Finding $\int_{-\infty}^\infty |f\ast f'|^2(x)\,dx$ using Plancherel’s theorem 
Suppose $G(\mathbb R)\ni f(x),\mathcal{F}[f](\omega)=\frac{1}{1+|w|^3}$ find the value of
  $$\int_{-\infty}^\infty |f\ast f'|^2(x)\,dx$$

I thought using Plancherel’s theorem
\begin{align}
\int_{-\infty}^\infty |f\ast f'|^2(x)\,dx
&=2\pi\int_{-\infty}^\infty|\mathcal{F}(f\ast f')|\,d\omega\\[2ex]
&=2\pi\int_{-\infty}^\infty(\frac{1}{1+|w|^3}\cdot
  \frac{i\omega}{1+|\omega|^3})^2\,d\omega\\[2ex]
&=-4\pi\int_0^\infty(\frac{\omega}{(1+\omega^3)^2})^2\,d\omega
\end{align}
but here I'm stuck because I can't find any Fourier series so that I'd be able to apply Parseval's identity or any other ways to simplify this integral. How can I simplify the last integral?

Hint: Substitute $u=\omega^3$.
