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

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Rubi [A]  time = 0.026015, antiderivative size = 50, normalized size of antiderivative = 1., 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 2637

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

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*}

Mathematica [A]  time = 0.0101918, 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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Maple [A]  time = 0.031, size = 32, normalized size = 0.6 \begin{align*}{\frac{\sin \left ( bx+a \right ) }{b}\ln \left ({\frac{\sin \left ( bx+a \right ) }{2}} \right ) }-{\frac{\sin \left ( bx+a \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)*ln(cos(1/2*a+1/2*b*x)*sin(1/2*a+1/2*b*x)),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 = 1.07812, size = 57, normalized size = 1.14 \begin{align*} \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} \end{align*}

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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Fricas [A]  time = 2.60892, size = 189, normalized size = 3.78 \begin{align*} \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} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [A]  time = 1.47756, size = 57, normalized size = 1.14 \begin{align*} \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} \end{align*}

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