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?
 A: Given $$\displaystyle \int\frac{1}{\sin^5 x+\cos^5 x}dx$$
First we will simplify $$\sin^5 x+\cos^5 x = \left(\sin^2 x+\cos^2 x\right)\cdot \left(\sin^3 x+\cos^3 x\right) - \sin ^2x\cdot \cos ^2x\left(\sin x+\cos x\right)$$
$$\displaystyle  \sin^5 x+\cos^5 x= (\sin x+\cos x)\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)$$
So Integral is $$\displaystyle \int\frac{1}{\sin^5 x+\cos^5 x}dx $$
$$\displaystyle = \int\frac{1}{(\sin x+\cos x)\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)}dx$$
$$\displaystyle = \int \frac{(\sin x+\cos x)}{(\sin x+\cos x)^2\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)}dx$$
$$\displaystyle  = \int \frac{(\sin x+\cos x)}{(1+\sin  2x)\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)}dx$$
Let $$(\sin x-\cos x) = t\;,$$ Then $$(\cos +\sin x)dx = dt$$ and $$(1-\sin 2x) = t^2\Rightarrow (1+\sin 2x) = (2-t^2)$$
So Integral Convert into $$\displaystyle  = 4\int\frac{1}{(2-t^2)\cdot(5-t^4)}dt = 4\int\frac{1}{(t^2-2)\cdot (t^2-\sqrt{5})\cdot (t^2+\sqrt{5})}dt$$
Now Using partial fraction, we get
$$\displaystyle = 4\int \left[\frac{1}{2-t^2}+\frac{1}{(2-\sqrt{5})\cdot 2\sqrt{5}\cdot (\sqrt{5}-t^2)}+\frac{1}{(2+\sqrt{5})\cdot 2\sqrt{5}\cdot (\sqrt{5}+t^2)}\right]dt$$
$$ = \displaystyle \sqrt{2}\ln \left|\frac{\sqrt{2}+t}{\sqrt{2}-t}\right|+\frac{1}{(2-\sqrt{5})\cdot 5^{\frac{3}{4}}}\cdot \ln \left|\frac{5^{\frac{1}{4}}+t}{5^{\frac{1}{4}}-t}\right|+\frac{2}{(2+\sqrt{5})\cdot 5^{\frac{3}{4}}}\cdot \tan^{-1}\left(\frac{t}{5^{\frac{1}{4}}}\right)+\mathbb{C}$$
where $$t=(\sin x-\cos x)$$
A: $\newcommand{\+}{^{\dagger}}%
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$\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}
A: For the second integral, decompose the integrand as follows
\begin{align}
&\frac52\int\frac{1}{\sin^5 x+\cos^5 x}dx\\
=&\int\frac{2}{\sin x+\cos x}-\frac{\sin x + \cos x}{2\sin x\cos x-\sec\frac{\pi}5}
-\frac{\sin x + \cos x}{2\sin x\cos x -\sec\frac{3\pi}5}\ dx\\
=&\sqrt2\tanh^{-1}\frac {\sin x- \cos x}{\sqrt2}
-\frac{1}{\sqrt{1-\sec\frac{3\pi}5}}\tanh^{-1}\frac{\sin x- \cos x}{\sqrt{1-\sec\frac{3\pi}5}}\\
&\hspace{4.5cm} +\frac{1}{\sqrt{\sec\frac{\pi}5-1}}\tan^{-1}\frac{\sin x- \cos x}{\sqrt{\sec\frac{\pi}5-1}}
\end{align}
Specifically
\begin{align}
&\int_0^{\pi/2}
\frac{1}{\sin^5 x+\cos^5 x}dx\\
=&\ \frac45\bigg(\sqrt2\coth^{-1}\sqrt2
-\frac{\coth^{-1}\sqrt{\sqrt5+2}}{\sqrt{\sqrt5+2}} +\frac{\cot^{-1}\sqrt{\sqrt5-2}}{\sqrt{\sqrt5-2}}\bigg)
\end{align}
A: I am not sure how you can continue either (the second term in the denominator can be expressed as $1-\sin(2 x)/2 - \sin^2(2x)/4,$ but I am not aware of any double angle formula for $\sin x + \cos x.$ The simplest approach to your integral is to use the feared $u = \tan \frac{x}2$ substitution, which reduces the integral to a rational function integral....
A: 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
$$
\begin{aligned}I&=4  \int \frac{d t}{\left(t^2-2\right)\left(t^4-4 t^2-1\right)}\\&=
\frac{4}{5} \int\left(\frac{-1}{t^2-2}+\frac{1}{t^2+\sqrt{5}-2}+\frac{1}{t^2-\sqrt{5}-2}\right) d t \\&=\frac{4}{5}\left[-\frac{1}{\sqrt{2}} \tanh ^{-1}\left(\frac{t}{\sqrt{2}}\right)+\frac{1}{\sqrt{\sqrt 5-2}} \tan ^{-1}\left(\frac{t}{\sqrt{\sqrt{5}-2}}\right)-\frac{1}{\sqrt{\sqrt{5}+2}} \tanh ^{-1}\left(\frac{t}{\sqrt{\sqrt{5}+2}}\right)\right]+C\\ &\boxed{I=\frac{4}{5}\left[-\frac{1}{\sqrt{2}} \tanh ^{-1}\left(\frac{\sin x-\cos x}{\sqrt{2}}\right)+\frac{1}{\sqrt{\sqrt 5-2}} \tan ^{-1}\left(\frac{\sin x-\cos x}{\sqrt{\sqrt{5}-2}}\right)\qquad\qquad  -\frac{1}{\sqrt{\sqrt{5}+2}} \tanh ^{-1}\left(\frac{\sin x-\cos x}{\sqrt{\sqrt{5}+2}}\right)\right]+C}\end{aligned}$$
A: $\displaystyle \begin{aligned}\int \frac{1}{\sin ^3 x+\cos ^3 x} d x & =\frac{1}{3} \int \left(\frac{2}{\sin x+\cos x}+\frac{\sin x+\cos x}{\sin ^2 x-\sin x \cos x+\cos ^2 x}\right) d x \\& =\frac{2}{3} J+\frac{1}{3} K \\\\J&=\int \frac{1}{\sin x+\cos x} d x \\& =\frac{1}{\sqrt{2}} \int \frac{d x}{\cos \left(x-\frac{\pi}{4}\right)} \\& =\frac{1}{\sqrt{2}} \ln\left|\sec \left(x-\frac{\pi}{4}\right)+\tan \left(x-\frac{\pi}{4}\right) \right|+c_1 \\\\K&=2 \int \frac{\sin x+\cos x}{2-2 \sin x \cos x} d x \\& =2 \int \frac{d(\sin x-\cos x)}{1+(\sin x-\cos x)^2} \\& =2 \tan ^{-1}(\sin x-\cos x)+c_2 \\\\\therefore \int \frac{1}{\sin ^3 x+\cos ^3 x} d x &=\frac{\sqrt{2}}{3} \ln \left|\sec \left(x-\frac{\pi}{4}\right)+\tan \left(x-\frac{\pi}{4}\right) \right|+2 \tan ^{-1}(\sin x-\cos x)+C \\& =\frac{\sqrt{2}}{3} \ln \left|\frac{\sqrt{2}+\sin x-\cos x}{\cos x+\sin x}\right|+\frac{2}{3}   \tan ^{-1}(\sin x-\cos x)+C \\&\end{aligned}\tag*{} $
