Difficult infinite integral involving a Gaussian, Bessel function and complex singularities I've come across the following integral in my work. $$\intop_{0}^{\infty}dk\, e^{-ak^{2}}J_{0}\left(bk\right)\frac{k^{3}}{c^{2}+k^{4}}
 $$
Where $a$,$b$,$c$ are all positive.
I've seen and evaluate similar integrals without the denominator using resources like Watson's Theory of Bessel Functions, but I've had no luck finding anything resembling this integral. It would get me a very awesome result if I were to evaluate this. Does anyone have any ideas on how to approach it?
edit: Some of my attempts so far include:
1)Using "Infinite integrals involving Bessel functions by an improved approach of contour integration and the residue theorem"  by Qiong-Gui Lin. This (like other residue theorem approaches) doesn't seem to work since the gaussian blows up on the imaginary axis.
2)I recall seeing expressions like $Z_u(ax)X_v(b\sqrt x)$ where Z,X are some variation of bessel functions in Gradshteyn, Ryzhik. This inspired me to write $e^{-ax^2}=\sum_{k=-\infty}^{\infty}I_{k}(-ax^{2})$ and substitute $t=x^2$, then integrate term by term. This hasn't gotten me anywhere either.
 A: 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]
