I want to deal with the integral of the form $$ F_{\nu} = \frac{1}{\pi}\int_0^{\pi}\frac{1}{\sqrt{\frac{a^2+b^2}{2}+\frac{\nu}{2}\sqrt{(a^2-b^2)^2+16ab\cos^2k}}}\;dk \\ \;\;\;\;\;\; = \frac{2}{\pi}\int_0^{\pi/2}\frac{1}{\sqrt{\frac{a^2+b^2}{2}+\frac{\nu}{2}\sqrt{(a^2-b^2)^2+16ab\cos^2k}}}\;dk $$ where $a$ and $b$ are two real integrals, $\nu=\pm1$. Specically, if $a=b$, the integral above is just $$ F_{\nu} = \frac{1}{\pi}\int_0^{\pi}\frac{1}{\sqrt{a^2+2\nu a\vert\cos k\vert}}\;dk \\ \;\;\;\;\;\; = \frac{2}{\pi}\int_0^{\pi/2}\frac{1}{\sqrt{a^2+2\nu a\cos k}}\;dk $$ and the summation of $F_++F_-$ relates to the elliptic integral of the first kind. But for general case where $a\neq b$, how to handle the integral? If a close-form exist? Any comment or solution will be welcomed.
EDIT (8 Dec. 2016):
In order to avoid the confusions, I shall introduce the equivalent integral (up to a constant coefficient) $$ F_{\nu} = \frac{1}{\pi}\int_0^{\pi}\frac{1}{\sqrt{1+\nu\sqrt{q^2+d^2\cos^2k}}}\;dk $$ where $q\in(-1,1)$, $d\in[0,1)$, and $\nu=\pm1$. It is alse assumed that $q^2+d^2<1$, so do not warry about the complex number.