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- #1

$$

\int_0^{\nu}\frac{d\theta}{(1 + e\cos\theta)^2}

$$

where $0<e<1$.

I tried using Residue Theory but that was messy and didn't come up as needed.

The highlights of that were:

$\cos z = \frac{z + \frac{1}{z}}{2}$

Denominator became $z + 2ze^2 + 2z^2e^2 + 2e^2 + 4z^2 + 4e$ and that factor to

$$

\left(z + \frac{1+2e^2+\sqrt{1-4e^2[15e(16+3e)]}}{4e(2+e)}\right)\left(z + \frac{1+2e^2-\sqrt{1-4e^2[15e(16+3e)]}}{4e(2+e)}\right)

$$

It is highly probably there were some errors with that mess. I wasn't sure if I could do:

$$

\cos 2z = \frac{2z + \frac{1}{2z}}{2}

$$

but I did.

Is there another way to integrate this? Any method real or complex is fine.