= Solution
Let $X=\partial^{-2}\zeta$, with an inverse spatial <Laplacian> defined by a boundary or zero-mode prescription. It depends on $\zeta$, not on its time derivative. Differentiation gives the <canonical momentum>
$$
\boxed{\pi=\dot\zeta+3\alpha\dot\zeta^2+2\gamma X\dot\zeta.}
$$
Count $\zeta,\pi,X$ as first order. Inverting perturbatively gives
$$
\dot\zeta=\pi-3\alpha\pi^2-2\gamma X\pi+O(\zeta^3).
$$
For the <Legendre transform in mechanics>, write $v=\dot\zeta=\pi+\Delta$ with $\Delta$ second order. The quadratic kinetic contribution is $\pi v-v^2/2=\pi^2/2-\Delta^2/2$, whose correction is fourth order; in the cubic terms one may consequently replace $v$ by $\pi$. The <Hamiltonian> density is therefore
$$
\boxed{\mathcal H=\tfrac12\pi^2+\tfrac12(\partial\zeta)^2-\alpha\pi^3-\beta\zeta(\partial\zeta)^2-\gamma(\partial^{-2}\zeta)\pi^2+O(\zeta^4).}
$$
The cubic <interaction Hamiltonian> density is
$$
\boxed{\mathcal H_{\rm int}=-\alpha\pi^3-\beta\zeta(\partial\zeta)^2-\gamma(\partial^{-2}\zeta)\pi^2.}
$$
In the free <interaction picture>, $\pi_I=\dot\zeta_I$. Thus here $\mathcal H_{\rm int}=-\mathcal L_3$ with the time derivatives interpreted as free fields. This equality follows from the <cubic Legendre transform for scalar derivative interactions>; at higher orders additional terms can appear.
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