I am interested in integrals of the following form: $$\text{I}(\mu,\sigma^2) = \int_{-\infty}^\infty dxH_n^2(x)e^{-(x-\mu)^2/2\sigma^2} $$ I know that in the case when $\mu = 0$ and $\sigma^2 = 1/2$ then $ I(0,1/2) = \sqrt{\pi}2^n n!$. I have tried following along the analysis by Jack D'Aurizio in Fourier transform of squared Gaussian Hermite polynomial, but using integrals of the form I am interested in. When doing this I end up with integrals of the form $$ \int_{-\pi}^\pi d\phi e^{a(cos(\phi)+b)^2}$$ which I believe have no solution in terms of elementary functions.
I have not been able to find a change of variables combined with properties of the Hermite Polynomials that simplify my integral into pieces with known solutions, but I am not very familiar with the properties of these polynomials.
(Not a complete answer, more a naive suggestion.) For Hermite polynomials there is the transation formula $$H_n(x+y)=\sum_{k,s=0}^n\frac{H_s(x)}{s!}\,\frac{H_{n-2k-s}(y)}{(n-2k-s)!}\,\frac1{k!},\tag{1}$$ and the scaling formula $$H_n(cx)=\sum_{k=0}^{\lfloor n/2\rfloor}\frac{n!\,(-1)^k}{k!(n-2k)!}{(1-c^2)}^k\,c^{n-2k}\,H_{n-2k}(x);\tag{2}$$ see (4.6.33) and (4.6.36) in Mourad Ismail's book “Classical and quantum orthogonal polynomials in one variable”. So one (arguably tedious) way to compute this could be as follows: