$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Note that
$\ds{{\arctan\pars{\xi} \over \xi} =
\int_{1}^{\infty}{\dd t \over t^{2} + \xi^{2}}}$.
\begin{align}
&\bbox[#ffd,10px]{\ds{\int_{-\infty}^{\infty}{b\arctan\pars{\root{x^{2} + a^{2}}/b} \over \pars{x^{2} + b^{2}}\root{x^{2} + a^{2}}}\,\dd x}} =
2\int_{0}^{\infty}{\arctan\pars{\root{x^{2} + a^{2}}/\verts{b}} \over \root{x^{2} + a^{2}}/\verts{b}}\,{\dd x \over x^{2} + b^{2}}
\\[5mm] = &\
2\int_{0}^{\infty}\int_{1}^{\infty}{\dd t \over
t^{2} + \pars{x^{2} + a^{2}}/b^{2}}\,{\dd x \over x^{2} + b^{2}} =
2b^{2}\int_{1}^{\infty}\int_{0}^{\infty}{\dd x \over
\pars{x^{2} + b^{2}t^{2} + a^{2}}\pars{x^{2} + b^{2}}}\,\dd t
\\[5mm] = &\
{2 \over \verts{b}}\int_{1}^{\infty}\int_{0}^{\infty}{\dd x \over
\pars{x^{2} + t^{2} + \mu^{2}}\pars{x^{2} + 1}}\,\dd t\,,
\qquad\qquad\qquad\mu \equiv {a \over b}
\end{align}
\begin{align}
&\bbox[#ffd,10px]{\ds{\int_{-\infty}^{\infty}{b\arctan\pars{\root{x^{2} + a^{2}}/b} \over \pars{x^{2} + b^{2}}\root{x^{2} + a^{2}}}\,\dd x}}
\\[5mm] = &\
{2 \over \verts{b}}\int_{1}^{\infty}{1 \over t^{2} + \mu^{2} - 1}
\int_{0}^{\infty}{\dd x \over x^{2} + 1}\,\dd t -
{2 \over \verts{b}}\int_{1}^{\infty}{1 \over t^{2} + \mu^{2} - 1}
\int_{0}^{\infty}{\dd x \over x^{2} + t^{2} + \mu^{2}}\,\dd t
\\[5mm] = &\
{\pi \over \verts{b}}\int_{1}^{\infty}{\dd t \over t^{2} + \mu^{2} - 1} -
{\pi \over \verts{b}}\int_{1}^{\infty}{\dd t \over
\pars{t^{2} + \mu^{2} - 1}\root{t^{2} + \mu^{2}}}
\\[5mm] = &\
\bbox[#ffe,10px,border:1px groove navy]{{\pi \over \verts{b}}\bracks{{\arctan\pars{\root{\mu^{2} - 1}} -
\mrm{arccot}\pars{\root{\mu^{2} - 1}} +
\mrm{arccot}\pars{\root{\mu^{4} - 1}} \over \root{\mu^{2} - 1}}}}\,,
\quad\mu \equiv {a \over b}
\end{align}
The second integral was evaluated as follows
\begin{align}
&\bbox[10px,#ffd]{\ds{\int_{1}^{\infty}{\dd t \over
\pars{t^{2} + \mu^{2} - 1}\root{t^{2} + \mu^{2}}}}}
\,\,\,\stackrel{t\ \mapsto\ 1/t}{=}\,\,\,
\int_{0}^{1}{t\,\dd t \over
\bracks{\pars{\mu^{2} - 1}t^{2} + 1}\root{1 + \mu^{2}t^{2}}}
\\[5mm] \stackrel{t^{2}\ \mapsto\ t}{=}\,\,\, &\
{1 \over 2}\int_{0}^{1}{\dd t \over
\bracks{\pars{\mu^{2} - 1}t + 1}\root{1 + \mu^{2}t}}
\,\,\,\stackrel{\root{1 + \mu^{2}t}\ \mapsto\ t}{=}\,\,\,
\int_{1}^{\root{1 + \mu^{2}}}{\dd t \over \pars{\mu^{2} - 1}t^{2} + 1}
\\[5mm] = &\
{\mrm{arccot}\pars{\root{\mu^{2} - 1}} -
\mrm{arccot}\pars{\root{\mu^{4} - 1}} \over \root{\mu^{2} - 1}}
\end{align}
