Integrate $\frac1{\sqrt{u^2+v^2} \left(1+u^2+v^2\right)}$ over $[-1,1]^2$?

Question

Problem

$$ \text{Evaluate} \quad I = \iint\limits_{[-1,1]^2} \frac{du\,dv}{\sqrt{u^2+v^2} \left(1+u^2+v^2\right)}$$

The integral is a special case of one that appeared in this recent post, in which I included a Wolfram cloud notebook containing the numerical and exact values of $I$ over the first quadrant.

Attempt

I tried evaluating $I$ in polar coordinates. A round of integration by parts, a substitution of $t=\cos\theta$, and an application of $\arccos t=\dfrac\pi2-\arcsin t$ later, I arrive at

$$\newcommand{dilog}[1]{\operatorname{Li}_2\left(#1\right)} \begin{align*} I &= \int_{-1}^1 \int_{-1}^1 \frac{du \, dv}{\sqrt{u^2+v^2} \left(1+u^2+v^2\right)} \\ &= 8 \int_0^1 \int_0^u \frac{du \, dv}{\sqrt{u^2+v^2} \left(1+u^2+v^2\right)} \\ &= 8 \int_0^\tfrac\pi4 \int_0^{\sec\theta} \frac{dr \, d\theta}{1+r^2} & (u,v)=(r\cos\theta,r\sin\theta) \\ &= 8 \int_0^\tfrac\pi4 \arctan(\sec\theta) \, d\theta \tag{$*$} \\ &= 2\pi \arctan\sqrt2 - 8 \int_0^\tfrac\pi4 \theta \frac{\sec\theta \tan\theta}{1 + \sec^2\theta} \, d\theta & \rm IBP \\ &= 2\pi \arctan\sqrt2 - 8 \int_\tfrac1{\sqrt2}^1 \frac{\arccos t}{1+t^2} \, dt & t=\cos\theta \\ &= \pi^2 - 2\pi\arctan\sqrt2 + 8 \int_\tfrac1{\sqrt2}^1 \frac{\arcsin t}{1+t^2} \, dt \tag{$\star$} \end{align*}$$

I've encountered a handful of integrals before with forms similar to that of $(*)$, i.e. composition of one trig function with another's inverse (example) that usually admitted tidy closed forms. (Granted, the integrals I have in mind usually were taken over $0\le\theta\le\pi/2$.)

As for the integral in $(\star)$, Mathematica produces a fairly digestible closed form (see the first output of the bottom-most section in the notebook) which I've been able to manually simplify and verify numerically to end up with

$$I = \pi^2 + 8 \, \Re \left[\dilog{\left(\sqrt2-1\right)e^{i3\pi/4}} - \dilog{\left(\sqrt2-1\right)e^{i\pi/4}}\right]$$

Questions

1. How can we arrive at this result?

2. Is this as simple as the closed form gets?

Edit: The answer to Q2 may actually be "no, we can do better", provided that the integrals below can be evaluated in closed form. With $\alpha=1-\dfrac1{\sqrt2}$, we have

$$\newcommand{\dilog}[1]{\operatorname{Li}_2\left(#1\right)} \begin{align*} \Re \dilog{\left(\sqrt2-1\right)e^{i\pi/4}} &= \Re \dilog{\alpha (1+i)} = I(-\alpha) \\ \\ \Re\dilog{\left(\sqrt2-1\right)e^{i3\pi/4}} &= \Re \dilog{\alpha (-1+i)} = I(\alpha) \end{align*}$$

where for $a\in\Bbb R$,

$$I(a) = -\frac12 \int_0^1 \frac{\log\left(1 + 2a t+2a^2t^2\right)}t \, dt$$

$$\implies I = \pi^2 + 4 \int_0^1 \log\left(\frac{1-2\alpha t+2\alpha^2t^2}{1+2\alpha t+2\alpha^2t^2}\right) \, \frac{dt}t$$

Answer 1

$$\int\frac{du}{\sqrt{u^2+v^2}\left(1+u^2+v^2\right)}=\frac{1}{\sqrt{v^2+1}}\tanh ^{-1}\left(\frac{u^2+v^2+1-u \sqrt{u^2+v^2}}{\sqrt{v^2+1}}\right)$$ Use the bounds and the logarithmic form of the hyperbolic arctangent $$\int_{-1}^{+1}\frac{du}{\sqrt{u^2+v^2}\left(1+u^2+v^2\right)}=\frac{1}{\sqrt{v^2+1}}\log \left(1+\frac{2}{v^2}\right)$$

Mathematica finds the (nasty) antiderivative writing (I suppose)

$$\log \left(1+\frac{2}{v^2}\right)=\log \left(v-i\sqrt{2}\right)+ \log \left(v+i\sqrt{2}\right)-2 \log (v)$$ For $$\int \frac{\log (v+a)}{\sqrt{v^2+1}}\,dv$$ have a look here

Using bounds and simplifying, the required integral is $$\frac{5 \pi ^2}{12}+2 \sinh ^{-1}(1)^2+\frac{1}{2} \text{Li}_2\left(577-408 \sqrt{2}\right)-\text{Li}_2\left(17-12\sqrt{2}\right)- 2\, \text{Li}_2\left(3-2 \sqrt{2}\right)$$