<s>[INST] calculation of $\int\frac{1}{\sin^3 x+\cos^3 x}dx$ and $\int\frac{1}{\sin^5 x+\cos^5x}dx$ 
Solve the following indefinite integrals:
$$
\begin{align}
&(1)\;\;\int\frac{1}{\sin^3 x+\cos^3 x}dx\\
&(2)\;\;\int\frac{1}{\sin^5 x+\cos^5 x}dx
\end{align}
$$

My Attempt for $(1)$:
$$
\begin{align}
I &= \int\frac{1}{\sin^3 x+\cos ^3 x}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(\sin^2 x+\cos ^2 x-\sin x \cos x\right)}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(1-\sin x\cos x\right)}\;dx\\
&= \frac{1}{3}\int \left(\frac{2}{\left(\sin x+\cos x\right)}+\frac{\left(\sin x+\cos x \right)}{\left(1-\sin x\cos x\right)}\right)\;dx\\
&= \frac{2}{3}\int\frac{1}{\sin x+\cos x}\;dx + \frac{1}{3}\int\frac{\sin x+\cos x}{1-\sin x\cos x}\;dx
\end{align}
$$
Using the identities
$$
\sin x = \frac{2\tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}},\;\cos x = \frac{1-\tan ^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}
$$
we can transform the integral to
$$I = \frac{1}{3}\int\frac{\left(\tan \frac{x}{2}\right)^{'}}{1-\tan^2 \frac{x}{2}+2\tan \frac{x}{2}}\;dx+\frac{2}{3}\int\frac{\left(\sin x- \cos x\right)^{'}}{1+(\sin x-\cos x)^2}\;dx
$$
The integral is easy to calculate from here.
My Attempt for $(2)$:
$$
\begin{align}
J &= \int\frac{1}{\sin^5 x+\cos ^5 x}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(\sin^4 x -\sin^3 x\cos x+\sin^2 x\cos^2 x-\sin x\cos^3 x+\cos^4 x\right)}\;dx\\
&= \int\frac{1}{(\sin x+\cos x)(1-2\sin^2 x\cos^2 x-\sin x\cos x+\sin^2 x\cos^2 x)}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(1-\sin x\cos x-\left(\sin x\cos x\right)^2\right)}\;dx
\end{align}
$$
How can I solve $(2)$ from this point?
 [/INST] By factorisation of
$$
\begin{aligned}
\sin ^5 x+\cos ^5 x= & \left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^3 x+\cos ^3 x\right) -\sin ^2 x \cos ^2 x(\sin x+\cos x) \\
= & (\sin x+\cos x)\left(1-\sin x \cos x-\sin ^2 x \cos ^2 x\right),
\end{aligned}
$$
we have $$
I=\int \frac{1}{\sin ^5 x+\cos ^5 x} d x=\int \frac{\sin x+\cos x}{(\sin x+\cos x)^2\left(1-\sin x \cos x-\sin ^2x \cos ^2 x\right)} d x
$$
Noting that letting $t=\sin x-\cos x$, then $dt=(\cos x+\sin x)dx$ and $\sin x \cos x=\frac{1-t^2}{2}$ yields
$$
\begin{aligned}I&=4  \int \frac{d t}{\left(t^2-2\right)\left(t^4-4 t^2-1\right)}\\&=
\frac{4}{5} \int\left(\frac{-1}{t^2-2}+\frac{1}{t^2+\sqrt{5}-2}+\frac{1}{t^2-\sqrt{5}-2}\right) d t \\&=\frac{4}{5}\left[-\frac{1}{\sqrt{2}} \tanh ^{-1}\left(\frac{t}{\sqrt{2}}\right)+\frac{1}{\sqrt{\sqrt 5-2}} \tan ^{-1}\left(\frac{t}{\sqrt{\sqrt{5}-2}}\right)-\frac{1}{\sqrt{\sqrt{5}+2}} \tanh ^{-1}\left(\frac{t}{\sqrt{\sqrt{5}+2}}\right)\right]+C\\ &\boxed{I=\frac{4}{5}\left[-\frac{1}{\sqrt{2}} \tanh ^{-1}\left(\frac{\sin x-\cos x}{\sqrt{2}}\right)+\frac{1}{\sqrt{\sqrt 5-2}} \tan ^{-1}\left(\frac{\sin x-\cos x}{\sqrt{\sqrt{5}-2}}\right)\qquad\qquad  -\frac{1}{\sqrt{\sqrt{5}+2}} \tanh ^{-1}\left(\frac{\sin x-\cos x}{\sqrt{\sqrt{5}+2}}\right)\right]+C}\end{aligned}$$
</s>