How to prove that the integral
$\int_{0}^{\infty} z^a (1+z)^{1-a} e^{-b z^2} dx; \quad a, b>0 $
is given by
$\frac{2^{a-3}}{\pi \Gamma(a-1)} \times G^{3,2}_{2,3}\left(b \middle| \begin{array}{c} \frac{1-a}{2},\frac{-a}{2} \\ -1,\frac{-1}{2},0 \\ \end{array} \right).$