3.394 \(\int \frac {1}{(1-\cos (3 x))^{3/2}} \, dx\)

Optimal. Leaf size=53 \[ -\frac {\sin (3 x)}{6 (1-\cos (3 x))^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sin (3 x)}{\sqrt {2} \sqrt {1-\cos (3 x)}}\right )}{6 \sqrt {2}} \]

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Rubi [A]  time = 0.03, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2650, 2649, 206} \[ -\frac {\sin (3 x)}{6 (1-\cos (3 x))^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sin (3 x)}{\sqrt {2} \sqrt {1-\cos (3 x)}}\right )}{6 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[(1 - Cos[3*x])^(-3/2),x]

[Out]

-ArcTanh[Sin[3*x]/(Sqrt[2]*Sqrt[1 - Cos[3*x]])]/(6*Sqrt[2]) - Sin[3*x]/(6*(1 - Cos[3*x])^(3/2))

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]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rubi steps

\begin {align*} \int \frac {1}{(1-\cos (3 x))^{3/2}} \, dx &=-\frac {\sin (3 x)}{6 (1-\cos (3 x))^{3/2}}+\frac {1}{4} \int \frac {1}{\sqrt {1-\cos (3 x)}} \, dx\\ &=-\frac {\sin (3 x)}{6 (1-\cos (3 x))^{3/2}}-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{2-x^2} \, dx,x,\frac {\sin (3 x)}{\sqrt {1-\cos (3 x)}}\right )\\ &=-\frac {\tanh ^{-1}\left (\frac {\sin (3 x)}{\sqrt {2} \sqrt {1-\cos (3 x)}}\right )}{6 \sqrt {2}}-\frac {\sin (3 x)}{6 (1-\cos (3 x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 61, normalized size = 1.15 \[ -\frac {\sin ^3\left (\frac {3 x}{2}\right ) \left (\csc ^2\left (\frac {3 x}{4}\right )-\sec ^2\left (\frac {3 x}{4}\right )-4 \log \left (\sin \left (\frac {3 x}{4}\right )\right )+4 \log \left (\cos \left (\frac {3 x}{4}\right )\right )\right )}{12 (1-\cos (3 x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - Cos[3*x])^(-3/2),x]

[Out]

-1/12*((Csc[(3*x)/4]^2 + 4*Log[Cos[(3*x)/4]] - 4*Log[Sin[(3*x)/4]] - Sec[(3*x)/4]^2)*Sin[(3*x)/2]^3)/(1 - Cos[
3*x])^(3/2)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{(1-\cos (3 x))^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[(1 - Cos[3*x])^(-3/2),x]

[Out]

Could not integrate

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fricas [B]  time = 0.92, size = 107, normalized size = 2.02 \[ \frac {{\left (\sqrt {2} \cos \left (3 \, x\right ) - \sqrt {2}\right )} \log \left (-\frac {{\left (\cos \left (3 \, x\right ) + 3\right )} \sin \left (3 \, x\right ) - 2 \, {\left (\sqrt {2} \cos \left (3 \, x\right ) + \sqrt {2}\right )} \sqrt {-\cos \left (3 \, x\right ) + 1}}{{\left (\cos \left (3 \, x\right ) - 1\right )} \sin \left (3 \, x\right )}\right ) \sin \left (3 \, x\right ) + 4 \, {\left (\cos \left (3 \, x\right ) + 1\right )} \sqrt {-\cos \left (3 \, x\right ) + 1}}{24 \, {\left (\cos \left (3 \, x\right ) - 1\right )} \sin \left (3 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(3*x))^(3/2),x, algorithm="fricas")

[Out]

1/24*((sqrt(2)*cos(3*x) - sqrt(2))*log(-((cos(3*x) + 3)*sin(3*x) - 2*(sqrt(2)*cos(3*x) + sqrt(2))*sqrt(-cos(3*
x) + 1))/((cos(3*x) - 1)*sin(3*x)))*sin(3*x) + 4*(cos(3*x) + 1)*sqrt(-cos(3*x) + 1))/((cos(3*x) - 1)*sin(3*x))

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giac [A]  time = 0.81, size = 59, normalized size = 1.11 \[ -\frac {\sqrt {2} {\left (\frac {2 \, \sqrt {\tan \left (\frac {3}{2} \, x\right )^{2} + 1}}{\tan \left (\frac {3}{2} \, x\right )^{2}} + \log \left (\sqrt {\tan \left (\frac {3}{2} \, x\right )^{2} + 1} + 1\right ) - \log \left (\sqrt {\tan \left (\frac {3}{2} \, x\right )^{2} + 1} - 1\right )\right )}}{24 \, \mathrm {sgn}\left (\tan \left (\frac {3}{2} \, x\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(3*x))^(3/2),x, algorithm="giac")

[Out]

-1/24*sqrt(2)*(2*sqrt(tan(3/2*x)^2 + 1)/tan(3/2*x)^2 + log(sqrt(tan(3/2*x)^2 + 1) + 1) - log(sqrt(tan(3/2*x)^2
+ 1) - 1))/sgn(tan(3/2*x))

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maple [A]  time = 0.14, size = 52, normalized size = 0.98

method	result	size
default	\(-\frac {\left (\frac {\cos \left (\frac {3 x}{2}\right )}{2}+\frac {\left (\ln \left (1+\cos \left (\frac {3 x}{2}\right )\right )-\ln \left (\cos \left (\frac {3 x}{2}\right )-1\right )\right ) \left (\sin ^{2}\left (\frac {3 x}{2}\right )\right )}{4}\right ) \sqrt {2}}{3 \sin \left (\frac {3 x}{2}\right ) \sqrt {2-2 \cos \left (3 x \right )}}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1-cos(3*x))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/6*(1/2*cos(3/2*x)+1/4*(ln(1+cos(3/2*x))-ln(cos(3/2*x)-1))*sin(3/2*x)^2)/sin(3/2*x)*2^(1/2)/(sin(3/2*x)^2)^(
1/2)

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maxima [B]  time = 1.22, size = 433, normalized size = 8.17 \[ \frac {4 \, {\left (\sin \left (6 \, x\right ) - 2 \, \sin \left (3 \, x\right )\right )} \cos \left (\frac {3}{2} \, \pi + \frac {3}{2} \, \arctan \left (\sin \left (3 \, x\right ), \cos \left (3 \, x\right )\right )\right ) - 4 \, {\left (\sin \left (6 \, x\right ) - 2 \, \sin \left (3 \, x\right )\right )} \cos \left (\frac {1}{2} \, \pi + \frac {1}{2} \, \arctan \left (\sin \left (3 \, x\right ), \cos \left (3 \, x\right )\right )\right ) + {\left (2 \, {\left (2 \, \cos \left (3 \, x\right ) - 1\right )} \cos \left (6 \, x\right ) - \cos \left (6 \, x\right )^{2} - 4 \, \cos \left (3 \, x\right )^{2} - \sin \left (6 \, x\right )^{2} + 4 \, \sin \left (6 \, x\right ) \sin \left (3 \, x\right ) - 4 \, \sin \left (3 \, x\right )^{2} + 4 \, \cos \left (3 \, x\right ) - 1\right )} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (3 \, x\right ), \cos \left (3 \, x\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (3 \, x\right ), \cos \left (3 \, x\right )\right )\right )^{2} + 2 \, \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (3 \, x\right ), \cos \left (3 \, x\right )\right )\right ) + 1\right ) - {\left (2 \, {\left (2 \, \cos \left (3 \, x\right ) - 1\right )} \cos \left (6 \, x\right ) - \cos \left (6 \, x\right )^{2} - 4 \, \cos \left (3 \, x\right )^{2} - \sin \left (6 \, x\right )^{2} + 4 \, \sin \left (6 \, x\right ) \sin \left (3 \, x\right ) - 4 \, \sin \left (3 \, x\right )^{2} + 4 \, \cos \left (3 \, x\right ) - 1\right )} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (3 \, x\right ), \cos \left (3 \, x\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (3 \, x\right ), \cos \left (3 \, x\right )\right )\right )^{2} - 2 \, \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (3 \, x\right ), \cos \left (3 \, x\right )\right )\right ) + 1\right ) - 4 \, {\left (\cos \left (6 \, x\right ) - 2 \, \cos \left (3 \, x\right ) + 1\right )} \sin \left (\frac {3}{2} \, \pi + \frac {3}{2} \, \arctan \left (\sin \left (3 \, x\right ), \cos \left (3 \, x\right )\right )\right ) + 4 \, {\left (\cos \left (6 \, x\right ) - 2 \, \cos \left (3 \, x\right ) + 1\right )} \sin \left (\frac {1}{2} \, \pi + \frac {1}{2} \, \arctan \left (\sin \left (3 \, x\right ), \cos \left (3 \, x\right )\right )\right )}{12 \, {\left (\sqrt {2} \cos \left (6 \, x\right )^{2} + 4 \, \sqrt {2} \cos \left (3 \, x\right )^{2} + \sqrt {2} \sin \left (6 \, x\right )^{2} - 4 \, \sqrt {2} \sin \left (6 \, x\right ) \sin \left (3 \, x\right ) + 4 \, \sqrt {2} \sin \left (3 \, x\right )^{2} - 2 \, {\left (2 \, \sqrt {2} \cos \left (3 \, x\right ) - \sqrt {2}\right )} \cos \left (6 \, x\right ) - 4 \, \sqrt {2} \cos \left (3 \, x\right ) + \sqrt {2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(3*x))^(3/2),x, algorithm="maxima")

[Out]

1/12*(4*(sin(6*x) - 2*sin(3*x))*cos(3/2*pi + 3/2*arctan2(sin(3*x), cos(3*x))) - 4*(sin(6*x) - 2*sin(3*x))*cos(
1/2*pi + 1/2*arctan2(sin(3*x), cos(3*x))) + (2*(2*cos(3*x) - 1)*cos(6*x) - cos(6*x)^2 - 4*cos(3*x)^2 - sin(6*x
)^2 + 4*sin(6*x)*sin(3*x) - 4*sin(3*x)^2 + 4*cos(3*x) - 1)*log(cos(1/2*arctan2(sin(3*x), cos(3*x)))^2 + sin(1/
2*arctan2(sin(3*x), cos(3*x)))^2 + 2*cos(1/2*arctan2(sin(3*x), cos(3*x))) + 1) - (2*(2*cos(3*x) - 1)*cos(6*x)
- cos(6*x)^2 - 4*cos(3*x)^2 - sin(6*x)^2 + 4*sin(6*x)*sin(3*x) - 4*sin(3*x)^2 + 4*cos(3*x) - 1)*log(cos(1/2*ar
ctan2(sin(3*x), cos(3*x)))^2 + sin(1/2*arctan2(sin(3*x), cos(3*x)))^2 - 2*cos(1/2*arctan2(sin(3*x), cos(3*x)))
+ 1) - 4*(cos(6*x) - 2*cos(3*x) + 1)*sin(3/2*pi + 3/2*arctan2(sin(3*x), cos(3*x))) + 4*(cos(6*x) - 2*cos(3*x)
+ 1)*sin(1/2*pi + 1/2*arctan2(sin(3*x), cos(3*x))))/(sqrt(2)*cos(6*x)^2 + 4*sqrt(2)*cos(3*x)^2 + sqrt(2)*sin(
6*x)^2 - 4*sqrt(2)*sin(6*x)*sin(3*x) + 4*sqrt(2)*sin(3*x)^2 - 2*(2*sqrt(2)*cos(3*x) - sqrt(2))*cos(6*x) - 4*sq
rt(2)*cos(3*x) + sqrt(2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\left (1-\cos \left (3\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1 - cos(3*x))^(3/2),x)

[Out]

int(1/(1 - cos(3*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (1 - \cos {\left (3 x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(3*x))**(3/2),x)

[Out]

Integral((1 - cos(3*x))**(-3/2), x)

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