I managed to convince myself it's guaranteed that a 10-mover won't cancel any moves if no 8-movers exist.
I'll try to explain my logic.
I started out trying to work out if conjugates are ever worth it. The first move of a conjugated comm will always be the same type as the last move (inverses of eachother for a 10-mover or e.g. R and R2 for a 9-mover), so if the comm can cancel from both sides then the moves in the skeleton would already cancel. So a 9-mover or more will only ever be able to cancel on one side. Also since it has to cancel at least 2*(number of setup moves) to match the effectiveness 8-mover, the setup has to completely cancel with the adjacent moves of the skeleton (i.e. the setup must be the inverse of the adjacent moves in the skeleton). For 10-movers, if this is the case, there is always an 8-mover that does the same job.
Imagine the sketon with an insertion X X X A' (A: comm) B X X X. Instead of inserting the 10-mover, you can insert the same comm without the setup one move earlier: X X X (comm) A' B X X X, which is exactly equivalent, so looking for 10+ movers is never worth it.
The same argument applies to 9-movers on the side where the setup doesn't cancel into the comm. On the side where it does, things are a little different. You can occasionally have decent cancelations with a 9-mover, e.g with the following: X X X A2 (A [A B A' C A B' A' C'] A') X X X, you can cancel 2 moves which you would not have been able to cancel with an 8-mover. But since you know that the setup has to cancel both the adjacent move on the comm and the adjacent move of the skeleton, it gives a very easy criteria for filtering the sorts of 9-movers you try insert.
Of course, if you go through your entire skeleton and there are no 8-movers at all, you may have to settle for something longer that doesn't cancel. That's really rare though.
Of course this logic doesn't apply to pure comms that are longer because of longer interchanges like Cale mentioned above.
