Here is a direct computation. For $\beta>0$, we need the following fact
$$\int_{\mathbb{R}}e^{-x^2\beta^2}dx=\frac{\sqrt{\pi}}{\beta},$$
and the so-called Error function
$$\pmb{\text{erf}}(s)=\frac2{\sqrt{\pi}}\int_0^se^{-y^2}dy.$$
Then, the complementary error function is defined as $\pmb{\text{erfc}}(s)=1-\pmb{\text{erf}}(s)$. Let $b^2:=-c, b>0$.

Now, write $x^2=\frac{1+b^2x^2-1}{b^2}$ so that
\begin{align} \int_{\mathbb{R}}e^{-x^2/2\sigma^2}\frac{x^2dx}{1+b^2x^2}
&=\frac1{b^2}\int_{\mathbb{R}}e^{-x^2/2\sigma^2}dx-\frac1{b^2}\int_{\mathbb{R}}
e^{-x^2/2\sigma^2}\frac{dx}{1+b^2x^2} \\
&=\frac{\sigma\sqrt{2\pi}}{b^2}-\frac1{b^3}\int_{\mathbb{R}}
e^{-u^2/2b^2\sigma^2}\frac{du}{1+u^2}.
\end{align}
Let $a:=1/2b^2\sigma^2$ and rewrite
$$\int_{\mathbb{R}}e^{-u^2a}\frac{du}{1+u^2}=e^a\int_{\mathbb{R}}e^{-(1+u^2)a}\frac{du}{1+u^2}:=e^ag(a).$$
Computing the derivative $\frac{d}{da}$, we find that
$$g'(a)=-\int_{\mathbb{R}}e^{-(1+u^2)a}du=-e^{-a}\int_{\mathbb{R}}e^{-u^2a}du
=-\sqrt{\pi}\frac{e^{-a}}a.$$
This differential equation and initial value $g(0)=\pi$ lead to
\begin{align} g(a)
&=-\sqrt{\pi}\int_0^a\frac{e^{-t}}tdt+\pi
=-2\sqrt{\pi}\int_0^{\sqrt{a}}e^{-y^2}dy+\pi \\
&=-\pi\left(\frac2{\sqrt{\pi}}\int_0^{\sqrt{a}}e^{-y^2}dy\right)+\pi
=-\pi\cdot\pmb{\text{erf}}(\sqrt{a})+\pi \\
&=\pi\cdot\pmb{\text{erfc}}(\sqrt{a}).
\end{align}
Finally, we combine all that we calculated so far to obtain
\begin{align} \frac1{\sqrt{2\sigma^2\pi}}\int_{\mathbb{R}}e^{-x^2/2\sigma^2}\frac{x^2dx}{1+b^2x^2}
&=\frac1{\sqrt{2\sigma^2\pi}}\left[\frac{\sigma\sqrt{2\pi}}{b^2}
-\frac{\pi}{b^3}e^{1/2b^2\sigma^2}\cdot\pmb{\text{erfc}}\left(\frac1{b\sigma\sqrt{2}}\right)\right] \\
&=\frac1{b^2}-\frac{\sqrt{\pi}}{b^3\sigma\sqrt{2}}\cdot e^{1/2b^2\sigma^2}\cdot\pmb{\text{erfc}}\left(\frac1{b\sigma\sqrt{2}}\right).
\end{align}
We made the assumption $\sigma>0$.

**REMARK.** The integral in the OP's question can be regarded, up to scaling, as the Weierstrass transform
$$H(y)=\frac1{\sqrt{4\pi}}\int_{\mathbb{R}}h(x)e^{-(y-x)^2/4}dx$$
of the function $h(x)=\frac1{1+b^2x^2}$, but evaluated at $y=0$; that is, $H(0)$.