Given $x\in \Re$, $a \in \Re$ where $-1 \le x \le 1$ and $0 \le a \le 4$, is it possible to reduce the following expression:
$\tanh(a \tanh^{-1}(x))$
E.g. to some kind of polynomium?
I know that if $a$ is an integer then the following holds:
$\tanh(0 \tanh^{-1}(x)) = 0$
$\tanh(1 \tanh^{-1}(x)) = x$
$\tanh(2 \tanh^{-1}(x)) = 2 x / (1 + x^2)$
$\tanh(3 \tanh^{-1}(x)) = (3 x + x^3) / (1 + 3 x^2)$
$\tanh(4 \tanh^{-1}(x)) = (4 x + 4 x^3) / (1 + 6 x^2 + x^4)$
Any suggestions?
$$\tanh^{-1}{x} = \frac12 \log{\left (\frac{1+x}{1-x} \right )}$$
$$\tanh{(a \tanh^{-1} x)} = \frac{e^{(2 a \tanh^{-1} x)}-1}{e^{(2 a \tanh^{-1} x)}+1} = \frac{(1+x)^a-(1-x)^a}{(1+x)^a+(1-x)^a}$$