Find $\int_0^\infty \sqrt{x} \left( \tan^{-1} \left(\frac{x+a}{c} \right)- \tan^{-1} \left(\frac{x-a}{c} \right) \right)dx$ How to compute the integral
\begin{align}
\int_{0}^{\infty}\sqrt{x}\left[\tan^{-1}\left(\frac{x+a}{c} \right)- \tan^{-1} \left(\frac{x-a}{c} \right) \right]\mathrm{d}x
\end{align}
for $c,a>0$.
If the it the integration is to difficult can we give a good upper bound? 
 A: By using integration by parts, the given integral equals
$$ \frac{2}{3}\int_{0}^{+\infty}\frac{4ac x^{5/2}}{\left(a^2+c^2-2 a x+x^2\right) \left(a^2+c^2+2 a x+x^2\right)}\,dx $$
and by substituting $x=z^2$, this integral becomes
$$ \frac{2}{3}\int_{-\infty}^{+\infty}\frac{4ac z^{6}}{\left(a^2+c^2-2 a z^2+z^4\right) \left(a^2+c^2+2 a z^2+z^4\right)}\,dz $$
where the integrand function is $O\left(\frac{1}{|z|^2}\right)$ as $|z|\to +\infty$. By the residue theorem, we just need to compute the residues of the integrand function at the poles in the upper half-plane, given by $\sqrt{\pm a+ ic}$ and $-\sqrt{\pm a-ic}$. It follows that the given integral equals:
$$\boxed{ \frac{2\pi}{3}\sqrt{c(3a^2-c^2)+(a^2+c^2)\sqrt{a^2+c^2}} }$$
and a simple upper bound (that is tight iff $a\approx 0$ or $a\approx c\sqrt{3}$) is given by $\color{red}{\frac{2\pi\sqrt{2}}{3}(a^2+c^2)^{3/4}}$.
A: With the short hand $b=\sqrt{a^2+c^2}$ and the substitution $x=t^2$, the integral is
\begin{align}
I=& \ 2\int_0^\infty t^2\left( \tan^{-1} \frac{t^2+a}{c} - \tan^{-1} \frac{t^2-a}{c} \right)dt\\
\overset{ibp}=& \ \frac{4c}3 \int_0^\infty
\frac{2at^2-b^2}{t^4-2at^2+b^2}+ \frac{2at^2+b^2}{t^4+2at^2+b^2}\ dt\\
=& \ \frac{4c}3 \bigg(\frac{(2a-b)\pi}{2\sqrt{2(b-a)}}+ \frac{(2a+b)\pi}{2\sqrt{2(b+a)}} \bigg)\\
= &\ \frac{2\pi}3\left(2a\sqrt{b+c}-b \sqrt{b-c} \right)
\end{align}
