Evaluating integral $\int_a^\infty dx\ \frac{x}{\sqrt{(1+\frac{x^2}{b^2})^5(x^2-a^2)}}$ Where $a,b\in\mathbb{R}$ and we may consider $a\ge 0$,
\begin{align}
\mathrm{I}\left(a\right) & =
\int_{a}^{\infty}{x \over
\,\sqrt{\,\left(\, 1 + x^{2}/b^{2}\,\right)^{5}
\left(\, x^{2} - a^{2}\,\right)}\,}\,\,\mathrm{d}x =
{2b \over 3\left(\, 1 + {a^{2}/b^{2}}\,\right)^{2}}
\end{align}
Mathematica provided the solution. Any ideas on how to solve by hand ?. 
I solved the case $\,\mathrm{I}\left(0\right) = 2b/3$ with some simple trig substitutions. I'm not sure how to approach the general solution. If the general solution is rather nasty, I'm open to cases where $a$ or $b$ or both are considered to be small parameters. My purpose is to expand my knowledge of integration techniques.

Let $t=\sqrt{x^2-a^2}$, then
$$dt=\frac{x}{\sqrt{x^2-a^2}}dx,\quad x^2=t^2+a^2.$$
It follows that
$$\begin{aligned}
\int_a^\infty\frac{x}{\sqrt{(1+\frac{x^2}{b^2})^5(x^2-a^2)}}dx&=\int_0^\infty\frac{1}{\left(1+\frac{t^2+a^2}{b^2}\right)^{\frac{5}{2}}}dt.
\end{aligned}$$
Then use the substitution 
$$t=\sqrt{a^2+b^2}\tan u,$$
then
$$dt=\sqrt{a^2+b^2}\sec^2udu,$$
so we have
$$\begin{aligned}\int_0^\infty\frac{1}{\left(1+\frac{t^2+a^2}
{b^2}\right)^{\frac{5}{2}}}dt&=\int_0^{\frac{\pi}{2}}\frac{\sqrt{a^2+b^2}\sec^2u}{\left(\frac{a^2+b^2}{b^2}\sec^2u\right)^{\frac{5}{2}}}du\\
&=\int_0^{\frac{\pi}{2}}\frac{b^5}{(a^2+b^2)^2}\cos^3udu\\
&=\frac{b^5}{(a^2+b^2)^2}\int_0^{\frac{\pi}{2}}\cos u\frac{1+\cos(2u)}{2}du\\
&=\frac{b^5}{(a^2+b^2)^2}\int_0^{\frac{\pi}{2}}\frac{\cos u}{2}+\frac{\cos(3u)+\cos(u)}{4}du\\
&=\frac{b^5}{(a^2+b^2)^2}\left.\left[\frac{\sin u}{2}+\frac{\sin(3u)}{12}+\frac{\sin u}{4}\right]\right|_0^{\frac{\pi}{2}}\\
&=\frac{b^5}{(a^2+b^2)^2}\cdot\frac{2}{3}\\
&=\frac{2b}{3\left(\frac{a^2}{b^2}+1\right)^2}
\end{aligned}$$
