Prove $\int_{\frac{\pi}{20}}^{\frac{3\pi}{20}} \ln \tan x\,\,dx= - \frac{2G}{5}$ Context: This question
asks to calculate a definite integral which turns out to be equal to $\displaystyle 4 \, \text{Ti}_2\left( \tan \frac{3\pi}{20} \right) -
4 \, \text{Ti}_2\left( \tan \frac{\pi}{20} \right),$ where $\text{Ti}_2(x) = \operatorname{Im}\text{Li}_2( i\, x)$ is the 
Inverse Tangent Integral function. 
The source for this integral is this question on brilliant.org.
In a comment, the OP
claims that the closed form can be further simplified to $-\dfrac\pi5 \ln\left( 124 - 55\sqrt5 + 2\sqrt{7625 - 3410\sqrt5} \right) + \dfrac85 G$.

How can we prove that?

I have thought about using the formula $$\text{Ti}_2(\tan x) = x \ln \tan x+ \sum_{n=0}^{\infty} \frac{\sin(2x(2n+1))}{(2n+1)^2}. \tag{1}$$
but that only mildly simplifies the problem.
Equivalent formulations include: 
$$\, \text{Ti}_2\left( \tan \frac{3\pi}{20} \right) -
 \, \text{Ti}_2\left( \tan \frac{\pi}{20} \right) \stackrel?= \frac{ \pi}{20} \ln \frac{  \tan^3( 3\pi/20)}{\tan ( \pi/20)}   + \frac{2 G}{5} \tag{2}$$
$$ \sum_{n=0}^{\infty} \frac{\sin \left(\frac{3\pi}{10}(2n+1) \right)- \sin \left(\frac{\pi}{10}(2n+1)\right)}{(2n+1)^2} \stackrel?=\
\frac{2G}{5} \tag{3}$$
$$\int_{\pi/20}^{3\pi/20} \ln \tan x\,\,dx \stackrel?= - \frac{2G}{5} \tag{4}$$
A related similar question is this one.

Let $J(a)=\int_{\frac{\pi}{20}}^{\frac{3\pi}{20}}\tanh^{-1}\frac{2a\cos2x}{1+a^{2}}dx$ and evaluate
\begin{align}
J’(a) &= \int_{\frac\pi{20}}^{\frac{3\pi}{20}}\frac{2(1-a^2)\cos2x}{a^4+1-2a^2\cos4x}dx=\frac1{2a}{\left.\tan^{-1}\frac{2a\sin2x}{1-a^2}\right|_{\frac{\pi}{20}}^{\frac{3\pi}{20} } }\\
&=\frac1{2a}\tan^{-1}\frac{a-a^5}{1+a^6}
= \frac1{2a}(\tan^{-1}a - \tan^{-1}a^5)
\end{align}
where $\sin\frac{3\pi}{10}-\sin\frac{\pi}{10}=\frac12$ and $\sin\frac{3\pi}{10}\sin\frac{\pi}{10}=\frac14$ are recognized. Then
\begin{align}
\int_\frac\pi{20}^\frac{3\pi}{20}\ln(\tan x)~dx
&= -\int_\frac\pi{20}^\frac{3\pi}{20}\tanh^{-1}(\cos2x)dx =-J(1)
=-\int_0^1 J’(a)da \\
& =-\frac12 \int_0^1\left(\frac{\tan^{-1}a}{a}\right.
-\underset{a^5\to a}{\left.\frac{\tan^{-1}a^5}{a}\right)}da\\
&=-\left(\frac12-\frac1{10}\right) \int_0^1\frac{\tan^{-1}a}{a}da=-\frac25 G
\end{align}
