Solving a first order differential equation in terms of Lambert W-function I am having great difficulty solving the following equation
$$\  \frac{ax}{(bx^2 + c)} = \frac{dx}{dt}  $$
I have re-edited the question. Any help is appreciated. Thank you.

For the equation
$$\frac{ax}{(bx^2 + c)} = \frac{dx}{dt}$$
it can be seen that
\begin{align}
\frac{dx}{dt} &= \frac{a}{2 b} \, \frac{2 b x}{b x^{2} + c} = \frac{a}{2 b} \, \frac{d}{dt} \, \ln(b x^{2} + c )
\end{align}
which yields 
$$x(t) = \frac{a}{2 b} \, \ln(b x^{2} + c) + c_{0}.$$
Solving for $x(t)$ requires much struggle to obtain
$$x^{2}(t) = \frac{c}{b} \, W\left(\frac{b}{c} \, e^{\frac{2 \, a}{c} \, (t + p_{0})} \right),$$
where $p_{0}$ is a constant, and $W(x)$ is the Lambert w-function. 
For verification use:
$$\frac{dW}{d z} = \frac{W(z)}{z \, (1 + W(z))}$$
for which
\begin{align}
2 \, x \, \frac{dx}{dt} &= \frac{c}{b} \, \frac{b}{c} \, \frac{2 a}{c} \, e^{\frac{2 \, a}{c} \, (t + p_{0})} \, W'(\cdot) \\
&= \frac{2 a}{c} \, e^{\frac{2 \, a}{c} \, (t + p_{0})} \, \frac{W(\cdot)}{\frac{b}{c} \, e^{\frac{2 \, a}{c} \, (t + p_{0})} \, (1 + W(\cdot))} \\
&= \frac{2 a}{b} \, \frac{W(\cdot)}{1 + W(\cdot)} \\
&= \frac{2 a}{b} \, \frac{\frac{b}{c} \, x^{2}}{1 + \frac{b}{c} \, x^{2}} \\
&= \frac{2 a \, x^{2}}{b \, x^2 + c} \\
\frac{dx}{dt} &= \frac{a \, x}{b \, x^{2} + c}.
\end{align}
This is the first order differential equation that was given.
