3.179 \(\int \cos (x) \log (\frac {1}{2} (1-\cos (2 x))) \, dx\)

Optimal. Leaf size=21 \[ \sin (x) \log \left (\frac {1}{2} (1-\cos (2 x))\right )-2 \sin (x) \]

[Out]

-2*sin(x)+ln(1/2-1/2*cos(2*x))*sin(x)

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Rubi [A]  time = 0.02, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2637, 2554, 12} \[ \sin (x) \log \left (\frac {1}{2} (1-\cos (2 x))\right )-2 \sin (x) \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]*Log[(1 - Cos[2*x])/2],x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \cos (x) \log \left (\frac {1}{2} (1-\cos (2 x))\right ) \, dx &=\log \left (\frac {1}{2} (1-\cos (2 x))\right ) \sin (x)-\int 2 \cos (x) \, dx\\ &=\log \left (\frac {1}{2} (1-\cos (2 x))\right ) \sin (x)-2 \int \cos (x) \, dx\\ &=-2 \sin (x)+\log \left (\frac {1}{2} (1-\cos (2 x))\right ) \sin (x)\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 13, normalized size = 0.62 \[ \sin (x) \log \left (\sin ^2(x)\right )-2 \sin (x) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]*Log[(1 - Cos[2*x])/2],x]

[Out]

-2*Sin[x] + Log[Sin[x]^2]*Sin[x]

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fricas [A]  time = 0.49, size = 17, normalized size = 0.81 \[ \log \left (-\cos \relax (x)^{2} + 1\right ) \sin \relax (x) - 2 \, \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(-cos(x)^2 + 1)*sin(x) - 2*sin(x)

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giac [A]  time = 0.18, size = 13, normalized size = 0.62 \[ \log \left (\sin \relax (x)^{2}\right ) \sin \relax (x) - 2 \, \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(sin(x)^2)*sin(x) - 2*sin(x)

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maple [C]  time = 0.60, size = 72, normalized size = 3.43 \[ \frac {i {\mathrm e}^{-i x} \ln \left (-2 \cos \left (2 x \right )+2\right )}{2}-\frac {i {\mathrm e}^{i x} \ln \left (-2 \cos \left (2 x \right )+2\right )}{2}-i \ln \relax (2) {\mathrm e}^{-i x}-i {\mathrm e}^{-i x}+i \ln \relax (2) {\mathrm e}^{i x}+i {\mathrm e}^{i x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)*ln(1/2-1/2*cos(2*x)),x)

[Out]

I*ln(2)*exp(I*x)-I*ln(2)*exp(-I*x)-1/2*I*exp(I*x)*ln(2-2*cos(2*x))+1/2*I*exp(-I*x)*ln(2-2*cos(2*x))+I*exp(I*x)
-I*exp(-I*x)

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maxima [A]  time = 0.45, size = 17, normalized size = 0.81 \[ \log \left (-\frac {1}{2} \, \cos \left (2 \, x\right ) + \frac {1}{2}\right ) \sin \relax (x) - 2 \, \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \[ \int \ln \left (\frac {1}{2}-\frac {\cos \left (2\,x\right )}{2}\right )\,\cos \relax (x) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(1/2 - cos(2*x)/2)*cos(x),x)

[Out]

int(log(1/2 - cos(2*x)/2)*cos(x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(1/2 - cos(2*x)/2)*cos(x), x)

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