Okay, so to satisfy requirements 5, 6, and 7, the soft-clipping function will have the form:

$$ f(x) = K \int\limits_{0}^{x} \left(1 - u^2 \right)^N \ du \quad \text{ for } |x|\le 1 $$

$K$ is a constant chosen to satisfy requirement 4.

The 1st derivative of $f(x)$ is

$$ f'(x) = K \left(1 - x^2 \right)^N $$

and the $n$th derivative is

$$ f^{(n)}(x) = K \left(1 - x^2 \right)^{N-n+1}g(x) $$

where $g(x)$ is the $(n-1)$th derivative of $-x^2$ and some other non-infinite stuff (I wouldn't mind if someone straightens this out).  Nonetheless, when $x = \pm 1$, then $f^{(n)}(x) = 0$ for $1 \le n < N-1$.


The integrand is a binomial and can be expressed as a power series using binomial expansion:

$$\begin{align}
\left(1 - u^2 \right)^N & = \sum\limits_{n=0}^{N} \binom{N}{n} (-u^2)^n (1)^{N-n} \\
  & = \sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} (-u^2)^n (1)^{N-n} \\
  & = \sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} (-1)^n u^{2n} \\
\end{align}$$

So the integral can be expressed as an integral of a power series:

$$\begin{align}
\int\limits_{0}^{x} \left(1 - u^2 \right)^N \ du & = \int\limits_{0}^{x} \sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} (-1)^n u^{2n} \ du \\
& = \sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} (-1)^n \int\limits_{0}^{x} u^{2n} \ du \\
& = \sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} \frac{(-1)^n}{2n+1} u^{2n+1} \Bigg|_0^x \\
& = \sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} \frac{(-1)^n}{2n+1} x^{2n+1} \\
& = x \sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} \frac{(-1)^n}{2n+1} \big(x^2 \big)^n \\
\end{align}$$

When $x = \pm 1$, we get

$$ \int\limits_{0}^{\pm 1} \left(1 - u^2 \right)^N \ du = \pm \sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} \frac{(-1)^n}{2n+1} $$

(there is a little improperness in the integral limit.)  So the scaler $K$ must be

$$ K = \frac{1}{ \sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} \frac{(-1)^n}{2n+1} }$$

This makes the entire soft-clipping function to be:

$$ f(x) = \begin{cases} 
-1 & x \le -1 \\
\frac{ x \sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} \frac{(-1)^n}{2n+1} \big(x^2 \big)^n}{ \sum\limits_{n=0}^{N} \frac{N!}{n!(N-n)!} \frac{(-1)^n}{2n+1} } \quad & -1 \le x \le +1 \\
+1 & +1 \le x \\
\end{cases} $$

and it appears that the odd-power (that is the power = $2n+1$) polynomial coefficients are

$$ a_n = \frac{ \frac{N!}{n!(N-n)!} \frac{(-1)^n}{2n+1}}{ \sum\limits_{i=0}^{N} \frac{N!}{i!(N-i)!} \frac{(-1)^i}{2i+1} } $$