It really depends on your input...

Anoter alternative for your regex would be: tr/x//d; (but ofcourse this only works when you want to remove some charachters and not a word).

What I noticed during benchmarking is that the less x's follows eachother the faster tr/// gets (or the slower s/// gets). Below you find the benchmarks I ran:

String: 'a' . ('x' x 1000) . 'b':

           Rate NoPlus     Tr   Plus
NoPlus   2465/s     --   -97%   -98%
Tr      94056/s  3715%     --   -15%
Plus   111285/s  4414%    18%     --
[download]

String: 'a' . ('x' x 100) . 'b':

           Rate NoPlus   Plus     Tr
NoPlus  22929/s     --   -91%   -93%
Plus   245996/s   973%     --   -25%
Tr     328951/s  1335%    34%     --
[download]

String: 'ax' x 1000:

          Rate NoPlus   Plus     Tr
NoPlus  2224/s     --    -6%   -95%
Plus    2367/s     6%     --   -95%
Tr     49416/s  2122%  1988%     --
[download]

String: 'ax' x 100:

           Rate NoPlus   Plus     Tr
NoPlus  21151/s     --    -4%   -93%
Plus    22038/s     4%     --   -93%
Tr     302771/s  1331%  1274%     --
[download]

Based on these results I would most likely use tr/x//d; ...