<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] $\newcommand{\+}{^{\dagger}}%
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$\large\tt\mbox{Just a hint:}$
Write
$$
\int{\cos\pars{x}\,\dd x \over \cos\pars{x}\sin^{3}\pars{x} + \cos^{4}\pars{x}}
=
\int{\dd z \over \root{1 - z^{2}}z^{3} + \bracks{1 - z^{2}}^{2}}
\quad\mbox{with}\quad z \equiv \sin\pars{x}
$$
Use an Euler substitution: $\root{1 - z^{2}} \equiv t + \ic z$ which yields
$1 - z^{2} = t^{2} + 2t\ic z - z^{2}$ such that
$\ds{z = {1 - t^{2} \over 2t\ic}}$:
\begin{align}
\root{1 - z^{2}}&=t + {1 - t^{2} \over 2t} = {1 + t^{2} \over 2t}
\\[3mm]
\dd z&= {\pars{-2t}\pars{2t\ic} - \pars{2\ic}\pars{1 - t^{2}} \over -4t^{2}}\,\dd t
=
\ic\,{t^{2} + 1 \over 2t^{2}}\,\dd t
\end{align}
\begin{align}
\int&=\int{1 \over
\bracks{\pars{1 + t^{2}}/2t}\bracks{\pars{1 - t^{2}}/2t}^{3}\pars{-1/\ic}
+
\bracks{\pars{1 + t^{2}}/2t}^{4}}
\,\ic\,{t^{2} + 1 \over 2t^{2}}\,\dd t
\\[3mm]&=-8\int{t^{2} \over -\pars{1 - t^{2}}^{3} + \ic\pars{1 + t^{2}}^{3}}\,\dd t
\end{align}
</s>