I have an integral (I) given by: $$I =\int _0^{1}\:d\sigma\int _0^{2\pi}\:d\phi\;\sigma \sqrt{1-\rho^2sin^2(\theta-\phi)-[\gamma\sigma\,+\,\rho\,cos(\theta-\phi)]^2}$$
Here, $\rho$ runs from 0 to 1 -$\gamma$ and $\theta$ from 0 to $2\pi$. How can the integral over $\sigma$ is analytically expressed in the form below: $$I = \frac{2\pi}{3\gamma^2}(1-\rho^2)^{1/2}\,\left(1-\frac{\rho^2}{4}\right) - A+B$$ where
$$A =\frac{1}{\gamma^2}\left[\frac{1}{3}\int _0^{2\pi}\:d\phi\;(1-\rho^2-\gamma^2 - 2 \rho\gamma\,cos\phi)^{3/2} + \frac{1}{2}\int _0^{2\pi}\:d\phi\;(\rho^2\cos\phi + \rho\gamma)\cos\phi\;(1-\rho^2-\gamma^2 - 2 \rho\gamma\,cos\phi)^{1/2}\right]$$
$$B =\frac{1}{2\gamma^2}\int _0^{2\pi}\:d\phi\;\rho\cos\phi\,(1-\rho^2\sin^2\phi)\left[\sin^{-1}\left(\frac{\rho\cos\phi}{\sqrt{1-\rho^2\sin^2\phi}}\right) - \sin^{-1}\left(\frac{\gamma + \rho\cos\phi}{\sqrt{1-\rho^2\sin^2\phi}}\right)\right]$$
I thought this integral as another form of Bushell's integral. An application of Green's theorem in polar coordinates. How can I write $I$ in terms of A and B and how can A and B be written in elliptic integral form? Some reference will be very helpful for me.