An ant on the move (noch nicht übersetzt)

On the Euclidean plane, an ant travels from point A(0, 1) to point B(d, 1) for an integer d.

In each step, the ant at point (x₀, y₀) chooses one of the lattice points (x₁, y₁) which satisfy x₁ ≥ 0 and y₁ ≥ 1 and goes straight to (x₁, y₁) at a constant velocity v. The value of v depends on y₀ and y₁ as follows:

• If y₀ = y₁, the value of v equals y₀.
• If y₀ ≠ y₁, the value of v equals (y₁ - y₀) / (ln(y₁) - ln(y₀)).

The left image is one of the possible paths for d = 4. First the ant goes from A(0, 1) to P₁(1, 3) at velocity (3 - 1) / (ln(3) - ln(1)) ≈ 1.8205. Then the required time is sqrt(5) / 1.8205 ≈ 1.2283.
From P₁(1, 3) to P₂(3, 3) the ant travels at velocity 3 so the required time is 2 / 3 ≈ 0.6667. From P₂(3, 3) to B(4, 1) the ant travels at velocity (1 - 3) / (ln(1) - ln(3)) ≈ 1.8205 so the required time is sqrt(5) / 1.8205 ≈ 1.2283.
Thus the total required time is 1.2283 + 0.6667 + 1.2283 = 3.1233.

The right image is another path. The total required time is calculated as 0.98026 + 1 + 0.98026 = 2.96052. It can be shown that this is the quickest path for d = 4.

Let F(d) be the total required time if the ant chooses the quickest path. For example, F(4) ≈ 2.960516287.
We can verify that F(10) ≈ 4.668187834 and F(100) ≈ 9.217221972.

Find F(10000). Give your answer rounded to nine decimal places.