A integral involving hyperbolic and trigonometric functions

I think that the faster would be to use the series expansion of $tanh(t)$ which gives $$ tanh(cos(x))=sum_{n=1}^infty frac{4^n left(4^n-1right) B_{2 n} }{(2 n)!}cos ^{2 n-1}(x)$$ which leads to $$I=int_0^1tanh(cos(x)),dx=$$ $$I=frac{1}{2}sum_{n=1}^infty left(4^n-1right) B_{2 n} left(frac{pi }{n Gamma left(n+frac{1}{2}right)^2}-frac{4^n B_{cos ^2(1)}left(n,frac{1}{2}right)}{(2n)!}right)$$

Computing the partial sums from $n=1$ to $n=p$, some results $$left( begin{array}{cc} p & text{result} \ 1 & 0.841471 \ 2 & 0.627183 \ 3 & 0.697668 \ 4 & 0.673052 \ 5 & 0.681934 \ 6 & 0.678660 \ 7 & 0.679885 \ 8 & 0.679422 \ 9 & 0.679598 \ 10 & 0.679531 \ 11 & 0.679557 \ 12 & 0.679547 \ 13 & 0.679551 \ 14 & 0.679549 \ 15 & 0.679550 \ 16 & 0.679549 end{array} right)$$

Composing the Taylor series in terms of $x$ would be a pure nightmare.

However, for a quick and dirty approximation, we could use the $[2n,2]$Padé approximant of the integrand. For example $$tanh(cos(x))sim frac {t+a x^2+b x^4}{1+c x^2}$$ with $$a=frac{15 t^2-61 t+15}{30 (6 t-1)} qquad b=frac{21 left(t^2-1right)}{40 (6 t-1)}qquad c=frac{-90 t^2+30 t+29}{30 (6 t-1)}$$ and $t=tanh(1)$. This would give as a result $$I sim frac {a+b}c-frac{b}{c^2}+frac{ c (c t-a)+b}{c^{5/2}}tan ^{-1}left(sqrt{c}right)sim 0.679076$$