How to find $\int \sqrt{x^8 + 2 + x^{-8}} \,\mathrm{d}x$? The integral to be solved is $\int \sqrt{x^8 + 2 + x^{-8}} \,\mathrm{d}x$. I tried substitution $t=x^8$ which got me to $\frac{1}{8} \int \sqrt{t + 2 + \frac{1}{t}} t^{-7/8} \,\mathrm{d}t$ and I'm stuck. Can you help?

\begin{align}
\int \sqrt{x^8 + 2 + x^{-8}} dx &= \int \sqrt{\left(x^4+\frac1{x^4}\right)^2} dx \\
&= \int \left|\underbrace{x^4+\frac1{x^4}}_{\ge0}\right| dx \\
&= \int \left(x^4+\frac1{x^4}\right) dx \\
&= \frac{x^5}5-\frac1{3x^3}+C
\end{align}
