Integral over Modified Bessel Functions I've stumbled upon this integral
\begin{equation}
\mathscr{I}_\nu^k(a,b)=\int_0^\infty dr\frac{r^k}{r^2+m^2}e^{-ar^2}I_\nu(br)
\end{equation}
(where $I_\nu(x)$ is the modified Bessel function) during some QFT research, but I cannot seem to crack it. I'm specifically interested in the integrals $\mathscr{I}_1^2$, $\mathscr{I}_1^4$, and $\mathscr{I}_2^3$, each of which have an odd integrand, so that the typical semicircular contours are seemingly inutile. I've tried Weierstrass transforms, Hankel transforms, Laplace transforms, and the like, but each made the evaluation even more complicated. Additionally, I cannot find any similar integrals in the literature. Does anyone know a solution or a plan of attack? Any input would be greatly appreciated.

We can transform
\begin{align}
\mathscr{I}_\nu^k(a,b)&=\int_0^\infty dr\frac{r^k}{r^2+m^2}e^{-ar^2}I_\nu(br)\\
&=e^{am^2}\int_0^\infty dr\frac{r^k}{r^2+m^2}e^{-a(r^2+m^2)}I_\nu(br)\\
&=e^{am^2}J_\nu^k(a,b)
\end{align} 
We have 
\begin{equation}
\frac{\partial J_\nu^k(a,b)}{\partial a}=-e^{-am^2}\int_0^\infty r^ke^{-ar^2}I_\nu(br)\,dr
\end{equation} 
with $\lim_{a\to\infty} J_\nu^k(a,b)=0$. When $k=\nu+1$, this integral is tabulated (DLMF):
\begin{equation}
\int_{0}^{\infty}t^{\nu+1}I_{\nu}\left(bt\right)\exp\left(-at^{2}\right)\,dt=\frac{b^\nu}{(2a)^{\nu+1}}\exp\left(\frac{b^{2}}{4a}\right)
\end{equation} 
then
\begin{equation}
\frac{\partial J_\nu^{\nu+1}(a,b)}{\partial a}=-\frac{b^\nu}{(2a)^{\nu+1}}\exp\left(\frac{b^{2}}{4a}-am^2\right)
\end{equation} 
We deduce
\begin{equation}
J_\nu^{\nu+1}(a,b)=\frac{b^{\nu}}{2^{\nu+1}}\int_a^\infty\exp\left(\frac{b^{2}}{4\alpha}-\alpha m^2\right)\frac{d\alpha}{\alpha^{\nu+1}}
\end{equation} 
Thus
\begin{align}
\mathcal{I}_\nu^{\nu+1}(a,b)&=\frac{b^{\nu}e^{am^2}}{2^{\nu+1}}\int_a^\infty\exp\left(\frac{b^{2}}{4\alpha}-\alpha m^2\right)\frac{d\alpha}{\alpha^{\nu+1}}\\
&=\frac{b^{\nu}e^{am^2}}{2^{\nu+1}}\int_0^{1/a}\exp\left(\frac{b^{2}x}{4}-\frac{m^2}{x}\right)x^{\nu-1}\,dx
\end{align}
For the cases of interest,
\begin{align}
\mathscr{I}_1^{2}(a,b)&=\frac{be^{am^2}}{4}\int_0^{1/a}\exp\left(\frac{b^{2}x}{4}-\frac{m^2}{x}\right)\,dx\\
\mathscr{I}_2^{3}(a,b)&=\frac{b^{2}e^{am^2}}{8}\int_0^{1/a}\exp\left(\frac{b^{2}x}{4}-\frac{m^2}{x}\right)x\,dx
\end{align}
and, as noted in @NinadMunshi answer,
\begin{equation}
\mathscr{I}_1^{4}(a,b)=-\partial_a\mathscr{I}_1^{2}(a,b)
\end{equation} 
Using derivations and another tabulated integral (DLMF), it seems possible to obtain  integral expressions for other values of $\nu$ and $k$. It should also be noticed that obtained integrals could be expressed in terms of the  incomplete Bessel functions
