A hard integral $\int\frac{(x^4-1)\sqrt{x^4+1}}{x^8+1} \mathrm{d}x$ This integtare is very hard for me，I do not known how to deal with it.
$$\int\frac{(x^4-1)\sqrt{x^4+1}}{x^8+1} \mathrm{d}x$$
I use “mathematica” and obtain this

next the “maple” ：

the “maple” tell me it's a Elementary function

Note
$$\frac2{x^8+1}=\frac1{x^4+1}\left( \frac1{x^4+\sqrt2x^2+1}+  \frac1{x^4-\sqrt2x^2+1}\right)
$$
Then
\begin{align}
&\int\frac{(x^4-1)\sqrt{x^4+1}}{x^8+1} \> dx \\
=& \frac12 \int\frac{\frac{x^4-1}{\sqrt{x^4+1}}dx}{x^4+\sqrt2x^2+1}
 +\frac12 \int\frac{\frac{x^4-1}{\sqrt{x^4+1}}dx}{x^4-\sqrt2x^2+1}\\
 =& \frac12\int\frac{d\sqrt{ x^2+\frac1{x^2}}}{x^2+\frac1{x^2}+\sqrt2}
+\frac12 \int\frac{d\sqrt{ x^2+\frac1{x^2}}}{x^2+\frac1{x^2}-\sqrt2}\\
=& \frac1{2\sqrt[4]2}\tan^{-1} \frac{\sqrt{ x^2+\frac1{x^2}} }{\sqrt[4]2 } - \frac1{2\sqrt[4]2}\coth^{-1} \frac{\sqrt{ x^2+\frac1{x^2}} }{\sqrt[4]2 }+C
\end{align}
