math: fix atanh (overflow and underflow issues)

in atanh exception handling was left to the called log functions,
but the argument to those functions could underflow or overflow.

use double_t and float_t to avoid some useless stores on x86

src/math/atanh.c

@@ -6,16 +6,24 @@ double atanh(double x)
 	union {double f; uint64_t i;} u = {.f = x};
 	unsigned e = u.i >> 52 & 0x7ff;
 	unsigned s = u.i >> 63;
+	double_t y;
 
 	/* |x| */
 	u.i &= (uint64_t)-1/2;
-	x = u.f;
+	y = u.f;
 
 	if (e < 0x3ff - 1) {
-		/* |x| < 0.5, up to 1.7ulp error */
-		x = 0.5*log1p(2*x + 2*x*x/(1-x));
+		if (e < 0x3ff - 32) {
+			/* handle underflow */
+			if (e == 0)
+				FORCE_EVAL((float)y);
+		} else {
+			/* |x| < 0.5, up to 1.7ulp error */
+			y = 0.5*log1p(2*y + 2*y*y/(1-y));
+		}
 	} else {
-		x = 0.5*log1p(2*x/(1-x));
+		/* avoid overflow */
+		y = 0.5*log1p(2*(y/(1-y)));
 	}
-	return s ? -x : x;
+	return s ? -y : y;
 }

src/math/atanhf.c

@@ -5,16 +5,24 @@ float atanhf(float x)
 {
 	union {float f; uint32_t i;} u = {.f = x};
 	unsigned s = u.i >> 31;
+	float_t y;
 
 	/* |x| */
 	u.i &= 0x7fffffff;
-	x = u.f;
+	y = u.f;
 
 	if (u.i < 0x3f800000 - (1<<23)) {
-		/* |x| < 0.5, up to 1.7ulp error */
-		x = 0.5f*log1pf(2*x + 2*x*x/(1-x));
+		if (u.i < 0x3f800000 - (32<<23)) {
+			/* handle underflow */
+			if (u.i < (1<<23))
+				FORCE_EVAL((float)(y*y));
+		} else {
+			/* |x| < 0.5, up to 1.7ulp error */
+			y = 0.5f*log1pf(2*y + 2*y*y/(1-y));
+		}
 	} else {
-		x = 0.5f*log1pf(2*x/(1-x));
+		/* avoid overflow */
+		y = 0.5f*log1pf(2*(y/(1-y)));
 	}
-	return s ? -x : x;
+	return s ? -y : y;
 }

src/math/atanhl.c

@@ -17,11 +17,18 @@ long double atanhl(long double x)
 	u.i.se = e;
 	x = u.f;
 
-	if (e < 0x3fff - 1) {
-		/* |x| < 0.5, up to 1.7ulp error */
-		x = 0.5*log1pl(2*x + 2*x*x/(1-x));
+	if (e < 0x3ff - 1) {
+		if (e < 0x3ff - LDBL_MANT_DIG/2) {
+			/* handle underflow */
+			if (e == 0)
+				FORCE_EVAL((float)x);
+		} else {
+			/* |x| < 0.5, up to 1.7ulp error */
+			x = 0.5*log1pl(2*x + 2*x*x/(1-x));
+		}
 	} else {
-		x = 0.5*log1pl(2*x/(1-x));
+		/* avoid overflow */
+		x = 0.5*log1pl(2*(x/(1-x)));
 	}
 	return s ? -x : x;
 }