$f(x)=\tanh(1+\tanh^{-1}(x))$ or $f:\tanh(x) \to \tanh(x+1)$ is a rational function? This is (again) more a recreational/incidental question.      
Playing with iteration of functions I considered the function $$ f(x) = \tanh(1+\tanh^{-1}(x)) \tag1$$ such that $$ f : \tanh(x) \to \tanh(x+1) \tag 2$$
Pari/GP is kind enough to provide the first few coefficients of the Taylor-series of $f(x)$ numerically.
$$ f(x) \sim  0.7615941559557649 + 0.4199743416140261 x - 0.3198500042246123 x^2 \\ + 0.2435958939998914 x^3 - 0.1855212092851373 x^4 + 0.1412918687974069 x^5  \\ - 0.1076070615601738 x^6 + 0.08195290922380060 x^7 - 0.06241485672841984 x^8  \\ + O(x^9) $$
Looking at the coefficients it seemed to me that they give just an alternating geometric series with quotient $q=-\tanh(1)$ and a scaling factor $a = \frac 1q - q$ such that -by that numerical heuristic-  the power-series of $f(x)$ is  $$f(x) \underset{\text{guessed}}{=} -q + a \sum_{k=1}^\infty q^k x^k \tag 3$$ which reduces then to the rational function $$ f(x) \underset{\text{guessed}}{=} { a \over 1-x\cdot q }- \frac 1q \tag 4$$
I'm surprised that this results in such a simple function - how would a proof for the algebraic identity (4) look like?
 A: $$\tanh(x+y) = \frac{\sinh(x+y)}{\cosh(x+y)} = \frac{\tanh x  + \tanh y }{1 + \tanh x \tanh y}$$
$$ \tanh(1 + x) = \frac{\tanh 1 + \tanh x}{1 + \tanh 1 \tanh x} $$
$$ f(\tanh x) = \tanh(1+x) \implies f(x) = \frac{\tanh 1 + x}{1 + x \tanh 1 } $$
A: Here, we use the alternative expressions
$$\mathrm{arctanh}(x) = \frac{1}{2} \log \left(\frac{1+x}{1-x}\right) \quad \text{and} \quad \tanh(x)=\frac{1-e^{-2x}}{1+e^{-2x}}.$$
The first is valid in $\vert x \vert < 1$, and the latter valid in all of $\mathbb{C}$.  Algbraic manipulation now gives
$$\tanh(1+\mathrm{arctanh}(x))=\frac{x+e^2 (x+1)-1}{-x+e^2 (x+1)+1}=\frac{x(e^2+1)+e^2-1}{x(e^2-1)+e^2+1},$$
at least for $\vert x \vert <1$.  This result may be analytically continued, of course.
A: If it is true for the ordinary tangent it is true for the hyperbolic tangent (due to $\tan(ix)=\pm i \tanh(x)$ (for some choice of the sign that I don't need to remember in order to answer the question).
And  $\tan(x+a)$ is indeed a rational function of $\tan x$ and $\tan a$, by the familiar addition formula for tangent.
This way we can foresee that a rational function expression will exist, and a bit of care about the signs will pin down what the formula is.  
