A hard integral $\int\frac{(x^4-1)\sqrt{x^4+1}}{x^8+1} \mathrm{d}x$

Question

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

Answer 1

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}

Answer 2

Hint:

$$\dfrac{(x^4-1)\sqrt{x^4+1}}{x^8+1}=x\cdot\dfrac{\left(1-\dfrac1{x^4}\right)\sqrt{x^2+\dfrac1{x^2}}}{x^4+\dfrac1{x^4}}$$

Set $x^2=y$ to find

$$\int\dfrac{(x^4-1)\sqrt{x^4+1}}{x^8+1}dx=\dfrac12\int\dfrac{\left(1-\dfrac1{y^2}\right)\sqrt{y+\dfrac1y}}{\left(y+\dfrac1y\right)^2-2}dy$$

Now choose $\sqrt{y+\dfrac1y}=u\implies y+\dfrac1y=u^2$