Hint 1

Use conservation of energy,

$${1\over 2}m\dot r^2-{GMm\over r}=-{GMm\over D},$$

to get a first-order differential equation for $r$. Integrating this, you should find that the time taken is

$$T={1\over\sqrt{2GM}}\int_{D/2}^D\left({1\over r}-{1\over D} \right)^{-1/2}dr.$$

The integral can be done by changing variable, with $r=D\cos^2 u$, and gives

$$T=\sqrt{{2D^3\over GM}}\,{\pi+2\over 8}.$$