An Elliptic Integral - What's the Simplest Answer?

Question

I have

$$ \int_{0}^{2\pi}d\theta\left(R^{2}-\epsilon^{2}\right)\sqrt{R^{2}-\epsilon^{2}\sin^{2}\left(\theta\right)} $$

which Mathematica thinks is

$$ 2R\left(R^{2}-\epsilon^{2}\right)\left(E\left(\frac{\epsilon}{R}\right)+\sqrt{\frac{R^{2}-\epsilon^{2}}{R^{2}}}E\left(\frac{\epsilon}{\sqrt{\epsilon^{2}-R^{2}}}\right)\right) $$

where $E(k)$ is the complete elliptic integral of the second kind, given by

$$ E\left(k\right)=\int_{0}^{\pi/2}d\theta\sqrt{1-k^{2}\sin^{2}\left(\theta\right)} $$

though I think if you do this 'by hand' there is a simpler answer Mathematica can't find due to the many relations between the elliptic integrals. Can anyone have some sort of attempt at finding it?

Answer 1

I tried first the antiderivative $$\int\left(R^{2}-\epsilon^{2}\right)\sqrt{R^{2}-\epsilon^{2}\sin^{2}\left(\theta\right)}d\theta=\frac{R^2 \left(R^2-\epsilon ^2\right) \sqrt{\frac{2 R^2+\epsilon ^2 \cos (2 \theta )-\epsilon ^2}{R^2}} E\left(\theta \left|\frac{\epsilon ^2}{R^2}\right.\right)}{\sqrt{2 R^2+\epsilon ^2 \cos (2 \theta )-\epsilon ^2}}$$ So $$\int_0^{2\pi}\left(R^{2}-\epsilon^{2}\right)\sqrt{R^{2}-\epsilon^{2}\sin^{2}\left(\theta\right)}d\theta=4 R (R^2-\epsilon^2 ) E\left(\frac{\epsilon ^2}{R^2}\right)$$

Hoping that this helps.

Answer 2

Using Maple I am obtaining directly

$$4\,R \left( {R}^{2}-{\epsilon}^{2} \right) {\it EllipticE} \left( { \frac {\epsilon}{R}} \right)$$