/* Definitions of some C99 math library functions, for those platforms
   that don't implement these functions already. */

#include "Python.h"
#include 
#include "_math.h"

/* The following copyright notice applies to the original
   implementations of acosh, asinh and atanh. */

/*
 * ====================================================
 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
 *
 * Developed at SunPro, a Sun Microsystems, Inc. business.
 * Permission to use, copy, modify, and distribute this
 * software is freely granted, provided that this notice
 * is preserved.
 * ====================================================
 */

#if !defined(HAVE_ACOSH) || !defined(HAVE_ASINH)
static const double ln2 = 6.93147180559945286227E-01;
static const double two_pow_p28 = 268435456.0; /* 2**28 */
#endif
#if !defined(HAVE_ASINH) || !defined(HAVE_ATANH)
static const double two_pow_m28 = 3.7252902984619141E-09; /* 2**-28 */
#endif
#if !defined(HAVE_ATANH) && !defined(Py_NAN)
static const double zero = 0.0;
#endif


#ifndef HAVE_ACOSH
/* acosh(x)
 * Method :
 *      Based on
 *            acosh(x) = log [ x + sqrt(x*x-1) ]
 *      we have
 *            acosh(x) := log(x)+ln2, if x is large; else
 *            acosh(x) := log(2x-1/(sqrt(x*x-1)+x)) if x>2; else
 *            acosh(x) := log1p(t+sqrt(2.0*t+t*t)); where t=x-1.
 *
 * Special cases:
 *      acosh(x) is NaN with signal if x= two_pow_p28) {        /* x > 2**28 */
        if (Py_IS_INFINITY(x)) {
            return x+x;
        }
        else {
            return log(x) + ln2;          /* acosh(huge)=log(2x) */
        }
    }
    else if (x == 1.) {
        return 0.0;                     /* acosh(1) = 0 */
    }
    else if (x > 2.) {                  /* 2 < x < 2**28 */
        double t = x * x;
        return log(2.0 * x - 1.0 / (x + sqrt(t - 1.0)));
    }
    else {                              /* 1 < x 2, else
 *               := sign(x)*log1p(|x| + x^2/(1 + sqrt(1+x^2)))
 */

double
_Py_asinh(double x)
{
    double w;
    double absx = fabs(x);

    if (Py_IS_NAN(x) || Py_IS_INFINITY(x)) {
        return x+x;
    }
    if (absx < two_pow_m28) {           /* |x| < 2**-28 */
        return x;                       /* return x inexact except 0 */
    }
    if (absx > two_pow_p28) {           /* |x| > 2**28 */
        w = log(absx) + ln2;
    }
    else if (absx > 2.0) {              /* 2 < |x| < 2**28 */
        w = log(2.0 * absx + 1.0 / (sqrt(x * x + 1.0) + absx));
    }
    else {                              /* 2**-28 =0.5
 *                  1              2x                          x
 *      atanh(x) = --- * log(1 + -------) = 0.5 * log1p(2 * -------)
 *                  2             1 - x                      1 - x
 *
 *      For x= 1 with signal;
 *      atanh(NaN) is that NaN with no signal;
 *
 */

double
_Py_atanh(double x)
{
    double absx;
    double t;

    if (Py_IS_NAN(x)) {
        return x+x;
    }
    absx = fabs(x);
    if (absx >= 1.) {                   /* |x| >= 1 */
        errno = EDOM;
#ifdef Py_NAN
        return Py_NAN;
#else
        return x / zero;
#endif
    }
    if (absx < two_pow_m28) {           /* |x| < 2**-28 */
        return x;
    }
    if (absx < 0.5) {                   /* |x| < 0.5 */
        t = absx+absx;
        t = 0.5 * m_log1p(t + t*absx / (1.0 - absx));
    }
    else {                              /* 0.5 =
       DBL_EPSILON/2 or the rounding-mode is some form of round-to-nearest
       then y-1-x will be exactly representable, and is computed exactly by
       (y-1)-x.

       If abs(x) < DBL_EPSILON/2 and the rounding mode is not known to be
       round-to-nearest then this method is slightly dangerous: 1+x could be
       rounded up to 1+DBL_EPSILON instead of down to 1, and in that case
       y-1-x will not be exactly representable any more and the result can be
       off by many ulps.  But this is easily fixed: for a floating-point
       number |x| < DBL_EPSILON/2., the closest floating-point number to
       log(1+x) is exactly x.
    */

    double y;
    if (fabs(x) < DBL_EPSILON / 2.) {
        return x;
    }
    else if (-0.5