A heavy uniform chain has length \(L\) and mass \(m\). The chain is inextensible and perfectly flexible. One end \(A\) of the chain is fixed to the ceiling. The other end, \(B\), is held at rest right next to \(A\) at the level of the ceiling (so that initially \(A\) and \(B\) are at the same height).

The second end \(B\) is released, and the chain starts to fall until the chain is fully extended with \(B\) vertically below \(A\). Note that there are no internal friction forces operating within the chain, and so energy is conserved throughout the motion. The horizontal component of the motion of the chain can be ignored.

It can be shown that the time \(T\) that it takes for the chain to become fully extended is \[ T \; = \; \sqrt{\frac{2L}{g}} \frac{\Gamma\big(\frac{A}{B}\big)^C}{\sqrt[D]{\pi}} \] where \(A,B,C,D\) are positive integers, with \(A,B\) coprime. Find the value of \(A+B+C+D\).**N.B.** The time \(T \approx 0.847\sqrt{\frac{2L}{g}}\) is less than the time \(\sqrt{\frac{2L}{g}}\) taken for a particle to fall a height of \(L\) from rest. The chain falls "faster than gravity".

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