I want to solve this equation for $t$: $$ \frac{1}{\kappa}\tanh^{-1}(\kappa t)+bt=c, $$ given that $\kappa$ is a constant, and $b$ and $c$ are functions not in terms of $t$.
I tried to convert it into the form of natural logarithms and transform it with the power of $e$ like below: $$ e^{bt}+\kappa te^{bt}+\kappa te^c=e^c $$
However, $t$ lies on both power of $e$ and the linear position, and I have been stuck at extracting $t$ for hours.
How can I get the solution to the equation? Is there any clever way to extract the solution?
$$\frac{1}{\kappa}\tanh^{-1}(\kappa t)+bt=c\tag 1$$ is an highly transcendental equation and then you will require some numerical method for obtainin the solution.
In fact, there is an analytical solution. Rewrite $(1)$ as $$\kappa t=\tanh (\kappa (c-b t)) \tag 2$$ Let $x=\kappa (c-b t)$ to make $$\frac{c \kappa -x}{b}=\tanh(x)= \frac {e^{2x}-1 } {e^{2x}+1 } \tag 3$$ which leads to $$e^{-2x}=-\frac{x+(b-c \kappa)}{x-(b+c \kappa)} \tag 4$$ which has solutions in terms of the generalized Lambert function (have a look at equation $(4)$ in the linked paper).
This being said, it is not very useful from a practical point of view.
Just think about numerical methods.
Edit
Equation $(2)$ can rewrite and look for the zero of function $$f(x)=x+b\tanh(x)-c \kappa\qquad \text{where} \qquad x=\kappa (c-b t)$$
Assuming $b>0$ and $c \kappa >0$, we face an increasing function since $$f'(x)=1+b \,\text{sech}^2(x)\quad > 0 \quad \forall x$$ Starting Newton method with $x_0=0$ will give as an estimate $$x_1=\frac{c \kappa }{b+1}$$
Trying with $b=\pi$ and $c \kappa=12.345$, Newton iterates will be $$\left( \begin{array}{cc} n & x_n \\ 0 & 0 \\ 1 & 2.9807374 \\ 2 & 9.0248926 \\ 3 & 9.2034074 \end{array} \right)$$
Doing the same with $b=\frac 1\pi$ $$\left( \begin{array}{cc} n & x_n \\ 0 & 0 \\ 1 & 9.3642626 \\ 2 & 12.026690 \end{array} \right)$$