Upper approximation of $\mathrm{atanh}(x)$? Is there are nice upper approximation of $\mathrm{atanh(x)}$?
For example, $\ln(x)$ is nicely approximated by $x-1$ for $x$ around $1$.

$\displaystyle \mathrm{atanh}(x) = \mathrm{asinh}(\sqrt{y})\qquad\text{, where } x ≥ 0\;,\;y=\frac{x^2}{1-x^2}$
If y is big, we can reduce it, with "half-angle" formula
$\displaystyle \mathrm{asinh}(\sqrt{y}) = 2\;\mathrm{asinh}
\left(\sqrt{\frac{y/2}{\sqrt{y+1}+1}} \right) $
asinh has taylor series of alternating signs, we can easily get both bounds.
$\displaystyle \mathrm{asinh}(\sqrt{y}) =
\sqrt{y}\;\left[ 1 - \left(\frac{1^2·y}{2·3}\right) \left[ 1 - \left(\frac{3^2·y}{4·5} \right)\left[ 1 - \left(\frac{5^2·y}{6·7}\right) \left[\,...\,
\right] \right]\right]\right]$
To make bounds even tighter, we flip "last" factor, $\displaystyle (1-ε) → \frac{1}{(1+ε)}$
Example, for upper bound of $\;\large \mathrm{atanh}(0.9)$
lua> x = 0.9
lua> y = x*x / (1-x*x)          --> 4.263, y too big
lua> y = y/2 / (sqrt(y+1)+1)    --> 0.647
lua> y = y/2 / (sqrt(y+1)+1)    --> 0.142
lua> y = y/2 / (sqrt(y+1)+1)    --> 0.034
lua> y = y/2 / (sqrt(y+1)+1)    --> 0.0085
lua> r = sqrt(y) * 2^4

lua> r
1.4742977995430115
lua> r * (1 - y/6 / (1 + 9/20*y))
1.4722194992970188
lua> r * (1 - y/6 * (1 - 9/20*y * (1 - 25/42*y / (1 + 49/72*y))))
1.472219489583329
lua> atanh(x)
1.4722194895832204

