<s>[INST] 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
 [/INST] 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 $\sqrt{y+\dfrac1y}=u\implies y+\dfrac1y=u^2$
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