z\right)}\right)}}\right]}\\
&=\arctan{\left[\frac{\left(\frac{1-c}{1+c}\right)\left(\frac{2z}{1+z^{2}}\right)}{1-\left(\frac{1-c}{1+c}\right)\left(\frac{1-z^{2}}{1+z^{2}}\right)}\right]}\\
&=\arctan{\left[\frac{2\left(1-c\right)z}{\left(1+c\right)\left(1+z^{2}\right)-\left(1-c\right)\left(1-z^{2}\right)}\right]}\\
&=\arctan{\left[\frac{\left(1-c\right)z}{c+z^{2}}\right]}\\
&=\arctan{\left(\frac{bz-az}{a+bz^{2}}\right)}\\
&=\arctan{\left(\frac{a^{-1}bz-z}{1+a^{-1}bz^{2}}\right)}\\
&=\arctan{\left(\frac{bz}{a}\right)}-\arctan{\left(z\right)}.\\
\end{align}$$
Putting everything together, for $a\in(0,1)\land p:=\sqrt{\frac{1-a}{1+a}}$ we have
$$\begin{align}
\mathcal{I}{\left(a\right)}
&=\mathcal{J}{\left(a\right)}-\mathcal{K}{\left(a\right)}\\
&=\frac12G{\left(1,p,1\right)}-\frac12G{\left(p,1,1\right)}\\
&~~~~~-\sqrt{1-a^{2}}\operatorname{artanh}{\left(a\right)}\arcsin{\left(a\right)}\\
&~~~~~+\frac{\sqrt{1-a^{2}}}{4}\left[G{\left(p^{2},1,p\right)}-G{\left(1,p^{2},p^{-1}\right)}\right]\\
&=-\sqrt{1-a^{2}}\operatorname{artanh}{\left(a\right)}\arcsin{\left(a\right)}\\
&~~~~~+\frac{\pi}{2}\ln{\left(\frac{1+p}{2}\right)}+\frac12\operatorname{Cl}_{2}{\left(\frac{\pi}{2}+2\arctan{\left(p\right)}\right)}-\frac12\operatorname{Cl}_{2}{\left(2\arctan{\left(p\right)}-\frac{\pi}{2}\right)}\\
&~~~~~-\left[\arctan{\left(p\right)}-\frac{\pi}{4}\right]\ln{\left(\frac{1-p}{1+p}\right)}\\
&~~~~~-\frac{\pi}{2}\ln{\left(\frac{p+1}{2}\right)}-\frac12\operatorname{Cl}_{2}{\left(\frac{3\pi}{2}-2\arctan{\left(p\right)}\right)}+\frac12\operatorname{Cl}_{2}{\left(\frac{\pi}{2}-2\arctan{\left(p\right)}\right)}\\
&~~~~~+\left[\frac{\pi}{4}-\arctan{\left(p\right)}\right]\ln{\left(\frac{1-p}{1+p}\right)}\\
&~~~~~+\frac{\sqrt{1-a^{2}}}{4}\bigg{[}4\ln{\left(\frac{p^{2}+1}{2}\right)}\arctan{\left(p\right)}-\operatorname{Cl}_{2}{\left(4\arctan{\left(p\right)}\right)}-\operatorname{Cl}_{2}{\left(\pi-4\arctan{\left(p\right)}\right)}\\
&~~~~~-\left[\pi-4\arctan{\left(p\right)}\right]\ln{\left(\frac{1-p^{2}}{1+p^{2}}\right)}\\
&~~~~~-2\ln{\left(\frac{1+p^{2}}{2}\right)}\left[\pi-2\arctan{\left(p\right)}\right]-\left[\pi-4\arctan{\left(p\right)}\right]\ln{\left(\frac{1-p^{2}}{1+p^{2}}\right)}\\
&~~~~~-\operatorname{Cl}_{2}{\left(4\arctan{\left(p\right)}\right)}-\operatorname{Cl}_{2}{\left(\pi-4\arctan{\left(p\right)}\right)}\bigg{]}\\
&=-\sqrt{1-a^{2}}\operatorname{artanh}{\left(a\right)}\arcsin{\left(a\right)}\\
&~~~~~+\left[\frac{\pi}{2}-2\arctan{\left(p\right)}\right]\ln{\left(\frac{1-p}{1+p}\right)}\\
&~~~~~+\operatorname{Cl}_{2}{\left(\frac{\pi}{2}+2\arctan{\left(p\right)}\right)}+\operatorname{Cl}_{2}{\left(\frac{\pi}{2}-2\arctan{\left(p\right)}\right)}\\
&~~~~~+\frac{\sqrt{1-a^{2}}}{4}\bigg{[}2\left[\pi-4\arctan{\left(p\right)}\right]\ln{\left(\frac{2}{1-p^{2}}\right)}\\
&~~~~~-2\operatorname{Cl}_{2}{\left(4\arctan{\left(p\right)}\right)}-2\operatorname{Cl}_{2}{\left(\pi-4\arctan{\left(p\right)}\right)}\bigg{]}\\
&=-\sqrt{1-a^{2}}\operatorname{artanh}{\left(a\right)}\arcsin{\left(a\right)}\\
&~~~~~-2\left[\frac{\pi}{2}-\arccos{\left(a\right)}\right]\operatorname{artanh}{\left(\sqrt{\frac{1-a}{1+a}}\right)}\\
&~~~~~+\operatorname{Cl}_{2}{\left(\frac{\pi}{2}+\arccos{\left(a\right)}\right)}+\operatorname{Cl}_{2}{\left(\frac{\pi}{2}-\arccos{\left(a\right)}\right)}\\
&~~~~~+\frac{\sqrt{1-a^{2}}}{4}\bigg{[}2\left[\pi-2\arccos{\left(a\right)}\right]\ln{\left(\frac{1+a}{a}\right)}\\
&~~~~~-2\operatorname{Cl}_{2}{\left(2\arccos{\left(a\right)}\right)}-2\operatorname{Cl}_{2}{\left(\pi-2\arccos{\left(a\right)}\right)}\bigg{]}\\
&=-\sqrt{1-a^{2}}\operatorname{artanh}{\left(a\right)}\arcsin{\left(a\right)}\\
&~~~~~-2\arcsin{\left(a\right)}\operatorname{artanh}{\left(\sqrt{\frac{1-a}{1+a}}\right)}\\
&~~~~~+\operatorname{Cl}_{2}{\left(\pi-\arcsin{\left(a\right)}\right)}+\operatorname{Cl}_{2}{\left(\arcsin{\left(a\right)}\right)}\\
&~~~~~+\frac{\sqrt{1-a^{2}}}{4}\bigg{[}4\arcsin{\left(a\right)}\ln{\left(\frac{1+a}{a}\right)}\\
&~~~~~-2\operatorname{Cl}_{2}{\left(\pi-2\arcsin{\left(a\right)}\right)}-2\operatorname{Cl}_{2}{\left(2\arcsin{\left(a\right)}\right)}\bigg{]},\\
\end{align}$$
and then, setting $\alpha:=\arcsin{\left(a\right)}\in\left(0,\frac{\pi}{2}\right)$,
$$\begin{align}
\mathcal{I}{\left(a\right)}
&=-\sqrt{1-a^{2}}\operatorname{artanh}{\left(a\right)}\arcsin{\left(a\right)}\\
&~~~~~-2\arcsin{\left(a\right)}\operatorname{artanh}{\left(\sqrt{\frac{1-a}{1+a}}\right)}\\
&~~~~~+\operatorname{Cl}_{2}{\left(\pi-\arcsin{\left(a\right)}\right)}+\operatorname{Cl}_{2}{\left(\arcsin{\left(a\right)}\right)}\\
&~~~~~+\frac{\sqrt{1-a^{2}}}{4}\bigg{[}4\arcsin{\left(a\right)}\ln{\left(\frac{1+a}{a}\right)}\\
&~~~~~-2\operatorname{Cl}_{2}{\left(\pi-2\arcsin{\left(a\right)}\right)}-2\operatorname{Cl}_{2}{\left(2\arcsin{\left(a\right)}\right)}\bigg{]}\\
&=\alpha\sqrt{1-a^{2}}\ln{\left(\frac{1+a}{a}\right)}-\alpha\sqrt{1-a^{2}}\operatorname{artanh}{\left(a\right)}\\
&~~~~~-2\alpha\operatorname{artanh}{\left(\sqrt{\frac{1-a}{1+a}}\right)}\\
&~~~~~+\operatorname{Cl}_{2}{\left(\alpha\right)}+\operatorname{Cl}_{2}{\left(\pi-\alpha\right)}\\
&~~~~~-\frac{\sqrt{1-a^{2}}}{2}\operatorname{Cl}_{2}{\left(2\alpha\right)}-\frac{\sqrt{1-a^{2}}}{2}\operatorname{Cl}_{2}{\left(\pi-2\alpha\right)}\\
&=\frac12\alpha\sqrt{1-a^{2}}\ln{\left(\frac{1-a^{2}}{a^{2}}\right)}-2\alpha\operatorname{artanh}{\left(\sqrt{\frac{1-a}{1+a}}\right)}\\
&~~~~~+2\operatorname{Cl}_{2}{\left(\alpha\right)}-\frac12\operatorname{Cl}_{2}{\left(2\alpha\right)}\\
&~~~~~-\frac{\sqrt{1-a^{2}}}{2}\operatorname{Cl}_{2}{\left(2\alpha\right)}-\frac{\sqrt{1-a^{2}}}{2}\operatorname{Cl}_{2}{\left(\pi-2\alpha\right)}\\
&=\sqrt{1-a^{2}}\ln{\left(\frac{\sqrt{1-a^{2}}}{a}\right)}\arcsin{\left(a\right)}-2\arcsin{\left(a\right)}\operatorname{artanh}{\left(\sqrt{\frac{1-a}{1+a}}\right)}\\
&~~~~~+2\operatorname{Cl}_{2}{\left(\arcsin{\left(a\right)}\right)}-\frac{1+\sqrt{1-a^{2}}}{2}\operatorname{Cl}_{2}{\left(2\arcsin{\left(a\right)}\right)}\\
&~~~~~-\frac{\sqrt{1-a^{2}}}{2}\operatorname{Cl}_{2}{\left(\pi-2\arcsin{\left(a\right)}\right)}.\blacksquare\\
\end{align}$$
It would seem then that $\mathcal{I}$ cannot be evaluated in terms of elementary functions, and in general will require some version of the dilogarithm.
