The following integral comes up for me when I'm computing a normalizing constant for a probability distribution:
$$\int_0^\infty r^{p-1}\left(1+\frac{\gamma}{\alpha}r^2\right)^{-1/2}\exp\left\{-\frac{\alpha}{2}r^2\right\}dr.$$
I know that it has a closed form solution, because you can type it into Mathematica and see the answer, but I can't find the right approach. If p=2, then I know the integral equals
$$\frac{1}{\sqrt{2\gamma}} \exp\left\{\frac{\alpha^2}{2\gamma}\right\} \Gamma\left(1/2, \frac{\alpha^2}{2\gamma}\right)$$ where $\Gamma$ is the incomplete gamma function (notation matches that of Wikipedia).
I was considering using the resulting knowledge that
$$\frac{\sqrt{2\gamma}\exp\left\{-\frac{\alpha^2}{2\gamma}\right\}}{\Gamma\left(1/2, \frac{\alpha^2}{2\gamma}\right)} \, r \, \left(1+\frac{\gamma}{\alpha}r^2\right)^{-1/2}\exp\left\{-\frac{\alpha}{2}r^2\right\}$$
is a (normalized) probability density function and think of the first integral as the p-1 moment of the random variable defined by this PDF, but then I couldn't derive the moment generating function or characteristic function.
I also considered whether I could do integration by parts or recursive integration by parts, but I couldn't find the right approach.
The closed form can be expressed with the Whittaker's confluent hypergeometric function, thanks to the Laplace transform from ref. $[1]$, page 139, Eq.(22) in : Tables of Integral Transforms, H.Bateman, Edit. McGraw-Hill (1954).