\frac{b}{m_p^2} \frac{(h+1)^4}{(6h)^2(2h+1)} R^2 \ln [R] - \frac{b}{m_p^2} \frac{(h+1)^4}{(6h)^2(2h+1)} \ln \left[ \frac{m_p^2 6h(2h+1)}{(h+1)^2}\right] R^2 \nonumber \\
&+ \frac{c}{m_p^2} \frac{(h+1)^4}{(6h)^2(2h+1)} R^2 ~~,\end{aligned}$$ where $m_p^2 = \frac{1}{8\pi G}$. This equation is a nonhomogeneous Euler differential equation and the solution can be written as $$\label{fR1}
f(R) = C_1 R^q + C_2 R^r + \delta R + \alpha R^{\frac{3}{2}} + \beta R^2 + \gamma R^2 \ln [R] ~~,$$ where $$\label{set1}
\left\{\begin{array}{ll} 
q = \frac{1}{4} \left[3 + h + \sqrt{h^2 - 10h + 1}~\right] \\\\
r = \frac{1}{4} \left[3 + h - \sqrt{h^2 - 10h + 1}~\right] \\\\
\delta = - \frac{n^2 (h+1)^2}{h^2} \\\\
\alpha = - \frac{1}{m_p} \frac{4a (h+1)^3}{(h+2)(6h)^{\frac{3}{2}} (2h+1)^{\frac{1}{2}}} \\\\
\beta = \frac{1}{m_p^2} \left[- \frac{b~(10+3h-h^2)(h+1)^4}{162~h^2(h+2)(2h+1)} - \frac{b~(h+1)^4}{54~h^2(2h+1)} \ln \left\{ \frac{m_p^2 6~h(2h+1)}{(h+1)^2}\right\} +  \frac{c~(h+1)^4}{54~h^2(2h+1)} \right] \\\\
\gamma = \frac{b}{m_p^2}\frac{(h+1)^4}{54~h^2(2h+1)}
\end{array} \right.$$ with $C_{1,2}$ as integration constants whose value can be predicted by the boundary conditions. The boundary conditions being $$f(R) \vert_{R=R_0} =- 2R_0  ~~~~~~~~~~     \text{and}  ~~~~~~~~~~  f'(R)\vert_{R=R_0} \sim 0 ~~,$$ where $R=R_0$ is the present value of $R$ which is a small constant. The value of $R_0$ is of the order of $(10^{-33}eV)^2$. If we apply the boundary conditions we get the values of $C_{1,2}$ as $$\begin{aligned}
C_1  =&  \frac{R_0^{1-q}}{2(q-r)} \Big[2(2r - \delta +  r 