Totalling Troublesome T's
JavaScript (ES6), 32 30 bytes

Saved 2 bytes thanks to @Shaggy

(w,h,x)=&gt;Q=n=&gt;--h?n+Q(n-x):n*w


Takes input in a curried format: f(W,H,x,y)(n)



let f =
(w,h,x)=&gt;Q=n=&gt;--h?n+Q(n-x):n*w;

console.log(
  f(3, 4, 9, 9)(32)
);




How?

First we note that the sum of the T starting at n with height H can be broken down into two sums:


n
The sum of the T starting one row higher with height H - 1


By repeatedly adding n to the total, moving one row up, and subtracting 1 from H until it reaches 1, we end up summing the vertical stem of the T. Moving one row up is accomplished by subtracting x from n, since it can be observed that the difference between any cell and the one above is x.

When H reaches 1, we now have only the crossbar of width W left to sum. At this point n represents the center of the crossbar. We could sum the crossbar recursively as we did with the stem, but instead we take advantage of a fairly simple formula:

sum(n - a : n + a) = (n - a) + (n - (a-1)) + ... + n + ... + (n + (a-1)) + (n + a)
                   = (n - a) + (n + a) + (n - (a-1)) + (n + (a-1)) + ... + n
                   = 2n + 2n + ... + n
                   = n * (2a + 1)


In this case, our a is (W - 1) / 2, which makes

n * (2a + 1) = n * (((W - 1) / 2) * 2 + 1)
             = n * ((W - 1) + 1)
             = n * W


which is the sum of the crossbar.