I am trying to find the closed form for the following integral: $$I=\int_{-\infty}^{\infty}e^{-a^2(x-b)^2+\frac{c}{1+x^2}}\ {\rm d}x$$ where $a,b,c\in\mathbb{R}$, and $a\ge0$. I know that the integral is gaussian at $c=0$, but I am trying to find it for $c>0$.
Edit: This integral appeared in the calculation of the average interaction of particles that follows Gaussian distribution $e^{-a^2(x-b)^2}$, where $a=\frac{1}{\sqrt{2}\sigma}$ and $b=\mu$, with a field $e^{\frac{c}{1+x^2}}$, where "$c$" controls the strength and width of the field. This integral is convergent. I looked through integrals in Table of Integrals, Series, and Products, but I could not find an answer. Also i look through these solutions 1, 2 and 3, but it does not help here.
My idea of solution was by expanding the field function using Taylor expansion, which leads to the following integral: $$I=\sum_{n=0}^{\infty}\int_{-\infty}^{\infty}\frac{c^n}{n!(1+x^2)^n}e^{-a^2(x-b)^2}\ {\rm d}x$$ the resultant integral is hard to solve.
Being myself a quantum mechanicist, I faced a quite similar problem and my approached was presented in a conference almost fifty years ago.
The idea is the same as your (expand as a series). The problem is that $$J= \int_{-\infty}^{+\infty}\frac{e^{-a^2(x-b)^2}}{(1+x^2)^n}\, dx$$ does not seem to show solution.
What I proposed was to expand now the integrand as $$\frac{e^{-a^2(x-b)^2}}{(1+x^2)^n}=\sum_{m=0}^\infty \frac{e^{-a^2x^2}}{(1+x^2)^n} d_m\,b^m$$ where $$d_0=e^{-a^2 x^2}\qquad d_1=2 a^2 x e^{-a^2 x^2}\quad \quad d_m=\frac 2 m a^2(x\, d_{m-1}-d_{m-2})$$
The $d_m$ being polynomials of degree $m$ in $x$, what is required is to compute only the even terms $$K_{2p}=\int_{-\infty}^{+\infty}\frac{e^{-a^2x^2}}{(1+x^2)^n}x^{2p}\, dx$$ which are $$K_{2p}=a^{(2 n-2 p-1)}\,\Gamma \left(p+\frac{1}{2}\right)\, U\left(n,n-p+\frac{1}{2},a^2\right)$$ where $U(.)$ is Tricomi's confluent hypergeometric function of the second kind (see here).
The calculations would be easy since (see here) $$0=a^2 (n+1) \,U\left(n+2,n-p+\frac{5}{2},a^2\right)+$$ $$\left(a^2-n+p-\frac{1}{2}\right)\, U\left(n+1,n-p+\frac{3}{2},a^2\right)-U\left(n,n-p+\frac{1}{2},a^2\right)$$with $$U\left(1,1-p+\frac{1}{2},a^2\right)=e^{a^2} E_{p+\frac{1}{2}}\left(a^2\right)$$ $$U\left(2,2-p+\frac{1}{2},a^2\right)=\frac{e^{a^2} \left(2 a^2+2 p-1\right) E_{p-\frac{1}{2}}\left(a^2\right)-2}{2 p-1}$$ where appears the exponential integral function.