I am confident with
$$A\cosh(\alpha x) + B\sinh(\alpha x) + C = \frac{A+B}{2} e^{\alpha x} (1-pe^{-\alpha x}) (1-qe^{-\alpha x}).$$
where $p$ and $q$ are solutions of the equation $ \frac{A+B}{2}t^2 + Ct + \frac{A-B}{2} = 0$.
I assumed that taking natural $\log$ (i.e. ln) yields:
\begin{align*} &\log\left(A\cosh(\alpha x) + B\sinh(\alpha x) + C\right) \\ &\hspace{6em}= \log(\frac{A+B}{2}) + \alpha x + \log(1-pe^{-\alpha x}) + \log(1-qe^{-\alpha x}). \end{align*}
However, wolfram says:
\begin{align*} &\log\left(A\cosh(\alpha x) + B\sinh(\alpha x) + C\right) \\ &\hspace{6em}= 2i\pi\lfloor(\frac{-(\arg(1-pe^{-ax})+\arg(1-qe^{-ax})-\pi)}{2\pi})\rfloor\\ &\hspace{6em}+\log(\frac{A+B}{2}) + \alpha x + \log(1-pe^{-\alpha x}) + \log(1-qe^{-\alpha x}). \end{align*}
Can someone please explain the appearance of those args. I do not get it.
PS: More confusion comes from the fact that the quantity that I'm measuring (i.e. $F(x)=A\cosh(\alpha x) + B\sinh(\alpha x) + C$) is all real, so having imaginary part in $\log$ result does not seem right to me at all.