How find this integral $I=\int\frac{1}{\sin^5{x}+\cos^5{x}}dx$ Question:
Find  the integral
$$I=\int\dfrac{1}{\sin^5{x}+\cos^5{x}}dx$$
my solution: since 
\begin{align*}\sin^5{x}+\cos^5{x}&=(\sin^2{x}+\cos^2{x})(\sin^3{x}+\cos^3{x})-\sin^2{x}\cos^2{x}(\sin{x}+\cos{x})\\
&=(\sin{x}+\cos{x})(1-\sin{x}\cos{x})-(\sin{x}+\cos{x})(\sin^2{x}\cos^2{x})\\
&=(\sin{x}+\cos{x})[1-\sin{x}\cos{x}-\sin^2{x}\cos^2{x})
\end{align*}
then let $$\sin{x}+\cos{x}=t\Longrightarrow \sin{x}\cos{x}=\dfrac{1}{2}(t^2-2)$$
then
$$\sin^5{x}+\cos^5{x}=t[1-\dfrac{1}{2}(t^2-2)-\dfrac{1}{4}(t^2-2)^2]=-\dfrac{1}{4}t(t^4-2t^2+4)$$
$$x+\dfrac{\pi}{4}=\arcsin{\dfrac{t}{\sqrt{2}}}\Longrightarrow dx=\dfrac{1}{\sqrt{2-t^2}}dt$$
so
$$I=-4\int\dfrac{dt}{t\sqrt{2-t^2}(t^4-2t^2+4)}=-2\int\dfrac{1}{u\sqrt{2-u^2}(u^2-2u+4)}du$$
where $u=t^2$
Then I can't,
because this wolf can't http://www.wolframalpha.com/input/?i=1%2F%28xsqrt%282-x%5E2%29%28x%5E2-2x%2B4%29%29dx
Thank you for  you help
 A: since 
\begin{align*}\sin^5{x}+\cos^5{x}&=(\sin^2{x}+\cos^2{x})(\sin^3{x}+\cos^3{x})-\sin^2{x}\cos^2{x}(\sin{x}+\cos{x})\\
&=(\sin{x}+\cos{x})(1-\sin{x}\cos{x})-(\sin{x}+\cos{x})(\sin^2{x}\cos^2{x})\\
&=(\sin{x}+\cos{x})[1-\sin{x}\cos{x}-\sin^2{x}\cos^2{x}]
\end{align*}
so
\begin{align*}\int\dfrac{1}{\sin^5{x}+\cos^5{x}}dx&=\dfrac{1}{(\sin{x}+\cos{x})(1-\cos{x}\sin{x}-\cos^2{x}\sin^2{x})}dx\\
&=\int\dfrac{1}{\sqrt{1-2y}(1+2y)(1-y-y^2)}dy (y=\cos{x}\sin{x})\\
&=\dfrac{4}{5}\int\dfrac{1}{\sqrt{1-2y}(1+2y)}dy+\dfrac{1}{5}\int\dfrac{1+2y}{\sqrt{1-2y}(1-y-y^2)}dy
\end{align*}
note
$$\int\dfrac{1}{\sqrt{1-2y}(1+2y)}=\int\dfrac{1}{z^2-2}dz=\dfrac{1}{2\sqrt{2}}\ln{\left|\dfrac{z-\sqrt{2}}{z+\sqrt{2}}\right|}+C_{1} (z=\sqrt{1-2y})$$
and we 
$$\int\dfrac{1+2y}{\sqrt{1-2y}(1-y-y^2)}dy=-4\int\dfrac{z^2-2}{z^4-4z^2+1}dz$$
and
\begin{align*}
\int\dfrac{z^2-2}{z^4-4z^2+1}dz&=\int\dfrac{z^2-1}{z^4-4z^2+1}dz-\int\dfrac{1}{z^4-4z^2+1}dz\\
&=\int\dfrac{d(z+\dfrac{1}{z})}{(z+\dfrac{1}{z})^2-6}-\int\dfrac{1}{z^4-4z^2+1}dz
\end{align*}
we can easy have
$$\int\dfrac{1}{z^4-4z^2+1}dz=\dfrac{1}{2}\left(\dfrac{1+z^2}{z^4-4z^2+1}+\int\dfrac{1-z^2}{z^4-4z^2+4}\right)$$
so
$$\dfrac{1}{5}\int\dfrac{1+2y}{\sqrt{1-2y}(1-y-y^2)}dy=\dfrac{1}{5}\left(\dfrac{1}{\sqrt{2}}\ln{\left|\dfrac{z^2-\sqrt{2}z+1}{z^2+\sqrt{2}z+1}\right|}-\dfrac{1}{\sqrt{6}}\ln{\left|\dfrac{z^2-\sqrt{6}z+1}{z^2+\sqrt{6}z+1}\right|}\right)+C_{2}$$
so
$$I=\dfrac{\sqrt{2}}{5}\ln{\left|\dfrac{z-\sqrt{2}}{z+\sqrt{2}}\right|}+\dfrac{1}{5}\left(\dfrac{1}{\sqrt{2}}\ln{\left|\dfrac{z^2-\sqrt{2}z+1}{z^2+\sqrt{2}z+1}\right|}-\dfrac{1}{\sqrt{6}}\ln{\left|\dfrac{z^2-\sqrt{6}z+1}{z^2+\sqrt{6}z+1}\right|}\right)+C$$
where $z=\cos{x}-\sin{x},C=\dfrac{4}{5}C_{1}+C_{2}$
