Indefinitely integrating $\log(\sinh(x)+a)$? How can we integrate this indefinitely?

$$\int\ln(\sinh(x)+a)\\=-\frac{x^2}2+x\ln(\sinh x+a)-x\ln(1-(-a+\sqrt{a^2+1})e^{-x})-x\ln(1-(-a-\sqrt{a^2+1})e^{-x})+{\rm Li}_2((-a+\sqrt{a^2+1})e^{-x})+{\rm Li_2}((-a-\sqrt{a^2+1})e^{-x})$$

I found two relevant images but I think it's hand waving. This and this. Better if someone could explain in better words?
 A: First, expand $\sinh(x)=\dfrac{e^x-e^{-x}}2$ to get
$$
\begin{align}
\int\log(\sinh(x)+a)\,\mathrm{d}x
&=\int\log\left(\frac{e^x-e^{-x}+2a}2\right)\,\mathrm{d}x\\
&=\int\log\left(1+\left(a+\sqrt{a^2+1}\right)e^{-x}\right)\,\mathrm{d}x\\
&+\int\log\left(1+\left(a-\sqrt{a^2+1}\right)e^{-x}\right)\,\mathrm{d}x\\
&+\int\log\left(\frac{e^x}2\right)\,\mathrm{d}x\tag{1}
\end{align}
$$
Then, as was verified in this answer,
$$
\int\log(1+ae^{bx})\,\mathrm{d}x
=-\frac1b\,\mathrm{Li}_2\!\left(-ae^{bx}\right)+C\tag{2}
$$
Applying $(2)$ to $(1)$ gives
$$
\begin{align}
\int\log(\sinh(x)+a)\,\mathrm{d}x
&=\mathrm{Li}_2\!\left(-\left(a+\sqrt{a^2+1}\right)e^{-x}\right)\\
&+\mathrm{Li}_2\!\left(-\left(a-\sqrt{a^2+1}\right)e^{-x}\right)\\
&+\frac{x^2}2-\log(2)x+C\tag{3}
\end{align}
$$
To evaluate $\mathrm{Li}_2(x)$ for arguments outside of $[-1,1]$, we can use the identities shown in this answer.

Applying the Identities from the Cited Answer
For all $a,x\in\mathbb{R}$, $-\left(a+\sqrt{a^2+1}\right)e^{-x}\lt0$; therefore, $\mathrm{Li}_2\!\left(-\left(a+\sqrt{a^2+1}\right)e^{-x}\right)\in\mathbb{R}$. For $x\lt-1$, we can evaluate $\mathrm{Li}_2(x)$ using the identity
$$
\mathrm{Li}_2(x)=-\frac{\pi^2}{6}-\mathrm{Li}_2(1/x)-\frac12\log(-x)^2\tag{4}
$$
However, if $\sinh(x)+a\lt0$, then $-\left(a-\sqrt{a^2+1}\right)e^{-x}\gt1$; therefore, for $\sinh(x)+a\lt0$, we have $\mathrm{Li}_2\!\left(-\left(a-\sqrt{a^2+1}\right)e^{-x}\right)\not\in\mathbb{R}$. For $x\gt1$, we can evaluate $\mathrm{Li}_2(x)$ using the identity
$$
\mathrm{Li}_2(x)=\frac{\pi^2}{3}-\frac12\log(x)^2-\mathrm{Li}_2(1/x)-i\pi\log(x)\tag{5}
$$
