3.157 \(\int \cos (a+b x) \log (x) \, dx\)

Optimal. Leaf size=35 \[ -\frac {\sin (a) \text {Ci}(b x)}{b}-\frac {\cos (a) \text {Si}(b x)}{b}+\frac {\log (x) \sin (a+b x)}{b} \]

[Out]

-cos(a)*Si(b*x)/b-Ci(b*x)*sin(a)/b+ln(x)*sin(b*x+a)/b

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Rubi [A]  time = 0.06, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {2637, 2554, 12, 3303, 3299, 3302} \[ -\frac {\sin (a) \text {CosIntegral}(b x)}{b}-\frac {\cos (a) \text {Si}(b x)}{b}+\frac {\log (x) \sin (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]*Log[x],x]

[Out]

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

Rule 12

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

Rule 2637

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

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 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]

Rubi steps

\begin {align*} \int \cos (a+b x) \log (x) \, dx &=\frac {\log (x) \sin (a+b x)}{b}-\int \frac {\sin (a+b x)}{b x} \, dx\\ &=\frac {\log (x) \sin (a+b x)}{b}-\frac {\int \frac {\sin (a+b x)}{x} \, dx}{b}\\ &=\frac {\log (x) \sin (a+b x)}{b}-\frac {\cos (a) \int \frac {\sin (b x)}{x} \, dx}{b}-\frac {\sin (a) \int \frac {\cos (b x)}{x} \, dx}{b}\\ &=-\frac {\text {Ci}(b x) \sin (a)}{b}+\frac {\log (x) \sin (a+b x)}{b}-\frac {\cos (a) \text {Si}(b x)}{b}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 30, normalized size = 0.86 \[ -\frac {\sin (a) \text {Ci}(b x)+\cos (a) \text {Si}(b x)-\log (x) \sin (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]*Log[x],x]

[Out]

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

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fricas [A]  time = 0.70, size = 38, normalized size = 1.09 \[ \frac {2 \, \log \relax (x) \sin \left (b x + a\right ) - {\left (\operatorname {Ci}\left (b x\right ) + \operatorname {Ci}\left (-b x\right )\right )} \sin \relax (a) - 2 \, \cos \relax (a) \operatorname {Si}\left (b x\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*log(x),x, algorithm="fricas")

[Out]

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

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giac [C]  time = 0.19, size = 108, normalized size = 3.09 \[ \frac {\log \relax (x) \sin \left (b x + a\right )}{b} + \frac {\Im \left (\operatorname {Ci}\left (b x\right ) \right ) \tan \left (\frac {1}{2} \, a\right )^{2} - \Im \left (\operatorname {Ci}\left (-b x\right ) \right ) \tan \left (\frac {1}{2} \, a\right )^{2} + 2 \, \operatorname {Si}\left (b x\right ) \tan \left (\frac {1}{2} \, a\right )^{2} - 2 \, \Re \left (\operatorname {Ci}\left (b x\right ) \right ) \tan \left (\frac {1}{2} \, a\right ) - 2 \, \Re \left (\operatorname {Ci}\left (-b x\right ) \right ) \tan \left (\frac {1}{2} \, a\right ) - \Im \left (\operatorname {Ci}\left (b x\right ) \right ) + \Im \left (\operatorname {Ci}\left (-b x\right ) \right ) - 2 \, \operatorname {Si}\left (b x\right )}{2 \, {\left (b \tan \left (\frac {1}{2} \, a\right )^{2} + b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x)*sin(b*x + a)/b + 1/2*(imag_part(cos_integral(b*x))*tan(1/2*a)^2 - imag_part(cos_integral(-b*x))*tan(1/2
*a)^2 + 2*sin_integral(b*x)*tan(1/2*a)^2 - 2*real_part(cos_integral(b*x))*tan(1/2*a) - 2*real_part(cos_integra
l(-b*x))*tan(1/2*a) - imag_part(cos_integral(b*x)) + imag_part(cos_integral(-b*x)) - 2*sin_integral(b*x))/(b*t
an(1/2*a)^2 + b)

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maple [C]  time = 1.11, size = 79, normalized size = 2.26 \[ \frac {i \Ei \left (1, -i b x \right ) {\mathrm e}^{-i a}}{2 b}-\frac {i \Ei \left (1, -i b x \right ) {\mathrm e}^{i a}}{2 b}-\frac {\Si \left (b x \right ) {\mathrm e}^{-i a}}{b}+\frac {\pi \,\mathrm {csgn}\left (b x \right ) {\mathrm e}^{-i a}}{2 b}+\frac {\ln \relax (x ) \sin \left (b x +a \right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(x)*sin(b*x+a)/b+1/2/b*exp(-I*a)*Pi*csgn(b*x)-1/b*exp(-I*a)*Si(b*x)+1/2*I/b*exp(-I*a)*Ei(1,-I*b*x)-1/2*I/b*e
xp(I*a)*Ei(1,-I*b*x)

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maxima [C]  time = 1.02, size = 55, normalized size = 1.57 \[ \frac {\log \relax (x) \sin \left (b x + a\right )}{b} + \frac {{\left (i \, E_{1}\left (i \, b x\right ) - i \, E_{1}\left (-i \, b x\right )\right )} \cos \relax (a) + {\left (E_{1}\left (i \, b x\right ) + E_{1}\left (-i \, b x\right )\right )} \sin \relax (a)}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*log(x),x, algorithm="maxima")

[Out]

log(x)*sin(b*x + a)/b + 1/2*((I*exp_integral_e(1, I*b*x) - I*exp_integral_e(1, -I*b*x))*cos(a) + (exp_integral
_e(1, I*b*x) + exp_integral_e(1, -I*b*x))*sin(a))/b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \cos \left (a+b\,x\right )\,\ln \relax (x) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)*log(x),x)

[Out]

int(cos(a + b*x)*log(x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(x)*cos(a + b*x), x)

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