3 H \dot{H} + \frac{M^2}{6 \left( 2\alpha+3\beta+2\beta \ln x \right)} H \ \nonumber \\
& - \frac{2\beta}{2\alpha+3\beta+2\beta \ln x } H^3 =  \left(\frac{M^2}{2\alpha+3\beta+2\beta \ln x} \right) \frac{\rho}{3 \bMp^2 H} \, . \label{dotR3} \end{aligned}$$ Now one can scrutinize the behaviors of $R$ and $H$ from Eqs.(\[dotR2\])-(\[dotR3\]). First, one can puts $\dot{R} = 0$ and $\rho = 0$ in the slow-roll region. Then one obtains $$\begin{aligned}
 & R_{i} \simeq 3 b H_{i}^2 \left( 1 + \sqrt{1 + \frac{2c}{3b^2 H_{i}^2}} \, \right) \label{Ri} \, , \\
& \text{where} \,\, b \equiv \frac{2\alpha+\beta+2\beta \ln x}{\alpha+\beta+\beta\ln x} \,\, , \,\, c \equiv \frac{M^2}{\alpha+\beta+\beta\ln x} \, . \label{bc} 
 \end{aligned}$$ As one expects $R_{i} \sim 12 H_{i}^2$ in this region, because $b \sim 2$ when $\alpha \gg \beta$. Even though $H$ is almost constant, it decreases almost linearly in time with a small slope. This can be obtained from Eq.(\[dotR3\]) $$\begin{aligned}
 \dot{H} &\simeq - \frac{M^2}{18 \left(2\alpha+3\beta+2\beta \ln x \right)} + \frac{2\beta}{3 \left(2\alpha+3\beta+2\beta \ln x \right) } H^2 \nonumber \\ 
& \simeq - \frac{M^2}{18 \left(2\alpha+3\beta+2\beta \ln x \right)} \label{SRdotH} \, , \end{aligned}$$ where we use the approximation $\beta \ll 1$ in the second equality. In order to obtain the finite period of inflation, $\left(2\alpha+3\beta+2\beta \ln x \right)$ should 