\(\int \cos (a+b x) \log (x) \, dx\) [157]

Optimal result

Integrand size = 9, antiderivative size = 35 \[ \int \cos (a+b x) \log (x) \, dx=-\frac {\operatorname {CosIntegral}(b x) \sin (a)}{b}+\frac {\log (x) \sin (a+b x)}{b}-\frac {\cos (a) \text {Si}(b x)}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {2717, 2634, 12, 3384, 3380, 3383} \[ \int \cos (a+b x) \log (x) \, dx=-\frac {\sin (a) \operatorname {CosIntegral}(b x)}{b}-\frac {\cos (a) \text {Si}(b x)}{b}+\frac {\log (x) \sin (a+b x)}{b} \]

[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 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]

Rule 3380

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 3383

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 3384

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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.86 \[ \int \cos (a+b x) \log (x) \, dx=-\frac {\operatorname {CosIntegral}(b x) \sin (a)-\log (x) \sin (a+b x)+\cos (a) \text {Si}(b x)}{b} \]

[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)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 1.02 (sec) , antiderivative size = 79, normalized size of antiderivative = 2.26

method	result	size
risch	\(\frac {\ln \left (x \right ) \sin \left (b x +a \right )}{b}+\frac {{\mathrm e}^{-i a} \pi \,\operatorname {csgn}\left (b x \right )}{2 b}-\frac {{\mathrm e}^{-i a} \operatorname {Si}\left (b x \right )}{b}+\frac {i {\mathrm e}^{-i a} \operatorname {Ei}_{1}\left (-i b x \right )}{2 b}-\frac {i {\mathrm e}^{i a} \operatorname {Ei}_{1}\left (-i b x \right )}{2 b}\)	\(79\)

[In]

int(cos(b*x+a)*ln(x),x,method=_RETURNVERBOSE)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.86 \[ \int \cos (a+b x) \log (x) \, dx=\frac {\log \left (x\right ) \sin \left (b x + a\right ) - \operatorname {Ci}\left (b x\right ) \sin \left (a\right ) - \cos \left (a\right ) \operatorname {Si}\left (b x\right )}{b} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(log(x)*cos(a + b*x), x)

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.57 \[ \int \cos (a+b x) \log (x) \, dx=\frac {\log \left (x\right ) \sin \left (b x + a\right )}{b} + \frac {{\left (i \, E_{1}\left (i \, b x\right ) - i \, E_{1}\left (-i \, b x\right )\right )} \cos \left (a\right ) + {\left (E_{1}\left (i \, b x\right ) + E_{1}\left (-i \, b x\right )\right )} \sin \left (a\right )}{2 \, b} \]

[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

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.31 (sec) , antiderivative size = 108, normalized size of antiderivative = 3.09 \[ \int \cos (a+b x) \log (x) \, dx=\frac {\log \left (x\right ) \sin \left (b x + a\right )}{b} + \frac {\Im \left ( \operatorname {Ci}\left (b x\right ) \right ) \tan \left (\frac {1}{2} \, a\right )^{2} - \Im \left ( \operatorname {Ci}\left (-b x\right ) \right ) \tan \left (\frac {1}{2} \, a\right )^{2} + 2 \, \operatorname {Si}\left (b x\right ) \tan \left (\frac {1}{2} \, a\right )^{2} - 2 \, \Re \left ( \operatorname {Ci}\left (b x\right ) \right ) \tan \left (\frac {1}{2} \, a\right ) - 2 \, \Re \left ( \operatorname {Ci}\left (-b x\right ) \right ) \tan \left (\frac {1}{2} \, a\right ) - \Im \left ( \operatorname {Ci}\left (b x\right ) \right ) + \Im \left ( \operatorname {Ci}\left (-b x\right ) \right ) - 2 \, \operatorname {Si}\left (b x\right )}{2 \, {\left (b \tan \left (\frac {1}{2} \, a\right )^{2} + b\right )}} \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \cos (a+b x) \log (x) \, dx=\int \cos \left (a+b\,x\right )\,\ln \left (x\right ) \,d x \]

[In]

int(cos(a + b*x)*log(x),x)

[Out]

int(cos(a + b*x)*log(x), x)