3.2.84 \(\int \sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)} \, dx\) [184]

Optimal. Leaf size=123 \[ \frac {2 \sqrt {a} \sqrt {c} \text {ArcTan}\left (\frac {\sqrt {a} \sqrt {c} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)}}\right )}{f}+\frac {2 \sqrt {a} \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)}}\right )}{f} \]

[Out]

2*arctan(a^(1/2)*c^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e))^(1/2))*a^(1/2)*c^(1/2)/f+2*arctanh
(a^(1/2)*d^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e))^(1/2))*a^(1/2)*d^(1/2)/f

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Rubi [A]
time = 0.21, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {4017, 4019, 209, 4065, 212} \begin {gather*} \frac {2 \sqrt {a} \sqrt {c} \text {ArcTan}\left (\frac {\sqrt {a} \sqrt {c} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a} \sqrt {c+d \sec (e+f x)}}\right )}{f}+\frac {2 \sqrt {a} \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a} \sqrt {c+d \sec (e+f x)}}\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Sec[e + f*x]]*Sqrt[c + d*Sec[e + f*x]],x]

[Out]

(2*Sqrt[a]*Sqrt[c]*ArcTan[(Sqrt[a]*Sqrt[c]*Tan[e + f*x])/(Sqrt[a + a*Sec[e + f*x]]*Sqrt[c + d*Sec[e + f*x]])])
/f + (2*Sqrt[a]*Sqrt[d]*ArcTanh[(Sqrt[a]*Sqrt[d]*Tan[e + f*x])/(Sqrt[a + a*Sec[e + f*x]]*Sqrt[c + d*Sec[e + f*
x]])])/f

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 4017

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)], x_Symbol] :> Dist[c
, Int[Sqrt[a + b*Csc[e + f*x]]/Sqrt[c + d*Csc[e + f*x]], x], x] + Dist[d, Int[Csc[e + f*x]*(Sqrt[a + b*Csc[e +
f*x]]/Sqrt[c + d*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 4019

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)], x_Symbol] :> Dist[-
2*(a/f), Subst[Int[1/(1 + a*c*x^2), x], x, Cot[e + f*x]/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*x]])],
x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 4065

Int[(csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)])/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.) +
(c_)], x_Symbol] :> Dist[-2*(b/f), Subst[Int[1/(1 - b*d*x^2), x], x, Cot[e + f*x]/(Sqrt[a + b*Csc[e + f*x]]*Sq
rt[c + d*Csc[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[
c^2 - d^2, 0]

Rubi steps

\begin {align*} \int \sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)} \, dx &=c \int \frac {\sqrt {a+a \sec (e+f x)}}{\sqrt {c+d \sec (e+f x)}} \, dx+d \int \frac {\sec (e+f x) \sqrt {a+a \sec (e+f x)}}{\sqrt {c+d \sec (e+f x)}} \, dx\\ &=-\frac {(2 a c) \text {Subst}\left (\int \frac {1}{1+a c x^2} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)}}\right )}{f}-\frac {(2 a d) \text {Subst}\left (\int \frac {1}{1-a d x^2} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)}}\right )}{f}\\ &=\frac {2 \sqrt {a} \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {c} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)}}\right )}{f}+\frac {2 \sqrt {a} \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)}}\right )}{f}\\ \end {align*}

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Mathematica [A]
time = 20.51, size = 240, normalized size = 1.95 \begin {gather*} -\frac {2 \cot (e+f x) \sqrt {a (1+\sec (e+f x))} \sqrt {c+d \sec (e+f x)} \left (-2 \sqrt {c} \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+c \cos (e+f x)}}{\sqrt {d} \sqrt {c-c \cos (e+f x)}}\right ) \sqrt {c (1+\cos (e+f x))} \sin ^2\left (\frac {1}{2} (e+f x)\right )+\text {ArcTan}\left (\frac {\sqrt {c (1+\cos (e+f x))} \sqrt {d+c \cos (e+f x)}}{\sqrt {c^2 \sin ^2(e+f x)}}\right ) \sqrt {c-c \cos (e+f x)} \sqrt {c^2 \sin ^2(e+f x)}\right )}{f \sqrt {c (1+\cos (e+f x))} \sqrt {c-c \cos (e+f x)} \sqrt {d+c \cos (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + a*Sec[e + f*x]]*Sqrt[c + d*Sec[e + f*x]],x]

[Out]

(-2*Cot[e + f*x]*Sqrt[a*(1 + Sec[e + f*x])]*Sqrt[c + d*Sec[e + f*x]]*(-2*Sqrt[c]*Sqrt[d]*ArcTanh[(Sqrt[c]*Sqrt
[d + c*Cos[e + f*x]])/(Sqrt[d]*Sqrt[c - c*Cos[e + f*x]])]*Sqrt[c*(1 + Cos[e + f*x])]*Sin[(e + f*x)/2]^2 + ArcT
an[(Sqrt[c*(1 + Cos[e + f*x])]*Sqrt[d + c*Cos[e + f*x]])/Sqrt[c^2*Sin[e + f*x]^2]]*Sqrt[c - c*Cos[e + f*x]]*Sq
rt[c^2*Sin[e + f*x]^2]))/(f*Sqrt[c*(1 + Cos[e + f*x])]*Sqrt[c - c*Cos[e + f*x]]*Sqrt[d + c*Cos[e + f*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1550\) vs. \(2(99)=198\).
time = 2.07, size = 1551, normalized size = 12.61

method	result	size
default	\(\text {Expression too large to display}\)	\(1551\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*sec(f*x+e))^(1/2)*(a+a*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/f*((d+c*cos(f*x+e))/cos(f*x+e))^(1/2)*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)*cos(f*x+e)*(-1+cos(f*x+e))*(-ln((
(-2*(d+c*cos(f*x+e))/(cos(f*x+e)+1))^(1/2)*(c-d)^(1/2)*sin(f*x+e)-c*cos(f*x+e)+d*cos(f*x+e)+c-d)/sin(f*x+e)/(c
-d)^(1/2))*2^(1/2)*(-d)^(1/2)*c^3+3*ln(((-2*(d+c*cos(f*x+e))/(cos(f*x+e)+1))^(1/2)*(c-d)^(1/2)*sin(f*x+e)-c*co
s(f*x+e)+d*cos(f*x+e)+c-d)/sin(f*x+e)/(c-d)^(1/2))*2^(1/2)*(-d)^(1/2)*c^2*d-3*ln(((-2*(d+c*cos(f*x+e))/(cos(f*
x+e)+1))^(1/2)*(c-d)^(1/2)*sin(f*x+e)-c*cos(f*x+e)+d*cos(f*x+e)+c-d)/sin(f*x+e)/(c-d)^(1/2))*2^(1/2)*(-d)^(1/2
)*c*d^2+ln(((-2*(d+c*cos(f*x+e))/(cos(f*x+e)+1))^(1/2)*(c-d)^(1/2)*sin(f*x+e)-c*cos(f*x+e)+d*cos(f*x+e)+c-d)/s
in(f*x+e)/(c-d)^(1/2))*2^(1/2)*(-d)^(1/2)*d^3+ln(((-2*(d+c*cos(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)-(c-d)^
(1/2)*cos(f*x+e)+(c-d)^(1/2))/sin(f*x+e))*2^(1/2)*(-d)^(1/2)*c^3-3*ln(((-2*(d+c*cos(f*x+e))/(cos(f*x+e)+1))^(1
/2)*sin(f*x+e)-(c-d)^(1/2)*cos(f*x+e)+(c-d)^(1/2))/sin(f*x+e))*2^(1/2)*(-d)^(1/2)*c^2*d+3*ln(((-2*(d+c*cos(f*x
+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)-(c-d)^(1/2)*cos(f*x+e)+(c-d)^(1/2))/sin(f*x+e))*2^(1/2)*(-d)^(1/2)*c*d^2
-ln(((-2*(d+c*cos(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)-(c-d)^(1/2)*cos(f*x+e)+(c-d)^(1/2))/sin(f*x+e))*2^(
1/2)*(-d)^(1/2)*d^3+ln(-2*(2^(1/2)*(-d)^(1/2)*(-2*(d+c*cos(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+c*cos(f*x+
e)-d*cos(f*x+e)-c*sin(f*x+e)-d*sin(f*x+e)-c+d)/(cos(f*x+e)-1-sin(f*x+e)))*(c-d)^(1/2)*c^2*d-2*ln(-2*(2^(1/2)*(
-d)^(1/2)*(-2*(d+c*cos(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c*sin(f*x+e)-d*sin(f
*x+e)-c+d)/(cos(f*x+e)-1-sin(f*x+e)))*(c-d)^(1/2)*c*d^2+ln(-2*(2^(1/2)*(-d)^(1/2)*(-2*(d+c*cos(f*x+e))/(cos(f*
x+e)+1))^(1/2)*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c*sin(f*x+e)-d*sin(f*x+e)-c+d)/(cos(f*x+e)-1-sin(f*x+e)))*
(c-d)^(1/2)*d^3-ln(-2*(2^(1/2)*(-d)^(1/2)*(-2*(d+c*cos(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)-c*cos(f*x+e)+d
*cos(f*x+e)-c*sin(f*x+e)-d*sin(f*x+e)+c-d)/(cos(f*x+e)-1+sin(f*x+e)))*(c-d)^(1/2)*c^2*d+2*ln(-2*(2^(1/2)*(-d)^
(1/2)*(-2*(d+c*cos(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)-c*cos(f*x+e)+d*cos(f*x+e)-c*sin(f*x+e)-d*sin(f*x+e
)+c-d)/(cos(f*x+e)-1+sin(f*x+e)))*(c-d)^(1/2)*c*d^2-ln(-2*(2^(1/2)*(-d)^(1/2)*(-2*(d+c*cos(f*x+e))/(cos(f*x+e)
+1))^(1/2)*sin(f*x+e)-c*cos(f*x+e)+d*cos(f*x+e)-c*sin(f*x+e)-d*sin(f*x+e)+c-d)/(cos(f*x+e)-1+sin(f*x+e)))*(c-d
)^(1/2)*d^3+2*(-(c-d)^4*c)^(1/2)*arctan((-1+cos(f*x+e))/(-2*(d+c*cos(f*x+e))/(cos(f*x+e)+1))^(1/2)/sin(f*x+e)*
(c-d)^2*c*2^(1/2)/(-(c-d)^4*c)^(1/2))*(c-d)^(1/2)*(-d)^(1/2))/(-2*(d+c*cos(f*x+e))/(cos(f*x+e)+1))^(1/2)/sin(f
*x+e)^2/(c-d)^(1/2)*2^(1/2)/(-d)^(1/2)/(c^2-2*c*d+d^2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sec(f*x+e))^(1/2)*(a+a*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*sec(f*x + e) + a)*sqrt(d*sec(f*x + e) + c), x)

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Fricas [A]
time = 12.18, size = 868, normalized size = 7.06 \begin {gather*} \left [\frac {\sqrt {a d} \log \left (\frac {2 \, \sqrt {a d} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (a c - a d\right )} \cos \left (f x + e\right )^{2} + 2 \, a d + {\left (a c + a d\right )} \cos \left (f x + e\right )}{\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )}\right ) + \sqrt {-a c} \log \left (\frac {2 \, a c \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-a c} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - a c + a d + {\left (a c + a d\right )} \cos \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{f}, -\frac {2 \, \sqrt {a c} \arctan \left (\frac {\sqrt {a c} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{a c \sin \left (f x + e\right )}\right ) - \sqrt {a d} \log \left (\frac {2 \, \sqrt {a d} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (a c - a d\right )} \cos \left (f x + e\right )^{2} + 2 \, a d + {\left (a c + a d\right )} \cos \left (f x + e\right )}{\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )}\right )}{f}, -\frac {2 \, \sqrt {-a d} \arctan \left (\frac {\sqrt {-a d} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{a d \sin \left (f x + e\right )}\right ) - \sqrt {-a c} \log \left (\frac {2 \, a c \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-a c} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - a c + a d + {\left (a c + a d\right )} \cos \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{f}, -\frac {2 \, {\left (\sqrt {a c} \arctan \left (\frac {\sqrt {a c} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{a c \sin \left (f x + e\right )}\right ) + \sqrt {-a d} \arctan \left (\frac {\sqrt {-a d} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{a d \sin \left (f x + e\right )}\right )\right )}}{f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sec(f*x+e))^(1/2)*(a+a*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[(sqrt(a*d)*log((2*sqrt(a*d)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) + d)/cos(f*x + e))*c
os(f*x + e)*sin(f*x + e) + (a*c - a*d)*cos(f*x + e)^2 + 2*a*d + (a*c + a*d)*cos(f*x + e))/(cos(f*x + e)^2 + co
s(f*x + e))) + sqrt(-a*c)*log((2*a*c*cos(f*x + e)^2 - 2*sqrt(-a*c)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqr
t((c*cos(f*x + e) + d)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) - a*c + a*d + (a*c + a*d)*cos(f*x + e))/(cos(f*
x + e) + 1)))/f, -(2*sqrt(a*c)*arctan(sqrt(a*c)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) +
d)/cos(f*x + e))*cos(f*x + e)/(a*c*sin(f*x + e))) - sqrt(a*d)*log((2*sqrt(a*d)*sqrt((a*cos(f*x + e) + a)/cos(
f*x + e))*sqrt((c*cos(f*x + e) + d)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + (a*c - a*d)*cos(f*x + e)^2 + 2*a
*d + (a*c + a*d)*cos(f*x + e))/(cos(f*x + e)^2 + cos(f*x + e))))/f, -(2*sqrt(-a*d)*arctan(sqrt(-a*d)*sqrt((a*c
os(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) + d)/cos(f*x + e))*cos(f*x + e)/(a*d*sin(f*x + e))) - sqrt
(-a*c)*log((2*a*c*cos(f*x + e)^2 - 2*sqrt(-a*c)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) +
d)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) - a*c + a*d + (a*c + a*d)*cos(f*x + e))/(cos(f*x + e) + 1)))/f, -2
*(sqrt(a*c)*arctan(sqrt(a*c)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) + d)/cos(f*x + e))*c
os(f*x + e)/(a*c*sin(f*x + e))) + sqrt(-a*d)*arctan(sqrt(-a*d)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c
*cos(f*x + e) + d)/cos(f*x + e))*cos(f*x + e)/(a*d*sin(f*x + e))))/f]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )} \sqrt {c + d \sec {\left (e + f x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sec(f*x+e))**(1/2)*(a+a*sec(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(a*(sec(e + f*x) + 1))*sqrt(c + d*sec(e + f*x)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sec(f*x+e))^(1/2)*(a+a*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(a*sec(f*x + e) + a)*sqrt(d*sec(f*x + e) + c), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}\,\sqrt {c+\frac {d}{\cos \left (e+f\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^(1/2)*(c + d/cos(e + f*x))^(1/2),x)

[Out]

int((a + a/cos(e + f*x))^(1/2)*(c + d/cos(e + f*x))^(1/2), x)

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