I am looking for any type of solution (closed form/recursive integral/special functions/...) or at least a good approximation for this definite integral:
$$
\int\limits_{\scriptsize 0}^{\scriptsize y}{\frac{1}{2}+\frac{1}{2}\,\tanh\left(a\cdot\,\operatorname{atanh}\left(2\,x-1\right)+b\right)}{\;\mathrm{d}x}
$$
where $a>0$, $x$ domain is from $0$ to $1$.
Other form of function after reduction is:
$$
\int_0^y{\frac{1}{1+c\,\left(\frac{1-x}{x}\right)^{a}}}{\;\mathrm{d}x}
$$
where $a>0$, $c=e^{-2b}$, $c>0$, $x$ domain is from $0$ to $1$.
I have been trying to solve it for 1 month right now, but I reached my knowledge threshold so I need to ask for help.
PS: Taylor series is not satisfying as it converges only in a small radius when $a$ is big value. Maclaurin series for function centered around $0$ also fails for some $b$ values.
Edit:
Most promising method I have tried so far was residue method, when I replace x so that bounds of integration goes from $-\infty$ to $\infty$.
$$
u=\operatorname{atanh}\left(\frac{2x}{y}-1\right)
$$
Only in this situation integral across arc of semi-cirle vanishes (acc. to Jordan's lemma) and we are left with integral of real $x$. So then I should just calculate the Residues to get the result, but... plot of the function shows that there are inifinite numbers of singularities in points $0+i(1/2\pi +k\pi )$. I think there is no way to converge inifite series of residues.
Another good attempt was to switch to inifite series: $$ \int_0^y{\frac{1}{1+c\,\left(\frac{1-x}{x}\right)^{a}}}{\;\mathrm{d}x}=\sum_n{(-c)^n\int_0^y{\left(\frac{1-x}{x}\right)^{an}}{\;\mathrm{d}x}}=\sum_n{(-c)^nB(y,1-an,1+an)} $$ So we end up with Incomplete Beta Function, which I don't know how to "roll back" to fraction or any other way to solve the series. Maybe you know if there exists a function "$f$" so: $$ B(y,1-an,1+an)=f^n(B(y,A,B)) $$