Consider a rectangle in the complex plane with vertices at $z=-R, z=R, z=R+ ik,$ and $z=-R+ik$.
Let $ f(z) = \displaystyle e^{-z^{2}}$ and integrate counterclockwise around the rectangle.
Then letting $R$ go to infinity,
$$ \int_{-\infty}^{\infty} e^{-x^{2}} \ dx + \lim_{R \to \infty} \int_{0}^{k} f(R+ it) \ i \ dt + \int_{\infty+ik}^{-\infty+ik} e^{-z^{2}} \ dz + \lim_{R \to \infty} \int_{k}^{0} f(-R+it) \ i \ dt = 0 . $$
If we can show that the second and fourth integrals vanish, then we have the result.
Notice that
$$ \begin{align} \Big| \int_{0}^{k} f(R+it) \ i \ dt \Big| & \le \int _{0}^{k} \Big|e^{-(R+it)^{2}} \Big| \ dt \\ &= e^{-R^{2}}\int_{0}^{k} e^{t^{2}} \ dt \end{align} $$
which vanishes as $ R \to \infty$ since $k$ is finite.
Similarly,
$$ \begin{align} \Big| \int_{0}^{k} f(-R+it) \ i \ dt \Big| &\le \int _{0}^{k} \Big| e^{-(-R+it)^{2}} \Big| dt \\ &= e^{-R^{2}}\int_{0}^{k} e^{t^{2}} \ dt \end{align}$$ which vanishes as $R \to \infty$ for the same reason.
I shall describe an expansion for this integral $\mathcal{I}(a,b,c)$ in powers of $c^{-2}$ . To do so I will make a few changes of parameters first. Observe that the substitution $x=bk$ yields
$$ \mathcal{I}(a,b,c) =\int_0^\infty dx \,e^{-a x^2/b^2} J_0(x)\frac{x^3}{b^{4} c^{2}+x^4}.$$ Defining $t=b^2/4a$, $\epsilon=1/b^4 c^2,$ and $I(\alpha,\epsilon) = b^4 c^2 \mathcal{I}(a,b,c)$, we have $$I(t,\epsilon) = \int_0^\infty dx\, e^{-x^2/4t} J_0(x)\frac{x^3}{1+\epsilon x^4}. \tag{1}$$
With this form in hand, we expand in powers of $\epsilon\sim c^{-2}$ to obtain $$I(t,\epsilon)=\sum_{k=0}^\infty (-\epsilon)^k \int_0^\infty dx\, x^{3+4k} e^{-x^2/4t}J_0(x).$$ The resulting term-by-term integration may be treated using formula 6.631.1 of Gradshteyn and Ryzhik (for reference, this is with $(\mu,\nu,\alpha,\beta)=(3+4k,0,1/4t,1))$:
$$\int_0^{\infty} dx\,x^{3+4k} e^{-x^2/4t}J_0(x) = \frac{1}{2}(4t)^{2k+2}(2k+1)!\,_1F_1(2k+2;1;-t)$$ where $ _1F_1(a;1;t)$ is Kummer's confluent hypergeometric series. This satisfies Kummer's transformation, allowing us to write
\begin{align}
_1F_1(2k+2;1;-t)
&=e^{-t}\,_1F_1(-1-2k;1;t)\\
&=e^{-t} \sum_{j=0}^{2k+1} \frac{(-1-2k)_j}{(j!)^2}t^j=\sum_{j=0}^{2k+1}\binom{2k+1}{j}\frac{t^j}{j!}e^{-t} \tag{2} \end{align} where the summation is terminated by the negative argument of the rising factorial $(x)_n$.
Recalling that the definition of the $n$th Laguerre polynomial is $L_n(x)=\sum_{k=0}^n\dfrac{(-x)^k}{k!}$, we may write equation $(2)$ as $e^{-t} L_{2k+1}(t)$. Hence we may express equation $(1)$ as
\begin{align}
I(t,\epsilon)
&=\sum_{k=0}^\infty (-\epsilon)^k \frac{1}{2}(4t)^{2k+2}(2k+1)! \, e^{-t} L_{2k+1}(t)\\
&=2t e^{-t} \sum_{m\text{ odd}}^\infty (-\epsilon)^{\frac{m-1}{2}} (4t)^m m! \, L_m(t)\
\end{align}
[to be continued]
Best Answer
Since $b$ is purely imaginary, this is a Fourier Transform.
Assuming $a>0$, $c>0$ ,and $\Re(b) = 0$; make the substitution $- \pi s = \Im (b)$ or equivalently $-i\pi s =b$ or $\pi s = ib$
$$\begin{align*}\displaystyle & \int_{-\infty}^{\infty} \frac{\exp\left[-a\left(x-b\right)^2\right]}{1+cx^2} dx\\ \\ &= \int_{-\infty}^{\infty} \frac{\exp\left[-a\left(x+i\pi s\right)^2\right]}{1+cx^2} dx\\ \\ &= \int_{-\infty}^{\infty} \frac{\exp\left[-ax^2+a(\pi s)^2-2\pi i axs\right]}{cx^2+1} dx\\ \\ &= \frac{a^2}{c}e^{a(\pi s)^2}\int_{-\infty}^{\infty} \frac{e^{-\frac{1}{a}(ax)^2}}{(ax)^2+\frac{a^2}{c}}e^{-2\pi i (ax)s} dx\\ \\ &= \frac{a}{c}e^{a(\pi s)^2}\int_{-\infty}^{\infty} \frac{e^{-\frac{1}{a}y^2}}{y^2+\frac{a^2}{c}}e^{-2\pi iys} dy\\ \\ &= \frac{a}{c}e^{a(\pi s)^2}(2\pi)^2\frac{1}{2\left(\frac{2\pi a}{\sqrt{c}}\right)}\int_{-\infty}^{\infty} \frac{2\left(\frac{2\pi a}{\sqrt{c}}\right)}{(2\pi y)^2+\left(\frac{2\pi a}{\sqrt{c}}\right)^2}\space e^{-\pi^2\left(\frac{y}{\pi\sqrt{a}}\right)^2}\space e^{-2\pi iys} dy\\ \\ &= \frac{\pi}{\sqrt{c}}e^{a(\pi s)^2}\mathscr{F}\left\{ \frac{2\left(\frac{2\pi a}{\sqrt{c}}\right)}{(2\pi y)^2+\left(\frac{2\pi a}{\sqrt{c}}\right)^2}\space e^{-\pi\left(\frac{y}{\sqrt{\pi a}}\right)^2}\right\} \\ \\ &= \frac{\pi}{\sqrt{c}}e^{a(\pi s)^2}\left[\mathscr{F}\left\{ \frac{2\left(\frac{2\pi a}{\sqrt{c}}\right)}{(2\pi y)^2+\left(\frac{2\pi a}{\sqrt{c}}\right)^2}\right\} * \mathscr{F}\left\{ e^{-\pi\left(\frac{y}{\sqrt{\pi a}}\right)^2}\right\} \right]\\ \\ &= \frac{\pi}{\sqrt{c}}e^{a(\pi s)^2}\left[e^{-\frac{2 a}{\sqrt{c}}|\pi s|} * \sqrt{\pi a} e^{-a\left(\pi s\right)^2}\right] \\ \\ &= \pi \sqrt{\frac{\pi a}{c}}e^{a(\pi s)^2}\int_{-\infty}^{\infty}e^{-\frac{2 a}{\sqrt{c}}|\pi \tau|} e^{-a\left(\pi s -\pi \tau\right)^2}\space d\tau \\ \\ &= \pi \sqrt{\frac{\pi a}{c}}e^{a(\pi s)^2}\left[\int_{-\infty}^{0}e^{\frac{2 a}{\sqrt{c}}\pi \tau} e^{-a\left(\pi s -\pi \tau\right)^2}\space d\tau +\int_{0}^{\infty}e^{-\frac{2 a}{\sqrt{c}}\pi \tau} e^{-a\left(\pi s -\pi \tau\right)^2}\space d\tau \right]\\ \\ &= \pi \sqrt{\frac{\pi a}{c}}\left(\int_{-\infty}^{0}\exp\left[-a\left([\pi\tau]^2-2\left[\pi s+\frac{1}{\sqrt{c}}\right]\pi \tau\right)\right]\space d\tau +\int_{0}^{\infty}\exp\left[-a\left([\pi\tau]^2-2\left[\pi s-\frac{1}{\sqrt{c}}\right]\pi \tau\right)\right]\space d\tau \right)\\ \\ &= \pi \sqrt{\frac{\pi a}{c}}\left(e^{a\left(\pi s+\frac{1}{\sqrt{c}}\right)^2}\int_{-\infty}^{0}\exp\left[-a\left(\pi\tau-\left[\pi s+\frac{1}{\sqrt{c}}\right]\right)^2\right]\space d\tau +e^{a\left(\pi s-\frac{1}{\sqrt{c}}\right)^2}\int_{0}^{\infty}\exp\left[-a\left(\pi\tau-\left[\pi s-\frac{1}{\sqrt{c}}\right]\right)^2\right]\space d\tau \right)\\ \\ &= \pi \sqrt{\frac{\pi a}{c}}\left(e^{a\left(\pi s+\frac{1}{\sqrt{c}}\right)^2}\frac{1}{\pi\sqrt{a}}\int_{-\infty}^{-\sqrt{a}\left(\pi s+\frac{1}{\sqrt{c}}\right)}e^{-u^2}\space du +e^{a\left(\pi s-\frac{1}{\sqrt{c}}\right)^2}\frac{1}{\pi\sqrt{a}}\int_{-\sqrt{a}\left(\pi s-\frac{1}{\sqrt{c}}\right)}^{\infty}e^{-u^2}\space du\right)\\ \\ &= \sqrt{\frac{\pi}{c}}\left(e^{a\left(\pi s+\frac{1}{\sqrt{c}}\right)^2}\int_{-\infty}^{-\sqrt{a}\left(\pi s+\frac{1}{\sqrt{c}}\right)}e^{-u^2}\space du +e^{a\left(\pi s-\frac{1}{\sqrt{c}}\right)^2}\int_{-\sqrt{a}\left(\pi s-\frac{1}{\sqrt{c}}\right)}^{\infty}e^{-u^2}\space du\right)\\ \\ &= \frac{\pi}{2}\frac{1}{\sqrt{c}}\left(e^{a\left(\pi s+\frac{1}{\sqrt{c}}\right)^2}\mathrm{erf}(u)\biggr{|}_{-\infty}^{-\sqrt{a}\left(\pi s+\frac{1}{\sqrt{c}}\right)} +e^{a\left(\pi s-\frac{1}{\sqrt{c}}\right)^2}\mathrm{erf}(u)\biggr{|}_{-\sqrt{a}\left(\pi s-\frac{1}{\sqrt{c}}\right)}^{\infty}\right)\\ \\ &= \frac{\pi}{2}\frac{1}{\sqrt{c}}\left(e^{a\left(\pi s+\frac{1}{\sqrt{c}}\right)^2}\left[\mathrm{erf}\left(-\sqrt{a}\left[\pi s+\frac{1}{\sqrt{c}}\right]\right)+1\right] +e^{a\left(\pi s-\frac{1}{\sqrt{c}}\right)^2}\left[1 -\mathrm{erf}\left(-\sqrt{a}\left[\pi s-\frac{1}{\sqrt{c}}\right]\right)\right]\right)\\ \\ &= \frac{\pi}{2}\frac{1}{\sqrt{c}}\left(e^{\left[-\sqrt{a}\left(ib+\frac{1}{\sqrt{c}}\right)\right]^2}\left[\mathrm{erf}\left(-\sqrt{a}\left[ib+\frac{1}{\sqrt{c}}\right]\right)+1\right] +e^{\left[-\sqrt{a}\left(ib-\frac{1}{\sqrt{c}}\right)\right]^2}\left[1 -\mathrm{erf}\left(-\sqrt{a}\left[ib-\frac{1}{\sqrt{c}}\right]\right)\right]\right)\\ \end{align*}$$