in much more complicated differential equations. This would be the subject of a future work.

From the definitions of $\Omega$ and $z$ we have $$\begin{aligned}
\Omega &=& H^{1+r}\dot H, \label{Eq1} \\
z &=& H^{2+r} \left(\frac{-3[1+(1-r)\gamma]H}{2(1-r)} +\frac
{1}{\alpha r}\right), \label{Eq2}\end{aligned}$$ Analogous to the solution of reduced Abel type canonical equation, $$\label{abel1}
y \frac{dy}{dx} - y = a x$$ (see Appendix C) we obtain the relation $\Omega = z/{\cal P}$. Therefore, from Eqs. (\[Eq1\]) and (\[Eq2\]) we obtain the following first order differential equation for Hubble parameter $H$, $$\label{init-polyn}
{\cal P} \dot H = \frac{-3[1+(1-r)\gamma]}{2(1-r)} H^2 +\frac{1}{\alpha r}H$$ with the solution $$H(t) = \frac{B}{\exp(-Bt/{\cal P})-A} \label{eq:mysolut1}$$ where $$\label{paramsab}
A=\frac{-3[1+(1-r)\gamma]}{2(1-r)}, \hspace*{1cm} B=\frac{1}{\alpha r},$$ and ${\cal P}$ is taken as a free parameter. We can assign any real value to ${\cal P}$. For the results presented in this work, we used a negative value. This negative sign is necessarily to overcome the sign from the integral limits. The geometric and thermodynamic quantities of the Universe read $$\begin{aligned}
 a(t) &=& a_0\left(\frac{\exp(-B t/{\cal P})}{\exp(-B
   t/{\cal P})-A}\right)^{{\cal P}/A} \label{approx-a}, \\
\rho(t) &=& 3\, H^2= 3 \left(\frac{B}{\exp(-Bt/{\cal P})-A}\right)^2 \label{approx-rho},\\
T(t) &=& \beta \rho^{r}=\beta \left(3 \frac{B^{2}}{[\exp(-Bt/{\cal P})-A]^{2}}\right)^r \label{approx-T},\\
\Pi(t) &=& -2\dot{H}-3\gamma H^{2}=-\frac{B^2}{{\cal P}} \left(\frac{2\exp(-Bt/{\cal P})+3\gamma{\cal P}}{[\exp(-Bt/{\cal P})-A]^2}\right) \label{approx-Pi},\\
 q(t)&=&\frac{d}{dt}H^{-1}-1=-\frac{1}{{\cal 