TITLE: Can this integral be calculated in closed form?
QUESTION [12 upvotes]: I'm trying to calculate the following integral:
$$
\int_{0}^{1}x\,\mathrm{J}_{2}\!\left(\,bx\,\right)
\sin\left(\,a\,\sqrt{\,1 - x^{2}\,}\,\right)\,\mathrm{d}x
$$where $a$ and $b$ are parameters ( independent of $x$ ).
Things I have tried so far, without success (but possibly not driven through far enough):

look up in tables: no joy. The ''closest'' match I have found is eq. (6.738.1) in Gradshteyn and Ryzhik, but that result is for the
first factor being $x^3$ instead of $x$;
use WolframAlpha (Standard): does not give values for general $a$
and $b$; only for assigned selected $a$ and $b$. Since I need results for a wide range of values for both $a$ and $b$, this (or any numerical quadrature for arbitrary $a$ and $b$) is not practical;
use an integral representation for Bessel functions: \begin{align}
J_2(bx) = \frac{1}{\pi} \int^\pi_0 \cos(2\theta - bx \sin\theta)
d\theta \end{align} then swap the order of integration. However, the outer integration (i.e., with respect to $\theta$) then becomes problematic;
use a recurrence relation for Bessel functions: \begin{align} J_2(bx) =
(2/(bx)) J_1(bx) - J_0(bx) \end{align} This does not seem to simplify matters, because the square root in sin() remains a difficulty.
attempt partial integration: since the integrand contains three
factors, the choice is not obvious. I tried grouping the first two
factors and use the partial integral result \begin{align} \int x^m J_n(x)
dx = -x^m J_{n-1}(x) + (m+n-1) \int x^{m-1} J_{n-1}(x) dx
\end{align} for $m=1$ , $n=2$, i.e., \begin{align} \int x J_2(x) dx
= -x J_{1}(x) - 2 J_{0}(x) + C\end{align} but the derivative of $\sin(a\sqrt{1-x^2})$ with respect to $x$ complicates the remaining
integration;
use a series representation of the Bessel function: this leads to a
double summation of integrals (one sum is semi-infinite) of the
form \begin{align} \int^1_0 u^{2m+1} \sin(a u) du \end{align}  but
this integral is a itself a difference of hypergeometric functions (or
alternatively an additional series representation). Such a double or triple summation is again
impractical for calculating for parametrized $a$ and $b$;
converting the original integral to \begin{align}
\int^1_0 y J_2 (b \sqrt{1-y^2}) \sin(a y) dx \end{align} where $y=\sqrt{1-x^2}$ and repeating the previous approaches; 
attempt
    a trigonometric substitution such as $x = \sin \alpha$ and repeat the
    previous approaches.

Can you find a solution or give further suggestions what could be attempted?

REPLY [2 votes]: This is not an answer either, but MSE wouldn't let me edit my comment.
Let $M(a,b)$ be your integral, and let $N(a,b)$ be the integral you found in the table with $x^3$.  Then $M$ and $N$ are related by the ODE: $M(a,b)-N(a,b)=\frac{\partial^2 M}{\partial a^2}(a,b)$, which can be solved for $M(a,b)$ in terms of $N$ (which is in your table), $M(0,b)=0$,  and $\frac{\partial M}{\partial a}(0,b)$ in the usual way. According to WolframAlpha, $$\frac{\partial M}{\partial a}(0,b)=\int_0^1xJ_2(bx)\sqrt{1-x^2}dx=\frac{b(2+\cos b)-3\sin b}{b^3}.$$  So if $N(a,b)$ is sufficiently nice, you might have a chance!