Объединение нескольких регулярных выражений в один

Я пытаюсь написать код для переноса строки в латинские стихи. Есть несколько ограничений, на которые я позаботился, однако я не получаю желаемого результата. Мой код приведен ниже:

<?php $string = "impulerittantaenanimis caelestibusirae"; $precedingC = precedingConsonant($string); $xrule = xRule($precedingC); $consonantc = consonantCT($xrule); $consonantp = consonantPT($consonantc); $cbv = CbetweenVowels($consonantp); $tv = twoVowels($cbv); echo $tv; function twoVowels($string) { return preg_replace('/([aeiou])([aeiou])/', '$1-$2', $string); } function CbetweenVowels($string) { return preg_replace('/([aeiou])([^aeiou])([aeiou])/', '$1-$2$3', $string); } function consonantPT($string) { return preg_replace('/([^aeiou]p)(t[aeiou])/', '$1-$2', $string); } function consonantCT($string) { return preg_replace('/([^aeiou]c)(t[aeiou])/', '$1-$2', $string); } function precedingConsonant($string) { $arr1 = str_split($string); $length = count($arr1); for($j=0;$j<$length;$j++) { if(isVowel($arr1[$j]) && !isVowel($arr1[$j+1]) && !isVowel($arr1[$j+2]) && isVowel($arr1[$j+3])) { $pc++; } } function strAppend2($string) { $arr1 = str_split($string); $length = count($arr1); for($i=0;$i<$length;$i++) { $check = $arr1[$i+1].$arr1[$i+2]; $check2 = $arr1[$i+1].$arr1[$i+2].$arr1[$i+3]; if($check=='br' || $check=='cr' || $check=='dr' || $check=='fr' || $check=='gr' || $check=='pr' || $check=='tr' || $check=='bl' || $check=='cl' || $check=='fl' || $check=='gl' || $check=='pl' || $check=='ch' || $check=='ph' || $check=='th' || $check=='qu' || $check2=='phl' || $check2=='phr') { if(isVowel($arr1[$i]) && !isVowel($arr1[$i+1]) && !isVowel($arr1[$i+2]) && isVowel($arr1[$i+3])) { $updatedString = substr_replace($string, "-", $i+1, 0); return $updatedString; } } else { if(isVowel($arr1[$i]) && !isVowel($arr1[$i+1]) && !isVowel($arr1[$i+2]) && isVowel($arr1[$i+3])) { $updatedString = substr_replace($string, "-", $i+2, 0); return $updatedString; } } } } $st1 = $string; for($k=0;$k<$pc;$k++) { $st1 = strAppend2($st1); } return $st1; } function xRule($string) { return preg_replace('/([aeiou]x)([aeiou])/', '$1-$2', $string); } function isVowel($ch) { if($ch=='a' || $ch=='e' || $ch=='i' || $ch=='o' || $ch=='u') { return true; } else { return false; } } function isConsonant($ch) { if($ch=='a' || $ch=='e' || $ch=='i' || $ch=='o' || $ch=='u') { return false; } else { return true; } } ?> 

Я считаю, что если объединить все эти функции, это приведет к желаемому результату. Однако я укажу свои ограничения ниже:

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 Rule 1 : When two or more consonants are between vowels, the first consonant is joined to the preceding vowel; for example - rec-tor, trac-tor, ac-tor, delec-tus, dic-tator, defec-tus, vic-tima, Oc-tober, fac-tum, pac-tus, Rule 2 : 'x' is joined to the preceding vowel; as, rex-i. However we give a special exception to the following consonants - br, cr, dr, fr, gr, pr, tr; bl, cl, fl, gl, pl, phl, phr, ch, ph, th, qu. These consonants are taken care by adding them to the later vowel for example - con- sola-trix n- sola-trix. Rule 3 : When 'ct' follows a consonant, that consonant and 'c' are both joined to the first vowel for example - sanc-tus and junc-tum Similarly for 'pt' we apply the same rule for example - scalp-tum, serp-tum, Redemp-tor. Rule 4 : A single consonant between two vowels is joined to the following vowel for example - ma-ter, pa-ter AND Z is joined to the following vowel. Rule 5 : When two vowels come together they are divided, if they be not a diphthong; as au-re-us. Diaphthongs are - "ae","oe","au" 

 $pattern = '~ (?<=[aeiou]) # each rule involves a vowel at the beginning (also called a # "preceding vowel") (?: # Rule 2: capture particular cases ( (?:[bcdfgpt]r | [bcfgp] l | ph [lr] | [cpt] h | qu ) [aeiou] x ) | [bcdfghlmnp-tx] (?: # Rule 3: When "ct" follows a consonant, that consonant and "c" are both # joined to the first vowel [cp] \K (?=t) | # Rule 1: When two or more consonants are between vowels, the first # consonant is joined to the preceding vowel \K (?= [bcdfghlmnp-tx]+ [aeiou] ) ) | # Rule 4: a single consonant between two vowels is joined to the following # vowel (?: \K (?= [bcdfghlmnp-t] [aeiou] ) | # Rule 2: "x" is joined to the preceding vowel x \K (?= [az] | (*SKIP)(*F) ) ) | # Rule 5: When two vowels come together they are divided, if they not be a # diphthong ("ae", "oe", "au") \K (?= [aeiou] (?<! a[eu] | oe ) ) ) ~xi'; 

 $string = <<<EOD Aeneadum genetrix, hominum diuomque uoluptas, alma Uenus, caeli subter labentia signa quae mare nauigerum, quae terras frugiferentis concelebras, per te quoniam genus omne animantum EOD; $result = preg_replace($pattern, '-$1', $string);