m_p^2 (h-1)^2}{2h(2h-1)}R + \frac{a m_p (h-1)^3}{(6h)^{\frac{3}{2}}(2h-1)^{\frac{3}{2}}} R^{\frac{3}{2}} + \frac{(h-1)^4 R^2}{(6h)^2(2h-1)^2}\left[b\ln\left\{\frac{6h(2h-1) m_p^2}{(h-1)^2 R} \right\} + c \right] ~~.$$ So for the quintessence scale factor eqn.(\[quin1\] and \[quin2\]) we can rewrite eqn.(\[maineq\]) with eqn.(\[rhoq\]) as $$\begin{aligned}
R^2 f''(R) &+ \frac{(h-1)}{2}Rf'(R) - \frac{(2h-1)}{2}f(R) = \frac{n^2 (h-1)^2}{2h}R  +\frac{a}{m_p} \frac{(h-1)^3}{(6h)^{\frac{3}{2}}(2h-1)^{\frac{1}{2}}} R^{\frac{3}{2}} \nonumber \\
&- \frac{b}{m_p^2} \frac{(h-1)^4}{(6h)^2(2h-1)} R^2 \ln [R] + \frac{b}{m_p^2} \frac{(h-1)^4}{(6h)^2(2h-1)} \ln \left[ \frac{m_p^2 6h(2h-1)}{(h-1)^2}\right] R^2 \nonumber \\
&+ \frac{c}{m_p^2} \frac{(h-1)^4}{(6h)^2(2h-1)} R^2 ~~,\end{aligned}$$ where $m_p^2 = \frac{1}{8\pi G}$. Similarly like the phantom case the solution can be written as $$\label{fR2}
f(R) = C_1 R^q + C_2 R^r + \delta R + \alpha R^{\frac{3}{2}} + \beta R^2 + \gamma R^2 \ln [R] ~~.$$ where $$\label{set2}
\left\{\begin{array}{ll} 
q = \frac{1}{4} \left[3 - h + \sqrt{h^2 + 10h + 1}~\right] \\\\
r = \frac{1}{4} \left[3 - h - \sqrt{h^2 + 10h + 1}~\right] \\\\
\delta = - \frac{n^2 (h-1)^2}{h^2} \\\\
\alpha =  \frac{2a}{3m_p} \frac{(h-1)^3}{h(2-h)(2h-1)^{\frac{1}{2}}} \\\\
\beta = \frac{1}{m_p^2} \left[ \frac{b~(10-3h-h^2)(h-1)^4}{162~h^2(2-h)(2h-1)} + \frac{b~(h-1)^4}{54~h^2(2h-1)} \ln \left\{ \frac{m_p^2 6~h(2h-1)}{(h-1)^2}\right\} +  \frac{c~(h-1)^4}{54~h^2(2h-1)} \right] \\\\
\gamma = - \frac{b}{m_p^2}\frac{(h-1)^4}{54~h^2(2h-1)}
\end{array} \right.$$ The boundary conditions will give $$\begin{aligned}
C_1  =&  \frac{R_0^{1-q}}{2(q-r)} \Big[2(2r - \delta +  r \delta) - \alpha \sqrt{R_0}~(3 - 2r) - 2\beta R_0~ (2 -r) - \nonumber \\
& 2\gamma R_0 