Context
This question is related to Integral $\int_{0}^{1}\frac{K\left(x\right)\ln\left(1-x^{2}\right)}{x^{2}}dx=-4\ln2$. But deals with the following generalization.
Being: $$F(\phi,k)=\int_{0}^{\phi}\frac{dx}{\sqrt{1-k^2\sin^2x}},$$ the incomplete elliptic of the first kind. Consider: $$I(\phi)=\int_{0}^{\pi/2}\frac{\cos{k}\log{(\cos{k})}F(\phi,\sin{k})}{\sin^2{k}} \tag{1}dk.$$ We have seen in the linked question: $$I(\pi/2)=-4\log{2}.\tag{2}$$ But we have also: $$I(\pi/3)=\frac{\pi}{6}-\log{(2+\sqrt{3})}\tag{3},$$
$$I(\pi/4)=\frac{\sqrt{2}\pi}{8}-\log{(1+\sqrt{2})}-\frac{\sqrt{2}\log{2}}{4},\tag{4}$$ $$I(\pi/5)=-\log{\left(\sqrt{\frac{5}{8}-\frac{\sqrt{5}}{8}}+1\right)}+\left(1-\sqrt{\frac{5}{8}-\frac{\sqrt{5}}{8}} \right)\log{(1+\sqrt{5})}+\left(\sqrt{\frac{5}{8}-\frac{\sqrt{5}}{8}} -2\right)\log{2}+\pi\sqrt{\frac{\sqrt{5}}{20}+\frac{1}{20}},\tag{5}$$ $$I(\pi/8) =\log{\left(\frac{\sqrt{2\sqrt{2}+4}-\sqrt{2}-2}{\sqrt{2}-\sqrt{2\sqrt{2}+4}}\right) -\sqrt{\frac{1}{8}-\frac{\sqrt{2}}{16}} }\log{(1+\sqrt{2})}-\sqrt{\frac{1}{32}-\frac{\sqrt{2}}{64}}\log{2}+\pi\sqrt{\frac{\sqrt{2}}{256}+\frac{1}{128}}.\tag{6}$$ Question
Does anybody know if $I(\phi)$ has series representations when $\phi \neq \pi/2$? The purpose is to obtain some series for these cases if its possible. Thanks for your cooperation!
Using
Mathematicanotations $$\int_{0}^{\phi}\frac{dx}{\sqrt{1-k^2\sin^2(x)}}=F\left(\phi \left|k^2\right.\right)$$ if$$0<\Re(\phi )\leq \frac{\pi }{2}\lor -\frac{\pi }{2}\leq \Re(\phi )<0$$
If you are concerned by small values of $\phi$, may be, you could use $$F\left(\phi \left|k^2\right.\right)=\phi+k^2\sum_{n=1}^\infty \frac{P_n(k)}{(2n+1)!}\, \phi^{2n+1}$$ where the first polynomials are $$\left( \begin{array}{cc} n & P_n(k) \\ 1 & 1 \\ 2 & 9 k^2-4 \\ 3 & 225 k^4-180 k^2+16 \\ 4 & 11025 k^6-12600 k^4+3024 k^2-64 \\ \end{array} \right)$$ and use $$\int_{0}^{\frac \pi 2}\frac{\cos(k)\log{(\cos(k))}}{\sin^2(k)} \,\,\sin^{2n}(k) \,dk=-\frac 1{2n-1 }\,\,H_{n-\frac{1}{2}}$$