integrating incomplete elliptic integral of first kind I am trying to integrate the integral $\displaystyle \int \dfrac{1}{\sqrt{1 - \beta \cos(\theta)}}d\theta$, where $0 \leq \beta \leq 1$.  
I believe that this is the incomplete elliptic integral of the first kind.  I have found several books on special functions that extensively covered the complete elliptic integral, but not so much for the incomplete elliptic integral.
I did recently stumble on the following formula online:
$\displaystyle \int \dfrac{1}{\sqrt{1 - \beta \cos(\theta)}}d\theta = \dfrac{2\sqrt{\dfrac{\beta \cos(\theta) -1}{\beta-1}}F\bigg(\dfrac{\theta}{2}, \dfrac{2\beta}{\beta - 1}\bigg)}{\sqrt{1 - \beta \cos(\theta)}}$
where $F\bigg(\dfrac{\theta}{2}, \dfrac{2\beta}{\beta - 1}\bigg)$ is the incomplete elliptic integral of the first kind.
My questions are:
(1) Is this formula correct, and where can I find more resources on this to further verify my integral?  
(2) When $\beta = \dfrac{1}{2}$ then $F\bigg(\dfrac{\theta}{2}, \dfrac{2\beta}{\beta -1}\bigg)$, then the $\dfrac{2\beta}{\beta -1}$ is a negative value.  Is that possible?
Any suggestion is much appreciated.
 A: Using
$$\cos\theta=1-2\sin^2\frac{\theta}{2}$$
we have
$$\int\frac1{\sqrt{1 - \beta \cos(\theta)}}\mathrm d\theta=\frac1{\sqrt{1-\beta}}\int\frac1{\sqrt{1 + \frac{2\beta}{1-\beta} \sin^2\left(\frac{\theta}{2}\right)}}\mathrm d\theta$$
from which the form of the incomplete elliptic integral of the first kind should already be apparent:
$$\frac1{\sqrt{1-\beta}}\int\frac1{\sqrt{1 + \frac{2\beta}{1-\beta} \sin^2\left(\frac{\theta}{2}\right)}}\mathrm d\theta=\frac2{\sqrt{1-\beta}}F\left(\frac{\theta}{2}\middle|\frac{2\beta}{\beta-1}\right)$$
(Note my use of the parameter instead of the modulus.)
Since you mention that $0\le\beta\le1$, we're not quite out of the woods yet, because in most practical computing environments, the modulus or parameter should be within $(0,1)$. To that end, apply the imaginary-modulus transformation to finally yield
$$\frac2{\sqrt{1-\beta}}F\left(\frac{\theta}{2}\middle|\frac{2\beta}{\beta-1}\right)=\frac2{\sqrt{1+\beta}}F\left(\arcsin\left(\sqrt{\frac{1+\beta}{1-\beta\cos\theta}}\sin\frac{\theta}{2}\right)\middle|\frac{2\beta}{1+\beta}\right)$$
