The integral:
I am curious if the following integral can be computed analytically.
$$J=\int_{-\infty}^{\infty}e^{-a(x-b)^2}\sqrt{x^2+c^2}dx,\hspace{1cm}a,b,c\in\mathbb{R}^{+}$$
What I've tried:
[1] Trig substitution:
With $x=c\tan(\theta)$ can write $J=\int_{-\pi/2}^{\pi/2}e^{-a(c\tan(\theta)-b)^2}\sec^3(\theta)d\theta$ but I'm not sure where to go from here (integration-by-parts?).
[2] Fourier Convolution
Using the fact that$$\int_{-\infty}^{\infty}g(y)f(x-y)dy=\int_{-\infty}^{\infty}g^{T}(t)f^{T}(t)e^{-ixt}dt,$$ where $g^{T}$ is the Fourier transform (FT) of $g$ I can replace the shift in the gaussian with a complex exponential. I know the FT of a gaussian is a gaussian and Mathematica gives the FT of $\sqrt{x^2+c^2}$ as $\sqrt{2/\pi}cK_1(c|t|)/|t|$, where $K_1$ is the modified Bessel function of the second kind.
I then have $J\propto\int_{-\infty}^{\infty}e^{-t^2/4a}K_1(c|t|)/|t|e^{-ixt}dx$.
I've considered contour integration and using the recurrence relation between the $K_\nu$'s but seem to get nowhere.
[3] Look up in table of integrals
I have looked through Gradshteyn and Ryzhik (7e) but to no avail.
Best Answer
As said in comments, for $b=0$, CAS produce different expressions. Another one is $$\int_{-\infty}^{\infty}e^{-ax^2}\sqrt{x^2+c^2}\,dx=\frac{1}{2} c^2 e^{\frac{a c^2}{2}} \left(K_0\left(\frac{a c^2}{2}\right)+K_1\left(\frac{a c^2}{2}\right)\right)$$
So, a possible idea would be to expand the exponential as a Taylor series built around $b=0$ such as $$e^{-a(x-b)^2}=$$ $$e^{-ax^2}\left(1+2 a b x+a b^2 \left(2 a x^2-1\right)+\frac{2}{3} a^2 b^3 x \left(2 a x^2-3\right)+\frac{1}{6} a^2 b^4 \left(4 a^2 x^4-12 a x^2+3\right)+\frac{1}{15} a^3 b^5 x \left(4 a^2 x^4-20 a x^2+15\right)+\frac{1}{90} a^3 b^6 \left(8 a^3 x^6-60 a^2 x^4+90 a x^2-15\right)+O\left(b^7\right) \right)$$ which makes that we would face integrals $$I_n=\int_{-\infty}^{\infty}e^{-ax^2}\sqrt{x^2+c^2}\,x^n\,dx$$ If $n$ is odd, the result is $I_{2n+1}=0$.
For even values, they write $$\color{blue}{I_{2n}=\frac { c^{2(2 n+1)}\, e^t} {2^{(2 n+1)}\, t^{2 n} }\Big[P_n(t)\, K_0(t) - Q_n(t)\, K_1(t) \Big]}\qquad \text{where} \qquad \color{red}{t=\frac{a c^2}{2}}$$
The first polynomials are
$$\left( \begin{array}{cc} n & P_n(t) \\ 0 & 1 \\ 1 & t \\ 2 & 4 t^3-9 t^2+12 t \\ 3 & 16 t^5-60 t^4+159 t^3-300 t^2+360 t \\ 4 & 64 t^7-336 t^6+1320 t^5-4095 t^4+9900 t^3-17640 t^2+20160 t \end{array} \right)$$
$$\left( \begin{array}{cc} n & Q_n(t) \\ 0 & -1 \\ 1 & t-2 \\ 2 & 4 t^3-11 t^2+18 t-24 \\ 3 & 16 t^5-68 t^4+195 t^3-408 t^2+600 t-720 \\ 4 & 64 t^7-368 t^6+1512 t^5-4905 t^4+12600 t^3-24840 t^2+35280 t-40320 \end{array} \right)$$
Edit
It could be simpler to rewrite $$\int e^{-a(x-b)^2}\sqrt{x^2+c^2}\,dx=c^2\,\int e^{-\alpha (y-\beta )^2} \sqrt{y^2+1}\,dy$$ where $\alpha=\frac a {c^2}$ and $\beta=\frac b c$.