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

The substitution $x=\tan^{1/4}t$ gives $$\int\frac{\sqrt{1+x^8}}{x^{13}}dx=\int\frac{1}{4}\sin^{-4}t\cos tdt=-\frac{1}{12}\sin^{-3}t+C=-\frac{1}{12}\bigg(\frac{x^8}{1+x^8}\bigg)^{-3/2}+C.$$

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

$-\frac{1}{12}(\frac{x^8}{1+x^8})^\frac{-3}{2}+C$

$x=\dfrac1y,dx=-\dfrac{dy}{y^2}$

$$\int\dfrac{\sqrt{1+x^8}}{x^{13}}dx=-\int\dfrac{y^{13}\sqrt{y^8+1}}{y^4\cdot y^2}dy$$

Set $\sqrt{y^8+1}=u $ or $y^8+1=v$