Inverse of $\tanh^{\prime\prime}(x)$

Question

I have problems finding the inverse function of the second derivative of the hyperbolic tangent. I know it is not invertible on the whole of $\mathbb{R}$, but having a closed form for the inverse on, lets say, $[-0.5,0.5]$ would be enough for me. Any help or references appreciated.
$$\tanh^{\prime\prime}(x)=-2\tanh(x)\mathop{\rm sech}\nolimits^2 (x) .$$

Answer 1

Converting to exponential form and plugging into Mathematica yields the three inverse functions

$ \frac{1}{2} \log \left(-\frac{-72 y-64}{6 y \sqrt[3]{-27 y^2-108 y+3 \sqrt{3} \sqrt{27 y^4-16 y^2}-64}}-\frac{3 y+8}{3 y}+\frac{2 \sqrt[3]{-27 y^2-108 y+3 \sqrt{3} \sqrt{27 y^4-16 y^2}-64}}{3 y}\right) $

$ \frac{1}{2} \log \left(\frac{\left(1+i \sqrt{3}\right) (-72 y-64)}{12 y \sqrt[3]{-27 y^2-108 y+3 \sqrt{3} \sqrt{27 y^4-16 y^2}-64}}-\frac{3 y+8}{3 y}-\frac{\left(1-i \sqrt{3}\right) \sqrt[3]{-27 y^2-108 y+3 \sqrt{3} \sqrt{27 y^4-16 y^2}-64}}{3 y}\right) $

$ \frac{1}{2} \log \left(\frac{\left(1-i \sqrt{3}\right) (-72 y-64)}{12 y \sqrt[3]{-27 y^2-108 y+3 \sqrt{3} \sqrt{27 y^4-16 y^2}-64}}-\frac{3 y+8}{3 y}-\frac{\left(1+i \sqrt{3}\right) \sqrt[3]{-27 y^2-108 y+3 \sqrt{3} \sqrt{27 y^4-16 y^2}-64}}{3 y}\right) $

which graphically seem to be correct for various ranges of $y$; no one of those inverses seems perfect for your desired interval.

Answer 2

Using the identity $\text{sech}^2(x)=1-\tanh^2(x)$, it can be seen that an equivalent expression for $\frac{d^2}{dx^2}\tanh(x)$ is $2\left(\tanh^3(x)-\tanh(x)\right)$, so solving the equation $y=\frac{d^2}{dx^2}\tanh(x)$ with $y\in\mathbb R$ amounts to solving the cubic $$\frac{y}{2}=t^3-t$$ and then applying $\tanh^{-1}$ to $t$. We can solve the cubic with Cardano's formula.