3.185 \(\int \cos (a+b x) \log (\cos (\frac {a}{2}+\frac {b x}{2}) \sin (\frac {a}{2}+\frac {b x}{2})) \, dx\)

Optimal. Leaf size=50 \[ \frac {\sin (a+b x) \log \left (\sin \left (\frac {a}{2}+\frac {b x}{2}\right ) \cos \left (\frac {a}{2}+\frac {b x}{2}\right )\right )}{b}-\frac {\sin (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {2637, 2554} \[ \frac {\sin (a+b x) \log \left (\sin \left (\frac {a}{2}+\frac {b x}{2}\right ) \cos \left (\frac {a}{2}+\frac {b x}{2}\right )\right )}{b}-\frac {\sin (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]*Log[Cos[a/2 + (b*x)/2]*Sin[a/2 + (b*x)/2]],x]

[Out]

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

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

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Mathematica [A]  time = 0.01, size = 33, normalized size = 0.66 \[ \frac {\sin (a+b x) \log \left (\frac {1}{2} \sin (a+b x)\right )}{b}-\frac {\sin (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]*Log[Cos[a/2 + (b*x)/2]*Sin[a/2 + (b*x)/2]],x]

[Out]

-(Sin[a + b*x]/b) + (Log[Sin[a + b*x]/2]*Sin[a + b*x])/b

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fricas [A]  time = 0.49, size = 65, normalized size = 1.30 \[ \frac {2 \, {\left (\cos \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) \log \left (\cos \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) \sin \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )\right ) \sin \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) - \cos \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) \sin \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )\right )}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*log(cos(1/2*a+1/2*b*x)*sin(1/2*a+1/2*b*x)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.93, size = 42, normalized size = 0.84 \[ \frac {\log \left (\cos \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) \sin \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )\right ) \sin \left (b x + a\right )}{b} - \frac {\sin \left (b x + a\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*log(cos(1/2*a+1/2*b*x)*sin(1/2*a+1/2*b*x)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.74, size = 32, normalized size = 0.64 \[ \frac {\ln \left (\frac {\sin \left (b x +a \right )}{2}\right ) \sin \left (b x +a \right )}{b}-\frac {\sin \left (b x +a \right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)*ln(cos(1/2*b*x+1/2*a)*sin(1/2*b*x+1/2*a)),x)

[Out]

ln(1/2*sin(b*x+a))/b*sin(b*x+a)-sin(b*x+a)/b

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maxima [A]  time = 0.54, size = 42, normalized size = 0.84 \[ \frac {\log \left (\cos \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) \sin \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )\right ) \sin \left (b x + a\right )}{b} - \frac {\sin \left (b x + a\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*log(cos(1/2*a+1/2*b*x)*sin(1/2*a+1/2*b*x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.52, size = 29, normalized size = 0.58 \[ -\frac {\sin \left (a+b\,x\right )-\ln \left (\frac {\sin \left (a+b\,x\right )}{2}\right )\,\sin \left (a+b\,x\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(cos(a/2 + (b*x)/2)*sin(a/2 + (b*x)/2))*cos(a + b*x),x)

[Out]

-(sin(a + b*x) - log(sin(a + b*x)/2)*sin(a + b*x))/b

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \log {\left (\sin {\left (\frac {a}{2} + \frac {b x}{2} \right )} \cos {\left (\frac {a}{2} + \frac {b x}{2} \right )} \right )} \cos {\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*ln(cos(1/2*a+1/2*b*x)*sin(1/2*a+1/2*b*x)),x)

[Out]

Integral(log(sin(a/2 + b*x/2)*cos(a/2 + b*x/2))*cos(a + b*x), x)

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