Evaluate the integral, $\int_0^{\infty} \frac{x^2}{(x+c)^{3/2}}\,e^{-x^2} dx$, where c>0

This integral came up in a research problem I'm working on, but I haven't had much luck calculating it. I suspect that the integral doesn't have a very clean form, but if anyone knows of an easy substitution or some elementary way to evaluate this integral, it would be very helpful.

Thanks!

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Mathematica does not find a solution at all, not even with Appell Hypergeometric functions in conjunction with error functions, I dont think you will either... I am very sorry; I am not saying its impossible, but very, VERY hard. – CBenni Jan 30 '13 at 0:03
Actually, Mathematica 9 gives an answer in terms of Bessel and generalized hypergeometric functions. Try Assuming[c>0,Integrate[x^2/(x+c)^(3/2) Exp[-x^2],{x,0,\[Infinity]}]]. – Ayman Hourieh Jan 30 '13 at 0:14
@AymanHourieh I overlooked the bounds >_< sorry. I will post the solution as an answer, because it is impossible in the comments. – CBenni Jan 30 '13 at 0:24
Of course you must assume $c \notin (-\infty, 0]$ for this to exist. – Robert Israel Jan 30 '13 at 0:25
@RobertIsrael Yes, you're absolutely right. Should have mentioned that c is a positive constant. – Sam Jan 30 '13 at 4:44

Denote $b[x]:=\; \text{BesselI}\left(x,\frac{c^2}{2}\right)$ and $h:=\text{HypergeometricPFQ}\left[\left\{\frac{3}{2},2\right\},\left\{\frac{5}{4},\frac{7}{4}\right\},-c^2\right]$. Then (according to mathematica)
$$\int\limits_0^\infty\frac{x^2}{(x+c)^{3/2}}e^{-x^2}=\\ \frac{e^{-\frac{c^2}{2}} \left(3 \pi \left(c^2 \left(-3+4 c^2\right) b[-1/4]+\left(1-7 c^2+4 c^4\right) b[1/4]-c^2 \left(-1+4 c^2\right) \left(b[3/4]+b[5/4]\right)\right)+64 c^2 e^{\frac{c^2}{2}} h\right)}{12 \sqrt{c}}$$