This integration takes forever in Mathematica. Does it mean that it is not solvable?

I have a quite complicated function f(x,y) as follows.

f(x,y) = -(2/(27 \[Pi]^2))*(-2 I + I Coth[1/2 (Log[4] + Log[2 Sin[x]^2 + 2 Sin[y]^2 - Sin[x + y]^2] + Log[-Sin[x]^2 + 2 (Sin[y]^2 + Sin[x + y]^2)] - 2 Log[1/2 (3 + Cos[2 x] - 5 Cos[2 y] + Cos[2 (x + y)] + 3 I Sin[2 x] + 3 I Sin[2 y] - 3 I Sin[2 (x + y)])])] - I Coth[1/2 (-Log[4] - Log[2 Sin[x]^2 - Sin[y]^2 + 2 Sin[x + y]^2] - Log[-Sin[x]^2 + 2 (Sin[y]^2 + Sin[x + y]^2)] + 2 Log[1/2 (3 + Cos[2 x] + Cos[2 y] - 5 Cos[2 (x + y)] - 3 I Sin[2 x] - 3 I Sin[2 y] + 3 I Sin[2 (x + y)])])]) (2 I + I Coth[1/2 (Log[4] + Log[2 Sin[x]^2 + 2 Sin[y]^2 - Sin[x + y]^2] + Log[-Sin[x]^2 + 2 (Sin[y]^2 + Sin[x + y]^2)] - 2 Log[1/2 (3 + Cos[2 x] - 5 Cos[2 y] + Cos[2 (x + y)] + 3 I Sin[2 x] + 3 I Sin[2 y] - 3 I Sin[2 (x + y)])])] - I Coth[1/2 (-Log[4] - Log[2 Sin[x]^2 - Sin[y]^2 + 2 Sin[x + y]^2] - Log[-Sin[x]^2 + 2 (Sin[y]^2 + Sin[x + y]^2)] + 2 Log[1/2 (3 + Cos[2 x] + Cos[2 y] - 5 Cos[2 (x + y)] - 3 I Sin[2 x] - 3 I Sin[2 y] + 3 I Sin[2 (x + y)])])]) (-I - 2 I Coth[1/2 (Log[4] + Log[2 Sin[x]^2 + 2 Sin[y]^2 - Sin[x + y]^2] + Log[-Sin[x]^2 + 2 (Sin[y]^2 + Sin[x + y]^2)] - 2 Log[1/2 (3 + Cos[2 x] - 5 Cos[2 y] + Cos[2 (x + y)] + 3 I Sin[2 x] + 3 I Sin[2 y] - 3 I Sin[2 (x + y)])])] - I Coth[1/2 (-Log[4] - Log[2 Sin[x]^2 - Sin[y]^2 + 2 Sin[x + y]^2] - Log[-Sin[x]^2 + 2 (Sin[y]^2 + Sin[x + y]^2)] + 2 Log[1/2 (3 + Cos[2 x] + Cos[2 y] - 5 Cos[2 (x + y)] - 3 I Sin[2 x] - 3 I Sin[2 y] + 3 I Sin[2 (x + y)])])]) (I - 2 I Coth[1/2 (Log[4] + Log[2 Sin[x]^2 + 2 Sin[y]^2 - Sin[x + y]^2] + Log[-Sin[x]^2 + 2 (Sin[y]^2 + Sin[x + y]^2)] - 2 Log[1/2 (3 + Cos[2 x] - 5 Cos[2 y] + Cos[2 (x + y)] + 3 I Sin[2 x] + 3 I Sin[2 y] - 3 I Sin[2 (x + y)])])] - I Coth[1/2 (-Log[4] - Log[2 Sin[x]^2 - Sin[y]^2 + 2 Sin[x + y]^2] - Log[-Sin[x]^2 + 2 (Sin[y]^2 + Sin[x + y]^2)] + 2 Log[1/2 (3 + Cos[2 x] + Cos[2 y] - 5 Cos[2 (x + y)] - 3 I Sin[2 x] - 3 I Sin[2 y] + 3 I Sin[2 (x + y)])])]) (-I - I Coth[1/2 (Log[4] + Log[2 Sin[x]^2 + 2 Sin[y]^2 - Sin[x + y]^2] + Log[-Sin[x]^2 + 2 (Sin[y]^2 + Sin[x + y]^2)] - 2 Log[1/2 (3 + Cos[2 x] - 5 Cos[2 y] + Cos[2 (x + y)] + 3 I Sin[2 x] + 3 I Sin[2 y] - 3 I Sin[2 (x + y)])])] - 2 I Coth[ 1/2 (-Log[4] - Log[2 Sin[x]^2 - Sin[y]^2 + 2 Sin[x + y]^2] - Log[-Sin[x]^2 + 2 (Sin[y]^2 + Sin[x + y]^2)] + 2 Log[1/2 (3 + Cos[2 x] + Cos[2 y] - 5 Cos[2 (x + y)] - 3 I Sin[2 x] - 3 I Sin[2 y] + 3 I Sin[2 (x + y)])])]) (I - I Coth[1/2 (Log[4] + Log[2 Sin[x]^2 + 2 Sin[y]^2 - Sin[x + y]^2] + Log[-Sin[x]^2 + 2 (Sin[y]^2 + Sin[x + y]^2)] - 2 Log[1/2 (3 + Cos[2 x] - 5 Cos[2 y] + Cos[2 (x + y)] + 3 I Sin[2 x] + 3 I Sin[2 y] - 3 I Sin[2 (x + y)])])] - 2 I Coth[1/2 (-Log[4] - Log[2 Sin[x]^2 - Sin[y]^2 + 2 Sin[x + y]^2] - Log[-[x]^2 + 2 (Sin[y]^2 + Sin[x + y]^2)] + 2 Log[1/2 (3 + Cos[2 x] + Cos[2 y] - 5 Cos[2 (x + y)] - 3 I Sin[2 x] - 3 I Sin[2 y] + 3 I Sin[2 (x + y)])])]) Csch[1/2 (Log[4] + Log[2 Sin[x]^2 + 2 Sin[y]^2 - Sin[x + y]^2] + Log[-Sin[x]^2 + 2 (Sin[y]^2 + Sin[x + y]^2)] - 2 Log[1/2 (3 + Cos[2 x] - 5 Cos[2 y] + Cos[2 (x + y)] + 3 I Sin[2 x] + 3 I Sin[2 y] - 3 I Sin[2 (x + y)])]) + 1/2 (-Log[4] - Log[2 Sin[x]^2 - Sin[y]^2 + 2 Sin[x + y]^2] - Log[-Sin[x]^2 + 2 (Sin[y]^2 + Sin[x + y]^2)] + 2 Log[1/2 (3 + Cos[2 x] + Cos[2 y] - 5 Cos[2 (x + y)] - 3 I Sin[2 x] - 3 I Sin[2 y] + 3 I Sin[2 (x + y)])])]^2 Sinh[1/2 (Log[4] + Log[2 Sin[x]^2 + 2 Sin[y]^2 - Sin[x + y]^2] + Log[-Sin[x]^2 + 2 (Sin[y]^2 + Sin[x + y]^2)] - 2 Log[1/2 (3 + Cos[2 x] - 5 Cos[2 y] + Cos[2 (x + y)] + 3 I Sin[2 x] + 3 I Sin[2 y] - 3 I Sin[2 (x + y)])])]^4 Sinh[1/2 (-Log[4] - Log[2 Sin[x]^2 - Sin[y]^2 + 2 Sin[x + y]^2] - Log[-Sin[x]^2 + 2 (Sin[y]^2 + Sin[x + y]^2)] + 2 Log[1/2 (3 + Cos[2 x] + Cos[2 y] - 5 Cos[2 (x + y)] - 3 I Sin[2 x] - 3 I Sin[2 y] + 3 I Sin[2 (x + y)])])]^4

There is a constraint x + y = Pi, and x and y are both positive real. I want to integrate f(x,y) over y from y = 0 to y = Pi - x in order to obtain f(x). However, the integration seems to take forever. Does it mean that it is not solvable in Mathematica? Or am I doing something wrong here? In general, is there anyway to check the progress of a long taking integration in Mathematica?