Let $\phi$ be a scalar field and $g_{\mu \nu} = \eta_{\mu \nu}+h_{\mu \nu}/M_p$ where $M_p$ is the Planck mass (so we assume we deal with perturbations). Let $\Lambda_2,\Lambda_3$ be energy scales such that $\Lambda_2 \gg \Lambda_3$. These are defined by $\Lambda^2_2 = M_p H_0$ and $\Lambda_3^3 = M_p H_0^2$. The [Horndeski action](https://en.wikipedia.org/wiki/Horndeski's_theory) is:

$$S = \int d^4 x \sqrt{-g} \sum^5_{i=2} \mathcal{L}_i,$$ 

where 

\begin{align}
\mathcal{L_2}&=\Lambda_2^4 G_2,\nonumber \\
\mathcal{L_3}&=\Lambda_2^4 G_3 [\Phi], \nonumber \\
\mathcal{L_4}&= M_p^2 G_4 R + \Lambda_2^4 G_{4,X}([\Phi]^2 - [\Phi^2]),\nonumber \\
\mathcal{L_5}&= M_p^2 G_5 G_{\mu \nu}\Phi^{\mu \nu} - \frac{1}{6}\Lambda_2^4 G_{5,X}([\Phi]^3 - 3[\Phi][\Phi^2] + 2[\Phi^3]), \nonumber
\end{align}

where $G_2,G_3,G_4,G_5$ are functions of $\phi$ and $X = -\frac{1}{2}\nabla^\mu \phi \nabla_\mu \phi /\Lambda_2^4$, $\Phi^{\mu}_{ \ \nu}:= \nabla^\mu \nabla_\nu \phi/\Lambda_3^3$ and square brackets indicate the trace, e.g. $[\Phi^2] = \nabla^\mu \nabla_\nu \phi \nabla^\nu \nabla_\mu \phi/\Lambda_3^6$ and $,$ denote partial derivatives.

In the paper [arXiv:1904.05874](https://arxiv.org/abs/1904.05874) they state the Feynman rules in the Appendix. In the paper they try to derive the Feynman rules from the above action around the vacuum expectation value $\langle \phi \rangle = 0$. It says for instance that the leading-order vertex for $\phi\phi\phi$ is given by:

$$-\frac{\bar{G}_{3,X} + 3\bar{G}_{4,\phi X}}{3\Lambda_3^3}\delta^{\mu \nu}_{\alpha \beta}\phi \phi^\nu_\mu \phi^\beta_\alpha,$$

where the bar indicates that the function is evaluated at $\langle \phi \rangle = 0$, $\delta$ is the generalized Kronecker delta and $\phi^\nu_\mu:= \nabla^\nu \nabla_\mu \phi$.

My question is how can I derive such an expression from the action?