TITLE: How to solve this set of definite integrals?
QUESTION [5 upvotes]: I'm looking to solve the integrals
$$
I_n=\int_0^\infty \frac{x^ne^{-2ax}}{\sqrt{x^4+1}}dx.
$$
for $a>0$.
This should be reduced to finding $I_0$ as numerically I find that
$$
I_n=\left(-2\right)^{-n}\frac{d^nI_0}{da^n}.
$$
Mathematica gives a result in terms of the MeijerG function $G^{51}_{15}\left(x|^{\frac{1}{4}}_{-\frac{1}{2},-\frac{1}{4},-\frac{1}{4},0,\frac{1}{4}}\right)$ but I cannot manage to prove it. I have looked for integral representations of this and hypergeometric functions in Gradshteyn but couldn't find anything useful.
Any help is much welcomed.
Thanks

REPLY [2 votes]: It's a general result that an integral over $[0, \infty)$ of two Meijer G-function of rational powers of $x$ is again a G-function:
$$\int_0^\infty \frac {x^n e^{-2 a x}} {\sqrt {x^4 + 1}} dx =
\frac 1 {\sqrt \pi} \int_0^\infty x^n
 G_{0, 1}^{1, 0} \left( 2 a x \middle| {- \atop 0} \right)
 G_{1, 1}^{1, 1} \left( x^4 \middle| {\frac 1 2 \atop 0} \right) dx = \\
\frac {2^{n - 3/2}} {\pi^2 a^{n + 1}}
 G_{5, 1}^{1, 5} \left( \frac {16} {a^4} \middle|
  {\frac {-n} 4, \frac {-n + 1} 4, \frac {-n + 2} 4, \frac {-n + 3} 4, \frac 1 2
   \atop 0} \right).$$
The resulting G-function is not reducible to a sum of hypergeometric functions because one of the four coefficients $(-n + i)/4$ is half-integral for any integral $n$.