Integral involving hypergeometric function I've worked out the projection of a spherically symmetric power law volume density profile $\rho(r)=br^a$, i.e. its surface density $\sigma(R)$, and am now trying to integrate this in a series of annular bins to get the total mass in each, i.e. $\int_A^B\sigma(R)R\,{\rm d}R$. (Actually the profile is a continuous series of power laws, but that's more of a footnote.)
I've worked just about everything out, but am stuck on an integral of the form:
$$I(r,R,a)=\int\left(\left(\frac{r}{R}\right)^2-1\right)^{\frac{1}{2}}R^{a+2} {}_2{\rm F}_1\left(\frac{1}{2},-\frac{a}{2};\frac{3}{2};-\left(\left(\frac{r}{R}\right)^2-1\right)\right){\rm d}R$$
${}_2{\rm F}_1$ is the hypergeometric function, as defined here. $a$ and $r$ are constants, in this context.
Given that the solution of this integral will be used to calculate the elements of a big-ish matrix and I have some iteration in mind, integrating numerically is... inconvenient, but I'm stumped on finding an analytic solution. Is there any hope here, or should I start thinking about efficient numerical solutions?
 A: In the following, we assume $R>r$.
\begin{equation}
 I(r,R,a)=\int\left(\frac{r^2}{R^2}-1\right)^{\frac{1}{2}}R^{a+2} {}_2{\rm F}_1\left(\frac{1}{2},-\frac{a}{2};\frac{3}{2};1-\frac{r^2}{R^2}\right){\rm d}R
\end{equation} 
changing $u=1-r^2/R^2$, we obtain
\begin{equation}
 I(r,R,a)=i\frac{r^{a+3}}{2}\int u^{\frac12}\left( 1-u \right)^{-\frac{a}{2}-\frac{5}{2}}{}_2{\rm F}_1\left(\frac{1}{2},-\frac{a}{2};\frac{3}{2};u\right){\rm d}u
\end{equation} 
where the principal determination for the square root was taken. Using the indefinite integration formula:
\begin{equation}
 \int  z^{c-1 }(1 - z)^{b - c-1}{}_2{\rm F}_1\left( a,b;c;z \right)= z^c(1 - z)^{b - c}  \frac{\Gamma(c)}{\Gamma(c+1)} {}_2{\rm F}_1\left( 1+a,b;1+c;z \right)
\end{equation} 
with $a=1/2,b=-a/2,c=3/2$, we find
\begin{equation}
  I(r,R,a)=-\frac{1}{3}R^{a+3}\left( \frac{r^2}{R^2}-1 \right)^{3/2} {}_2{\rm F}_1\left(\frac{3}{2},-\frac{a}{2};\frac{5}{2};1-\frac{r^2}{R^2}\right)
\end{equation} 
which seems to be numerically correct.
