3.159 \(\int \cos ^3(a+b x) \log (x) \, dx\)

Optimal. Leaf size=88 \[ -\frac{3 \sin (a) \text{CosIntegral}(b x)}{4 b}-\frac{\sin (3 a) \text{CosIntegral}(3 b x)}{12 b}-\frac{3 \cos (a) \text{Si}(b x)}{4 b}-\frac{\cos (3 a) \text{Si}(3 b x)}{12 b}-\frac{\log (x) \sin ^3(a+b x)}{3 b}+\frac{\log (x) \sin (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.468172, antiderivative size = 88, 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, 4430} \[ -\frac{3 \sin (a) \text{CosIntegral}(b x)}{4 b}-\frac{\sin (3 a) \text{CosIntegral}(3 b x)}{12 b}-\frac{3 \cos (a) \text{Si}(b x)}{4 b}-\frac{\cos (3 a) \text{Si}(3 b x)}{12 b}-\frac{\log (x) \sin ^3(a+b x)}{3 b}+\frac{\log (x) \sin (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^3*Log[x],x]

[Out]

(-3*CosIntegral[b*x]*Sin[a])/(4*b) - (CosIntegral[3*b*x]*Sin[3*a])/(12*b) + (Log[x]*Sin[a + b*x])/b - (Log[x]*
Sin[a + b*x]^3)/(3*b) - (3*Cos[a]*SinIntegral[b*x])/(4*b) - (Cos[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 4430

Int[Cos[(c_.) + (d_.)*(x_)]^(q_.)*((e_.) + (f_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Int[E
xpandTrigReduce[(e + f*x)^m, Sin[a + b*x]^p*Cos[c + d*x]^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[
p, 0] && IGtQ[q, 0]

Rubi steps

\begin{align*} \int \cos ^3(a+b x) \log (x) \, dx &=\frac{\log (x) \sin (a+b x)}{b}-\frac{\log (x) \sin ^3(a+b x)}{3 b}-\int \frac{(5+\cos (2 (a+b x))) \sin (a+b x)}{6 b x} \, dx\\ &=\frac{\log (x) \sin (a+b x)}{b}-\frac{\log (x) \sin ^3(a+b x)}{3 b}-\frac{\int \frac{(5+\cos (2 (a+b x))) \sin (a+b x)}{x} \, dx}{6 b}\\ &=\frac{\log (x) \sin (a+b x)}{b}-\frac{\log (x) \sin ^3(a+b x)}{3 b}-\frac{\int \left (\frac{5 \sin (a+b x)}{x}+\frac{\cos (2 a+2 b x) \sin (a+b x)}{x}\right ) \, dx}{6 b}\\ &=\frac{\log (x) \sin (a+b x)}{b}-\frac{\log (x) \sin ^3(a+b x)}{3 b}-\frac{\int \frac{\cos (2 a+2 b x) \sin (a+b x)}{x} \, dx}{6 b}-\frac{5 \int \frac{\sin (a+b x)}{x} \, dx}{6 b}\\ &=\frac{\log (x) \sin (a+b x)}{b}-\frac{\log (x) \sin ^3(a+b x)}{3 b}-\frac{\int \left (-\frac{\sin (a+b x)}{2 x}+\frac{\sin (3 a+3 b x)}{2 x}\right ) \, dx}{6 b}-\frac{(5 \cos (a)) \int \frac{\sin (b x)}{x} \, dx}{6 b}-\frac{(5 \sin (a)) \int \frac{\cos (b x)}{x} \, dx}{6 b}\\ &=-\frac{5 \text{Ci}(b x) \sin (a)}{6 b}+\frac{\log (x) \sin (a+b x)}{b}-\frac{\log (x) \sin ^3(a+b x)}{3 b}-\frac{5 \cos (a) \text{Si}(b x)}{6 b}+\frac{\int \frac{\sin (a+b x)}{x} \, dx}{12 b}-\frac{\int \frac{\sin (3 a+3 b x)}{x} \, dx}{12 b}\\ &=-\frac{5 \text{Ci}(b x) \sin (a)}{6 b}+\frac{\log (x) \sin (a+b x)}{b}-\frac{\log (x) \sin ^3(a+b x)}{3 b}-\frac{5 \cos (a) \text{Si}(b x)}{6 b}+\frac{\cos (a) \int \frac{\sin (b x)}{x} \, dx}{12 b}-\frac{\cos (3 a) \int \frac{\sin (3 b x)}{x} \, dx}{12 b}+\frac{\sin (a) \int \frac{\cos (b x)}{x} \, dx}{12 b}-\frac{\sin (3 a) \int \frac{\cos (3 b x)}{x} \, dx}{12 b}\\ &=-\frac{3 \text{Ci}(b x) \sin (a)}{4 b}-\frac{\text{Ci}(3 b x) \sin (3 a)}{12 b}+\frac{\log (x) \sin (a+b x)}{b}-\frac{\log (x) \sin ^3(a+b x)}{3 b}-\frac{3 \cos (a) \text{Si}(b x)}{4 b}-\frac{\cos (3 a) \text{Si}(3 b x)}{12 b}\\ \end{align*}

Mathematica [A]  time = 0.150755, size = 66, normalized size = 0.75 \[ -\frac{9 \sin (a) \text{CosIntegral}(b x)+\sin (3 a) \text{CosIntegral}(3 b x)+9 \cos (a) \text{Si}(b x)+\cos (3 a) \text{Si}(3 b x)-9 \log (x) \sin (a+b x)-\log (x) \sin (3 (a+b x))}{12 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^3*Log[x],x]

[Out]

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

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Maple [C]  time = 0.118, size = 162, normalized size = 1.8 \begin{align*}{\frac{3\,\ln \left ( x \right ) \sin \left ( bx+a \right ) }{4\,b}}+{\frac{\ln \left ( x \right ) \sin \left ( 3\,bx+3\,a \right ) }{12\,b}}+{\frac{{{\rm e}^{-3\,ia}}\pi \,{\it csgn} \left ( bx \right ) }{24\,b}}-{\frac{{{\rm e}^{-3\,ia}}{\it Si} \left ( 3\,bx \right ) }{12\,b}}+{\frac{{\frac{i}{24}}{{\rm e}^{-3\,ia}}{\it Ei} \left ( 1,-3\,ibx \right ) }{b}}+{\frac{3\,{{\rm e}^{-ia}}\pi \,{\it csgn} \left ( bx \right ) }{8\,b}}-{\frac{3\,{{\rm e}^{-ia}}{\it Si} \left ( bx \right ) }{4\,b}}+{\frac{{\frac{3\,i}{8}}{{\rm e}^{-ia}}{\it Ei} \left ( 1,-ibx \right ) }{b}}-{\frac{{\frac{3\,i}{8}}{{\rm e}^{ia}}{\it Ei} \left ( 1,-ibx \right ) }{b}}-{\frac{{\frac{i}{24}}{{\rm e}^{3\,ia}}{\it Ei} \left ( 1,-3\,ibx \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C]  time = 1.25374, size = 146, normalized size = 1.66 \begin{align*} -\frac{{\left (\sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )\right )} \log \left (x\right )}{3 \, b} + \frac{{\left (i \, E_{1}\left (3 i \, b x\right ) - i \, E_{1}\left (-3 i \, b x\right )\right )} \cos \left (3 \, a\right ) +{\left (9 i \, E_{1}\left (i \, b x\right ) - 9 i \, E_{1}\left (-i \, b x\right )\right )} \cos \left (a\right ) +{\left (E_{1}\left (3 i \, b x\right ) + E_{1}\left (-3 i \, b x\right )\right )} \sin \left (3 \, a\right ) + 9 \,{\left (E_{1}\left (i \, b x\right ) + 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(cos(b*x+a)^3*log(x),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.60629, size = 302, normalized size = 3.43 \begin{align*} \frac{8 \,{\left (\cos \left (b x + a\right )^{2} + 2\right )} \log \left (x\right ) \sin \left (b x + a\right ) -{\left (\operatorname{Ci}\left (3 \, b x\right ) + \operatorname{Ci}\left (-3 \, b x\right )\right )} \sin \left (3 \, a\right ) - 9 \,{\left (\operatorname{Ci}\left (b x\right ) + \operatorname{Ci}\left (-b x\right )\right )} \sin \left (a\right ) - 2 \, \cos \left (3 \, a\right ) \operatorname{Si}\left (3 \, b x\right ) - 18 \, \cos \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(cos(b*x+a)^3*log(x),x, algorithm="fricas")

[Out]

1/24*(8*(cos(b*x + a)^2 + 2)*log(x)*sin(b*x + a) - (cos_integral(3*b*x) + cos_integral(-3*b*x))*sin(3*a) - 9*(
cos_integral(b*x) + cos_integral(-b*x))*sin(a) - 2*cos(3*a)*sin_integral(3*b*x) - 18*cos(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 )} \cos ^{3}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(x)*cos(a + b*x)**3, x)

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Giac [C]  time = 1.21928, size = 668, normalized size = 7.59 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(sin(b*x + a)^3 - 3*sin(b*x + a))*log(x)/b + 1/24*(imag_part(cos_integral(3*b*x))*tan(3/2*a)^2*tan(1/2*a)
^2 + 9*imag_part(cos_integral(b*x))*tan(3/2*a)^2*tan(1/2*a)^2 - 9*imag_part(cos_integral(-b*x))*tan(3/2*a)^2*t
an(1/2*a)^2 - imag_part(cos_integral(-3*b*x))*tan(3/2*a)^2*tan(1/2*a)^2 + 2*sin_integral(3*b*x)*tan(3/2*a)^2*t
an(1/2*a)^2 + 18*sin_integral(b*x)*tan(3/2*a)^2*tan(1/2*a)^2 - 18*real_part(cos_integral(b*x))*tan(3/2*a)^2*ta
n(1/2*a) - 18*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)*tan(1/2*a)^2 - 2*real_part(cos_integral(-3*b*x))*tan(3/2*a)*tan(1/2*a)^2 + imag_part(cos_integral(3*b*x))*
tan(3/2*a)^2 - 9*imag_part(cos_integral(b*x))*tan(3/2*a)^2 + 9*imag_part(cos_integral(-b*x))*tan(3/2*a)^2 - im
ag_part(cos_integral(-3*b*x))*tan(3/2*a)^2 + 2*sin_integral(3*b*x)*tan(3/2*a)^2 - 18*sin_integral(b*x)*tan(3/2
*a)^2 - imag_part(cos_integral(3*b*x))*tan(1/2*a)^2 + 9*imag_part(cos_integral(b*x))*tan(1/2*a)^2 - 9*imag_par
t(cos_integral(-b*x))*tan(1/2*a)^2 + imag_part(cos_integral(-3*b*x))*tan(1/2*a)^2 - 2*sin_integral(3*b*x)*tan(
1/2*a)^2 + 18*sin_integral(b*x)*tan(1/2*a)^2 - 2*real_part(cos_integral(3*b*x))*tan(3/2*a) - 2*real_part(cos_i
ntegral(-3*b*x))*tan(3/2*a) - 18*real_part(cos_integral(b*x))*tan(1/2*a) - 18*real_part(cos_integral(-b*x))*ta
n(1/2*a) - imag_part(cos_integral(3*b*x)) - 9*imag_part(cos_integral(b*x)) + 9*imag_part(cos_integral(-b*x)) +
imag_part(cos_integral(-3*b*x)) - 2*sin_integral(3*b*x) - 18*sin_integral(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)