$\int^{\infty}_{-\infty}e^{-\alpha x^{2}}\frac{P(x)}{Q(x)}dx$ What would be the proper way to integrate the expression:
$\int^{\infty}_{-\infty}e^{-\alpha x^{2}}\frac{P(x)}{Q(x)}dx$
If everything is fine, i.e. $\alpha$ is positive, polynomial $Q(x)$ has no real zeroes and the order of $P(x)$ is less than the order of $Q(x)$?
In particular:
$P(x)=1+x$
$Q(x)=(1+(x-a)^2)^s$
$s>0$ is integer, $a$ is real.
 A: Let's try using substitution to get the integral into some recognizable form.
$$x=y+a$$
$$I_s(\alpha,a)= e^{- \alpha a^2} \int_{-\infty}^\infty e^{- \alpha y^2} e^{- 2\alpha a y} \frac{1+y+a}{(1+y^2)^s} dy$$
Clearly, we can consider two separate integrals:
$$J_s(\alpha,a)=\int_{-\infty}^\infty e^{- \alpha y^2} \frac{e^{- 2\alpha a y}}{(1+y^2)^s} dy$$
$$Y_s(\alpha,a)=\int_{-\infty}^\infty e^{- \alpha y^2} \frac{e^{- 2\alpha a y} y}{(1+y^2)^s} dy$$
1) Let's start with the first one. Expand the exponential as a series:
$$J_s(\alpha,a)=\sum_{n=0}^\infty \frac{(-1)^n (2 \alpha a)^n}{n!} \int_{-\infty}^\infty e^{- \alpha y^2} \frac{y^n}{(1+y^2)^s} dy$$
The integrals inside the series will be nonzero only for $n=2m$. Then we can write:
$$2\int_0^\infty e^{- \alpha y^2} \frac{y^{2m}}{(1+y^2)^s} dy=\int_0^\infty e^{- \alpha z} \frac{z^{m-1/2}}{(1+z)^s} dz=\Gamma\left(m+\frac12 \right) \mathrm U \left(m+\frac12,m-s+\frac32,\alpha \right)$$
Where $\mathrm U$ is the Tricomi hypergeometric function.
Let's not forget:
$$\Gamma\left(m+\frac12 \right)= 2 \sqrt{\pi} \frac{\Gamma(2m)}{\Gamma(m) 4^m}$$
Simplifying, we obtain:
$$J_s(\alpha,a)=\sqrt{\pi} \sum_{m=0}^\infty \frac{(\alpha a)^{2m}}{m!} \mathrm U \left(m+\frac12,m-s+\frac32,\alpha \right)$$
2) Doing the same series expansion for the second integral:
$$Y_s(\alpha,a)=\sum_{n=0}^\infty \frac{(-1)^n (2 \alpha a)^n}{n!} \int_{-\infty}^\infty e^{- \alpha y^2} \frac{y^{n+1}}{(1+y^2)^s} dy$$
Now we have to take $n=2m+1$, which gives us:
$$2 \int_0^\infty e^{- \alpha y^2} \frac{y^{2m+2}}{(1+y^2)^s} dy=\int_0^\infty e^{- \alpha z} \frac{z^{m+1/2}}{(1+z)^s} dz=\Gamma\left(m+\frac32 \right) \mathrm U \left(m+\frac32,m-s+\frac52,\alpha \right)$$
After simplifications:
$$Y_s(\alpha,a)=-\sqrt{\pi} \sum_{m=0}^\infty \frac{(\alpha a)^{2m+1}}{m!} \mathrm U \left(m+\frac32,m-s+\frac52,\alpha \right)$$
Finally we obtain:

$$I_s(\alpha,a)= \\ =  \sqrt{\pi} e^{- \alpha a^2} \sum_{m=0}^\infty \frac{(\alpha a)^{2m}}{m!} \left((1+a) \mathrm U \left(m+\frac12,m-s+\frac32,\alpha \right)- \\ - \alpha a \mathrm U \left(m+\frac32,m-s+\frac52,\alpha \right)  \right)$$

Note that the series should converge for all reasonable parameters, including $\alpha a>1$. Numerical experiments confirm that the series and the integral are equal.
The Tricomi function in Wolfram language is HypergeometricU[a,b,z].
As an example, let's take $\alpha=3$, $a=2$, $s=1/2$:
NIntegrate[Exp[12-3 x^2] (1+x)/(1+(x-2)^2)^(1/2)/Sqrt[Pi],{x,-Infinity,Infinity},WorkingPrecision->20]
gives 45951.232606275117969.
Sum[36^m/m! (3 HypergeometricU[m+1/2,m+1,3]-6 HypergeometricU[m+3/2,m+2,3]),{m,0,50}]
gives 45951.232606275096333.
The accuracy is good.
Obviously, this is not very useful for actual integral evaluation, because we use special functions, but this is all I managed to do.
