Integrating $\int \frac{1}{\sqrt[4]{1+x^4}} dx $ How to integrate 
$$\int \frac{1}{\sqrt[4]{1+x^4}} dx $$
 A: $$ \begin{aligned}
\int\dfrac{\mathrm{d}x}{\sqrt[\scriptstyle{4}]{x^4 + 1}} &=\int\frac{1}{x\,\sqrt[\scriptstyle{4}]{1 + 1/x^4}}\,\mathrm{d}x
\end{aligned}
$$
Now, set $ u=\sqrt[\scriptstyle{4}]{1+1/x^4} $. Then,
$$ u^4 - 1 = \frac{1}{x^4} $$
$$ 4u^3\,\mathrm{d}u = -4\frac{1}{x^4}\frac{1}{x}\,\mathrm{d}x $$
$$ \frac{\mathrm{d}x}{x} = -\frac{u^3}{u^4 - 1}\,\mathrm{d}u $$
$$
\begin{aligned}
\int\frac{\mathrm{d}x}{\sqrt[\scriptstyle{4}]{x^4+1}}&=-\int\frac{u^2}{u^4 - 1}\,\mathrm{d}u\\
&=-\int\frac{u^2 + 1 - 1}{\left(u^2 - 1\right)\!\!\left(u^2 + 1\right)}\,\mathrm{d}u\\
&=\int\frac{1}{1-u^2}\,\mathrm{d}u -\frac{1}{2} \int\frac{1}{u^2+1}-\frac{1}{u^2-1}\,\mathrm{d}u\\
&=\frac{1}{2}\int\frac{1}{1-u^2}\,\mathrm{d}u - \frac{1}{2}\arctan u + C\\
&=\frac{1}{4}\ln\left|\frac{u+1}{u-1}\right| - \frac{1}{2}\arctan u + C\\
&=\frac{1}{4}\ln\left|\frac{\sqrt[\scriptstyle{4}]{1+x^4}+x}{\sqrt[\scriptstyle{4}]{1+x^4}-x}\right| - \frac{1}{2}\arctan\!\left( \frac{\sqrt[\scriptstyle{4}]{1+x^4}}{x}\right) + C
\end{aligned}
$$
A: Maple says $ x {\rm hypergeom}([1/4, 1/4], [5/4], -x^4)$.  This can be converted to MeijerG or JacobiP or various Heun functions, but nothing elementary.
