3.154 \(\int \log (x) \sin (a+b x) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0780172, antiderivative size = 35, normalized size of antiderivative = 1., 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 = {2638, 2554, 12, 3303, 3299, 3302} \[ \frac{\cos (a) \text{CosIntegral}(b x)}{b}-\frac{\sin (a) \text{Si}(b x)}{b}-\frac{\log (x) \cos (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Log[x]*Sin[a + b*x],x]

[Out]

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

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[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]

Rule 12

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

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]

Rubi steps

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

Mathematica [A]  time = 0.0423173, size = 30, normalized size = 0.86 \[ \frac{\cos (a) \text{CosIntegral}(b x)-\sin (a) \text{Si}(b x)-\log (x) \cos (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[x]*Sin[a + b*x],x]

[Out]

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

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Maple [C]  time = 0.107, size = 80, normalized size = 2.3 \begin{align*} -{\frac{\cos \left ( bx+a \right ) \ln \left ( x \right ) }{b}}+{\frac{{\frac{i}{2}}{{\rm e}^{-ia}}\pi \,{\it csgn} \left ( bx \right ) }{b}}-{\frac{i{{\rm e}^{-ia}}{\it Si} \left ( bx \right ) }{b}}-{\frac{{{\rm e}^{-ia}}{\it Ei} \left ( 1,-ibx \right ) }{2\,b}}-{\frac{{{\rm e}^{ia}}{\it Ei} \left ( 1,-ibx \right ) }{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-cos(b*x+a)*ln(x)/b+1/2*I/b*exp(-I*a)*Pi*csgn(b*x)-I/b*exp(-I*a)*Si(b*x)-1/2/b*exp(-I*a)*Ei(1,-I*b*x)-1/2/b*ex
p(I*a)*Ei(1,-I*b*x)

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Maxima [C]  time = 1.15679, size = 77, normalized size = 2.2 \begin{align*} -\frac{\cos \left (b x + a\right ) \log \left (x\right )}{b} - \frac{{\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 (i \, b x\right ) - i \, E_{1}\left (-i \, b x\right )\right )} \sin \left (a\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.28442, size = 149, normalized size = 4.26 \begin{align*} \frac{{\left (\operatorname{Ci}\left (b x\right ) + \operatorname{Ci}\left (-b x\right )\right )} \cos \left (a\right ) - 2 \, \cos \left (b x + a\right ) \log \left (x\right ) - 2 \, \sin \left (a\right ) \operatorname{Si}\left (b x\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((cos_integral(b*x) + cos_integral(-b*x))*cos(a) - 2*cos(b*x + a)*log(x) - 2*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{\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),x)

[Out]

Integral(log(x)*sin(a + b*x), x)

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Giac [C]  time = 1.30149, size = 138, normalized size = 3.94 \begin{align*} -\frac{\cos \left (b x + a\right ) \log \left (x\right )}{b} - \frac{\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 )^{2} + 2 \, \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 ) + 4 \, \operatorname{Si}\left (b x\right ) \tan \left (\frac{1}{2} \, a\right ) - \Re \left ( \operatorname{Ci}\left (b x\right ) \right ) - \Re \left ( \operatorname{Ci}\left (-b x\right ) \right )}{2 \,{\left (b \tan \left (\frac{1}{2} \, a\right )^{2} + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-cos(b*x + a)*log(x)/b - 1/2*(real_part(cos_integral(b*x))*tan(1/2*a)^2 + real_part(cos_integral(-b*x))*tan(1/
2*a)^2 + 2*imag_part(cos_integral(b*x))*tan(1/2*a) - 2*imag_part(cos_integral(-b*x))*tan(1/2*a) + 4*sin_integr
al(b*x)*tan(1/2*a) - real_part(cos_integral(b*x)) - real_part(cos_integral(-b*x)))/(b*tan(1/2*a)^2 + b)