There is a python script that does prime factorization. It's very fast and runs in less than a second. But there are some functions for php that run very slow. It take one parameter (a long integer) like 1278426847636566097 and calculate prime factorization. Returns an array with 2 indexes. The result for this number is: 
Array ( [0] => 1233387599 [1] => 1036516703 )
The python script: (getpq.py)
#!/usr/bin/env python
from __future__ import print_function
import prime
import sys
import json
def eprint(*args, **kwargs):
    print(*args, file=sys.stderr, **kwargs)
pq = prime.primefactors(long(sys.argv[1]))
sys.stdout.write(json.dumps(pq))
sys.stdout.flush()
prime.py:
# NOTICE!!! This is copied from https://stackoverflow.com/questions/4643647/fast-prime-factorization-module
import random
def primesbelow(N):
    # http://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
    #""" Input N>=6, Returns a list of primes, 2 <= p < N """
    correction = N % 6 > 1
    N = {0:N, 1:N-1, 2:N+4, 3:N+3, 4:N+2, 5:N+1}[N%6]
    sieve = [True] * (N // 3)
    sieve[0] = False
    for i in range(int(N ** .5) // 3 + 1):
        if sieve[i]:
            k = (3 * i + 1) | 1
            sieve[k*k // 3::2*k] = [False] * ((N//6 - (k*k)//6 - 1)//k + 1)
            sieve[(k*k + 4*k - 2*k*(i%2)) // 3::2*k] = [False] * ((N // 6 - (k*k + 4*k - 2*k*(i%2))//6 - 1) // k + 1)
    return [2, 3] + [(3 * i + 1) | 1 for i in range(1, N//3 - correction) if sieve[i]]
smallprimeset = set(primesbelow(100000))
_smallprimeset = 100000
def isprime(n, precision=7):
    # http://en.wikipedia.org/wiki/Miller-Rabin_primality_test#Algorithm_and_running_time
    if n == 1 or n % 2 == 0:
        return False
    elif n < 1:
        raise ValueError("Out of bounds, first argument must be > 0")
    elif n < _smallprimeset:
        return n in smallprimeset
    d = n - 1
    s = 0
    while d % 2 == 0:
        d //= 2
        s += 1
    for repeat in range(precision):
        a = random.randrange(2, n - 2)
        x = pow(a, d, n)
        if x == 1 or x == n - 1: continue
        for r in range(s - 1):
            x = pow(x, 2, n)
            if x == 1: return False
            if x == n - 1: break
        else: return False
    return True
# https://comeoncodeon.wordpress.com/2010/09/18/pollard-rho-brent-integer-factorization/
def pollard_brent(n):
    if n % 2 == 0: return 2
    if n % 3 == 0: return 3
    y, c, m = random.randint(1, n-1), random.randint(1, n-1), random.randint(1, n-1)
    g, r, q = 1, 1, 1
    while g == 1:
        x = y
        for i in range(r):
            y = (pow(y, 2, n) + c) % n
        k = 0
        while k < r and g==1:
            ys = y
            for i in range(min(m, r-k)):
                y = (pow(y, 2, n) + c) % n
                q = q * abs(x-y) % n
            g = gcd(q, n)
            k += m
        r *= 2
    if g == n:
        while True:
            ys = (pow(ys, 2, n) + c) % n
            g = gcd(abs(x - ys), n)
            if g > 1:
                break
    return g
smallprimes = primesbelow(10000) # might seem low, but 1000*1000 = 1000000, so this will fully factor every composite < 1000000
def primefactors(n, sort=False):
    factors = []
    limit = int(n ** .5) + 1
    for checker in smallprimes:
        if checker > limit: break
        while n % checker == 0:
            factors.append(checker)
            n //= checker
            limit = int(n ** .5) + 1
            if checker > limit: break
    if n < 2: return factors
    while n > 1:
        if isprime(n):
            factors.append(n)
            break
        factor = pollard_brent(n) # trial division did not fully factor, switch to pollard-brent
        factors.extend(primefactors(factor)) # recurse to factor the not necessarily prime factor returned by pollard-brent
        n //= factor
    if sort: factors.sort()
    return factors
def factorization(n):
    factors = {}
    for p1 in primefactors(n):
        try:
            factors[p1] += 1
        except KeyError:
            factors[p1] = 1
    return factors
totients = {}
def totient(n):
    if n == 0: return 1
    try: return totients[n]
    except KeyError: pass
    tot = 1
    for p, exp in factorization(n).items():
        tot *= (p - 1)  *  p ** (exp - 1)
    totients[n] = tot
    return tot
def gcd(a, b):
    if a == b: return a
    while b > 0: a, b = b, a % b
    return a
def lcm(a, b):
    return abs(a * b) // gcd(a, b)
As I don't know python, is there any way to do this is php?
NOTE: I found a very bad algorithm, implemented in php, takes about 600 seconds:
public function primefactor($num) {
    $sqrt = sqrt($num);
    for ($i = 2; $i <= $sqrt; $i++) {
        if ($num % $i == 0) {
            return array_merge($this->primefactor($num/$i), array($i));
        }
    }
    return array($num);
}
 
    