How about this:
Inserting the last F2L pair while making the cross on top
and permuting two of the LL edges.

Kind of complicated, but it could mean a 1 look LL. There would be 6 possibilities for permuting the last 2 edges, excluding when they're already solved. So, 40 CLLs (I think) x 6 = 240. The cases would probably be slightly different, due to the mispermuted edges meaning mispermuted corners. However, it would probably still be 240 new cases to learn.
I hope that hasn't already been done before.