Evaluating $\int_0^1 \frac{z \log ^2\left(\sqrt{z^2+1}-1\right)}{\sqrt{1-z^2}} \, dz$

Question

What kind of real analysis tools would you employ for this integral?

$$\int_0^1 \frac{z \log ^2\left(\sqrt{1+z^2}-1\right)}{\sqrt{1-z^2}} \, dz$$

EDIT: Here is a supplementary question, the cubic log integral version $$\int_0^1 \frac{z \log ^3\left(\sqrt{1+z^2}-1\right)}{\sqrt{1-z^2}} \, dz$$

Answer 1

1. Change of variables $y=\sqrt{1-z^2}$ transforms the integral into $$\mathcal{I}=\int_0^1\ln^2\left(\sqrt{2-y^2}-1\right)dy.$$
2. Further change of variables $y=\sqrt2 \sin\varphi$ leads to $$\mathcal{I}=\sqrt2 \int_{0}^{\pi/4}\cos\varphi\ln^2\left(\sqrt2 \cos\varphi-1\right)d\varphi.$$
3. Next we can integrate once by parts to kill one of the logarithms : \begin{align}\mathcal{I}=\ln^2\left(\sqrt2-1\right)+2\sqrt2 \int_{0}^{\pi/4}\frac{\sqrt2\sin\varphi-1}{\sqrt{2}\cos\varphi-1}\sin\varphi\ln\left(\sqrt2 \cos\varphi-1\right)d\varphi. \end{align}
4. Finally, pass to a rational parametrization by setting $t=\tan\frac{\varphi}{2}$: $$\mathcal{I}=\ln^2\left(\sqrt2-1\right)+8\left(2-\sqrt2\right) \int_{0}^{\sqrt2-1}\frac{t\left(t-1-\sqrt2\right)}{\left(1+t^2\right)^2\left(t-1+\sqrt2\right)}\ln\left(\sqrt2\, \frac{1-t^2}{1+t^2}-1\right)dt.$$
5. Any antiderivative of the form $\displaystyle\int R_1(t)\ln R_2(t)\,dt$ with rational $R_{1,2}(t)$ can be expressed in terms of dilogarithms. In particular, this is true in our case (use e.g. Mathematica). A fearless reader is invited to simplify the resulting combination of dilogarithm values.

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6. After a simplification of the dilogarithmic expressions, one finally obtains $$\boxed{\quad\displaystyle\mathcal{I}=2\mathbf{K}-\frac{13\pi^2}{48}+\frac{\pi}{2}+\frac{\ln^22}{4}+\frac{\pi\ln 2}{4}-\ln 2+2\quad}$$ where $\mathbf{K}$ denotes the Catalan's constant.