I hope it won't be categorized as a trivial question, I solved this integral and arrived at the following form: $${\int_{0}^{\frac{\pi}{2}}\ln\left(1+\alpha^n\sin(x)^{2n}\right)\mathrm{d}x=\frac{n\pi}{2}\ln\left(\frac{\alpha}{4}\right)+\pi\sum_{k=1}^{n}\text{acosh}\left(\sqrt{\frac{1}{2\alpha}\left(\alpha+\sqrt{\alpha^2-2\cos\left(\frac{2k-1}{n}\pi\right)\alpha+1}+1\right)}\right)}$$ But it seems a bit long and too "complicated" (nested roots and hyperbolic inverse functions)
Would anyone be able to manipulate this solution to arrive at a more compact form?
I "played" around with it a bit and came to this form:
$${\int_{0}^{\frac{\pi}{2}}\ln\left(1+\alpha^n\sin(x)^{2n}\right)\mathrm{d}x=\pi\sum_{k=1}^{n}\ln\left(\frac{\sqrt{z_k+1+\alpha}+\sqrt{z_k+1-\alpha}}{2\sqrt{2}}\right)\qquad\text{where }z_k:=\sqrt{\alpha^{2}-2\cos\left(\frac{2k-1}{n}\pi\right)\alpha+1}}$$
For $\alpha\in\mathbb{R}$
Which however is just as long (technically has 2 nested radicals).
(Mainly I would like to remove those nested radicals, then the type of functions you use is indifferent, as long as it is a little more compact)
In terms of generalized hypergeometric functions $$I_n=\displaystyle\int_0^{\frac{\pi}{2}}\log\big(1+\alpha^n\sin(x)^{2n}\big)\,dx$$ for $n>1$ $$\large\color{blue}{I_n=\pi\, \frac{a_n}{b_n}\,\alpha^n\,J_n}$$ Coefficients $a_n$ $$\{16,32,256,512,2048,4096,65536\}$$
form sequence $A101926$ (with an offset of $2$).
Coefficients $b_n$ $$\{3,5,35,63,231,429,6435\}$$ form sequence $A001790$ (with an offset of $2$).
Concerning the $J_n$, the pattern is very clear $$J_2=\, _4F_3\left(1,1,\frac{5}{4},\frac{7}{4};\frac{3}{2},2,2;-\alpha ^2\right)$$ $$J_3=\, _5F_4\left(1,1,\frac{7}{6},\frac{9}{6},\frac{11}{6};\frac{4}{3},\frac{5}{3},2,2;-\alpha ^3\right)$$ $$J_4=\, _6F_5\left(1,1,\frac{9}{8},\frac{11}{8},\frac{13}{8},\frac{15}{8};\frac{5}{4}, \frac{6}{4},\frac{7}{4},2,2;-\alpha ^4\right)$$ $$J_5=\, _7F_6\left(1,1,\frac{11}{10},\frac{13}{10},\frac{15}{10},\frac{17}{10},\frac{19}{ 10};\frac{6}{5},\frac{7}{5},\frac{8}{5},\frac{9}{5},2,2;-\alpha ^5\right)$$ $$J_6=\, _8F_7\left(1,1,\frac{13}{12},\frac{15}{12},\frac{17}{12},\frac{19}{12},\frac{21}{12 },\frac{23}{12};\frac{7}{6},\frac{8}{6},\frac{9}{6},\frac{10}{6},\frac{11}{6},2,2;-\alpha ^6\right)$$