The following integrals, composed of a Gaussian measure and a positive fraction term, come up in my research. Neither the integration literature nor online integral calculators could help.
$$F_0=\int \frac{dx}{\sqrt{2\pi s^2}}e^{-(x-m)^2/2s^2} \frac{1}{a+bx^2}$$ $$F_2=\int \frac{dx}{\sqrt{2\pi s^2}}e^{-(x-m)^2/2s^2} \frac{x^2}{a+bx^2}$$
where $a,b,s\in\mathbb{R^+}, m\in\mathbb{R}$. Those are easy to calculate numerically, but I wonder if there is an analytic expression for any of them.
Preliminary note. First things first, let's notice that one of these two integrals can be deduced from the other one, because they satisfy the following relation : $$ aF_0 + bF_2 = \int_\Bbb{R} \frac{\mathrm{d}x}{\sqrt{2\pi s^2}}\, e^{-(x-m)^2/2s^2} = 1 $$
First method : differentiation. You may recast the computation of $F_0$ as the solution to an ODE by defining $$ F_0(t) = \int_\Bbb{R}\frac{\mathrm{d}x}{\sqrt{2\pi s^2}} \frac{1}{a+bx^2} \,\exp\left(-\frac{tx^2-2mx+m^2}{2s^2}\right) $$ for instance (other choices are possible), hence $$ 2s^2bF_0'(t) + aF_0(t) = \frac{1}{\sqrt{t}} \exp\left(-\frac{m^2}{2s^2}\left(1-\frac{1}{t}\right)\right) $$ Then, the solution is given by $F_0 = F_0(1)$.
Second method : substitution. You may use the so-called Hubbard–Stratonovich transformation (based on Fourier transforms), i.e. $$ \frac{1}{\sqrt{2\pi s^2}}e^{-(x-m)^2/2s^2} = \int_\Bbb{R}\frac{\mathrm{d}t}{2\pi}\, e^{-\frac{1}{2}s^2t^2-it(x-m)}, $$ before switching the integrals thanks to Fubini and integrating with respect to $x$ with the help of the following Fourier transform : $$ \int_\Bbb{R} \frac{e^{-itx}}{a+bx^2} \,\mathrm{d}x = \frac{1}{\sqrt{ab}} \exp\left(-|t|\sqrt{\frac{a}{b}}\right) $$ Note that this last result wouldn't be the same for values if $\operatorname{sgn}(a) \neq \operatorname{sgn}(b)$. Then, you will get a gaussian integral whose domain has to be split between the positive and negative values of $t$.
Third method : another substitution. Another natural substitution is given by Gérard Letac in his answer, namely $$ \frac{1}{a+bx^2} = \int_0^{\infty} e^{-t(a+bx^2)} \,\mathrm{d}t. $$ Note that it works positive $a,b$ only. Then, as said by Gérard Letac, Fubini allows to switch the integration in order to obtain gaussian integrals.
Final note. You won't be able to find closed form solutions, except for particular values of the parameters $(a,b,m,s)$, and the result involves special functions such as the error function.
EDITED: fixed a typo in the second method
Writing $$\frac{1}{a+bx^2}=\int_0^{\infty}e^{-t(a+bx^2)}dt$$ will lead to another integral using Fubini, but the result will not be more elementary.
