<s>[INST] A closed form for the antiderivative of $\frac 1{\sin^5 x +\cos^5 x} $ Does there exist a closed form expression for  $$\int \dfrac {dx}{\sin^5 x +\cos^5 x}? $$ 
 [/INST] Utilize the identity
$$\sin^5 (t+\frac\pi4) +\cos^5(t+\frac\pi4)
=\sqrt2 \cos t\left(\frac{5}4-\cos^4 t\right)
$$
to integrate
\begin{align}
&\int \dfrac {1}{\sin^5 (t+\frac\pi4) +\cos^5(t+\frac\pi4)}dt\\
=& \frac{2\sqrt2}5 \int\bigg(\frac1{\cos^2 t}-\frac{1}{2\cos^2 t+\sqrt5}
-\frac{1}{2\cos^2 t-\sqrt5} \bigg)\cos t \>dt \\
=& \frac25\bigg(\sqrt2\tanh^{-1}\sin t
-\frac{\tanh^{-1}\frac{\sqrt2\sin t}{\sqrt{\sqrt5+2}}}{\sqrt{\sqrt5+2}}
+\frac{\tan^{-1}\frac{\sqrt2 \sin t}{\sqrt{\sqrt5-2}}}{\sqrt{\sqrt5-2}}\bigg)+C
\end{align}
Then, replace $t=x-\frac\pi4$ to arrive at the original integral.
In particular
\begin{align}
&\int_0^{\frac\pi2}\frac {dx}{\sin^5 x +\cos^5 x}
=\int_{-\frac\pi4}^{\frac\pi4} \dfrac {1}{\sin^5 (t+\frac\pi4) +\cos^5(t+\frac\pi4)}dt
\\
=&\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}
</s>