Notation for (complete) elliptic integrals:
$$
K(k):=\int_0^1\frac{1}{\sqrt{(1-t^2)(1-k^2t^2)}}dt,
\qquad
E(k):=\int_0^1\frac{\sqrt{1-k^2t^2}}{\sqrt{1-t^2}}dt.
$$
By simple algebra we also have
$$
\int_0^1\frac{t^2}{\sqrt{(1-t^2)(1-k^2t^2)}}dt=\frac{K(k)-E(k)}{k^2}.
$$
Integrating by parts we get
$$
I:=\int_{\sqrt 3}^2\sqrt{(y^2-3)(4-y^2)}dy=\int_{\sqrt 3}^2\frac{2y^4-7y^2}{\sqrt{(y^2-3)(4-y^2)}}dy.
$$
We have
$$
2y^4-7y^2=-2(y^2-3)(4-y^2)+7y^2-24
$$
hence
$$
I=-2I+\int_{\sqrt 3}^2\frac{7y^2-24}{\sqrt{(y^2-3)(4-y^2)}}dy,
$$
i.e.,
$$
I=\frac 13\int_{\sqrt 3}^2\frac{7y^2-24}{\sqrt{(y^2-3)(4-y^2)}}dy.
$$
It is convenient to change integration variable by $y=2t$ to obtain
\begin{align*}
I&=\frac{4i}{3\sqrt 3}\int_{\sqrt 3/2}^1\frac{7t^2-6}{\sqrt{(1-t^2)(1-\frac 43t^2)}}
\\
&=\Re\left(\frac{4i}{3\sqrt 3}\int_{0}^1\frac{7t^2-6}{\sqrt{(1-t^2)(1-\frac 43t^2)}}\right)
\\
&=\Re\left(\frac{4i}{3\sqrt 3}\left(7\frac{K(\tfrac 2{\sqrt3})-E(\tfrac 2{\sqrt3})}{4/3}-6K(\tfrac 2{\sqrt3})\right)\right)
\\
&=\Im\left(\frac{K(\tfrac 2{\sqrt3})+7E(\tfrac 2{\sqrt3})}{\sqrt 3}\right).
\end{align*}
One can further simplify the result by splitting real and imaginary parts (see also Separate elliptic integrals into real and imaginary parts). In particular we use DLMF 19.7.3
\begin{align*}
K(k^{-1})&=k\left(K(k)-i K(k')\right),\\
E(k^{-1})&=k^{-1}\left(E(k)+iE(k')-{k'}^2K(k)-i k^2K(k')\right),
\end{align*}
where $k':=\sqrt{1-k^2}$ with $k=\sqrt 3/2$, $k'=1/2$ to obtain
$$
\Im(K(2/\sqrt 3))=-\frac{\sqrt 3} 2K(1/2),\qquad \Im(E(2/\sqrt 3))=\frac 2{\sqrt 3}E(1/2)-\frac{\sqrt 3}2K(1/2)
$$
so that finally
$$
I=\Im\left(\frac{K(\tfrac 2{\sqrt3})+7E(\tfrac 2{\sqrt3})}{\sqrt 3}\right)=\frac {14}3E(1/2)-4K(1/2).
$$
This agrees with the linked computation in Wolfram Alpha (up to the factor 4 omitted in the integral and up to the different convention in the notation for elliptic integrals, which in WA are expressed as functions of $k^2$ rather than $k$, explaining the argument $1/4$ rather than $1/2$ in the elliptic integrals).
