Divide numerator and denominator by $x^{15}$:

$\displaystyle\int\dfrac{2x^{-3}+5x^{-6}}{(1+x^{-2}+x^{-5})^3}\,dx$

Let $t=1+x^{-2}+x^{-5}$. Then $dt=(-2x^{-3}-5x^{-6})dx$, so the numerator (divided by $x^{15}$) gives $-dt$:

$=-\displaystyle\int\dfrac{dt}{t^3} \cdot \dfrac{1}{x^{15}/x^{15}}$

Wait — let me redo with $t = x^{-5}+x^{-3}+1$... Actually substituting directly: $t=1+x^{-2}+x^{-5}$, $-dt=(2x^{-3}+5x^{-6})dx$.

Integral $= -\int t^{-3}dt = \dfrac{1}{2t^2}+C = \dfrac{x^{10}}{2(x^5+x^3+1)^2}+C$