Indefinite integral $\int \frac{1}{1+\sin^4(x)} \, \mathrm dx$ I'm a bit lost in this integral: $$\int \frac{1}{1+\sin^4(x)} \, \mathrm dx$$
I have tried solving with Wolfram, but I was getting a cosecant solution which doesn't seem as the correct method.
Do you have any ideas? :)
EDIT:
Do you please have step-by-step solution, because I am now somewhat lost. Using the substitution $t=\tan(x)$, I got to
$$\int \left(\frac{t^2}{2t^4+2t^2+1}+\frac{1}{2t^4+2t^2+1}\right)\mathrm dt$$
By expanding with 1:
$$\int \frac{1}{1+\sin^4x}\cdot \frac{\frac{1}{\cos^4x}}{\frac{1}{\cos^4x}}\mathrm dx$$
$$\int \:\frac{1}{\frac{1}{\cos^4x}\cdot \frac{\sin^4x}{\cos^4x}}\cdot \frac{1}{\cos^4x} \mathrm dx$$
$$\int \:\frac{1}{\left(\frac{1}{\cos^2x}\right)^2\cdot \tan^4x}\cdot \frac{1}{\cos^2x}\cdot \frac{1}{\cos^2x}\mathrm dx$$
And using the substitution: $t=\tan\left(x\right)$
$$\mathrm dt=\frac{1}{\cos^2x}\mathrm dx$$
$$t^2=\tan^2\left(x\right)$$
$$t^2=\frac{\sin^2x}{\cos^2x}$$
$$t^2=\frac{1-\cos^2x}{\cos^2x}$$
$$t^2=\frac{1}{\cos^2x}-\frac{\cos^2x}{\cos^2x}=\frac{1}{\cos^2x}-1$$
$$t^2+1=\frac{1}{\cos^2x}$$
Using it:
$$\int \:\frac{t^2+1}{2t^4+2t^2+1}\mathrm dt$$
I don't think I got to the expected result but I can't seem to be able to find why…

Double-check my arithmetic, but here's the strategy.
@Bernard's suggested substitution $t=\tan x$ gives
\begin{align}
& \int\left(1-\frac{t^4}{(t^2\sqrt{2}+1)^2-2(\sqrt{2}-1)t^2}\right) dt \\
= {} & t-\int\frac{t^4}{(t^2\sqrt{2}+ct+1)(t^2\sqrt{2}-ct+1)} \, dt\end{align}
with $c:=\sqrt{2\sqrt{2}-1}$. You can do the rest with partial fractions.
