Pieces of a Latin Square

Place the numbers 1-4 in the 16 squares of a 4x4 grid, such that each row and column contains one of each number.  Then divide the grid into five pieces along the gridlines, such that:

1. Each piece has the same sum (8)
2. Each piece is unique (no pieces have the same numbers in the same shape, through rotations or reflections.)

Source: Original.

Each of our 5 pieces must contain at least 3 squares, since 4+4 cannot exist as a piece. 5 pieces of size 3 use 15 squares - 1 spare. Therefore, we need to cut into pieces of size {3,3,3,3,4}

At least 2 of the '1' cells will be part of a size 3 piece.

If we have '1's in 3 of our size 3 pieces, all values are forced: {1,3,4},{1,3,4},{1,3,4},{2,2,4} and {2,2,1,3}. Now, since we are not allowed pieces containing the same digits to be in identical shapes (under rotation/reflection), we have a problem, since there are only 2 possible tri-omino shapes.

Therefore, we can only have '1's in 2 of the size 3 pieces - the other 2 appearing in the size 4. So we have:
{1,1,3,3},{1,3,4},{1,3,4},{2,2,4},{2,2,4} or
{1,1,2,4},{1,3,4},{1,3,4},{2,2,4},{2,3,3}

Now, notice the first set has 2 size 3 pieces which both have 2x'2's. These cannot be drawn as different shaped tri-ominos, so this set is impossible to draw.

So, our pieces can ONLY be made of the following numbers:
{1,1,2,4},{1,3,4},{1,3,4},{2,2,4},{2,3,3}
Note that the only straight piece is one of the {1,3,4} pieces. ALL the others are bent. Also note that the straight {1,3,4} can only go on an edge of the 4x4 grid, otherwise a straight piece would be required to fit next to it. The same goes for the long edge of the size 4....

+-----+-+
|2 4 1|3|
| +---| |
|1|2 3|4|
+-| +-| |
|4|3|2|1|
| +-| +-+
|3 1|4 2|
+---+---+