How to solve $\int \frac{\,dx}{(x^3 + x + 1)^3}$? 
How to solve $$\int \frac{\,dx}{(x^3 + x + 1)^3}$$ ?  

Wolfram Alpha gives me something I am not familiar with. I thought that the idea was using partial fractions because $x^3$ and $x$ are bijections, there must be a real root but it seems that Wolfram Alpha is using numerical methods to approximate the root so it's not a "nice" number. I can't thing of any substitution which could help me nor any formula to transform this denominator. The formula $\int u\,dv = uv - \int v\,du$ also yields a more complicated integral:
$$ \,dv = \,dx \implies v = x \\
u = \frac{1}{(x^3 + x + 1)^3} \implies du = -3\frac{3x^2 + 1}{(x^3 + x + 1)^4} \,dx \\
\int \frac{\,dx}{(x^3 + x + 1)^3} = \frac{x}{(x^3 + x + 1)^3} + 3\int \frac{3x^3 + x}{(x^3 + x + 1)^4} \,dx \\ 
= \frac{x}{(x^3 + x + 1)^3} + 3\int \frac{3x^3 + x \pm 2x \pm 3}{(x^3 + x + 1)^4}\,dx  \\
= \frac{x}{(x^3 + x + 1)^3} + 3\int \frac{3x^3 + 3x + 3}{(x^3 + x + 1)^4}\,dx + 3\int \frac{- 2x - 3}{(x^3 + x + 1)^4}\,dx \\
= \frac{x}{(x^3 + x + 1)^3} + 9\int \frac{\,dx}{(x^3 + x + 1)^3} - 3\int \frac{2x + 3}{(x^3 + x + 1)^4} \,dx  \\
I = \int \frac{\,dx}{(x^3 + x + 1)^3} \implies \\
-8I = \frac{x}{(x^3 + x + 1)^3} - 3\int \frac{2x + 3}{(x^3 + x + 1)^4} \,dx$$
Is this one really as complicated as Wolfram Alpha "tells" me or is there some sort of "trick" which can be applied? 

Hint: the polynomial $x^3+x+1$ has one real root, say $\alpha$. Then $x^3+x+1=(x-\alpha)(x^2+\alpha x+\alpha^2+1)$ and then apply integration techniques of rational expressions of polynomials with repeated factors, see for example https://math.la.asu.edu/~surgent/mat271/parfrac.pdf
