Partial fraction decomposition is the key. Let $\xi=\exp\left(\frac{\pi i}{5}\right)$ be a primitive tenth root of unity.
Then $x^5+1 = (x-\xi)(x-\xi^3)(x-\xi^5)(x-\xi^7)(x-\xi^9)$ and:
$$\text{Res}\left(\frac{x^3}{1+x^5},x=\xi^{2k+1}\right)=\frac{1}{5}\xi^{10-(2k+1)}\tag{1}$$ so: $$ \frac{x^3}{1+x^5} = \frac{1}{5}\sum_{k=0}^{4}\frac{\xi^{9-2k}}{x-\xi^{2k+1}}\tag{2}$$ as well as: $$ \int\frac{x^3}{1+x^5}\,dx = \frac{1}{5}\sum_{k=0}^{4} \xi^{9-2k}\log\left(x-\xi^{2k+1}\right). \tag{3}$$