grep: regex only for match anything between parenthesis

Here are a few options, all of which print the desired output:

Using grep with the -o flag (only print matching part of line) and Perl compatible regular expressions (-P) that can do lookarounds:
```
printf "this is (test.com)\n" | grep -Po '(?<=\().*(?=\))'
```
That regex might need some explaining:
- (?<=\() : This is a positive lookbehind, the general format is (?<=foo)bar and that will match all cases of bar found right after foo. In this case, we are looking for an opening parenthesis, so we use \( to escape it.
- (?=\)) : This is a positive lookahead and simply matches the closing parenthesis.
The -o option to grep causes it to only print the matched part of any line, so we look for whatever is in parentheses and then delete them with sed:
```
printf "this is (test.com)\n" | grep -o '(.*)' | sed 's/[()]//g'
```

Parse the whole thing with Perl:

printf "this is (test.com)\n" | perl -pe 's/.*\((.+?)\)/$1/'

Parse the whole thing with sed:

printf "this is (test.com)\n" | sed 's/.*(\(.*\))/\1/'

One approach would be to use PCRE - Perl Compatible Regular Expressions with grep:

$ echo "this is (test.com)" | grep -oP '(?<=\().*(?=\))'
test.com

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