I am trying to find a general solution of integrals with the following form $$I= \int_{-\infty}^{\infty}\frac{e^{-(x-\mu)^2/\sigma^2}}{(1+a x^2)^b} dx$$ where $\{a,b,\mu, \sigma\}\in\mathbb{R}$, $a,b,\sigma>0 $ and $-\infty <\mu< \infty $. I tried to write $\frac{1}{(1+a x^2)^b}$ as a series expantion using the hypergeometric function 2F1 definition as $$\frac{1}{(1+a x^2)^b}={}_2F_1(1,b;1;-ax^2)= \sum_{k=0}^{\infty}\frac{(r)_k}{k!}(-ax^2)^{-k}$$ but later i figured out that this expansion is convergent only for $|ax^2|<1$, therefore this approach is not applicable.
Could you help me to find an approach to solve such integral or to find an a convergent expansion for $\frac{1}{(1+a x^2)^b}$?
If you want to do the expansion into a hypergeometric function, you can use the following way to handle the convergence. Basically, you get to do twice the amount of calculations. Set $y=ax^2$ for the moment, then
$\frac{1}{(1+y)^b} = \frac{1}{y^b}\frac{1}{(1+y^{-1})^b} = \frac{1}{y^b}{}_2F_{1}(b;;-y^{-1}) = \frac{1}{(ax^2)^b}{}_2F_{1}(b;;-(ax^2)^{-1})$
So this converges for $|ax^2|>1$. You have to check $|ax^2|=1$ separately.
I haven't though through if it is a good way to evaluate the integral though...