Counting symmetries but not inversions as the same case there are in total 270 different positions for this case. The perfect solutions for all these are well known, thanks to Bernhard Helmstetter, and average 12.08 moves.
The most successful such system is the Friedrich 2 look system, as described here. The first "look" (OLL) orients all pieces, and requires memorization of 57 algs. The second (PLL) places all pieces, using 21 algs. The average length of each step is around 10 moves [anyone know the exact numbers?], resulting in a total move count of around 20 for this approach.
The idea is to learn only a reasonably small number of algorithms - certainly less than the 78 needed for Friedrich, and for each of the 270 distinct positions, find the fastest combination of two of those algs for each of the 270 positions, and memorize those. Or to be precise, if you learn 40 algs, you will use a pair of algorithms for 230 positions, and only one for the positions the 40 solve.
Learning all those is still a fairly big memorization undertaking, but much smaller than memorizing 270 individual algorithms. My gut feeling number is that it's 10% of the effort. And for "muscle memory", you only need to keep 40 known fast algs up to speed, which is much easier than 270.
I think that the 2-3 more moves used compared to the perfect 12.08 is more than compensated by these factors in terms of execution speed.
I started by putting in all edge orientation preserving algs of 11 or less moves, as listed by Helmstetter (The "BH54" codes refers to position 54 in the Bernhard Helmstetter list on his 5+6+7 page.). I then tried to determine which ones was least useful. The color coding below is an attempt to show which are the core algs you need to learn (green), the more advanced one that are useful (yellow), and the even more advanced (red). It needs a lot more work, and is not all that reliable, but it's a start.
These are pretty much useless according to the computer, but I've always used them and I'm not about to stop now.
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