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)
$$\int\limits_0^\infty\frac{x^2}{(x+c)^{3/2}}e^{-x^2}=\\
\frac{e^{-\frac{c^2}{2}} \left(3 \pi \left(c^2 \left(-3+4 c^2\right) b[-1/4]+\left(1-7 c^2+4 c^4\right) b[1/4]-c^2 \left(-1+4 c^2\right) \left(b[3/4]+b[5/4]\right)\right)+64 c^2 e^{\frac{c^2}{2}} h\right)}{12 \sqrt{c}}
$$
I am absolutely sure this does not help you at all. I am very sorry to disappoint you, but I dont know how this can be done. I know improper Integrals are usually solved by integrating over semi-circles in complex space, however that is not easily done for this function.
Assuming[c>0,Integrate[x^2/(x+c)^(3/2) Exp[-x^2],{x,0,\[Infinity]}]]. – Ayman Hourieh Jan 30 '13 at 00:14