Integration of $\frac{1}{\sqrt{x^8+1}} $ How can we integrate $\frac{1}{\sqrt{x^8+1}}$ thanks for help.
I couldn't find any way. I tried to factor, substituted with $\tan(a)$ but it's not working :'(

Given $\displaystyle \int \frac{1}{\sqrt{x^8+1}}dx\;,$ Now Let $x^4 = t\;,$ Then $\displaystyle 4x^3dx = dt\Rightarrow dx = \frac{1}{4x^3}dt = \frac{1}{4}t^{-\frac{3}{4}}dt$
So Integral Convert into $$\displaystyle \frac{1}{4}\int t^{-\frac{3}{4}}\cdot \left(1+t^2\right)^{-\frac{1}{2}}dt$$
Now Using Integrals of differential binomials and Chebyshev’s criterion

Integration of $$\displaystyle \int x^{m}\cdot (a+bx^n)^{p}dx$$ in terms of elementry function is Possible only, When
$$\displaystyle p\;,\frac{m+1}{n}\;,\frac{m+1}{n}+p$$ is an Integer, Where $m,n,p\in \mathbb{Q}$.

So above we have $\displaystyle m=-\frac{3}{4}\notin \mathbb{Z}$ and $\displaystyle \frac{m+1}{n} = \frac{\left(-\frac{3}{4}+1\right)}{2} = \frac{1}{8}\notin  \mathbb{Z}$ and
$\displaystyle \frac{m+1}{n}+p = \frac{1}{8}-\frac{1}{2} = -\frac{3}{8}\notin  \mathbb{Z}$
So We can Not express the Integral $$\displaystyle \int\frac{1}{\sqrt{x^8+1}}dx$$ in terms of  Elementry function..
