Let us consider the following function $f(a,k)$ in the interval $a,k\in (0,1]$ :
$$f(a,k)=\frac{2 \sqrt{1-a^2} \sqrt{a^2-k^2}}{\sqrt{a^2}}\Pi\left(a^2,k^2\right)$$
where $\Pi\left(a^2,k^2\right)$ is the complete elliptic integral of the 3rd kind:
$$\Pi\left(a^2,k^2\right)=\int_0^{\frac{\pi }{2}} \frac{1}{\left(1-a^2\text{sin}^2(x)\right)\sqrt{1-k^2\text{sin}^2(x)}} \, dx$$
Since the integral is symmetric in $(x \leftrightarrow -x)$ we can spread it from $-\pi/2$ to $\pi/2$ and divide the result by two, so that we have:
$$ f(a,k)=\frac{\sqrt{1-a^2} \sqrt{a^2-k^2}}{\sqrt{a^2}}\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}} \frac{ dx}{\left(1+\frac{a^2}{4}\left(e^{i 2 x}-2+e^{-i 2 x}\right)\right)\sqrt{1+\frac{k^2}{4}\left(e^{i 2 x}-2+e^{-i 2 x}\right)}} $$
Considering that $e^{i 2 x}$ goes through one closed circle when $x$ goes from $-\pi/2$ to $\pi/2$, we can make the substitution:
$$ e^{i 2 x}=z~\text{,}~~~~dx =\frac{1}{2 i z} dz $$
and obtain a closed contour integral in the complex plane:
$$ f(a,k)=\oint _{|z|=1}\frac{\sqrt{1-a^2} \sqrt{a^2-k^2}\sqrt{z}}{2i\sqrt{a^2}\left(z+\frac{a^2}{4}\left(z^2-2z+1\right)\right)\sqrt{z+\frac{k^2}{4}\left(z^2-2z+1\right)}} dz $$
Now, since the contour is closed, the integral should only be different from zero if a singularity with a non-vanishing residuum is enclosed by the contour. The residuum associated with the vanishing of the square root term in the denominator is zero. The other terms in the denominator $\left(z+\frac{a^2}{4}\left(z^2-2z+1\right)\right)$ vanish for:
$$ z_\pm = \frac{a^2-2 \left(1 \pm \sqrt{1-a^2}\right)}{a^2} $$
and the corresponding residuum for the integral is $\text{Res}(z_\pm)=\pm \frac{i}{2}$. However, it turns out that $z_+ \leq -1$ with $z_+ = -1$ only for $a=1$. In the same way we have $0 > z_- \geq-1$ with $z_-=-1$ only for $a=1$. Therefore, at regular points on the interval in question only the residue at $z_-$ contributes, so that the integral equals to:
$$f(a,k)=2\pi i \left(-\frac{i}{2} \right)=\pi$$
Now, obviously this result is not entirely correct, since the function $f(a,k)$ is not even constant over the region when plotted e.g. in Mathematica. I would like to know what kind of mistakes I did in the computation, and how I should deal with residual contributions to the elliptic integrals in general? Also, the point $(a=1,k=1)$ seems to be very special, since there $z_+=z_-=-1$ which is $on$ the integration contour (but $z_-$ approaches this spot from within the contour and $z_+$ from outside of the contour). It would be nice to know how to treat this special point.