Evaluate $\int_0^\infty e^{-b\left(\frac{r^2}{a^2}+1\right)^\frac{-\gamma}{2}}\left (\frac{r^2}{a^2}+1\right)^\frac{-\gamma}{2} r^2 dr $ I'm working on research in astrophysics related to determining the ages of stellar nurseries. I've got the numerical solution, but need an analytic solution to the integral below in order to better illustrate the behavior of the key parameters. 
The problem is, after 3 days of trying, I'm no closer to a solution! Can anyone help me solve this?
$$
\int_0^\infty e^{-b\left(\frac{r^2}{a^2}+1\right)^\frac{-\gamma}{2}} \left(\frac{r^2}{a^2}+1\right)^\frac{-\gamma}{2} r^2 dr
$$
where $a,b,\gamma$ are real, positive constants and $\gamma > 4$.
$\gamma$ is the key here. I would love to have a solution for general values of $\gamma$, but it would be enough for me to just have solutions for some two specific values of $\gamma$ for comparison.
If anyone can help, I'd appreciate it very much. 
Or if anyone can say with certainty that the integral is not solvable, then at least I can move on. 
 A: Define:
$$f(a,b,\gamma)=\int _{0}^{\infty }\!{{\rm e}^{-b \left( {\frac {{r}^{2}}{{a}^{2}}}+1
 \right) ^{-\frac{1}{2}\gamma\, n}}} \left( {\frac {{r}^{2}}{{a}^{2}}}+1 \right) ^{\frac{1}{2}\gamma\, n}{r}^{2}{dr}$$
then after the variable change $R=ra$ you have:
$$
\begin{aligned}
f(a,b,\gamma)&={a}^{3}{\frac {\partial }{\partial b}}\int _{0}^{
\infty }\!-\mathrm{exp}\left[{-b \left( {R}^{2}+1 \right) ^{-\frac{1}{2}\gamma\, }}\right]{R}^{2}{dR}\\
&=-\sum_{n=0}^{\infty}\frac{{a}^{3}}{n!}{\frac {\partial }{\partial b}}\int _{0}^{
\infty }\!(-b)^n{\left( {R}^{2}+1 \right) ^{-\frac{1}{2}\gamma\, n}}{R}^{2}{dR}\\
&=-\sum_{n=0}^{\infty}(-b)^{n-1}\frac{{a}^{3}}{(n-1)!}\int _{0}^{
\infty }\!{\left( {R}^{2}+1 \right) ^{-\frac{1}{2}\gamma\, n}}{R}^{2}{dR}\\
\end{aligned}
$$
and the integral can be recognised as the beta function:
$$
\begin{aligned}
\int _{0}^{\infty }\! \left( {R}^{2}+1 \right) ^{-\frac{1}{2}\gamma\, n}{R}^{2}{dR}&=\frac{1}{2}\,\int _{0}^{\infty }\!{\frac {\sqrt {s}}{ \left( s+1 \right) ^{\frac{1}{2}\,\gamma\, n}}}{ds}\quad:R=\sqrt{s}\\
&=\frac{1}{2}\,\beta \!\left( \frac{3}{2},-\frac{3}{2}+\frac{1}{2}\,\gamma\,n \right) \\
&=\frac{1}{4}\,{\frac {\sqrt {\pi }\Gamma  \left( -\frac{3}{2}+\frac{1}{2}\,\gamma n \right) }{\Gamma 
 \left( \frac{1}{2}\,\gamma\, n \right) }}
\end{aligned}
$$
hence:

$$\int _{0}^{\infty }\!{{\rm e}^{-b \left( {\frac {{r}^{2}}{{a}^{2}}}+1
 \right) ^{-\frac{\gamma}{2}}}} \left( {\frac {{r}^{2}}{{a}^{2}}}+1 \right) ^{-\frac{\gamma}{2}}{r}^{2}{dr}=\frac{{a}^{3}\sqrt{\pi}}{4}\sum _{n=0}^{\infty }{\frac { \,\Gamma  \left( -\frac{3}{2}+\frac{1}{2}\gamma n \right) }{
\Gamma  \left( \frac{1}{2}\gamma\,n \right) \Gamma(n)}}\left( -b
 \right) ^{n-1}$$

An example for $\gamma=4$ will give closed form in terms of the generalized Hypergeometric function:
$$\int _{0}^{\infty }\!{{\rm e}^{- \left( {r}^{2}+1 \right) ^{-2}}}{r}^{
2} \left( {r}^{2}+1 \right) ^{-2}{dr}=a^3\frac{\pi}{4}\,
{\mbox{$_2$F$_2$}\!\left(\frac{1}{4},\frac{3}{4};\,1,\frac{3}{2};\,-b\right)}$$
