Integral $ \int_a^\infty \frac{dr}{r^5 \sqrt{r-a}}$ 
$$ \int_a^\infty \frac{dr}{r^5 \sqrt{r-a}}$$

By taking $r-a=t^2$ we are getting 
 $$ \int_0^\infty \frac{2dt}{(a+t^2)^5}$$
Now I am stuck after this. Thanks.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}}}
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\begin{align}
2\int_{0}^{\infty}{\dd t \over \pars{t^{2} + a}^{5}} & =
{1 \over 12}\,\totald[4]{}{a}\int_{0}^{\infty}{\dd t \over t^{2} + a} =
{\pi \over 24}\,\totald[4]{\pars{a^{-1/2}}}{a} =
{\pi \over 24}\,{105 \over 16\,a^{9/2}} =
\bbx{{35\pi \over 128}\,{1 \over a^{9/2}}}
\end{align}
A: Let $t = \sqrt{a}\tan(\theta), dt = \sqrt{a}\sec^2(\theta)$. Then your integral becomes
$$2\int \frac{\sqrt{a}\sec^2(\theta)}{(a+a\tan^2(\theta))^5} d \theta = \frac{2\sqrt a}{a^5} \int \cos^8(\theta)\, d\theta$$
$$ = \frac{2}{a^{9/2}} \int \left( \frac{1+\cos(2\theta)}{2}\right)^4 d\theta$$$$ = \frac{1}{8a^{9/2}}\int 1+4\cos(2\theta)+6\cos^2(2\theta)+4\cos^3(2\theta)+\cos^4(2\theta) d\theta$$
