So the 'standard' method seems to be to solve the inners and middle then solve U and D as a domino. It's about time I explained the method that me and Dan Cohen have been getting sub30 with.
The steps are 3x3x3 -> Columns -> L4E
The first step is to solve it as a 3x3x3. This allows all layers to be turned freely and just leaves the inner slices to solve. It's probably quite similar to the other method only you're solving the outer layers instead of the inner ones. You can use any method here, but some have delicious advantages that are 3x3x5 specific :3. I hear that F2L recognition is a bit annoying.
After this step is done, you want to orient your cube so that the 'top' and 'bottom' faces are on the L and R sides so you can use r and l moves for the inner layers.
Next step is columns (usually). The aim is to solve the edges and centres (the two types of inner layer pieces) in the first two layers leaving four edges left to solve. This is accomplished with a technique called columns.
The idea of the technique is to blockbuild the pieces together. Generally, I'll first solve one of the two 1x2x3 'blocks' then move onto the other. This is usually done by creating a 1x2x2 centre-edge-centre block and then adding the centre-edge pair. This leaves you with another block to do the same with on the other side.
The second block is a little harder. The first 1x2x2 goes in just fine, but the final centre-edge pair presents an interesting problem. It cannot be easily created as the other columns have been because you are restricted in where you can store pieces (all the empty space is in use). From here there are a few things you can do. At first I was just placing the edge and then using [TPerm, l] to comm the centre in. These days it mostly varies on the case. You can put the edge in and use some variation on r U2 r U2 r2 U2 r2 U2 r U2 r' to solve the centre or put the centre in and use some commutator to solve the edge. (With the latter technique you might get lucky and get the chance to force a L4E skip.)
I forgot to mention that the final pair may be easy to create. When that happens it feels like you're skipping a step or something.
When you get good, you can treat the last step as last 5 edges and three centres - looking for the best way to solve from there instead of moving to L4E. (However, L4E is often the best solution anyway.)
Finally, here's an alg for every L4E case;
[l, U' R U x U R2 U' x']
or l' U2 r' D2 r U2 r' D2 r l
2x2 Cycle Diag (PLL Parity)
R2 F2 U2 r2 U2 F2 R2
2x2 Cycle Adj (2 Flip)
U2 r' U2 l r U2 r' U2 r U2 l' r' U2 r
2x2 Cycle Opp (PLL Parity + 2 Flip)
l U2 r D2 r' U2 l' r U2 l D2 l' U2 r'
2 Cycle Diag
r U2 r' U2 r' U2 l U2 r' U2 r U2 F2 r2 F2 l'
2 Cycle Adj
r' U2 x l U2 l' U2 x' r2 U2 r U2 r' U2 F2 r2 F2
2 Cycle Opp
r U2 r U2 x U2 r U2 l' x' U2 l U2 r2
4 Cycle Clockwise
r U2 r2 U2 r' U2 r U2 r' U2 r2 U2 r
4 Cycle Anticlockwise
r' U2 r2 U2 r U2 r' U2 r U2 r2 U2 r'
4 Cycle X (British Parity)
r2 B2 r' U2 r' U2 B2 r' B2 r B2 r' B2 r2 B2
4 Cycle Z
r U2 r l' U2 r2 U2 r' U2 r U2 r' U2 r' U2 l U2 r'
The badasses that know K4LL should already know most of these. You'll have to mirror/invert them to cover everything.