$$x(t)\cdot r+x'(t)\cdot l+a\cdot\ln\left(1+\frac{x(t)}{b}\right)=0\space\Longleftrightarrow\space x(t)=\dots$$
This is a separable ODE.
$$x'(t)=\frac{dx}{dt}=-\frac{1}{l}\left(rx(t)+a\ln\left(1+\frac{x(t)}{b}\right)\right)$$
$$dt=-\frac{l}{rx+a\ln\left(1+\frac{x}{b}\right)}dx$$
$$t=-l\int\frac{dx}{rx+a\ln\left(1+\frac{x}{b}\right)}$$
With the condition $x(0)=x_0$ :
$$t(x)=-l\int_{x_0}^x\frac{d\xi}{r\xi+a\ln\left(1+\frac{\xi}{b}\right)}$$
As far as I know there is no closed form of this integral in terms of a finite number of standard functions. The same for the inverse function $x(t)$.
So, the analytic solution is a function defined by an integral. This is very common in practice. One have to proceed with numerical integration.
It is very easy to draw $x(t)$ thanks to usual numerical integration : Draw $t(x)$ from the above integral. Plot the points $(t,x)$ instead of $(x,t)$ , i.e. with $t$ on horizontal axis and $x$ on vertical axis.
To compute the value of $x(t)$ at a given value $t$ , proceed to the numerical integration of $-l\int_{x_0}\frac{d\xi}{r\xi+a\ln\left(1+\frac{\xi}{b}\right)}$ with $\xi$ increasing up to reach the specified value $t$ . The value of $\xi$ at this point gives the value of $x(t)$ .
