In a recent answer, I found that this definite integral takes on an exact value of
$$\begin{align*}
& \frac{4(a-1)}{\sqrt{(a-1)^2+1}} \left(a \sinh^{-1} \sqrt{a-1} – \sinh^{-1}(a-1)\right) \\
& = \frac{4(a-1)}{\sqrt{(a-1)^2+1}} \left(a \log\left(\sqrt{a-1}+\sqrt a\right) – \log\left(\sqrt{(a-1)^2+1} + a – 1\right)\right) \tag1
\end{align*}$$
for $a=\sqrt2=1+\tan\dfrac\pi8$. Numerical evidence {(1), (2)} suggests this is equivalent to the closed form given in the OP,
$$\begin{align*}
& \log\left[\left(2a – 1 + 2\sqrt{2-a}\right)^\sqrt{2+a} \left(2a + 1 – 2\sqrt{2+a}\right)^\sqrt{2-a}\right] \\
&= \sqrt{a^2+a} \log\left(a^3-1+a^2\sqrt{a^2-a}\right) + \sqrt{a^2-a} \log\left(a^3+1-a^2\sqrt{a^2+a}\right) \tag2
\end{align*}$$
However, establishing the identity $(1)=(2)$ seems like a nontrivial task. It's not true for all real $a$ (plotted on the domain of $(2)$). Solving the underlying transcendental equation is hopeless as far as I can tell.
Question: How can one show that $(1)=(2)$ when $a=\sqrt2$?
(attempt/previous edit rolled up into self-answer)
Best Answer
Since $a=\sqrt2$, we may simplify
$$\begin{cases} \frac{4(a-1)}{\sqrt{(a-1)^2+1}} \equiv \frac{a^5-a^4}{\sqrt{a^4-a^3}} = \sqrt{a^6-a^5} \\ \sqrt{\frac{a^2+a}{a^2-a}} \equiv 1+a \\ 1+a-a^3\equiv1-a & (*) \end{cases}$$
The problem thus boils down to showing
$$a^3 \log\left(\sqrt{a-1} + \sqrt a\right) - a^2 \log\left(a - 1 + a \sqrt{a^2-a}\right) \\ \equiv (a+1) \log\left(a^3-1+a^2\sqrt{a^2-a}\right) + \log\left(a^3+1-a^2\sqrt{a^2+a}\right) \tag{$\diamondsuit$}$$
We may also rewrite
$$\begin{align*} & a^3 \log\left(\sqrt{a-1} + \sqrt a\right) \\ &\equiv a \log\left(2a-1 + 2 \sqrt{a^2-a}\right) \\ &\equiv a \log \left(a^3 - 1 + a^2 \sqrt{a^2-a}\right) \end{align*}$$
which further reduces $(\diamondsuit)$ to
$$- a^2 \log\left(a - 1 + a \sqrt{a^2-a}\right) \\ \equiv \log\left(a^3-1+a^2\sqrt{a^2-a}\right) + \log\left(a^3+1-a^2\sqrt{a^2+a}\right)$$
The LHS can be rewritten
$$\begin{align*} & -a^2 \log\left(a-1+a\sqrt{a^2-a}\right) \\ &\equiv \log\frac1{(a-1)^2 + 2(a^2-a) \sqrt{a^2-a} + a^4-a^3} \\ &\equiv \log\left(-a^5 + a^4 + a^2 + 1 - (a^4-a^3) \sqrt{a^2-a}\right) \end{align*}$$
and joining the logs on the RHS yields an argument of
$$\begin{align*} & \left(a^3-1+a^2\sqrt{a^2-a}\right) \left(a^3+1-a^2\sqrt{a^2+a}\right) \\ &\equiv a^6-a^5-1 + a^2 (a^3+1) \sqrt{a^2-a} - a^2 (a^3-1) \sqrt{a^2+a} \\ &\equiv a^6 - a^5 - 1 + \left((a^2-a^5) (a+1) + a^2+a^5\right) \sqrt{a^2-a} \\ &\equiv a^6 - a^5 - 1 + a^3 (1 + a - a^3) \sqrt{a^2-a} \tag{$*$} \\ &\equiv a^6 - a^5 - 1 - (a^4-a^3) \sqrt{a^2-a} \end{align*}$$
$(1)=(2)$ then follows from the fact that
$$a^4+a^2+1 \equiv a^6-1=7$$