closed form for $\int_0^1x^{a+1}(1-x^2)^bJ_a(cx)dx$ $$\int_0^1x^{a+1}(1-x^2)^bJ_a(cx)dx$$
my friend posted the integral on the fb and I tried to solve it but I faild because I have little information about bessel function 
so could some one help ? 
 A: Using the series expansion for the Bessel function the integral is as follows:
\begin{align}
I &= \int_{0}^{1} x^{a+1} (1-x^{2})^{b} J_{a}(cx) \ dx \\
&= \sum_{n=0}^{\infty} \frac{(-1)^{n} (c/2)^{2n+a}}{n!(n+a)!} \ \int_{0}^{1} x^{2n+2a+1} (1-x^{2})^{b} \ dx.
\end{align}
Making the substitution $x = \sqrt{t}$ the integral now becomes
\begin{align}
I &= \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n} (c/2)^{2n+a}}{n!(n+a)!} \ \int_{0}^{1} t^{n+a} (1-t)^{b} \ dt \\
&= \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n} (c/2)^{2n+a}}{n!(n+a)!} \ B(n+a+1, b+1) \\
&= \frac{b!}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n} (c/2)^{2n+a}}{n!(n+a+b+1)!} \\
&= \frac{b!}{2} \left(\frac{2}{c}\right)^{b+1} \ J_{a+b+1}(c).
\end{align}
Hence
\begin{align}
\int_{0}^{1} x^{a+1} (1-x^{2})^{b} J_{a}(cx) \ dx = \frac{b!}{2} \left(\frac{2}{c}\right)^{b+1} \ J_{a+b+1}(c).
\end{align}
A: One of the ways is to write the bessel function as infinite series
\begin{equation}
J_a\left(cx\right) = \sum_{m=0}^{\infty}\frac{\left(-1\right)^m}{m!\Gamma\left(m+a+1\right)}\left(\frac{cx}{2}\right)^{2m+a}
\end{equation}
 and then evaluate the integral term by term using following formula:
\begin{equation}
\int_0^r x^m\left(r^n-x^n\right)^pdx = \frac{r^{m+1+np}\Gamma\left[(m+1)/n\right]\Gamma\left[p+1\right]}{n\Gamma\left[(m+1)/n+p+1\right]}
\end{equation}
