Evaluate an integral using power series I'm trying to develop power series solutions for the following integrals
$$\int_\phi^\pi\frac{(1-\nu \cos\theta)\,d\theta}{(1-2\nu \cos\theta+\nu^2)^{3/2}}$$
and 
$$\int_\phi^\pi\frac{(\cos\theta-\nu)\,d\theta}{(1-2\nu \cos\theta+\nu^2)^{3/2}}$$
in terms of $\nu$, where $\phi=0$ if $0<\nu<1$ or some small value $\epsilon$ if $\nu=1$   (i.e. so that $1-\nu\cos(\theta)$ is not equal to $0$).
These integrals arise in connection with  the gravitational field of a ring, and appear in a mathpages.com article entitled "Gravity of a Torus". The author (Dr. Kevin S. Brown) gives solutions that converge rapidly for all values of $\nu<1$. Presumably these could be extended if $\nu=1$ with $\phi = \epsilon$. 
I'm unable to replicate his results.
Dr. Brown, a prolific author, protects his privacy diligently, so any help would be greatly appreciated. 
To paraphrase his paper slightly he states:
We can also express (the gravitational forces) as power series. Expanding the integral in powers of $\nu$ and dividing out the geometric series (?) gives (for the first integral)
$$ I=\frac {\kappa}{(1-\nu^2)}\left(1-{\nu^2\over4}-{3\nu^4\over64}-{5\nu^6\over256}-{175\nu^8\over16384} +...\right)$$
$$=\frac {\kappa}{(1-\nu^2)}\sum_{k=0}^\infty \frac{1}{(2k-1)4^{2k}}\left(\frac{(2k)!}{k!k!}\right)^2\nu^{2k}$$
where $\kappa$ is a constant involving $\pi$ etc (it's getting rather late). How is he getting this result? Please excuse my typos and formatting - new to Latex.
 A: As David H commented, I suppose that you are considering $$f(\nu)=\int_\phi^\pi\frac{(1-\nu \cos\theta)\,d\theta}{(1-2\nu \cos\theta+\nu^2)^{3/2}}$$ and not $f(\theta)$ which is no meaning since you integrate over $\theta$.
As you know, the antiderivative involves elliptic functions. Using a CAS, I obtained $$I=\int_0^\pi\frac{(1-\nu \cos\theta)\,d\theta}{(1-2\nu \cos\theta+\nu^2)^{3/2}}=\frac{(\nu -1) K\left(\frac{4 \nu }{(\nu +1)^2}\right)-(\nu +1) E\left(\frac{4 \nu
   }{(\nu +1)^2}\right)}{(\nu^2 -1) }$$ provided $$\left(\sqrt{1-\Re(\nu )^2}>\Im(\nu )\land \Im(\nu )+\sqrt{1-\Re(\nu
   )^2}>0\right)\lor \sqrt{1-\Re(\nu )^2}<\Im(\nu )\lor \Im(\nu )+\sqrt{1-\Re(\nu
   )^2}<0\lor \Re(\nu )>1\lor \Re(\nu )+1<0 $$ The corresponding Taylor series built around $\nu=0$ is $$f(\nu)=\pi\left(1+\frac{3 }{4}\nu ^2+\frac{45}{64}\nu ^4+\frac{175 }{256}\nu ^6+\frac{11025 
   }{16384}\nu ^8+\frac{43659 }{65536}\nu ^{10}+\frac{693693 }{1048576}\nu ^{12}+O\left(\nu ^{14}\right) \right)$$ which is a quite good approximation for $0 \leq \nu \leq 0.8$. For sure, adding more terms makes the fit better and better.
Concerning the coefficients, the numerators and denominators correspond respectively to sequences $A069955$ and $A056982$ of $OEIS$.
Edit
Taking into account your last edit, what has been done is  a Taylor series expansion of $(1-\nu^2)f(\nu)$ and not of $f(\nu)$ as I did in my answer. Doing the same, what I obtain for this expression is $$\pi\left(1-\frac{1}{4}\nu ^2-\frac{3 }{64}\nu ^4-\frac{5 }{256}\nu ^6-\frac{175 
   }{16384}\nu ^8-\frac{441 }{65536}\nu ^{10}-\frac{4851 }{1048576}\nu ^{12} +O\left(\nu ^{14}\right) \right)$$ which is exactly what Dr. Brown produced.
This finally write
$$f(\nu)=-\frac {\pi}{(1-\nu^2)}\sum_{k=0}^\infty \frac{1}{(2k-1)4^{2k}}\left(\frac{(2k)!}{k!\,k!}\right)^2\nu^{2k}$$ which means that $\color{red}{\kappa=-\pi}$.
From a computational point of view, noting $$a_k=\frac{1}{(2k-1)4^{2k}}\left(\frac{(2k)!}{k!\,k!}\right)^2$$ you could notice the simple recurrence relation $$a_{k+1}=\frac{4 k^2-1}{4 (k+1)^2}a_k$$ where $a_0=-1$. This would make any coding quite unexpensive.
