3.156 \(\int \log (x) \sin ^3(a+b x) \, dx\)

Optimal. Leaf size=89 \[ \frac{3 \cos (a) \text{CosIntegral}(b x)}{4 b}-\frac{\cos (3 a) \text{CosIntegral}(3 b x)}{12 b}-\frac{3 \sin (a) \text{Si}(b x)}{4 b}+\frac{\sin (3 a) \text{Si}(3 b x)}{12 b}+\frac{\log (x) \cos ^3(a+b x)}{3 b}-\frac{\log (x) \cos (a+b x)}{b} \]

[Out]

(3*Cos[a]*CosIntegral[b*x])/(4*b) - (Cos[3*a]*CosIntegral[3*b*x])/(12*b) - (Cos[a + b*x]*Log[x])/b + (Cos[a +
b*x]^3*Log[x])/(3*b) - (3*Sin[a]*SinIntegral[b*x])/(4*b) + (Sin[3*a]*SinIntegral[3*b*x])/(12*b)

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Rubi [A]  time = 0.517726, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 8, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.727, Rules used = {2633, 2554, 12, 6742, 3303, 3299, 3302, 3312} \[ \frac{3 \cos (a) \text{CosIntegral}(b x)}{4 b}-\frac{\cos (3 a) \text{CosIntegral}(3 b x)}{12 b}-\frac{3 \sin (a) \text{Si}(b x)}{4 b}+\frac{\sin (3 a) \text{Si}(3 b x)}{12 b}+\frac{\log (x) \cos ^3(a+b x)}{3 b}-\frac{\log (x) \cos (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Log[x]*Sin[a + b*x]^3,x]

[Out]

(3*Cos[a]*CosIntegral[b*x])/(4*b) - (Cos[3*a]*CosIntegral[3*b*x])/(12*b) - (Cos[a + b*x]*Log[x])/b + (Cos[a +
b*x]^3*Log[x])/(3*b) - (3*Sin[a]*SinIntegral[b*x])/(4*b) + (Sin[3*a]*SinIntegral[3*b*x])/(12*b)

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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 12

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

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rubi steps

\begin{align*} \int \log (x) \sin ^3(a+b x) \, dx &=-\frac{\cos (a+b x) \log (x)}{b}+\frac{\cos ^3(a+b x) \log (x)}{3 b}-\int \frac{\cos (a+b x) \left (-3+\cos ^2(a+b x)\right )}{3 b x} \, dx\\ &=-\frac{\cos (a+b x) \log (x)}{b}+\frac{\cos ^3(a+b x) \log (x)}{3 b}-\frac{\int \frac{\cos (a+b x) \left (-3+\cos ^2(a+b x)\right )}{x} \, dx}{3 b}\\ &=-\frac{\cos (a+b x) \log (x)}{b}+\frac{\cos ^3(a+b x) \log (x)}{3 b}-\frac{\int \left (-\frac{3 \cos (a+b x)}{x}+\frac{\cos ^3(a+b x)}{x}\right ) \, dx}{3 b}\\ &=-\frac{\cos (a+b x) \log (x)}{b}+\frac{\cos ^3(a+b x) \log (x)}{3 b}-\frac{\int \frac{\cos ^3(a+b x)}{x} \, dx}{3 b}+\frac{\int \frac{\cos (a+b x)}{x} \, dx}{b}\\ &=-\frac{\cos (a+b x) \log (x)}{b}+\frac{\cos ^3(a+b x) \log (x)}{3 b}-\frac{\int \left (\frac{3 \cos (a+b x)}{4 x}+\frac{\cos (3 a+3 b x)}{4 x}\right ) \, dx}{3 b}+\frac{\cos (a) \int \frac{\cos (b x)}{x} \, dx}{b}-\frac{\sin (a) \int \frac{\sin (b x)}{x} \, dx}{b}\\ &=\frac{\cos (a) \text{Ci}(b x)}{b}-\frac{\cos (a+b x) \log (x)}{b}+\frac{\cos ^3(a+b x) \log (x)}{3 b}-\frac{\sin (a) \text{Si}(b x)}{b}-\frac{\int \frac{\cos (3 a+3 b x)}{x} \, dx}{12 b}-\frac{\int \frac{\cos (a+b x)}{x} \, dx}{4 b}\\ &=\frac{\cos (a) \text{Ci}(b x)}{b}-\frac{\cos (a+b x) \log (x)}{b}+\frac{\cos ^3(a+b x) \log (x)}{3 b}-\frac{\sin (a) \text{Si}(b x)}{b}-\frac{\cos (a) \int \frac{\cos (b x)}{x} \, dx}{4 b}-\frac{\cos (3 a) \int \frac{\cos (3 b x)}{x} \, dx}{12 b}+\frac{\sin (a) \int \frac{\sin (b x)}{x} \, dx}{4 b}+\frac{\sin (3 a) \int \frac{\sin (3 b x)}{x} \, dx}{12 b}\\ &=\frac{3 \cos (a) \text{Ci}(b x)}{4 b}-\frac{\cos (3 a) \text{Ci}(3 b x)}{12 b}-\frac{\cos (a+b x) \log (x)}{b}+\frac{\cos ^3(a+b x) \log (x)}{3 b}-\frac{3 \sin (a) \text{Si}(b x)}{4 b}+\frac{\sin (3 a) \text{Si}(3 b x)}{12 b}\\ \end{align*}

Mathematica [A]  time = 0.105273, size = 66, normalized size = 0.74 \[ \frac{9 \cos (a) \text{CosIntegral}(b x)-\cos (3 a) \text{CosIntegral}(3 b x)-9 \sin (a) \text{Si}(b x)+\sin (3 a) \text{Si}(3 b x)-9 \log (x) \cos (a+b x)+\log (x) \cos (3 (a+b x))}{12 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[x]*Sin[a + b*x]^3,x]

[Out]

(9*Cos[a]*CosIntegral[b*x] - Cos[3*a]*CosIntegral[3*b*x] - 9*Cos[a + b*x]*Log[x] + Cos[3*(a + b*x)]*Log[x] - 9
*Sin[a]*SinIntegral[b*x] + Sin[3*a]*SinIntegral[3*b*x])/(12*b)

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Maple [C]  time = 0.129, size = 162, normalized size = 1.8 \begin{align*} -{\frac{3\,\cos \left ( bx+a \right ) \ln \left ( x \right ) }{4\,b}}+{\frac{\ln \left ( x \right ) \cos \left ( 3\,bx+3\,a \right ) }{12\,b}}-{\frac{{\frac{i}{24}}{{\rm e}^{-3\,ia}}\pi \,{\it csgn} \left ( bx \right ) }{b}}+{\frac{{\frac{i}{12}}{{\rm e}^{-3\,ia}}{\it Si} \left ( 3\,bx \right ) }{b}}+{\frac{{{\rm e}^{-3\,ia}}{\it Ei} \left ( 1,-3\,ibx \right ) }{24\,b}}+{\frac{{\frac{3\,i}{8}}{{\rm e}^{-ia}}\pi \,{\it csgn} \left ( bx \right ) }{b}}-{\frac{{\frac{3\,i}{4}}{{\rm e}^{-ia}}{\it Si} \left ( bx \right ) }{b}}-{\frac{3\,{{\rm e}^{-ia}}{\it Ei} \left ( 1,-ibx \right ) }{8\,b}}-{\frac{3\,{{\rm e}^{ia}}{\it Ei} \left ( 1,-ibx \right ) }{8\,b}}+{\frac{{{\rm e}^{3\,ia}}{\it Ei} \left ( 1,-3\,ibx \right ) }{24\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(x)*sin(b*x+a)^3,x)

[Out]

-3/4*cos(b*x+a)*ln(x)/b+1/12*ln(x)/b*cos(3*b*x+3*a)-1/24*I/b*exp(-3*I*a)*Pi*csgn(b*x)+1/12*I/b*exp(-3*I*a)*Si(
3*b*x)+1/24/b*exp(-3*I*a)*Ei(1,-3*I*b*x)+3/8*I/b*exp(-I*a)*Pi*csgn(b*x)-3/4*I/b*exp(-I*a)*Si(b*x)-3/8/b*exp(-I
*a)*Ei(1,-I*b*x)-3/8/b*exp(I*a)*Ei(1,-I*b*x)+1/24/b*exp(3*I*a)*Ei(1,-3*I*b*x)

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Maxima [C]  time = 1.25428, size = 149, normalized size = 1.67 \begin{align*} \frac{{\left (\cos \left (b x + a\right )^{3} - 3 \, \cos \left (b x + a\right )\right )} \log \left (x\right )}{3 \, b} + \frac{{\left (E_{1}\left (3 i \, b x\right ) + E_{1}\left (-3 i \, b x\right )\right )} \cos \left (3 \, a\right ) - 9 \,{\left (E_{1}\left (i \, b x\right ) + E_{1}\left (-i \, b x\right )\right )} \cos \left (a\right ) -{\left (i \, E_{1}\left (3 i \, b x\right ) - i \, E_{1}\left (-3 i \, b x\right )\right )} \sin \left (3 \, a\right ) -{\left (-9 i \, E_{1}\left (i \, b x\right ) + 9 i \, E_{1}\left (-i \, b x\right )\right )} \sin \left (a\right )}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x)*sin(b*x+a)^3,x, algorithm="maxima")

[Out]

1/3*(cos(b*x + a)^3 - 3*cos(b*x + a))*log(x)/b + 1/24*((exp_integral_e(1, 3*I*b*x) + exp_integral_e(1, -3*I*b*
x))*cos(3*a) - 9*(exp_integral_e(1, I*b*x) + exp_integral_e(1, -I*b*x))*cos(a) - (I*exp_integral_e(1, 3*I*b*x)
- I*exp_integral_e(1, -3*I*b*x))*sin(3*a) - (-9*I*exp_integral_e(1, I*b*x) + 9*I*exp_integral_e(1, -I*b*x))*s
in(a))/b

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Fricas [A]  time = 2.27604, size = 304, normalized size = 3.42 \begin{align*} -\frac{{\left (\operatorname{Ci}\left (3 \, b x\right ) + \operatorname{Ci}\left (-3 \, b x\right )\right )} \cos \left (3 \, a\right ) - 9 \,{\left (\operatorname{Ci}\left (b x\right ) + \operatorname{Ci}\left (-b x\right )\right )} \cos \left (a\right ) - 8 \,{\left (\cos \left (b x + a\right )^{3} - 3 \, \cos \left (b x + a\right )\right )} \log \left (x\right ) - 2 \, \sin \left (3 \, a\right ) \operatorname{Si}\left (3 \, b x\right ) + 18 \, \sin \left (a\right ) \operatorname{Si}\left (b x\right )}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x)*sin(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/24*((cos_integral(3*b*x) + cos_integral(-3*b*x))*cos(3*a) - 9*(cos_integral(b*x) + cos_integral(-b*x))*cos(
a) - 8*(cos(b*x + a)^3 - 3*cos(b*x + a))*log(x) - 2*sin(3*a)*sin_integral(3*b*x) + 18*sin(a)*sin_integral(b*x)
)/b

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \log{\left (x \right )} \sin ^{3}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(x)*sin(b*x+a)**3,x)

[Out]

Integral(log(x)*sin(a + b*x)**3, x)

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Giac [C]  time = 1.33762, size = 613, normalized size = 6.89 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x)*sin(b*x+a)^3,x, algorithm="giac")

[Out]

1/3*(cos(b*x + a)^3/b - 3*cos(b*x + a)/b)*log(x) + 1/24*(real_part(cos_integral(3*b*x))*tan(3/2*a)^2*tan(1/2*a
)^2 - 9*real_part(cos_integral(b*x))*tan(3/2*a)^2*tan(1/2*a)^2 - 9*real_part(cos_integral(-b*x))*tan(3/2*a)^2*
tan(1/2*a)^2 + real_part(cos_integral(-3*b*x))*tan(3/2*a)^2*tan(1/2*a)^2 - 18*imag_part(cos_integral(b*x))*tan
(3/2*a)^2*tan(1/2*a) + 18*imag_part(cos_integral(-b*x))*tan(3/2*a)^2*tan(1/2*a) - 36*sin_integral(b*x)*tan(3/2
*a)^2*tan(1/2*a) + 2*imag_part(cos_integral(3*b*x))*tan(3/2*a)*tan(1/2*a)^2 - 2*imag_part(cos_integral(-3*b*x)
)*tan(3/2*a)*tan(1/2*a)^2 + 4*sin_integral(3*b*x)*tan(3/2*a)*tan(1/2*a)^2 + real_part(cos_integral(3*b*x))*tan
(3/2*a)^2 + 9*real_part(cos_integral(b*x))*tan(3/2*a)^2 + 9*real_part(cos_integral(-b*x))*tan(3/2*a)^2 + real_
part(cos_integral(-3*b*x))*tan(3/2*a)^2 - real_part(cos_integral(3*b*x))*tan(1/2*a)^2 - 9*real_part(cos_integr
al(b*x))*tan(1/2*a)^2 - 9*real_part(cos_integral(-b*x))*tan(1/2*a)^2 - real_part(cos_integral(-3*b*x))*tan(1/2
*a)^2 + 2*imag_part(cos_integral(3*b*x))*tan(3/2*a) - 2*imag_part(cos_integral(-3*b*x))*tan(3/2*a) + 4*sin_int
egral(3*b*x)*tan(3/2*a) - 18*imag_part(cos_integral(b*x))*tan(1/2*a) + 18*imag_part(cos_integral(-b*x))*tan(1/
2*a) - 36*sin_integral(b*x)*tan(1/2*a) - real_part(cos_integral(3*b*x)) + 9*real_part(cos_integral(b*x)) + 9*r
eal_part(cos_integral(-b*x)) - real_part(cos_integral(-3*b*x)))/(b*tan(3/2*a)^2*tan(1/2*a)^2 + b*tan(3/2*a)^2
+ b*tan(1/2*a)^2 + b)