Three-cushion billiards challenge.

The key trick is

flipping the table!
(╯°□°)╯︵ ┻━┻

Diagram:

By reflecting the pentagon across a side 3 times, the path forms a line connecting the centers of the farthest pentagons. Half its length can be computed using the law of cosines with an angle measuring $144^\circ$ (double the $72^\circ$ central angle of the pentagon) and sides of length $r$ and $2r$ where $r$ is the inradius of the pentagon.

Math:

$$\begin{align*}&\mathrel{\phantom=}2\sqrt{r^2+(2r)^2-2\cdot r\cdot2r\cdot\cos{144^\circ}}\\&=2\sqrt{5r^2-4r^2\left(-\frac{1+\sqrt5}4\right)}\\&=2r\sqrt{6+\sqrt5}\\&=2\cdot\frac l{2\sqrt{5-2\sqrt{5}}}\cdot\sqrt{6+\sqrt5}\\&=l\sqrt{\frac{6+\sqrt5}{5-2\sqrt{5}}}\\&=l\sqrt{\frac{6+\sqrt5}{5-2\sqrt{5}}\cdot\frac{2+\sqrt5}{2+\sqrt5}}\\&=l\sqrt{\frac{17+8\sqrt5}{\sqrt5}}\\&=\boxed{l\sqrt{8+\frac{17}{\sqrt5}}}\end{align*}$$

Update

Pranay found a shorter path. Using the same technique,

half its length can be computed using the law of cosines with an angle measuring $72^\circ$ (equal to the central angle) and sides of length $r$ and $2r$.

Very similar math:

$$\begin{align*}&\mathrel{\phantom=}2\sqrt{r^2+(2r)^2-2\cdot r\cdot2r\cdot\cos{72^\circ}}\\&=2\sqrt{5r^2-4r^2\cdot\frac{\sqrt5-1}4}\\&=2r\sqrt{6-\sqrt5}\\&=2\cdot\frac l{2\sqrt{5-2\sqrt{5}}}\cdot\sqrt{6-\sqrt5}\\&=l\sqrt{\frac{6-\sqrt5}{5-2\sqrt{5}}}\\&=l\sqrt{\frac{6-\sqrt5}{5-2\sqrt{5}}\cdot\frac{2+\sqrt5}{2+\sqrt5}}\\&=l\sqrt{\frac{7+4\sqrt5}{\sqrt5}}\\&=\boxed{l\sqrt{4+\frac{7}{\sqrt5}}}\end{align*}$$