An equivalent of : $f(x)=\int_0^{+\infty}\frac{e^{-xt}}{(1+t^3)^{1/3}} dt$ $\forall\ x\ \in\ \left]0,+\infty\right[\ $ we put:
$$
{\rm f}\left(x\right)
=
\int_{0}^{\infty}{{\rm e}^{-xt} \over \left(1 + t^{3}\right)^{1/3}}\,{\rm d}t
$$
The question is the question is to find an equivalent of $\,\,{\rm f}\left(x\right)$ when $x \to 0^{+}$.
That means find a simple function $g$ such that when $x \to 0^+$ we have : $f(x)\sim g(x)$ that means : $$\lim_{x \to 0^+} \frac{f(x)}{g(x)} =1$$

Mathematica gives 
$$
\frac{G_{1,4}^{4,1}\left(\frac{s^3}{27}|
\begin{array}{c}
 \frac{2}{3} \\
 0,0,\frac{1}{3},\frac{2}{3} \\
\end{array}
\right)}{2 \sqrt{3} \pi  \Gamma \left(\frac{1}{3}\right)}$$ for the integral, and 
expanding this in a power series at $0,$ gets
$$
\frac{-3 \sqrt{3} \Gamma \left(\frac{1}{3}\right) \Gamma
   \left(\frac{2}{3}\right) \log (s)-2 \sqrt{3} \gamma  \Gamma
   \left(\frac{1}{3}\right) \Gamma \left(\frac{2}{3}\right)+\sqrt{3} \log (27)
   \Gamma \left(\frac{1}{3}\right) \Gamma \left(\frac{2}{3}\right)+\sqrt{3}
   \Gamma \left(\frac{1}{3}\right) \Gamma \left(\frac{2}{3}\right) \psi
   ^{(0)}\left(\frac{2}{3}\right)}{6 \pi }+\frac{4 \pi ^2 s}{27 \Gamma
   \left(\frac{1}{3}\right) \Gamma \left(\frac{4}{3}\right)^2}+O\left(s^2\right),
$$
so the dominant term is 
$$
\frac{-3 \sqrt{3} \Gamma \left(\frac{1}{3}\right) \Gamma
   \left(\frac{2}{3}\right)}{6\pi} \log (s)
$$
Running FullSimplify on the power series, gets one
$$
\left(-\log (s)+\frac{\pi }{6 \sqrt{3}}-\gamma +\frac{\log
   (3)}{2}\right)+\frac{4 \pi ^2 s}{27 \Gamma \left(\frac{1}{3}\right) \Gamma
   \left(\frac{4}{3}\right)^2}+O\left(s^2\right),
$$
which agrees with Robert Israel's comment.
