3.393 \(\int \frac {1}{\sqrt {1-\cos (2 x)}} \, dx\)

Optimal. Leaf size=30 \[ -\frac {\tanh ^{-1}\left (\frac {\sin (2 x)}{\sqrt {2} \sqrt {1-\cos (2 x)}}\right )}{\sqrt {2}} \]

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Rubi [A]  time = 0.01, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2649, 206} \[ -\frac {\tanh ^{-1}\left (\frac {\sin (2 x)}{\sqrt {2} \sqrt {1-\cos (2 x)}}\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[1 - Cos[2*x]],x]

[Out]

-(ArcTanh[Sin[2*x]/(Sqrt[2]*Sqrt[1 - Cos[2*x]])]/Sqrt[2])

Rule 206

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

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {1-\cos (2 x)}} \, dx &=-\operatorname {Subst}\left (\int \frac {1}{2-x^2} \, dx,x,\frac {\sin (2 x)}{\sqrt {1-\cos (2 x)}}\right )\\ &=-\frac {\tanh ^{-1}\left (\frac {\sin (2 x)}{\sqrt {2} \sqrt {1-\cos (2 x)}}\right )}{\sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 33, normalized size = 1.10 \[ -\frac {\sin (x) \left (\log \left (\cos \left (\frac {x}{2}\right )\right )-\log \left (\sin \left (\frac {x}{2}\right )\right )\right )}{\sqrt {1-\cos (2 x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[1 - Cos[2*x]],x]

[Out]

-(((Log[Cos[x/2]] - Log[Sin[x/2]])*Sin[x])/Sqrt[1 - Cos[2*x]])

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {1-\cos (2 x)}} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[1/Sqrt[1 - Cos[2*x]],x]

[Out]

Could not integrate

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fricas [B]  time = 0.84, size = 58, normalized size = 1.93 \[ \frac {1}{4} \, \sqrt {2} \log \left (-\frac {{\left (\cos \left (2 \, x\right ) + 3\right )} \sin \left (2 \, x\right ) - 2 \, {\left (\sqrt {2} \cos \left (2 \, x\right ) + \sqrt {2}\right )} \sqrt {-\cos \left (2 \, x\right ) + 1}}{{\left (\cos \left (2 \, x\right ) - 1\right )} \sin \left (2 \, x\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(2*x))^(1/2),x, algorithm="fricas")

[Out]

1/4*sqrt(2)*log(-((cos(2*x) + 3)*sin(2*x) - 2*(sqrt(2)*cos(2*x) + sqrt(2))*sqrt(-cos(2*x) + 1))/((cos(2*x) - 1
)*sin(2*x)))

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giac [A]  time = 0.64, size = 16, normalized size = 0.53 \[ \frac {\sqrt {2} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{2 \, \mathrm {sgn}\left (\sin \relax (x)\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(2*x))^(1/2),x, algorithm="giac")

[Out]

1/2*sqrt(2)*log(abs(tan(1/2*x)))/sgn(sin(x))

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maple [A]  time = 0.14, size = 17, normalized size = 0.57

method	result	size
default	\(-\frac {\sin \relax (x ) \arctanh \left (\cos \relax (x )\right ) \sqrt {2}}{\sqrt {2-2 \cos \left (2 x \right )}}\)	\(17\)
risch	\(\frac {\sqrt {2}\, \ln \left ({\mathrm e}^{i x}-1\right ) \sin \relax (x )}{\sqrt {-\left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}-\frac {\sqrt {2}\, \ln \left ({\mathrm e}^{i x}+1\right ) \sin \relax (x )}{\sqrt {-\left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1-cos(2*x))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*sin(x)*arctanh(cos(x))*2^(1/2)/(sin(x)^2)^(1/2)

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maxima [B]  time = 1.17, size = 101, normalized size = 3.37 \[ -\frac {1}{4} \, \sqrt {2} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, x\right ), \cos \left (2 \, x\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, x\right ), \cos \left (2 \, x\right )\right )\right )^{2} + 2 \, \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, x\right ), \cos \left (2 \, x\right )\right )\right ) + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, x\right ), \cos \left (2 \, x\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, x\right ), \cos \left (2 \, x\right )\right )\right )^{2} - 2 \, \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, x\right ), \cos \left (2 \, x\right )\right )\right ) + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(2*x))^(1/2),x, algorithm="maxima")

[Out]

-1/4*sqrt(2)*log(cos(1/2*arctan2(sin(2*x), cos(2*x)))^2 + sin(1/2*arctan2(sin(2*x), cos(2*x)))^2 + 2*cos(1/2*a
rctan2(sin(2*x), cos(2*x))) + 1) + 1/4*sqrt(2)*log(cos(1/2*arctan2(sin(2*x), cos(2*x)))^2 + sin(1/2*arctan2(si
n(2*x), cos(2*x)))^2 - 2*cos(1/2*arctan2(sin(2*x), cos(2*x))) + 1)

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mupad [B]  time = 0.24, size = 28, normalized size = 0.93 \[ -\frac {\sqrt {2}\,\sin \left (2\,x\right )\,\mathrm {atanh}\left (\sqrt {{\cos \relax (x)}^2}\right )}{2\,\sqrt {1-{\cos \left (2\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1 - cos(2*x))^(1/2),x)

[Out]

-(2^(1/2)*sin(2*x)*atanh((cos(x)^2)^(1/2)))/(2*(1 - cos(2*x)^2)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {1 - \cos {\left (2 x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(2*x))**(1/2),x)

[Out]

Integral(1/sqrt(1 - cos(2*x)), x)

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