How to integrate $\int \frac{1}{\sin^4x + \cos^4 x} \,dx$? 
How to integrate 
  $$\int \frac{1}{\sin^4x + \cos^4 x} \,dx$$ 

I tried the following approach:
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \int \frac{1}{\sin^4x + (1-\sin^2x)^2} \,dx = \int \frac{1}{\sin^4x + 1- 2\sin^2x + \sin^4x} \,dx \\
= \frac{1}{2}\int \frac{1}{\sin^4x - \sin^2x + \frac{1}{2}} \,dx = \frac{1}{2}\int \frac{1}{(\sin^2x - \frac{1}{2})^2 + \frac{1}{4}} \,dx$$
The substitution $t = \tan\frac{x}{2}$ yields 4th degree polynomials and a $\sin$ substitution would produce polynomials and expressions with square roots while Wolfram Alpha's solution doesn't look that complicated. Another approach:
$\sin^4x + \cos^4 x = (\sin^2 x + \cos^2x)(\sin^2 x + \cos^2 x) - 2\sin^2 x\cos^2 x = 1 - 2\sin^2 x\cos^2 x  = (1-\sqrt2\sin x \cos x)(1+\sqrt2\sin x \cos x)$
 and then I tried substituting: $t = \sin x \cos x$ and got 
$$\int\frac{t\,dt}{2(1-2t^2)\sqrt{1-4t^2}}$$
Another way would maybe be to make two integrals:
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \int \frac{1}{(1-\sqrt2\sin x \cos x)(1+\sqrt2\sin x \cos x)} \,dx = \\ \frac{1}{2}\int \frac{1}{1-\sqrt2\sin x \cos x} \,dx +  \frac{1}{2}\int\frac{1}{1+\sqrt2\sin x \cos x} \,dx$$ 
... and again I tried $t = \tan\frac{x}{2}$ (4th degree polynomial) and $t=\sqrt2 \sin x \cos x$ and I get $\frac{\sqrt 2}{2} \int \frac{\,dt}{(1-t)\sqrt{1-2t^2}}$ for the first one.
Any hints?

$$
\begin{aligned}
\sin ^{4} x+\cos ^{4} x &=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x \\
&=1-2 \sin ^{2} x \cos ^{2} x \\
\int \frac{d x}{\sin ^{4} x+\cos ^{4} x} &=\int \frac{d x}{1-2 \sin ^{2} x \cos ^{2} x} \\
&=\int \frac{\sec ^{4} x}{\sec ^{4} x-2 \tan ^{2} x} d x \\
&\stackrel{t=\tan x}{=} \int \frac{1+t^{2}}{\left(1+t^{2}\right)^{2}-2 t^{2}} d t, \\
&=\int \frac{1+t^{2}}{t^{4}+1} d t \\
&=\int \frac{1+\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}} d t \\
&=\int \frac{d\left(1-\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)^{2}+2} \\
&=\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)+C \\
&=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\tan x-\cot x}{\sqrt{2}}\right)+C
\end{aligned}
$$
