TITLE: A better approximation than mine for $\int_0^1\log\left(1+\operatorname{gd}(x)\right)\,dx$, where $\operatorname{gd}(x)$ is the Gudermannian function
QUESTION [1 upvotes]: Let $\operatorname{gd}(x)$ the Gudermannian function, defined as in this MathWorld's article, and implemented in Wolfram Language as Gudermannian[x]. 
This afternoon I've spent an hour playing with Wolfram Alpha online calculator about integrals, for instance, like this $$I=\int_0^1\log\left(1+\operatorname{gd}(x)\right)\,dx.$$
Notice that the plot in previous article of MathWorld tell us that for $0<x<1$ one has $0<\operatorname{gd}(x)<1$. 
My calculations to get an approximation were $$I\approx\int_0^1 \operatorname{gd}(x)\,dx=-\frac{\pi}{2}+i\left(\operatorname{Li_2}(-ie)-\operatorname{Li_2}(ie)\right)-2C,$$ where I've combined with the indefinite integral of the Gudermannian functions that tell us MathWorld. Here $C$ is Catalan's constant. But this quantity is about $\approx 0.464065$.
But WA calculated $I$ as $0.365619$, play this code: 
integrate log(1+(gd(x)))dx, from x=0 to x=1

Question. Can you provide us a better approximation, than mine, for our integral $$\int_0^1\log\left(1+\operatorname{gd}(x)\right)\,dx?$$
  Please provide also the more important justifications in your calculations. Also if your approach uses the numerical analysis and/or calculations with your computer provide us some detail of your calculation/method. Thanks in advance.

REPLY [2 votes]: For sure, you could use Taylor expansion
$$\log (1+\text{gd}(x))=x-\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^4}{12}+\frac{3 x^5}{40}-\frac{x^6}{18}+\frac{169
   x^7}{5040}-\frac{19 x^8}{840}+\frac{2171 x^9}{120960}+O\left(x^{10}\right)$$ and integrate which would lead to $$\int_0^1 \log (1+\text{gd}(x))\,dx=\frac{443161}{1209600}\approx 0.366370$$ Continuing the expansion to $O\left(x^{20}\right)$ would give $$\int_0^1 \log (1+\text{gd}(x))\,dx=\frac{35582823156912797}{97316080327065600}\approx 0.365642$$ May be, better would be to use Padé approximants; a very simple would be 
$$\log (1+\text{gd}(x))\approx \frac{x-\frac{x^2}{2}}{1-\frac{x^2}{6}}$$ and the integration would lead to  $$\int_0^1 \log (1+\text{gd}(x))\,dx\approx 3+3 \log \left(\frac{6}{5}\right)-3 \sqrt{6} \tanh ^{-1}\left(\frac{1}{\sqrt{6}}\right)\approx 0.361349$$ Using the next one
$$\log (1+\text{gd}(x))\approx \frac{-\frac{17 x^3}{60}+\frac{6 x^2}{5}+x}{\frac{2 x^2}{5}+\frac{17 x}{10}+1}$$ we can still integrate and get a bunch of logarithms for a value $\approx 0.364644$.
Just for the fun, I put in the next table the decimal value of the integral of the $[n,2]$ Padé approximant
$$\left(
\begin{array}{cc}
 n & \text{result} \\
 2 & 0.3613491513 \\
 3 & 0.3646438085 \\
 4 & 0.3661395229 \\
 5 & 0.3657879040 \\
 6 & 0.3654692466 \\
 7 & 0.3655746142 \\
 8 & 0.3656449849 \\
 9 & 0.3656298359 \\
 10 & 0.3656072339
\end{array}
\right)$$
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