Evaluate $\int \frac{\sqrt{1+x^8} dx}{x^{13}}$ 
Evaluate: $\displaystyle\int \frac{\sqrt{1+x^8}}{x^{13}}dx$

My attempt:
I have tried substituting $1+x^8=z^2$ but that did not work. I also tried writing $x^{13}$ in the denominator as $x^{16}.x^{-3}$ hoping that it would bring the integrand into some form but that too did not work.

Let $x^4=\tan u$ so that $dx=\frac{\sec^2 u}{4x^3}du$
so our integral becomes
$\int \frac{\sqrt{1+\tan^2 u}}{x^{13}}.\frac{\sec^2 u}{4x^3}du$
or rather
$\frac{1}{4}\int \frac{\sec^3 u}{\tan^4 u}du$
which is
$\frac{1}{4}\int \cos u\sin^{-4} u du$
which equals
$-\frac{1}{12}\sin^{-3}u +C$
now substitute back from $u$ to $x$ which I think is
$-\frac{1}{12}(\frac{x^8}{1+x^8})^\frac{-3}{2}+C$
