#include "Python.h"

#ifndef HAVE_HYPOT

double hypot(double x, double y)

{

	double yx;

	x = fabs(x);

	y = fabs(y);

	if (x < y) {

		double temp = x;

		x = y;

		y = temp;

	}

	if (x == 0.)

		return 0.;

	else {

		yx = y/x;

		return x*sqrt(1.+yx*yx);

	}

}

#endif /* HAVE_HYPOT */

#ifndef HAVE_COPYSIGN

static double

copysign(double x, double y)

{

	/* use atan2 to distinguish -0. from 0. */

	if (y > 0. || (y == 0. && atan2(y, -1.) > 0.)) {

		return fabs(x);

	} else {

		return -fabs(x);

	}

}

#endif /* HAVE_COPYSIGN */

#ifndef HAVE_LOG1P

#include <float.h>

double

log1p(double x)

{

	/* For x small, we use the following approach.  Let y be the nearest

	   float to 1+x, then

	     1+x = y * (1 - (y-1-x)/y)

	   so log(1+x) = log(y) + log(1-(y-1-x)/y).  Since (y-1-x)/y is tiny,

	   the second term is well approximated by (y-1-x)/y.  If abs(x) >=

	   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 <= x && x <= 1.) {

		/* WARNING: it's possible than an overeager compiler

		   will incorrectly optimize the following two lines

		   to the equivalent of "return log(1.+x)". If this

		   happens, then results from log1p will be inaccurate

		   for small x. */

		y = 1.+x;

		return log(y)-((y-1.)-x)/y;

	} else {

		/* NaNs and infinities should end up here */

		return log(1.+x);

	}

}

#endif /* HAVE_LOG1P */

/*

 * ====================================================

 * 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.

 * ====================================================

 */

static const double ln2 = 6.93147180559945286227E-01;

static const double two_pow_m28 = 3.7252902984619141E-09; /* 2**-28 */

static const double two_pow_p28 = 268435456.0; /* 2**28 */

static const double zero = 0.0;

/* asinh(x)

 * Method :

 *	Based on 

 *		asinh(x) = sign(x) * log [ |x| + sqrt(x*x+1) ]

 *	we have

 *	asinh(x) := x  if  1+x*x=1,

 *		 := sign(x)*(log(x)+ln2)) for large |x|, else

 *		 := sign(x)*log(2|x|+1/(|x|+sqrt(x*x+1))) if|x|>2, else

 *		 := sign(x)*log1p(|x| + x^2/(1 + sqrt(1+x^2)))  

 */

#ifndef HAVE_ASINH

double

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 <= |x| < 2= */

		double t = x*x;

		w = log1p(absx + t / (1.0 + sqrt(1.0 + t)));

	}

	return copysign(w, x);

}

#endif /* HAVE_ASINH */

/* 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<1.

 *      acosh(NaN) is NaN without signal.

 */

#ifndef HAVE_ACOSH

double

acosh(double x)

{

	if (Py_IS_NAN(x)) {

		return x+x;

	}

	if (x < 1.) {			/* x < 1;  return a signaling NaN */

		errno = EDOM;

#ifdef Py_NAN

		return Py_NAN;

#else

		return (x-x)/(x-x);

#endif

	}

	else 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 */

		double t = x - 1.0;

		return log1p(t + sqrt(2.0*t + t*t));

	}

}

#endif /* HAVE_ACOSH */

/* atanh(x)

 * Method :

 *    1.Reduced x to positive by atanh(-x) = -atanh(x)

 *    2.For x>=0.5

 *		  1	      2x			  x

 *      atanh(x) = --- * log(1 + -------) = 0.5 * log1p(2 * --------)

 *		  2	     1 - x		      1 - x

 *

 *      For x<0.5

 *      atanh(x) = 0.5*log1p(2x+2x*x/(1-x))

 *

 * Special cases:

 *      atanh(x) is NaN if |x| >= 1 with signal;

 *      atanh(NaN) is that NaN with no signal;

 *

 */

#ifndef HAVE_ATANH

double

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 * log1p(t + t*absx / (1.0 - absx));

	} 

	else {				/* 0.5 <= |x| <= 1.0 */

		t = 0.5 * log1p((absx + absx) / (1.0 - absx));

	}

	return copysign(t, x);

}

#endif /* HAVE_ATANH */