During my research I came upon the following definite integral involving Bessel functions, which I have no idea how to solve.
$\int_{-1}^{1}{dx I_{0}(a\cdot sin(bx) \cdot \sqrt{1-x^2}})$
Where $a$ is complex and $b$ is real.
The first observation is that the integral is even, therefore we can multiply by two while changing the limit to be between 0 to 1.
From here I used a series expansion for the Bessel function in which I also used the bionomial theorem, allowing me to use a known integral for $\int{x^n\cdot sin^m(x) dx} $, this causes another series expansion which is combined with integral which is solvabale. The problem is ,I am left with hypergeometric function which I have been able to write explicitly summed with a term which holds 6 sums, which I don’t seem to be able to solve.
where in the image below you can see what I got. $a$ and $b$ are switched for $\xi$ and $\rho$.
$a$ and 
If you use the series expansion of the Bessel function
$$I_0\left(a \sin (b x) \sqrt{1-x^2} \right)=\sum_{n=0}^\infty\frac{ a^{2n}}{4^n \,(n!)^2 }\,\Big[\left(1-x^2\right) \sin ^2(b x)\Big]^n$$ and the problem reduces to the computation of $$J_n=\int \Big[\left(1-x^2\right) \sin ^2(b x)\Big]^n\, dx$$ which does not make problem except finding its general formula $$12b ^3\,J_1=8 b^3+6 b \cos (2 b)-3 \sin (2 b)$$ $$2560b^5\,J_2=1024 b^5+640 \left(4 b^2-3\right) \sin (2 b)+5 \left(3-16 b^2\right) \sin (4 b)+$$ $$3840 b \cos (2 b)-60 b \cos (4 b)$$
I have the feeling that, whatever you could do with the integrand in $J_n$, you will face a lot of difficulties.
For example, suppose that $b$ is small $$\left(1-x^2\right) \sin ^2(b x)=b^2 +\sum_{k=1}^\infty (-1)^{k+1}\frac{2^{2 k-3} b^{2 k-2} \left(4 b^2+2 k (2 k-1)\right)}{\Gamma (2 k+1)}\, x^{2k}$$ but now we need to handle its $n^{\text{th}}$ power and here will come some nasty hypergeomtric functions but, may be (?), it could be simpler than all the summations you have.