Can we evaluate $\int \frac{1}{\sin ^ {2n+1} x+\cos ^ {2n+1} x} d x?$ 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 $n\geq 3.$

Here is a generalized approach, which decomposes the integrand with $a_k=\frac{\pi k}{2n+1}$
\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 decomposed terms on the RHS can be integrated individually. For example
\begin{align}
&\int \frac1{\sin^{7} x+\cos^{7} x}dx\\
 =&\ \frac{8}7 \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.
