\(\int \cos (a+b x) \log (\cos (\frac {a}{2}+\frac {b x}{2}) \sin (\frac {a}{2}+\frac {b x}{2})) \, dx\) [185]

Optimal result

Integrand size = 35, antiderivative size = 50 \[ \int \cos (a+b x) \log \left (\cos \left (\frac {a}{2}+\frac {b x}{2}\right ) \sin \left (\frac {a}{2}+\frac {b x}{2}\right )\right ) \, dx=-\frac {\sin (a+b x)}{b}+\frac {\log \left (\cos \left (\frac {a}{2}+\frac {b x}{2}\right ) \sin \left (\frac {a}{2}+\frac {b x}{2}\right )\right ) \sin (a+b x)}{b} \]

[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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {2717, 2634} \[ \int \cos (a+b x) \log \left (\cos \left (\frac {a}{2}+\frac {b x}{2}\right ) \sin \left (\frac {a}{2}+\frac {b x}{2}\right )\right ) \, dx=\frac {\sin (a+b x) \log \left (\sin \left (\frac {a}{2}+\frac {b x}{2}\right ) \cos \left (\frac {a}{2}+\frac {b x}{2}\right )\right )}{b}-\frac {\sin (a+b x)}{b} \]

[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 2634

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 2717

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

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.66 \[ \int \cos (a+b x) \log \left (\cos \left (\frac {a}{2}+\frac {b x}{2}\right ) \sin \left (\frac {a}{2}+\frac {b x}{2}\right )\right ) \, dx=-\frac {\sin (a+b x)}{b}+\frac {\log \left (\frac {1}{2} \sin (a+b x)\right ) \sin (a+b x)}{b} \]

[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

Maple [A] (verified)

Time = 7.66 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.60

method	result	size
default	\(\frac {\ln \left (\frac {\sin \left (b x +a \right )}{2}\right ) \sin \left (b x +a \right )-\sin \left (b x +a \right )}{b}\)	\(30\)
risch	\(\text {Expression too large to display}\)	\(1389\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.30 \[ \int \cos (a+b x) \log \left (\cos \left (\frac {a}{2}+\frac {b x}{2}\right ) \sin \left (\frac {a}{2}+\frac {b x}{2}\right )\right ) \, dx=\frac {2 \, {\left (\cos \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) \log \left (\cos \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) \sin \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )\right ) \sin \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) - \cos \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) \sin \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )\right )}}{b} \]

[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

Sympy [F]

\[ \int \cos (a+b x) \log \left (\cos \left (\frac {a}{2}+\frac {b x}{2}\right ) \sin \left (\frac {a}{2}+\frac {b x}{2}\right )\right ) \, dx=\int \log {\left (\sin {\left (\frac {a}{2} + \frac {b x}{2} \right )} \cos {\left (\frac {a}{2} + \frac {b x}{2} \right )} \right )} \cos {\left (a + b x \right )}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.84 \[ \int \cos (a+b x) \log \left (\cos \left (\frac {a}{2}+\frac {b x}{2}\right ) \sin \left (\frac {a}{2}+\frac {b x}{2}\right )\right ) \, dx=\frac {\log \left (\cos \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) \sin \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )\right ) \sin \left (b x + a\right )}{b} - \frac {\sin \left (b x + a\right )}{b} \]

[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

Giac [F(-2)]

Exception generated. \[ \int \cos (a+b x) \log \left (\cos \left (\frac {a}{2}+\frac {b x}{2}\right ) \sin \left (\frac {a}{2}+\frac {b x}{2}\right )\right ) \, dx=\text {Exception raised: TypeError} \]

[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]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [B] (verification not implemented)

Time = 1.70 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.58 \[ \int \cos (a+b x) \log \left (\cos \left (\frac {a}{2}+\frac {b x}{2}\right ) \sin \left (\frac {a}{2}+\frac {b x}{2}\right )\right ) \, dx=-\frac {\sin \left (a+b\,x\right )-\ln \left (\frac {\sin \left (a+b\,x\right )}{2}\right )\,\sin \left (a+b\,x\right )}{b} \]

[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