I am stuck evaluating an integral that appears in a simplified theory of nuclear binding energy. The nucleus is modelled as a sphere of radius $R$ with a continuous charge distribution, and the integral gives the force experienced by a new proton of radius $r_0$. Here's the geometry:
geometry of nucleus-proton Coulomb interaction
The force along the line joining the proton-nucleus centers is $$ F = \frac{3e^2}{32\pi^2\epsilon_0r_0^3}\int \frac{\cos\alpha}{d^2}d\boldsymbol{r} $$
I approached this by using the cosine rule for $\cos\alpha = (r^2 - R'^2 - d^2)/(2R'd)$ and $d=r^2+R'^2 - 2rR'\cos\theta$ but I can't do the resulting integral, which i think is $$ 2\pi \int_0^\pi \int_0^R \frac{(r\cos\theta-R')r^2\sin\theta}{(r^2 + R'^2 - 2rR'\cos\theta)^{5/2}}\,\mathrm{d}r\mathrm{d}\theta $$ This integral leads me down a mess of algebra in $r$ if I try to substitute $u=\cos\theta$. Can anyone provide some hints or correct my working?
EDIT: I tried $u^2 = r^2 + R'^2 - 2rR'\cos\theta$ and got the $\theta$ integral out to: $$ \frac{r}{2R'^2}\left[ \frac{1}{u} - \frac{r^2-R'^2}{3u^3} \right]_{R'-r}^{R'+r} $$ but I'm really struggling to put this function of $r$ into a tidy form I can integrate.