Computing the Integral $\int r^2 \text{J}_0(\alpha r) \text{I}_1(\beta r)\text{d}r$ 
I encountered the following integral in a physical problem
  $$I=\int r^2 \text{J}_0(\alpha r) \text{I}_1(\beta r)\text{d}r$$
  where $\text{J}_0$ is the Bessel function of first kind of order $0$ and $\text{I}_1$ is the modified Bessel function of order $1$. Also, $\alpha$ and $\beta$ are arbitrary real numbers.

It seems that MAPLE and WOLFRAM are not able to find the primitive. However, I think that there should be a tidy one in terms of Bessel functions.

My Thought
My instinct in integration tells me to use integration by parts and the recursive relations for Bessel functions. But I couldn't get any where yet.
 A: First note that
$$ r\text{I}_1(\beta r)=\frac{\partial}{\partial\beta} \text{I}_0\left(\beta r\right).$$ 
Then we can write
\begin{align}
\mathcal{I}(t):=&\int_0^t r^2 \text{J}_0\left(\alpha r\right)\text{I}_1\left(\beta r\right) dr\\
=&\frac{\partial}{\partial\beta}\int_0^t r\text{J}_0\left(\alpha r\right)\text{I}_0\left(\beta r\right)dr\\
=& \frac{\partial}{\partial\beta}\left[\frac{t\left(\alpha \text{J}_1(\alpha t)\text{I}_0(\beta t)+\beta \text{J}_0(\alpha t)\text{I}_1(\beta t)\right)}{\alpha^2+\beta^2}\right].
\end{align}
The last line follows from the formula 1.11.5.2 in Vol. II of Prudnikov-Brychkov-Marychev (and can of course be verified by direct differentiation).
A: According to the Start Wearing Purple's answer, I just took the computations and wanted to put the final answer here
$$\eqalign{
  & I = {1 \over {{\alpha ^2} + {\beta ^2}}}\,\,\,\,\left( {\alpha {\text{I}_1}\left( {\beta r} \right){\text{J}_1}\left( {\alpha r} \right) + \beta {\text{I}_0}\left( {\beta r} \right){\text{J}_0}\left( {\alpha r} \right)} \right){r^2}  \cr 
  & \,\,\, - {{2\beta } \over {{{\left( {{\alpha ^2} + {\beta ^2}} \right)}^2}}}\left( {\alpha {\text{I}_0}\left( {\beta r} \right){\text{J}_1}\left( {\alpha r} \right) + \beta {\text{I}_1}\left( {\beta r} \right){\text{J}_0}\left( {\alpha r} \right)} \right)r \cr} $$
Hope that it may be useful for future readers. There may still be other heuristic ways to compute this integral so do not hesitate to write a new answer with a different approach. :)
