Replace [^a-zA-Z0-9 -] with an empty string.

Regex rgx = new Regex("[^a-zA-Z0-9 -]");
str = rgx.Replace(str, "");

Removing non-alphanumeric chars

The following is the/a correct regex to strip non-alphanumeric chars from an input string:

input.replace(/\W/g, '')

Note that \W is the equivalent of [^0-9a-zA-Z_] - it includes the underscore character. To also remove underscores use e.g.:

input.replace(/[^0-9a-z]/gi, '')

The input is malformed

Since the test string contains various escaped chars, which are not alphanumeric, it will remove them.

A backslash in the string needs escaping if it's to be taken literally:

"\\test\\red\\bob\\fred\\new".replace(/\W/g, '')
"testredbobfrednew" // output

Handling malformed strings

If you're not able to escape the input string correctly (why not?), or it's coming from some kind of untrusted/misconfigured source - you can do something like this:

JSON.stringify("\\test\red\bob\fred\new").replace(/\W/g, '')
"testredbobfrednew" // output

Note that the json representation of a string includes the quotes:

JSON.stringify("\\test\red\bob\fred\new")
""\\test\red\bob\fred\new""

But they are also removed by the replacement regex.

The Java Pattern class, which is Java's implementation of regex, supports Unicode Categories, e.g. \p{Lu}. Since you want alphanumeric, that would be Categories L (Letter) and N (Number).

Since your example shows you also want to keep spaces, you need to include that. Let's use the Predefined Character Class \s, so you also get to keep newlines and tabs.

To find anything but the specified characters, use a Negation Character Class: [^abc]

All-in-all, that means [^\s\p{L}\p{N}]:

String output = input.replaceAll("[^\\s\\p{L}\\p{N}]+", "");

Where What is that an animal No It is a plane
Dónde Qué es eso un animal No Es un avión
Onde O que é isso um animal Não É um avião

Or see regex101.com for demo.

Of course, there are multiple ways to do it.

You could alternatively use the POSIX Character Class \p{Alnum}, and then enable UNICODE_CHARACTER_CLASS, using (?U).

String output = input.replaceAll("(?U)[^\\s\\p{Alnum}]+", "");

Where What is that an animal No It is a plane
Dónde Qué es eso un animal No Es un avión
Onde O que é isso um animal Não É um avião

Now, if you didn't want spaces, that could be simplified by using \P{xx} instead:

String output = input.replaceAll("(?U)\\P{Alnum}+", "");

WhereWhatisthatananimalNoItisaplane
DóndeQuéesesounanimalNoEsunavión
OndeOqueéissoumanimalNãoÉumavião

This is actually fairly straightforward.

Assuming str is the string you're cleaning up:

str = str.replace(/[^a-z0-9+]+/gi, '+');

The ^ means "anything not in this list of characters". The + after the [...] group means "one or more". /gi means "replace all of these that you find, without regard to case".

So any stretch of characters that are not letters, numbers, or '+' will be converted into a single '+'.

To remove parenthesized substrings (as requested in the comments), do this replacement first:

str = str.replace(/\(.+?\)/g, '');

function replacer() {

  var str = document.getElementById('before').value.
    replace(/\(.+?\)/g, '').
    replace(/[^a-z0-9+]+/gi, '+');

  document.getElementById('after').value = str;
}

document.getElementById('replacem').onclick = replacer;

<p>Before:
  <input id="before" value="Durarara!!x2 Ten" />
</p>

<p>
  <input type="button" value="replace" id="replacem" />
</p>

<p>After:
  <input id="after" value="" readonly />
</p>