After investigating the integral
$$\boxed{\quad \int \frac{1}{\sin ^5 x+\cos ^5x} d x \\=\frac{4}{5}\left[-\frac{1}{\sqrt{2}} \tanh ^{-1}\left(\frac{\sin x-\cos x}{\sqrt{2}}\right)+\frac{1}{\sqrt{\sqrt 5-2}} \tan ^{-1}\left(\frac{\sin x-\cos x}{\sqrt{\sqrt{5}-2}}\right)\\ \qquad\ -\frac{1}{\sqrt{\sqrt{5}+2}} \tanh ^{-1}\left(\frac{\sin x-\cos x}{\sqrt{\sqrt{5}+2}}\right)\right]+C}$$
in my post, I try to evaluate $$\int \frac{1}{\sin ^ {7} x+\cos ^ {7} x} d x$$ using the factorisation
\begin{aligned} \sin ^{7} x+\cos ^7 x&=\left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^5 x+\cos ^5 x\right)-\sin ^2 x \cos ^2 x\left(\sin ^3 x+\cos ^3 x\right) \\ &=\left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^3 x+\cos ^3 x\right) -\sin ^2 x \cos ^2 x(\sin x+\cos x) \\ &\quad -\sin ^2 x \cos ^2 x\left(\sin ^3 x+\cos ^3 x\right) \\ &=(\sin x+\cos x)\left(1-\sin x \cos x-2 \sin ^2 x \cos ^2 x+\sin^3x\cos^3x\right) \end{aligned}
Let $t=\sin x-\cos x$, then $d t=(\cos x+\sin x) d x$ and $\sin x\cos x= \frac{1-t^2}{2}$ and yields $$ \begin{aligned} I =\int \frac{8}{\left(t^2-2\right)\left(t^6+t^4-9 t^2-1\right)} d t \end{aligned} $$ where I was stuck in the last integrand with power $6$. Your help and comments are highly appreciated.
My Question:
Is it difficult go further with $$\int \frac{1}{\sin ^ {2n+1} x+\cos ^ {2n+1} x} d x?$$ where $n\geq 3.$
With $a_k=\frac{\pi k}{2n+1}$, decompose the integrand as follows \begin{align} &\frac{2n+1}{\sin^{2n+1}x+\cos^{2n+1}x}\\ =& \ \frac{2^n}{\sin x+\cos x}+2^{n+1}\sum_{k=1}^n(-1)^{k}\frac{\cos^{n}2a_k}{\sec a_k }\frac{\sin x+\cos x}{1+\cos 2a_k\sin2x} \\ \end{align} Then, the RHS can be integrated piece-wise. For example
\begin{align} &\frac78\int \frac1{\sin^{7} x+\cos^{7} x}dx\\ =&\int \frac{1}{2-t^2} - \frac{2\cos\frac\pi7 \cos^2\frac{2\pi}7}{{\sec\frac{2\pi}7+1-t^2}} +\frac{2\cos\frac{2\pi}7 \cos^2\frac{4\pi}7}{{\sec\frac{4\pi}7+1-t^2}} -\frac{2\cos\frac{3\pi}7 \cos^2\frac{6\pi}7}{{\sec\frac{6\pi}7+1-t^2}}\ dt \end{align} where the substitution $t=\sin x-\cos x$ is made.