This integral (in a slightly different form) appeared in the book of Borwein and Devlin with a remark that no closed form exists for it so far:
$$I=\int_0^1 \frac{dy}{1+y^2} \tanh^{-1} \frac{1}{\sqrt{3+y^2}}$$
It was even named among "impossible integrals" (page 58).
I have spent several hours on this integral and found no closed form of course, but I don't think it's hopeless.
With the most obvious series expansion and some work (and liberal use of Mathematica), I have obtained the following explicit series for the integral:
$$I= \log(1+\sqrt{2})\arctan \frac{1}{\sqrt{2}}- \frac{1}{6} \sum_{n=0}^\infty \frac{1}{(2n+1)2^n}\sum_{l=0}^{n-1} \left(\frac{2}{3} \right)^l \sum_{k=0}^{l} \frac{(-1)^k}{(2k+1)4^k} \binom{l}{k}$$
The series has good convergence, by the way:
Is there a more simple series expression for the integral, or, who knows, maybe even a closed form was found after the book was published?
Recall the definition of the area hyperbolic tangent in terms of the natural logarithm as
$$\operatorname{artanh}{\left(z\right)}:=\frac12\ln{\left(\frac{1+z}{1-z}\right)};~~~\small{z\in\left(-1,1\right)},$$
and observe that
$$\forall\left(a,y\right)\in\left(0,1\right)\times\mathbb{R}:0<\frac{1}{\sqrt{a^{-2}+y^{2}}}<1.$$
Then, define the function $\mathcal{I}:\left(0,1\right)\rightarrow\mathbb{R}_{>0}$ via the definite integral
$$\mathcal{I}{\left(a\right)}:=\int_{0}^{1}\mathrm{d}y\,\frac{2}{1+y^{2}}\operatorname{artanh}{\left(\frac{1}{\sqrt{a^{-2}+y^{2}}}\right)}.$$
Note in particular that the OP's integral $I$ can be written as $I=\frac12\mathcal{I}{\left(\frac{1}{\sqrt{3}}\right)}$. We show that this quantity can be reduced to Clausen functions and elementary functions.
Suppose $a\in\left(0,1\right)$. Then, $0<\sqrt{2}-1<\sqrt{1+a^{2}}-a<1$. Also,
$$0<\frac{1+a-\sqrt{1+a^{2}}}{1-a+\sqrt{1+a^{2}}}=\frac{\sqrt{1+a^{2}}-1}{a}=\frac{a}{1+\sqrt{1+a^{2}}}<1.$$
Rescaling the integral and applying the appropriate Euler substitution, we find
$$\begin{align} \mathcal{I}{\left(a\right)} &=\int_{0}^{1}\mathrm{d}y\,\frac{2}{1+y^{2}}\operatorname{artanh}{\left(\frac{1}{\sqrt{a^{-2}+y^{2}}}\right)}\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{2a^{2}}{a^{2}+a^{2}y^{2}}\operatorname{artanh}{\left(\frac{a}{\sqrt{1+a^{2}y^{2}}}\right)}\\ &=\int_{0}^{a}\mathrm{d}x\,\frac{2a}{a^{2}+x^{2}}\operatorname{artanh}{\left(\frac{a}{\sqrt{1+x^{2}}}\right)};~~~\small{\left[ay=x\right]}\\ &=\int_{1}^{\sqrt{1+a^{2}}-a}\mathrm{d}t\,\frac{(-1)\left(1+t^{2}\right)}{2t^{2}}\cdot\frac{2a}{a^{2}+\left(\frac{1-t^{2}}{2t}\right)^{2}}\\ &~~~~~\times\operatorname{artanh}{\left(\frac{2at}{1+t^{2}}\right)};~~~\small{\left[\sqrt{1+x^{2}}=x+t\right]}\\ &=\int_{\sqrt{1+a^{2}}-a}^{1}\mathrm{d}t\,\frac{4a\left(1+t^{2}\right)}{4a^{2}t^{2}+\left(1-t^{2}\right)^{2}}\operatorname{artanh}{\left(\frac{2at}{1+t^{2}}\right)}\\ &=\int_{\sqrt{1+a^{2}}-a}^{1}\mathrm{d}t\,\frac{2a\left(1+t^{2}\right)}{4a^{2}t^{2}+\left(1-t^{2}\right)^{2}}\ln{\left(\frac{1+2at+t^{2}}{1-2at+t^{2}}\right)}\\ &=\int_{\sqrt{1+a^{2}}-a}^{1}\mathrm{d}t\,\frac{2a\left(1+t^{2}\right)}{1-2\left(1-2a^{2}\right)t^{2}+t^{4}}\ln{\left(\frac{1+2at+t^{2}}{1-2at+t^{2}}\right)}.\\ \end{align}$$
Set $b:=\sqrt{1-a^{2}}$. Then, $0<b<1$, and
$$\begin{align} \mathcal{I}{\left(a\right)} &=\int_{\sqrt{1+a^{2}}-a}^{1}\mathrm{d}t\,\frac{2a\left(1+t^{2}\right)}{1-2\left(1-2a^{2}\right)t^{2}+t^{4}}\ln{\left(\frac{1+2at+t^{2}}{1-2at+t^{2}}\right)}\\ &=\int_{\sqrt{1+a^{2}}-a}^{1}\mathrm{d}t\,\frac{2a\left(1+t^{2}\right)}{1-2\left(2b^{2}-1\right)t^{2}+t^{4}}\ln{\left(\frac{1+2at+t^{2}}{1-2at+t^{2}}\right)}\\ &=a\int_{\sqrt{1+a^{2}}-a}^{1}\mathrm{d}t\,\frac{2\left(1+t^{2}\right)}{\left(1-2bt+t^{2}\right)\left(1+2bt+t^{2}\right)}\ln{\left(\frac{1+2at+t^{2}}{1-2at+t^{2}}\right)}\\ &=a\int_{\sqrt{1+a^{2}}-a}^{1}\mathrm{d}t\,\left[\frac{1}{\left(1+2bt+t^{2}\right)}+\frac{1}{\left(1-2bt+t^{2}\right)}\right]\ln{\left(\frac{1+2at+t^{2}}{1-2at+t^{2}}\right)}\\ &=a\int_{\frac{1+a-\sqrt{1+a^{2}}}{1-a+\sqrt{1+a^{2}}}}^{0}\mathrm{d}u\,\frac{(-2)}{\left(1+u\right)^{2}}\cdot\frac12\left[\frac{\left(1+u\right)^{2}}{\left(1+b\right)+\left(1-b\right)u^{2}}+\frac{\left(1+u\right)^{2}}{\left(1-b\right)+\left(1+b\right)u^{2}}\right]\\ &~~~~~\times\ln{\left(\frac{\left(1+a\right)+\left(1-a\right)u^{2}}{\left(1-a\right)+\left(1+a\right)u^{2}}\right)};~~~\small{\left[t=\frac{1-u}{1+u}\right]}\\ &=a\int_{0}^{\frac{1+a-\sqrt{1+a^{2}}}{1-a+\sqrt{1+a^{2}}}}\mathrm{d}u\,\left[\frac{1}{\left(1+b\right)+\left(1-b\right)u^{2}}+\frac{1}{\left(1-b\right)+\left(1+b\right)u^{2}}\right]\\ &~~~~~\times\ln{\left(\frac{\left(1+a\right)+\left(1-a\right)u^{2}}{\left(1-a\right)+\left(1+a\right)u^{2}}\right)}\\ &=\frac{a}{\left(1+b\right)}\int_{0}^{\frac{a}{1+\sqrt{1+a^{2}}}}\mathrm{d}u\,\left[\frac{1}{1+\left(\frac{1-b}{1+b}\right)u^{2}}+\frac{1}{\left(\frac{1-b}{1+b}\right)+u^{2}}\right]\ln{\left(\frac{1+\left(\frac{1-a}{1+a}\right)u^{2}}{\left(\frac{1-a}{1+a}\right)+u^{2}}\right)}.\\ \end{align}$$
Set $A:=\sqrt{\frac{1-a}{1+a}}\land B:=\sqrt{\frac{1-b}{1+b}}\land Z:=\frac{a}{1+\sqrt{1+a^{2}}}$. Then, $0<A<1\land0<B<1\land0<Z<1\land\frac{a}{1+b}=B$, and
$$\begin{align} \mathcal{I}{\left(a\right)} &=\frac{a}{\left(1+b\right)}\int_{0}^{\frac{a}{1+\sqrt{1+a^{2}}}}\mathrm{d}u\,\left[\frac{1}{1+\left(\frac{1-b}{1+b}\right)u^{2}}+\frac{1}{\left(\frac{1-b}{1+b}\right)+u^{2}}\right]\ln{\left(\frac{1+\left(\frac{1-a}{1+a}\right)u^{2}}{\left(\frac{1-a}{1+a}\right)+u^{2}}\right)}\\ &=B\int_{0}^{Z}\mathrm{d}u\,\left[\frac{1}{1+B^{2}u^{2}}+\frac{1}{B^{2}+u^{2}}\right]\ln{\left(\frac{1+A^{2}u^{2}}{A^{2}+u^{2}}\right)}\\ &=\int_{0}^{Z}\mathrm{d}u\,\frac{B\ln{\left(\frac{1+A^{2}u^{2}}{A^{2}+u^{2}}\right)}}{1+B^{2}u^{2}}+\int_{0}^{Z}\mathrm{d}u\,\frac{B\ln{\left(\frac{1+A^{2}u^{2}}{A^{2}+u^{2}}\right)}}{B^{2}+u^{2}}\\ &=\int_{0}^{BZ}\mathrm{d}x\,\frac{\ln{\left(\frac{B^{2}+A^{2}x^{2}}{A^{2}B^{2}+x^{2}}\right)}}{1+x^{2}};~~~\small{\left[u=B^{-1}x\right]}\\ &~~~~~+\int_{0}^{B^{-1}Z}\mathrm{d}x\,\frac{\ln{\left(\frac{1+A^{2}B^{2}x^{2}}{A^{2}+B^{2}x^{2}}\right)}}{1+x^{2}};~~~\small{\left[u=Bx\right]}\\ &=\int_{0}^{BZ}\mathrm{d}x\,\frac{\ln{\left(\frac{B^{2}+A^{2}x^{2}}{A^{2}B^{2}+x^{2}}\right)}}{1+x^{2}}+\int_{0}^{B^{-1}Z}\mathrm{d}x\,\frac{\ln{\left(\frac{1+A^{2}B^{2}x^{2}}{A^{2}+B^{2}x^{2}}\right)}}{1+x^{2}}.\\ \end{align}$$
Define the function $\mathcal{J}:\mathbb{R}_{>0}^{2}\rightarrow\mathbb{R}$ via the definite integral
$$\mathcal{J}{\left(a,z\right)}:=\int_{0}^{z}\mathrm{d}x\,\frac{\ln{\left(a^{2}+x^{2}\right)}}{1+x^{2}}.$$
Now, suppose $\left(a,z\right)\in\mathbb{R}_{>0}^{2}\land a\neq1$, and set
$$\theta=2\arctan{\left(z\right)}\land\omega=\arctan{\left(\frac{z}{a}\right)}-\arctan{\left(z\right)}.$$
Then, it can be shown that
$$\begin{align} \mathcal{J}{\left(a,z\right)} &=\int_{0}^{z}\mathrm{d}x\,\frac{\ln{\left(a^{2}+x^{2}\right)}}{1+x^{2}}\\ &=\theta\ln{\left(1+a\right)}-\omega\ln{\left(\left|\frac{1-a}{1+a}\right|\right)}+\frac12\operatorname{Cl}_{2}{\left(2\theta+2\omega\right)}-\frac12\operatorname{Cl}_{2}{\left(2\omega\right)}-\operatorname{Cl}_{2}{\left(\theta\right)}.\\ \end{align}$$
Suppose $a=\frac{1}{\sqrt{3}}$. Then, $Z=A^{2}\land b=\sqrt{\frac23}\land A=\sqrt{2-\sqrt{3}}\land B=\sqrt{5-2\sqrt{6}}$. Note $0<B<A<1$. Also, observe that
$$\arctan{\left(A^{2}B^{-1}\right)}+\arctan{\left(A^{2}B\right)}=\frac{\pi}{4},$$
$$\arctan{\left(A^{2}B^{-1}\right)}-\arctan{\left(A^{2}B\right)}=\arcsin{\left(\frac{1}{\sqrt{3}}\right)}.$$
Then, setting
$$\theta:=\arctan{\left(A^{2}B\right)}\in\left(0,\frac{\pi}{4}\right),$$
$$\psi:=\arctan{\left(\frac{A^{2}}{B}\right)}\in\left(0,\frac{\pi}{4}\right),$$
$$\alpha:=\arctan{\left(A\right)}\in\left(0,\frac{\pi}{4}\right),$$
$$\gamma:=\arctan{\left(A^{3}\right)}\in\left(0,\frac{\pi}{4}\right),$$
we have
$$\begin{align} \mathcal{I}{\left(\frac{1}{\sqrt{3}}\right)} &=\int_{0}^{A^{2}B}\mathrm{d}x\,\frac{\ln{\left(\frac{B^{2}+A^{2}x^{2}}{A^{2}B^{2}+x^{2}}\right)}}{1+x^{2}}+\int_{0}^{A^{2}B^{-1}}\mathrm{d}x\,\frac{\ln{\left(\frac{1+A^{2}B^{2}x^{2}}{A^{2}+B^{2}x^{2}}\right)}}{1+x^{2}}\\ &=\int_{0}^{A^{2}B}\mathrm{d}x\,\frac{\ln{\left(A^{2}\right)}+\ln{\left(A^{-2}B^{2}+x^{2}\right)}-\ln{\left(A^{2}B^{2}+x^{2}\right)}}{1+x^{2}}\\ &~~~~~+\int_{0}^{A^{2}B^{-1}}\mathrm{d}x\,\frac{\ln{\left(A^{2}\right)}+\ln{\left(A^{-2}B^{-2}+x^{2}\right)}-\ln{\left(A^{2}B^{-2}+x^{2}\right)}}{1+x^{2}}\\ &=\int_{0}^{A^{2}B}\mathrm{d}x\,\frac{\ln{\left(A^{2}\right)}}{1+x^{2}}+\int_{0}^{A^{2}B^{-1}}\mathrm{d}x\,\frac{\ln{\left(A^{2}\right)}}{1+x^{2}}\\ &~~~~~+\int_{0}^{A^{2}B}\mathrm{d}x\,\frac{\ln{\left(A^{-2}B^{2}+x^{2}\right)}}{1+x^{2}}-\int_{0}^{A^{2}B}\mathrm{d}x\,\frac{\ln{\left(A^{2}B^{2}+x^{2}\right)}}{1+x^{2}}\\ &~~~~~+\int_{0}^{A^{2}B^{-1}}\mathrm{d}x\,\frac{\ln{\left(A^{-2}B^{-2}+x^{2}\right)}}{1+x^{2}}-\int_{0}^{A^{2}B^{-1}}\mathrm{d}x\,\frac{\ln{\left(A^{2}B^{-2}+x^{2}\right)}}{1+x^{2}}\\ &=\ln{\left(A^{2}\right)}\arctan{\left(A^{2}B\right)}+\ln{\left(A^{2}\right)}\arctan{\left(A^{2}B^{-1}\right)}\\ &~~~~~-\mathcal{J}{\left(AB,A^{2}B\right)}+\mathcal{J}{\left(\frac{B}{A},A^{2}B\right)}-\mathcal{J}{\left(\frac{A}{B},\frac{A^{2}}{B}\right)}+\mathcal{J}{\left(\frac{1}{AB},\frac{A^{2}}{B}\right)}\\ &=2\left(\theta+\psi\right)\ln{\left(A\right)}\\ &~~~~~-\mathcal{J}{\left(AB,A^{2}B\right)}+\mathcal{J}{\left(\frac{B}{A},A^{2}B\right)}-\mathcal{J}{\left(\frac{A}{B},\frac{A^{2}}{B}\right)}+\mathcal{J}{\left(\frac{1}{AB},\frac{A^{2}}{B}\right)},\\ \end{align}$$
where
$$\begin{align} \mathcal{J}{\left(AB,A^{2}B\right)} &=2\arctan{\left(A^{2}B\right)}\ln{\left(1+AB\right)}\\ &~~~~~-\left[\arctan{\left(A\right)}-\arctan{\left(A^{2}B\right)}\right]\ln{\left(\left|\frac{1-AB}{1+AB}\right|\right)}\\ &~~~~~+\frac12\operatorname{Cl}_{2}{\left(2\arctan{\left(A\right)}+2\arctan{\left(A^{2}B\right)}\right)}\\ &~~~~~-\frac12\operatorname{Cl}_{2}{\left(2\arctan{\left(A\right)}-2\arctan{\left(A^{2}B\right)}\right)}\\ &~~~~~-\operatorname{Cl}_{2}{\left(2\arctan{\left(A^{2}B\right)}\right)}\\ &=2\theta\ln{\left(1+AB\right)}-\left(\alpha-\theta\right)\ln{\left(\frac{1-AB}{1+AB}\right)}\\ &~~~~~+\frac12\operatorname{Cl}_{2}{\left(2\alpha+2\theta\right)}-\frac12\operatorname{Cl}_{2}{\left(2\alpha-2\theta\right)}-\operatorname{Cl}_{2}{\left(2\theta\right)},\\ \end{align}$$
and
$$\begin{align} \mathcal{J}{\left(\frac{B}{A},A^{2}B\right)} &=2\arctan{\left(A^{2}B\right)}\ln{\left(1+\frac{B}{A}\right)}\\ &~~~~~-\left[\arctan{\left(A^{3}\right)}-\arctan{\left(A^{2}B\right)}\right]\ln{\left(\left|\frac{1-\frac{B}{A}}{1+\frac{B}{A}}\right|\right)}\\ &~~~~~+\frac12\operatorname{Cl}_{2}{\left(2\arctan{\left(A^{3}\right)}+2\arctan{\left(A^{2}B\right)}\right)}\\ &~~~~~-\frac12\operatorname{Cl}_{2}{\left(2\arctan{\left(A^{3}\right)}-2\arctan{\left(A^{2}B\right)}\right)}\\ &~~~~~-\operatorname{Cl}_{2}{\left(2\arctan{\left(A^{2}B\right)}\right)}\\ &=2\theta\ln{\left(1+\frac{B}{A}\right)}-\left(\gamma-\theta\right)\ln{\left(\frac{A-B}{A+B}\right)}\\ &~~~~~+\frac12\operatorname{Cl}_{2}{\left(2\gamma+2\theta\right)}-\frac12\operatorname{Cl}_{2}{\left(2\gamma-2\theta\right)}-\operatorname{Cl}_{2}{\left(2\theta\right)},\\ \end{align}$$
and
$$\begin{align} \mathcal{J}{\left(\frac{A}{B},\frac{A^{2}}{B}\right)} &=2\arctan{\left(\frac{A^{2}}{B}\right)}\ln{\left(1+\frac{A}{B}\right)}\\ &~~~~~-\left[\arctan{\left(A\right)}-\arctan{\left(\frac{A^{2}}{B}\right)}\right]\ln{\left(\left|\frac{1-\frac{A}{B}}{1+\frac{A}{B}}\right|\right)}\\ &~~~~~+\frac12\operatorname{Cl}_{2}{\left(2\arctan{\left(A\right)}+2\arctan{\left(\frac{A^{2}}{B}\right)}\right)}\\ &~~~~~-\frac12\operatorname{Cl}_{2}{\left(2\arctan{\left(A\right)}-2\arctan{\left(\frac{A^{2}}{B}\right)}\right)}\\ &~~~~~-\operatorname{Cl}_{2}{\left(2\arctan{\left(\frac{A^{2}}{B}\right)}\right)}\\ &=2\psi\ln{\left(1+\frac{A}{B}\right)}-\left(\alpha-\psi\right)\ln{\left(\frac{A-B}{A+B}\right)}\\ &~~~~~+\frac12\operatorname{Cl}_{2}{\left(2\alpha+2\psi\right)}-\frac12\operatorname{Cl}_{2}{\left(2\alpha-2\psi\right)}-\operatorname{Cl}_{2}{\left(2\psi\right)},\\ \end{align}$$
and
$$\begin{align} \mathcal{J}{\left(\frac{1}{AB},\frac{A^{2}}{B}\right)} &=2\arctan{\left(\frac{A^{2}}{B}\right)}\ln{\left(1+\frac{1}{AB}\right)}\\ &~~~~~-\left[\arctan{\left(A^{3}\right)}-\arctan{\left(\frac{A^{2}}{B}\right)}\right]\ln{\left(\left|\frac{1-\frac{1}{AB}}{1+\frac{1}{AB}}\right|\right)}\\ &~~~~~+\frac12\operatorname{Cl}_{2}{\left(2\arctan{\left(A^{3}\right)}+2\arctan{\left(\frac{A^{2}}{B}\right)}\right)}\\ &~~~~~-\frac12\operatorname{Cl}_{2}{\left(2\arctan{\left(A^{3}\right)}-2\arctan{\left(\frac{A^{2}}{B}\right)}\right)}\\ &~~~~~-\operatorname{Cl}_{2}{\left(2\arctan{\left(\frac{A^{2}}{B}\right)}\right)}\\ &=2\psi\ln{\left(1+\frac{1}{AB}\right)}-\left(\gamma-\psi\right)\ln{\left(\frac{1-AB}{1+AB}\right)}\\ &~~~~~+\frac12\operatorname{Cl}_{2}{\left(2\gamma+2\psi\right)}-\frac12\operatorname{Cl}_{2}{\left(2\gamma-2\psi\right)}-\operatorname{Cl}_{2}{\left(2\psi\right)}.\\ \end{align}$$
Then,
$$\begin{align} \mathcal{I}{\left(\frac{1}{\sqrt{3}}\right)} &=2\left(\theta+\psi\right)\ln{\left(A\right)}\\ &~~~~~-\mathcal{J}{\left(AB,A^{2}B\right)}+\mathcal{J}{\left(\frac{B}{A},A^{2}B\right)}-\mathcal{J}{\left(\frac{A}{B},\frac{A^{2}}{B}\right)}+\mathcal{J}{\left(\frac{1}{AB},\frac{A^{2}}{B}\right)}\\ &=2\left(\theta+\psi\right)\ln{\left(A\right)}\\ &~~~~~-\bigg{[}2\theta\ln{\left(1+AB\right)}-\left(\alpha-\theta\right)\ln{\left(\frac{1-AB}{1+AB}\right)}\\ &~~~~~+\frac12\operatorname{Cl}_{2}{\left(2\alpha+2\theta\right)}-\frac12\operatorname{Cl}_{2}{\left(2\alpha-2\theta\right)}-\operatorname{Cl}_{2}{\left(2\theta\right)}\bigg{]}\\ &~~~~~+\bigg{[}2\theta\ln{\left(1+\frac{B}{A}\right)}-\left(\gamma-\theta\right)\ln{\left(\frac{A-B}{A+B}\right)}\\ &~~~~~+\frac12\operatorname{Cl}_{2}{\left(2\gamma+2\theta\right)}-\frac12\operatorname{Cl}_{2}{\left(2\gamma-2\theta\right)}-\operatorname{Cl}_{2}{\left(2\theta\right)}\bigg{]}\\ &~~~~~-\bigg{[}2\psi\ln{\left(1+\frac{A}{B}\right)}-\left(\alpha-\psi\right)\ln{\left(\frac{A-B}{A+B}\right)}\\ &~~~~~+\frac12\operatorname{Cl}_{2}{\left(2\alpha+2\psi\right)}-\frac12\operatorname{Cl}_{2}{\left(2\alpha-2\psi\right)}-\operatorname{Cl}_{2}{\left(2\psi\right)}\bigg{]}\\ &~~~~~+\bigg{[}2\psi\ln{\left(1+\frac{1}{AB}\right)}-\left(\gamma-\psi\right)\ln{\left(\frac{1-AB}{1+AB}\right)}\\ &~~~~~+\frac12\operatorname{Cl}_{2}{\left(2\gamma+2\psi\right)}-\frac12\operatorname{Cl}_{2}{\left(2\gamma-2\psi\right)}-\operatorname{Cl}_{2}{\left(2\psi\right)}\bigg{]}\\ &=2\left(\theta+\psi\right)\ln{\left(A\right)}\\ &~~~~~-2\theta\ln{\left(1+AB\right)}+\left(\alpha-\theta\right)\ln{\left(\frac{1-AB}{1+AB}\right)}\\ &~~~~~-\frac12\operatorname{Cl}_{2}{\left(2\alpha+2\theta\right)}+\frac12\operatorname{Cl}_{2}{\left(2\alpha-2\theta\right)}\\ &~~~~~+2\theta\ln{\left(A+B\right)}-2\theta\ln{\left(A\right)}-\left(\gamma-\theta\right)\ln{\left(\frac{A-B}{A+B}\right)}\\ &~~~~~+\frac12\operatorname{Cl}_{2}{\left(2\gamma+2\theta\right)}-\frac12\operatorname{Cl}_{2}{\left(2\gamma-2\theta\right)}\\ &~~~~~-2\psi\ln{\left(A+B\right)}+2\psi\ln{\left(B\right)}+\left(\alpha-\psi\right)\ln{\left(\frac{A-B}{A+B}\right)}\\ &~~~~~-\frac12\operatorname{Cl}_{2}{\left(2\alpha+2\psi\right)}+\frac12\operatorname{Cl}_{2}{\left(2\alpha-2\psi\right)}\\ &~~~~~+2\psi\ln{\left(1+AB\right)}-2\psi\ln{\left(A\right)}-2\psi\ln{\left(B\right)}-\left(\gamma-\psi\right)\ln{\left(\frac{1-AB}{1+AB}\right)}\\ &~~~~~+\frac12\operatorname{Cl}_{2}{\left(2\gamma+2\psi\right)}-\frac12\operatorname{Cl}_{2}{\left(2\gamma-2\psi\right)}\\ &=2\left(\psi-\theta\right)\ln{\left(\frac{1+AB}{A+B}\right)}+\left(\alpha-\gamma+\psi-\theta\right)\ln{\left(\frac{1-AB}{1+AB}\right)}\\ &~~~~~+\left(\alpha-\gamma+\theta-\psi\right)\ln{\left(\frac{A-B}{A+B}\right)}\\ &~~~~~-\frac12\operatorname{Cl}_{2}{\left(2\alpha+2\theta\right)}+\frac12\operatorname{Cl}_{2}{\left(2\alpha-2\theta\right)}+\frac12\operatorname{Cl}_{2}{\left(2\gamma+2\theta\right)}-\frac12\operatorname{Cl}_{2}{\left(2\gamma-2\theta\right)}\\ &~~~~~-\frac12\operatorname{Cl}_{2}{\left(2\alpha+2\psi\right)}+\frac12\operatorname{Cl}_{2}{\left(2\alpha-2\psi\right)}+\frac12\operatorname{Cl}_{2}{\left(2\gamma+2\psi\right)}-\frac12\operatorname{Cl}_{2}{\left(2\gamma-2\psi\right)},\\ \end{align}$$
and then, noting that $A+B=1-AB\land\alpha+\gamma+\theta=\psi$, we have
$$\begin{align} \mathcal{I}{\left(\frac{1}{\sqrt{3}}\right)} &=2\left(\psi-\theta\right)\ln{\left(\frac{1+AB}{A+B}\right)}+\left(\alpha-\gamma+\psi-\theta\right)\ln{\left(\frac{1-AB}{1+AB}\right)}\\ &~~~~~+\left(\alpha-\gamma+\theta-\psi\right)\ln{\left(\frac{A-B}{A+B}\right)}\\ &~~~~~-\frac12\operatorname{Cl}_{2}{\left(2\alpha+2\theta\right)}+\frac12\operatorname{Cl}_{2}{\left(2\alpha-2\theta\right)}+\frac12\operatorname{Cl}_{2}{\left(2\gamma+2\theta\right)}-\frac12\operatorname{Cl}_{2}{\left(2\gamma-2\theta\right)}\\ &~~~~~-\frac12\operatorname{Cl}_{2}{\left(2\alpha+2\psi\right)}+\frac12\operatorname{Cl}_{2}{\left(2\alpha-2\psi\right)}+\frac12\operatorname{Cl}_{2}{\left(2\gamma+2\psi\right)}-\frac12\operatorname{Cl}_{2}{\left(2\gamma-2\psi\right)}\\ &=2\gamma\ln{\left(\frac{1+AB}{A-B}\right)}\\ &~~~~~-\frac12\operatorname{Cl}_{2}{\left(2\alpha+2\theta\right)}+\frac12\operatorname{Cl}_{2}{\left(2\alpha-2\theta\right)}+\frac12\operatorname{Cl}_{2}{\left(2\gamma+2\theta\right)}-\frac12\operatorname{Cl}_{2}{\left(2\gamma-2\theta\right)}\\ &~~~~~-\frac12\operatorname{Cl}_{2}{\left(2\alpha+2\psi\right)}+\frac12\operatorname{Cl}_{2}{\left(-2\gamma-2\theta\right)}\\ &~~~~~+\frac12\operatorname{Cl}_{2}{\left(4\psi-2\theta-2\alpha\right)}-\frac12\operatorname{Cl}_{2}{\left(-2\alpha-2\theta\right)}\\ &=4\left(\psi-\theta-\alpha\right)\ln{\left(1+\sqrt{2}\right)}\\ &~~~~~+\frac12\operatorname{Cl}_{2}{\left(2\alpha-2\theta\right)}-\frac12\operatorname{Cl}_{2}{\left(2\psi-4\theta-2\alpha\right)}\\ &~~~~~-\frac12\operatorname{Cl}_{2}{\left(2\alpha+2\psi\right)}+\frac12\operatorname{Cl}_{2}{\left(4\psi-2\theta-2\alpha\right)}\\ &=\left(\pi-8\theta-4\alpha\right)\ln{\left(1+\sqrt{2}\right)}\\ &~~~~~+\frac12\operatorname{Cl}_{2}{\left(\frac{\pi}{4}\right)}-\frac12\operatorname{Cl}_{2}{\left(\frac{3\pi}{4}\right)}\\ &~~~~~-\frac12\operatorname{Cl}_{2}{\left(\frac{\pi}{2}-6\theta-2\alpha\right)}+\frac12\operatorname{Cl}_{2}{\left(\pi-6\theta-2\alpha\right)}\\ &=\left(\frac{\pi}{2}-12\theta\right)\ln{\left(1+\sqrt{2}\right)}\\ &~~~~~+\frac12\operatorname{Cl}_{2}{\left(\frac{\pi}{4}\right)}-\frac12\operatorname{Cl}_{2}{\left(\frac{3\pi}{4}\right)}\\ &~~~~~-\frac12\operatorname{Cl}_{2}{\left(\frac{\pi}{4}-8\theta\right)}+\frac12\operatorname{Cl}_{2}{\left(\frac{3\pi}{4}-8\theta\right)}\\ &=\frac14\operatorname{Cl}_{2}{\left(\frac{\pi}{2}\right)}+\left(6\sigma-\pi\right)\ln{\left(1+\sqrt{2}\right)}\\ &~~~~~+\frac12\operatorname{Cl}_{2}{\left(4\sigma-\frac{\pi}{4}\right)}-\frac12\operatorname{Cl}_{2}{\left(4\sigma-\frac{3\pi}{4}\right)};~~~\small{\left[\sigma:=\arcsin{\left(\frac{1}{\sqrt{3}}\right)}\right]}\\ &=\frac14C+\arctan{\left(\frac{10\sqrt{2}}{23}\right)}\ln{\left(1+\sqrt{2}\right)}\\ &~~~~~+\frac12\operatorname{Cl}_{2}{\left(\tau+\frac{\pi}{2}\right)}-\frac12\operatorname{Cl}_{2}{\left(\tau\right)};~~~\small{\left[4\sigma-\frac{3\pi}{4}=:\tau\in\left(0,\frac{\pi}{2}\right)\right]},\\ \end{align}$$
where here $C$ denotes the Catalan constant.