Integrate $\int_0^x \frac{d\theta}{(1+\epsilon\cos{\theta})^2}$ for $\epsilon>0$ If I can solve this integral below in analytical form, then I will be able to provide the time dependant orbit for a two-body problem, 
$$\int_0^x \frac{d\theta}{(1+\epsilon\cos{\theta})^2}$$
for $\epsilon>0$.
 A: Observe that,
$$\left(\frac{\sin\theta}{1+\epsilon\cos{\theta}} \right)'=
\frac{\cos\theta+\epsilon}{(1+\epsilon\cos{\theta})^2}
= \frac1\epsilon\left[\frac{1}{1+\epsilon\cos{\theta}}+\frac{\epsilon^2-1}{(1+\epsilon\cos{\theta})^2} \right]$$
and decompose the integral
$$I=\int_0^x \frac{d\theta}{(1+\epsilon\cos{\theta})^2}$$
$$=\frac{1}{\epsilon^2-1}\left[\int_0^x d\left(\frac{\epsilon\sin\theta}{1+\epsilon\cos{\theta}}\right)-\int_0^x \frac{d\theta}{1+\epsilon\cos{\theta}}\right]$$
$$=\frac{\epsilon}{\epsilon^2-1}\frac{\sin x}{1+\epsilon\cos{x}}-\frac{1}{\epsilon^2-1}\int_0^x \frac{d\theta}{1+\epsilon\cos{\theta}}$$
The second integral is better known
$$\int_0^x \frac{d\theta}{1+\epsilon\cos{\theta}}
=\int_0^x \frac{ 2d(\tan\frac{\theta}{2}) }{(1-\epsilon)\tan^2\frac{\theta}{2}+(1+\epsilon)} = \frac{2}{\sqrt{1-\epsilon^2}} \tan^{-1} \left(\sqrt{\frac{1-\epsilon}{1+\epsilon}} \tan\frac x2 \right) $$
Thus, 
$$I=-\frac{\epsilon}{1-\epsilon^2}\frac{\sin x}{1+\epsilon\cos{x}}+\frac{2}{{(1-\epsilon^2})^{3/2}} \tan^{-1} \left(\sqrt{\frac{1-\epsilon}{1+\epsilon}} \tan\frac x2 \right) $$
