# Can $\tanh(\arctan(x))$ be simplified?

Can $$\tanh(\arctan(x))$$ be simplified or re-expressed in terms of algebraic functions (or maybe a log here and there), kind of like the forward-inverse identities? (perhaps using the myriad of trig and hyperbolic formulas)

The only thing I found is the Gudermannian function, which in some sense yields the opposite of what I need

$$gd(x) = 2 \arctan\left(\tanh\left(\frac{x}{2}\right)\right)$$

Also, when I let Mathematica spit out the MacLaurin series, it seems that $$\tan(\arctan(x))$$ and $$\tanh(\arctan(x))$$ have similar expansions (differing only by the usual alternating sign). Is there a place where these kind of relationships between the trigonometric and hyperbolic functions are studied? Anything known about the properties of the coefficients of such series?

How do you feel about $$\tanh \arctan x = \frac{(1-\mathrm{i}x)^\mathrm{i} - (1+\mathrm{i}x)^\mathrm{i}}{(1-\mathrm{i}x)^\mathrm{i} + (1+\mathrm{i}x)^\mathrm{i}} \text{?}$$
This is obtained from $$\tanh y = \frac{\mathrm{e}^{2 y}-1}{\mathrm{e}^{2 y}+1}$$ and $$\arctan x = \frac{\mathrm{i}}{2} \log \frac{1-\mathrm{i} x}{1+\mathrm{i} x}$$, followed by some simplifying.
• How do you get that $i$ symbol like that? Using a \text block seems clumsy. – Oscar Lanzi Oct 21 '19 at 23:36
• @OscarLanzi : Even worse: "\mathrm{e}" Similarly for the "$\mathrm{i}$". It's so common on MSE, I type it automatically now. (I even forget in my own writing that I long ago created macros in my personal set for these.) – Eric Towers Oct 21 '19 at 23:47
• @EricTowers Bless you for putting $\rm e$ and $\rm i$ upright like the true numbers they are – gen-ℤ ready to perish Oct 22 '19 at 0:23