First step for $\int\frac{\operatorname{gd}\left(\log x\right)}{\sqrt{1-x^2}}dx$, where $\operatorname{gd}(x)$ denotes the Gudermannian function

In this post we denote the Gudermannian function as $\operatorname{gd}(x)$, I am saying the function from this Wikipedia.

I believe that it is possible to get the indefinite integral of $$\int\frac{\operatorname{gd}\left(\log x\right)}{\sqrt{1-x^2}}dx.\tag{1}$$

Question. What do you think that should be the first step in the calculation of such indefinite integral $(1)$? I am not asking about the full deduction of such integral, only the first step (or the first and second ones) to calculate the indefinite integral. That is what is the idea to start to compute such indefinite integral. Many thanks.

• That I evoke with this Question is the following situation: imagine that we need to calculate such kind of integrals, what should be a first good step that favors the solution of the problem of calculation of such indefinite integral? – user243301 May 12 '18 at 15:00
• What is your ultimate goal? Mathematica gives an answer involving arcsin, arctan, log, and trigamma which has several complicated terms and is unlikely to simplify much. The power series for it begins $\; -\pi x/2 + x^2 + (12\ln(x) - 11)x^4/48+ (45\ln(x)^2 -60\ln(x)+82)x^6/1080 + O(x^8).$ – Somos May 12 '18 at 16:37
• My Question arises when I was trying variations of my post On a combination between Ahmed's Integral and the Gudermannian function (that is motivated by the fact that MathWorld aritcle for Ahmed's integral, tell us that such integral is a particular case of other integral, I was trying to get different versions of my Question in the mentioned post). Thus my goal with this post is to know how to attack this kind of integrals. Many thanks for your help @Somos – user243301 May 12 '18 at 16:43
• I am sorry but you don't "attack" integrals. In my previous comment I told you that according to Mathematica, the integral is a complicated expression with many terms that almost surely has no simple form. What would you do with such a complicated expression? – Somos May 12 '18 at 16:47
• @Somos in this post I do not ask about such complicated expression (that is the full elaboration of the indefinite integral). My intention and question were about the first step to attack the integral. – user243301 May 12 '18 at 16:54

Famously, $\operatorname{gd}y=2\arctan\exp y -\pi/2$. So your integral is $\int (2\arctan x -\pi/2)\arcsin' x dx$. By parts, you just need to obtain $\int\frac{\arcsin x dx}{1+x^2}$. Wolfram Alpha makes that look messy.