Evaluate $\int \frac{\sqrt{1+x^4}}{1-x^4}dx$ How to evaluate

$$\int \frac{\sqrt{1+x^4}}{1-x^4}dx.$$

If we substitute $x=\sqrt{\tan{\theta}}$, it becomes
$$\int \frac{1}{2\cos{\theta}\cos{2\theta}\sqrt{\tan{\theta}}}d\theta.$$
What can I do next ?
Edit
Are these steps correct ?
$$=\int \frac{1}{2\cos{\theta}\cos{2\theta}\sqrt{\frac{\sin\theta}{\cos\theta}}}d\theta$$
$$=\int \frac{1}{2\cos{2\theta}\sqrt{\cos^2 \theta\frac{\sin\theta}{\cos\theta}}}d\theta$$
$$=\int \frac{1}{2\cos{2\theta}\sqrt{\cos^2 \theta\frac{\sin\theta}{\cos\theta}}}d\theta$$
$$=\int \frac{1}{\sqrt{2}\cos{2\theta}\sqrt{\sin{2 \theta}}}d\theta$$
Now If we substitute $t=\sin{2\theta}$, it becomes
$$=\int \frac{1}{\sqrt{2}(1-t^2)\sqrt{t}}dt$$
$$=\int \frac{\frac{1}{t^2}}{\sqrt{2}\frac{(1-t^2)}{t}\sqrt{\frac{t}{t^2}}}dt$$
$$=\int \frac{\frac{1}{t^2}}{\sqrt{2}(\frac{1}{t} -t)\sqrt{\frac{1}{t}}}dt$$
Now If we substitute $u^2=\frac{1}{t}$, it becomes
$$=\int \frac{-2u}{\sqrt{2}(u^2 -\frac{1}{u^2})u}du$$
$$=\int \frac{-u^2}{\sqrt{2}(u^4 -1^2)}du$$
$$=\int \frac{-u^2}{\sqrt{2}(u^2 +1)(u^2-1)}du$$
$$=-\frac{1}{2\sqrt{2}}\int \frac{u^2+1+u^2-1}{(u^2 +1)(u^2-1)}du$$
This can be solved easily now and we will get an elementary solution.
But Wolfram alpha gives a solution in non elementary functions.
Therefore I am confused. I guess the above solution is correct only for some restricted values of x.
 A: Alternatively, substitute $t=\frac{\sqrt2 x}{\sqrt{1+x^4}}$. Then
$$\frac{\sqrt{1+x^4}}{1-x^4}=\frac{xt}{\sqrt2(t^2-x^2)}, \>\>\>\>\>dx =\frac{x}{t(1-t^2x^2)}dt$$
and
\begin{align}
&\int \frac{\sqrt{1+x^4}}{1-x^4}dx\\
=&\ \frac1{\sqrt2}\int \frac{x^2}{(t^2-x^2)(1-t^2 x^2)}dt
= \frac1{\sqrt2}\int \frac1{t^2(\frac1{x^2}+x^2)-1-t^4}dt\\
=& \frac1{\sqrt2}\int \frac1{1-t^4}dt= \frac1{2\sqrt2}\int \frac1{1-t^2}+ \frac1{1+t^2}\ dt\\
=& \ \frac1{2\sqrt2}\left(\tanh^{-1}t +\tan^{-1}t\right)+C
\end{align}
A: Maple also gives a non-elementary answer to the original integral in terms of elliptic integral functions.  However, for the integral over $\theta$ it does give an elementary answer, which translated back to the original integral becomes
$$ \frac{\sqrt{2}\, \mathrm{arctanh}\! \left(\sqrt{\sin\left(2 \arctan \! \left(x^{2}\right)\right)}\right)}{4}+\frac{\sqrt{2}\, \arctan \! \left(\sqrt{\sin\left(2 \arctan \! \left(x^{2}\right)\right)}\right)}{4}$$
which is correct for $x > 0$ (for $x < 0$ you want to multiply that by $-1$).
I believe both Maple and Wolfram Alpha use some form of the Risch algorithm to decide whether an integral is elementary, however in the algebraic case, which the original integral is, the implementation seems not to be complete.
