<s>[INST] A general method for integration of rational function. $\int\frac {x^3}{1+x^5}$
ATTEMPT:
I did the following substitution: 
Let $x=\frac{1}{t}.$
$dx=\frac{-1}{t^2}dt.$
substituting back: 
$I=\int\frac{-1}{1+t^5}dt$ which doesn't seems a simpler integration.
Next  i substituted $x=p^\frac{2}{5}.$
$dx=\frac{2}{5}\frac{p^{3/5}}{1+p^2}.$
Now let $p=tan\theta.$
$dp=sec^2\theta$$d\theta.$
substituting back:
$I=\frac{2}{5}\int tan^{3/5}\theta$$d\theta$ which i couldn't integrate further even by trying By parts method.
Is there a general approach for problems of the form (not the algebraic twins):
$\int \frac{x^m}{1+x^n}dx$ where $n-m \ne1 $ 
 [/INST] What you already made is $$I=\int\frac {x^3}{1+x^5}dx=-\int\frac {dt}{1+t^5}$$ Now $$1+t^5=\prod_{i=1}^5 (t-\alpha_i)$$ where $\alpha_i=-(-1)^{\frac i5}$ which means that, using partial fraction decomposition, $$\frac {1}{1+t^5}=\sum_{i=1}^5 \frac{\beta_i}{t-\alpha_i}$$ $$\int\frac {dt}{1+t^5}=\sum_{i=1}^5 \beta_i \log({t-\alpha_i})$$ which is simple, except that the $\alpha_i$'s and $\beta_i$'s are complex numbers.
But the real part of $$(a+i b)\log(c+id)=\frac a2\, \log(c^2+d^2)-b\, \tan^{-1} \big(\frac d c\big)$$ and this has to be applied to each term (noticing that four of the roots $\alpha_i$ are conjugated by pairs). So, isolating the case of the real root, we need to combine by pairs the logarithms and the inverse tangents.
This is how,  in "Table of Integrals, Series, and Products" by I.S. Gradshteyn and I.M. Ryzhik, they arrive to equation $2.142$.
$$\frac{1}{20} \left(4 \log (t+1)+\left(\sqrt{5}-1\right) \log \left(t^2+\frac{1}{2}
   \left(\sqrt{5}-1\right) t+1\right)-\left(1+\sqrt{5}\right) \log
   \left(t^2-\frac{1}{2} \left(1+\sqrt{5}\right) t+1\right)-2 \sqrt{10-2
   \sqrt{5}} \tan ^{-1}\left(\frac{-4 t+\sqrt{5}+1}{\sqrt{10-2 \sqrt{5}}}\right)+2
   \sqrt{ \left(10+2\sqrt{5}\right)} \tan ^{-1}\left(\frac{4 t+\sqrt{5}-1}{\sqrt{
   \left(10+2\sqrt{5}\right)}}\right)\right)$$
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