^2\pi }{a}+\frac{4 \dot{\pi}  H \partial _i\partial _j^2\pi
   \partial _i\dot{\pi}}{a}+\frac{H \partial _j\pi  \partial _i\dot{\pi}
   \partial _i\partial _j\dot{\pi}}{a}+\frac{5 H \partial
   _i\partial _j\pi \partial _i\dot{\pi} \partial _j\dot{\pi}}{a}+\frac{H \partial _j^2\pi  \left(\partial _i\dot{\pi}\right){}^2}{a}+\frac{H \partial _i \pi \partial _i\dot{\pi} \partial
   _j^2\dot{\pi}}{a}-\frac{H^3 \left(\partial _i\pi \right){}^2 \partial _j^2\pi
   }{a}+3 \dot{\pi}  a H^4 \left(\partial _i\pi \right){}^2-6 \dot{\pi}  a H^3 \partial
   _i\pi  \partial _i\dot{\pi}+12 \dot{\pi}^2 a H^3 \partial _i^2\pi
   -2 \dot{\pi} a H^2 \left(\partial _i\dot{\pi}\right){}^2-\frac{2 \dot{\pi}  (\partial
   _i\partial _j^2\pi )^2
   }{a^3}-\frac{\partial _j\pi  \partial _i\partial _k^2\pi \partial
   _i\partial _j\dot{\pi}}{a^3}-\frac{5 \partial _i\partial _j\pi  \partial _i\partial _k^2\pi \partial _j\dot{\pi}}{a^3}-\frac{\partial _j^2\pi  \partial _i\partial _k^2\pi 
   \partial _i\dot{\pi}}{a^3}-\frac{\partial _j\pi  \partial _i^2\partial
   _j\pi  \partial _k^2\dot{\pi}}{a^3}+18 \dot{\pi}^3 a^3 H^4
\end{dmath}

\section{More Details about the Numerical Calculations of Bispectrum and the Variable Transformations} \label{numericalcalculations}
To calculate different contributions to $\left<\zeta(\textbf{k}_1)\zeta(\textbf{k}_2)\zeta(\textbf{k}_3)\right>$, we first substituted $\zeta$ for $\pi$ using $\pi \rightarrow \zeta = -H\pi$ and then transformed from cosmic time to conformal time $dt\rightarrow a\; d\tau$. For the $\dot\pi(\partial\pi)^2$ term, this results in factoring factors of $H$ and $a$ into the coefficients in the following way, 
\begin{equation}
   C_{\dot{\pi} (\partial\pi)^2}\dot{\pi} (\partial\pi)^2 \xrightarrow{}-\frac{1}{H^3}C_{\dot{\pi} (\partial\pi)^2}\dot\zeta(\partial\zeta)^2 \xrightarrow{} -\frac{1}{aH^3}C_{\dot{\pi} (\partial\pi)^2}\zeta'(\partial\zeta)^2.
\end{equation}
Therefore, we took track of these factors by defining
\begin{equation}
    C_{\zeta'(\partial\zeta)^2} = -\frac{1}{aH^3}C_{\dot{\pi} (\partial\pi)^2}=-\frac{1}{H^3}[-\delta_3H^4 +9\bar{M}_4H^3-2M_2^4]. 
\end{equation}
After evaluating all the commutation relations between the operators in the interaction terms and the external vertices, we found the contribution to the bispectrum to be
\begin{align}
   \left<\zeta(\textbf{k}_1)\zeta(\textbf{k}_2)\zeta(\textbf{k}_3)\right>_{\dot\pi(\partial\pi)^2}=-i\int_{-\infty}^{\tau_f} a(\tau')d\tau' C_{\zeta'(\partial\zeta)^2}&\;2(2\pi)^3\delta^3(\textbf{k}_3+\textbf{k}_2+\textbf{k}_1)\times\nonumber\\f_{k_1}(\tau_f){f_{k_1}^*}'(\tau')f_{k_2}(\tau_f){f_{k_2}^*}(\tau')&f_{k_3}(\tau_f){f_{k_3}^*}(\tau'){\textbf{k}}_2\cdot{\textbf{k}}_3+ \text{\text{cycl.}} + \text{c.c.},
\end{align} where $f$ as we had defined in \eqref{mfunction} represents the mode function for $\zeta(\textbf{k})$. Next, we substitute for $f$ and $f'$ in terms of the canonical variable $u_k(\tau)$ which satisfies the Wronskian condition and had been computed numerically from quadratic action. The substitution relations are given by
\begin{align*}
\sqrt{A_1}f &= -\frac{H}{a}u\\
\sqrt{A_1}f'&= -\frac{H}{a}\frac{du}{d\tau} +H^2u
\end{align*} 
to obtain
\begin{align}
   \left<\zeta(\textbf{k}_1)\zeta(\textbf{k}_2)\zeta(\textbf{k}_3)\right>_{\dot\pi(\partial\pi)^2}&=&-i\int_{-\infty}^{\tau_f}a(\tau')d\tau'\left(\frac{H^6}{a(\tau_f)^3a(\tau')^3 A_1^3}\right)\;C_{\zeta'(\partial\zeta)^2}\;2(2\pi)^3\delta^3(\textbf{k}_3+\textbf{k}_2+\textbf{k}_1)\nonumber\\
   &&u_{k_1}(\tau_f){u_{k_1}^*}'(\tau')u_{k_2}(\tau_f){u_{k_2}^*}(\tau')u_{k_3}(\tau_f){u_{k_3}^*}(\tau'){\textbf{k}}_2\cdot{\textbf{k}}_3+ \text{cycl.} + \text{c.c.}\nonumber\\
    &&+i\int_{-\infty}^{\tau_f}a(\tau')d\tau'\left(\frac{H^{7}}{a(\tau_f)^3a(\tau')^2 A_1^3}\right)C_{\zeta'(\partial\zeta)^2}\;2(2\pi)^3\delta^3(\textbf{k}_3+\textbf{k}_2+\textbf{k}_1)\nonumber\\
   &&u_{k_1}(\tau_f){u_{k_1}^*}(\tau')u_{k_2}(\tau_f){u_{k_2}^*}(\tau')u_{k_3}(\tau_f){u_{k_3}^*}(\tau'){\textbf{k}}_2\cdot{\textbf{k}}_3+ \text{cycl.} + \text{c.c.}
\end{align} 
Next, changing the integration variable to the dimensionless variable $x_i=c_sk_i\tau$, and substituting for $a=-1/H\tau= -c_sk_i/Hx_i$ and defining $\tilde{u}(x)= \sqrt{c_s k}\,u_k(\tau)$\footnote{This is due to the symmetry of our ODE and normalization under this change of variable. } as well as $x_f=c_sk_1\tau_f$, we arrive at. 
\begin{align}
   \left<\zeta(\textbf{k}_1)\zeta(\textbf{k}_2)\zeta(\textbf{k}_3)\right>_{\dot\pi(\partial\pi)^2}=&i2(2\pi)^3\delta^3(\textbf{k}_3+\textbf{k}_2+\textbf{k}_1)\frac{1}{\sqrt{c_s^6k_1^2k_2^2k_3^2}}\frac{H^{11}x_f^3}{(c_sk_1)^5 A_1^3} \left [\int_{-\infty}^{x_{f}}(x_1')^2 dx_1'\; \right. \nonumber\\&
  C_{\zeta'(\partial\zeta)^2}\; \tilde{u}(x_f){\tilde{u}^*}_{,x}(x_1')\tilde{u}(\frac{k_2}{k_1}x_f){\tilde{u}^*}(\frac{k_2}{k_1}x_3')\tilde{u}(\frac{k_3}{k_1}x_f){\tilde{u}^*}(\frac{k_2}{k_1} x_1')