A more elegant version of this spiral graph

Question

The problem

I set out to create a polar graph of a spiral with the following properties: (Listed in order of priority)

1. It should be performant to calculate (A household computer should be able to easily calculate 12k points per second)
2. At all times the spiral should be "smooth" with no bumps or sudden turns
3. It should be expressed in closed form
4. The radius should start near 1 and descend slowly at first
5. Near to $\pi$ it should descend more quickly, arriving at $\theta = \pi, r \approx 0.5$ (Arriving later is better than sooner)
6. After that, the rate of descent should slow down again and the spiral should make a few more ($\approx3$) revolutions before reaching 0 (If it ever does, it doesn't have to)
7. Nice to have: Be able to adjust the "inner radius" (currently 0.5) by changing a variable

My solution

After a lot of tweaking I managed to come up with this: $^\$$

$$\begin{align} F(\theta) &= \frac{\sin(\theta^3\pi - \frac \pi2) + 1}{2}\\ G(\theta) &= \frac{\sin(\theta\pi - \frac \pi2)+1.1}{2.2}\\ H(\theta) &= \frac{\sin(2\pi\theta^2)}{40 G(\theta)} \end{align} $$

and finally:

$$r = F(\frac {\theta}{5\pi}) + H(\frac {\theta}{5\pi})$$

This produces the following graph:

It's not perfect, but it is the closest I've been able to come to meeting all of the requirements.

I am happy with the fact that at the outermost point the tangent is almost straight up and it hits $r = 0.5$ right when I wanted it to.

The question

What I would like help with:

1. Make the function more elegant, as it stands coding this would be a mess and I only arrived at it by clumsily stumbling around
2. It would be nice (though not required) for the inner revolutions to be slightly flatter, with my solution the innermost revolution is about half the size it should be

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$^\$$ I am aware these functions are "backwards" but it's trivial to resolve that so I'm not concerned