I just managed to solve this with around 30 minutes to go on Saturday. I found it very difficult to get my hooks into it. Just wondering did others struggle with it. Seems to have fewer solvers than usual.

It was tough.

That puzzle (link will expire) does seem quite difficult. While I can't offer a full analytical solution, here are some of the observations that helped me solve it. (I did use "pencil" numbers for this one )

(A) The sum of the numbers in cells f2 and f3 is 5, so those two cells must contain either [1,4] or [2,3]. (In the former case, 4 must be in f3 because the 10 cage in that quadrant can't contain [1,4,5] while 1 is also in f3.)

(B) In column d, the sum of the numbers in the four cells outside the 15 and 19 cages is 45 - 15 - 19 = 11, which means that those cells must contain [1,2,3,5].

(C) If the 10 cage in cells c3, c4 and d4 contains the number 1, it must be in d4 (because the other cells are "covered" by the 1 in c5). If not, the numbers in that cage must be [2,3,5].

(D) In the upper left quadrant, the sum of all the cages that are wholly or partially in the quadrant is 10 + 11 + 12 + 13 + 14 = 60, of which 45 is inside the quadrant and 15 in the combined four "outie" cells of the cages 10, 13 and 14 (cells a4, c4, d4 and d1, of which the latter two are among the four cells mentioned in (B).

(E) I tried to determine where the number 9 could go within the same quadrant. Its upper row is "covered" by the 9 in i1, and c3 obviously can't contain 9 either. When I placed 9 in a2 or c2, there was no correct way to complete rows 1 and 2. That left the 11 cage (cells b2 and b3) and cell a3 as the only possible positions of 9 within that quadrant.

The latter two observations made me think that the rather small sum of the four "outies" combined with the knowledge that a4 had to contain a comparatively high number (5, 6, 8 or 9) might be enough of a constraint to permit some inferences. I tried putting 9 in that cell, which soon turned out not to permit any correct solution to columns c and d (the 21 cage at the bottom).

Keeping my focus on the number 9, I then tried putting it in a3, with 5 in a4. Apropos (C), I don't remember if I managed to prove (using columns c and d) that the 10 cage couldn't contain 1 under this hypothesis, or if [2,3,5] in that cage was just a sub-hypothesis I was checking, but at any rate it led to a correct solution to the puzzle.

Even assuming that [2,3,5] in the 10 cage was the result of a valid inference, I would of course still have to check and exclude both variants of [6,8] in the 14 cage in cells a3 and a4 (with [2,9] in the 11 cage within the upper left quadrant) to make sure that the solution I'd found was unique.

Working through the branches of trial and error this way after you've found a correct solution may give you a better understanding of the set of conditions/constraints that force the unique solution to the puzzle. You may find (or at least approach) a full analytical solution instead of a gamebook-style catalog of all the branches (and twigs) of trial and error where all paths execpt one lead to a dead end

But I'm trying to cut back on the puzzling so I won't attempt any further analysis of this one

For reference, a screenshot of the puzzle:

After your step (D) you wanted to find the cell that has 9 in the upper left 3x3 area. We can find that by using your total of 15 for cells a4,c4,d1,d4. First we can find that d1=5: eliminate 1 and 2 from d1 by referring to cages e1 and f1, and cell f3, then eliminate 3 by using your total of 15 for the 4 "outie" cells and the total of 11 for cells d1,d4,d8,d9 (step(B)). Still using the total of 15 we can now solve c3=5, then a4=5 and that gives us a3=9. From this point the puzzle is getting close to being solved.

Thanks a lot, michaele

Even using your strong hint, I struggled a bit with this step but finally managed: 3 in d1 -> 1 in d4 (2 and 5 don't work when considering (D)) -> [2,5] in d8-d9 -> [6,8] in c8-c9 -> no valid way to get the sum 10 in c1-c2

There are quite a number of conditions that can be deduced and used while solving the problem.
They are obtained by combining the totals of the cages with rectangles with a known total.
Rectangles can be 3x3, 3x6, nx9.
Here's a list of useful relations used solving the puzzle:

C1 + C2 + A3 + C3 = 22
D1 + A4 + C4 + D4 = 15
D1 + F1 + E3 + F3 = 20
G1 + I2 + H3 + I3 = 14
F1 + I4 + H4 = 13
A4 + C4 + A6 + C6 = 21
A3 + C3 + D4 + A7 + C7 = 22
G4 + H4 + I4 + G6 + I6 = 24
A7 + C7 + C8 + C9 = 23
A6 + C6 + D8 + D9 = 17
D7 + F7 + D8 + F8 + D9 + F9 = 27
G7 + I7 + G8 = 12
I6 + F8 + F9 = 13
D1 + D4 + A7 + C7 = 13
C5 = 1
D1 + D4 + D8 + D9 = 11
F2 + F3 = 5
I1 + I5 + I8 + I9 = 23
B4 + A5 + B5 + B6 = 23
B4 + A5 + B5 + B6 = 23
F2 + F3 = 5
I1 + I5 + I8 + I9 = 23
G5 + G9 = 15
B2 + D2 + G2 + I2 = 25
B7 + E7 + G7 + H7 = 21

For example C5=1 is based on cols A,B. The cages of this region total 91 and C5 is the only external point.
F2 + F3 = 5 is similarly based on the region consisting of columns A-E.