Integrating nasty gaussian over square root

Question

TLDR: trying to solve, $$\int_1^\infty \exp\left(-\frac{x^2}{2\omega^2}\right) \frac{1}{\sqrt{ax^2+bx-1}}dx$$

After doing some reading and looking at some other questions 1, 2 (and even going through a few integral tables) I realized there's probably no closed-form solution here, and that expressing the integral as a series of special functions is probably the best approach here,

$$ \sum_{k=0}^\infty \frac{(2\omega^2)^{-k}}{k!} \int_{1}^{\infty}\frac{x^{2k}}{\sqrt{(c-x)(x-d)}}dx$$

However, I wasn't able to find a solution even in this reduced form. Any ideas?

More details:

I encountered this integral during my research, trying to find the product of two functions $y_1, y_2$ of two independent random variables $p,q\sim N(\mu,\sigma^2)$, where

$$y_1(p,q) = \frac{1}{\sqrt{1+p^2+q^2}}$$ and $$y_2(p,q) = ap+bq+c$$

Answer 1

Let's factor the quadratic as $(x-c)(x-d)$, where I'll assume $|c|, |d| < 1$. For $x\ge 1$ we have $$ \frac{1}{\sqrt{x-c}} = \sum_{k=0}^\infty \frac{(2k)!}{k!^2} (c/4)^k x^{-(1/2+k)}$$ and similarly for $1/\sqrt{x-d}$. Thus $$ \eqalign{\frac{1}{\sqrt{(x-c)(x-d)}}&= \sum_{k=0}^\infty \sum_{j=0}^k \frac{(2j)! (2k-2j)!}{j!^2 (k-j)!^2} \left(\frac{c}{4}\right)^j \left(\frac{d}{4}\right)^{k-j} x^{-1-k}\cr &= \sum_{k-0}^\infty \frac{(2k)!}{k!^2} (d/4)^k {}_2F_1\left(\frac{1}{2},-k; -k+1/2; \frac{c}{d}\right) x^{-1-k}}$$ so that $$ \int_1^\infty e^{-x^2/(2\omega^2)} \frac{dx}{\sqrt{(x-c)(x-d)}} = \sum_{k=0}^\infty \frac{(2k)!}{k!^2} (d/4)^k {}_2F_1\left(\frac{1}{2},-k; -k+1/2; \frac{c}{d}\right) 2^{-1-k/2} \omega^{-k} \Gamma\left(-\frac{k}{2},\frac{1}{2\omega^2}\right) $$