Antisymmetric asymptotic curve with only simple binary arithmetic? I'm looking for an s-curve formula with similar properties to $Sigmoid$ or $\tan^{-1}$, but without 'expensive' unary functions or their binary generalizations (e.g. $^x\log y$). The only allowed operations are: addition, subtraction, multiplication, division, exponentiation-with-integer (and by extension, negation and reciprocation).
The formal properties should be as follows:
Antisymmetric around $x=0$: $$f(x) - f(0) = f(0) - f(-x)$$
Asymptotic: $$f(\infty) < \infty$$
Increasing: (derivative is non-negative): $$f'(x) \ge 0$$
Derivative is decreasing away from origin: $$f'(\left|x\right|) > f'(\left|x\right|+\delta)$$
Derivative is never zero except at infinity: $$\left|f'(x)\right| \neq 0, \text{ if } \left|x\right| \neq \infty$$$$\left|f'(x)\right| = 0, \text{ if } \left|x\right| = \infty$$
Derivative at zero is not infinite: $$f'(0) < \infty$$
[edit] Forgot a semi-important constraint (@Meelo's answer reminded me)
Derivative is smooth: $$f''(0) = 0$$
 A: If you don't care about splitting into cases, you can just define a nice rational function that satisfies the conditions for positive arguments, and just mirror it; for instance, we could use
$$f(x)=\frac{x}{|x|+1}$$
which works and can be calculated quickly, but is not smooth (though it is continuously differentiable) and, implicitly, requires casework for the absolute value.
Otherwise, we're just talking about a rational function (since products, quotients, sums, and exponentiation-with-integer of a rational functions are still rational functions) and it is impossible. For instance, let
$$f(x)=\frac{a(x)}{b(x)}$$
for two polynomials $a$ and $b$. The two must be of the same degree, because if $\operatorname{deg}(a)>\operatorname{deg}(b)$, then the function tends towards $\pm \infty$ as $x$ goes towards $\pm \infty$. If $\operatorname{deg}(a)<\operatorname{deg}(b)$, then the function would tend towards $0$ as $x$ goes to $\pm\infty$, which would mean that it can't be strictly increasing between. Since the degrees are the same, it must be that
$$\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow-\infty} f(x)=\frac{a_n}{b_n}$$
where $a_n$ and $b_n$ are the coefficients of the highest degree terms (coefficients of $x^n$ where $n=\deg(a)=\deg(b)$), since eventually $x^n$ will dominate every other term. This means that no rational function can satisfy your constraints.
One might put all of the above more succinctly by noting that rational functions can be extended to a continuous function in the one point compactification of the reals, but no function satisfying your constraints can be so extended.
