Trying to evaluate $\int_{0}^{\infty}\frac{\ln(1+x^3)}{1+x^3}\frac{dx}{1+x^3}$ I would like to work this out:

$$I=\large\int_{0}^{\infty}\frac{\ln(1+x^3)}{1+x^3}\frac{\mathrm dx}{1+x^3}$$

Making a sub: $u=x^3$, $dx=\frac{du}{3x^2}$
$$I=\frac{1}{3}\int_{0}^{\infty}\frac{\ln(1+u)}{u^{2/3}(1+u)^2}\mathrm du$$
Making a sub: $u=\tan^2(y)$, $du=\sec^2(y)dy$
$$I=\frac{2}{3}\int_{0}^{\pi/2}\frac{\ln\sec(y)}{\sec^2(y)\sqrt[3]{\tan^4(y)}}\mathrm dy$$
$$I=\frac{2}{3}\int_{0}^{\pi/2}\frac{\cot(y)\cos^2(y)\ln\sec(y)}{\sqrt[3]{\tan(y)}}\mathrm dy$$
I can't continue.
Maybe there is another alternative way to simplify $I$

In this solution we will not use the beta function or the digamma function.
Put
\begin{equation*}
f(s)=\int_{0}^{\infty}\dfrac{\ln(1+s^3x^3)}{(1+x^3)^2}\, \mathrm{d}x.
\end{equation*}
We want to calculate $f(1)$. However,
\begin{equation*}
f(1) = f(1)-f(0) = \int_{0}^{1}f'(s)\, \mathrm{d}s\tag{1}
\end{equation*}
and
\begin{equation*}
f'(s)=\int_{0}^{\infty}\dfrac{3s^2x^3}{(1+s^3x^3)(1+x^3)^2}\, \mathrm{d}x =[x^3=z] = \int_{0}^{\infty}\dfrac{s^2z^{1/3}}{(1+s^3z)(1+z)^2}\, \mathrm{d}z .
\end{equation*}
For $0<s<1$ we use keyhole contour integration and get
\begin{equation*}
f'(s) = \dfrac{2\pi\sqrt{3}}{9}\dfrac{2s+1}{(s^2+s+1)^2}\cdot s^2 .
\end{equation*}
Integration by parts in (1) will finally give us
\begin{equation*}
f(1) = \dfrac{2\pi}{27}\left(\sqrt{3}(3\ln 3 -1)-\pi\right).
\end{equation*}
