Google Sheets – How to Extract Image Link from Spreadsheet Cell

After seeing Rubén's answer I decided to write a different custom function for this task, with the following features:

1. The parameter is provided as a range, not as a string: that is, =linkURL(C2) instead of =linkURL("C2"). This is consistent with how parameters usually work, and makes the references more robust: they will be maintained if someone adds a new row on top.
2. Arrays are supported: =linkURL(B2:D5) returns the URLs of all hyperlink commands found in this range (and blank cells for other places).

To achieve 1, I do not use the argument passed by the sheet (which would be the text content of the target cell), but instead parse the formula =linkURL(...) itself and extract range notation from there.

/** 
 * Returns the URL of a hyperlinked cell, if it's entered with hyperlink command. 
 * Supports ranges
 * @param {A1}  reference Cell reference
 * @customfunction
 */
function linkURL(reference) {
  var sheet = SpreadsheetApp.getActiveSheet();
  var formula = SpreadsheetApp.getActiveRange().getFormula();
  var args = formula.match(/=\w+\((.*)\)/i);
  try {
    var range = sheet.getRange(args[1]);
  }
  catch(e) {
    throw new Error(args[1] + ' is not a valid range');
  }
  var formulas = range.getFormulas();
  var output = [];
  for (var i = 0; i < formulas.length; i++) {
    var row = [];
    for (var j = 0; j < formulas[0].length; j++) {
      var url = formulas[i][j].match(/=hyperlink\("([^"]+)"/i);
      row.push(url ? url[1] : '');
    }
    output.push(row);
  }
  return output
}

I found a solution... I ended up reconstructing the URL using the REGEXMATCH formula inside a copy recipe I made in IFTTT. That throws the formula with the raw data directly into the Google Sheet cell and the process to extract the URL is happening on the fly. The main obstacle is the limited API handlers offered by IFTTT to begin with. I found that zapier offers greater API access, especially when comes to Twitter, but it costs.