I propose a slight variation of SSC. Let's have a look at this example solve:
F' L D' B' D2 F //EO + line
R2 //separate RB/RG
U' R2 // RB pair
U2 D R // RG pair
D' R // CO = 2001 on U/D
U' D2 R // CO + belt
D' R2 D R2 // CTL
U' R2 D' R2 D R2 D' R2 D' R2 D2 R2 //CPBL
D2 //DLDR
M2 U2 M2 U //ULUR
D2 M' D2 M //M perm
39 ATM
Steps 3 and 4 are the same as in the original method although I'm pretty sure we can make use of a wider variety of algs for CPBL to influence LEE during that step.
Instead of EO + 3 edges followed by orienting 3 corners on D, setup and do WV/SLS I propose doing only the EOline (on left) as the first step (which is easier to see in inspection and has less moves) and then solve the remaining 2 edges of the belt and CO of both layers in one intuitive step. I believe the latter can always be done in 10 moves ATM but will take significantly less moves in most of the cases. With the first step being a bit simpler it's maybe possible to predict the location of the unsolved belt edges as well as CO on D so recognition would be fine (only required to look at U).
I'm not yet sure if that step can be performed fast using the intuitive way. I basically observed that by using SLS we always break up the 3rd belt edge anyway and that the algs finish in the same manner all the time so we can save a lot of moves by not putting it in the first place.
As for how to perform that intuitive step, we would first of all separate the two edges into the U and D layers and then manipulate the CO (in relation to those edges) in both layers by setup <U,D> followed by <R>. Directly before solving belt and CO there is always a pretty easily recognizable pattern, that is, by starting from the edge and going clockwise: ccw, solved, solved, cw (hence CO = 2001 in the comments of the example solve by assigning numbers modulo 3 to the respective corner orientation).
So this comes down to the quest of getting this pattern in 6 moves ATM. For now I'm not able to figure it out fast but I suspect we can find some tricks, orders and standard cases similar to how LSE is solved in Roux nowadays.
My example solve has almost the worst case for the first two steps and yet only 17 moves as opposed to the projected movecount of 23 for the original method.