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 $ 

There is a general closed form for such integrals but it's not elementary or pretty at all. 
$$\int \frac{x^m}{1+x^n} \, \mathrm{d}x = \frac{x^{m+1} \, _2F_1\left(1, \frac{m+1}{n};\frac{m+n+1}{n}; -x^n\right)}{m+1} + \mathrm{C}.$$
Where $_2F_1\left(a, b;c; x\right)$ is the hypergeometric function.
With regards to your specific problem, try partial fractions (although I suspect you may have mistyped, given how ugly this is)
$$\frac{x^3}{1+x^5} = \frac{\sqrt{5}x - 5x + \sqrt{5} + 5}{5\sqrt{5}(2x^2 + \sqrt{5}x - x + 2)} + \frac{\sqrt{5}x + 5x + \sqrt{5} - 5}{5\sqrt{5}(2x^2 - \sqrt{5}x - x + 2)} - \frac{1}{5(x+1)}.$$
