3.158 \(\int \cos ^2(a+b x) \log (x) \, dx\)

Optimal. Leaf size=66 \[ -\frac {\sin (2 a) \text {Ci}(2 b x)}{4 b}-\frac {\cos (2 a) \text {Si}(2 b x)}{4 b}+\frac {\log (x) \sin (a+b x) \cos (a+b x)}{2 b}-\frac {x}{2}+\frac {1}{2} x \log (x) \]

[Out]

-1/2*x+1/2*x*ln(x)-1/4*cos(2*a)*Si(2*b*x)/b-1/4*Ci(2*b*x)*sin(2*a)/b+1/2*cos(b*x+a)*ln(x)*sin(b*x+a)/b

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Rubi [A]  time = 0.12, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 8, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {2635, 8, 2554, 12, 3327, 3303, 3299, 3302} \[ -\frac {\sin (2 a) \text {CosIntegral}(2 b x)}{4 b}-\frac {\cos (2 a) \text {Si}(2 b x)}{4 b}+\frac {\log (x) \sin (a+b x) \cos (a+b x)}{2 b}-\frac {x}{2}+\frac {1}{2} x \log (x) \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2*Log[x],x]

[Out]

-x/2 + (x*Log[x])/2 - (CosIntegral[2*b*x]*Sin[2*a])/(4*b) + (Cos[a + b*x]*Log[x]*Sin[a + b*x])/(2*b) - (Cos[2*
a]*SinIntegral[2*b*x])/(4*b)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

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]

Rule 3327

Int[(u_)^(m_.)*((a_.) + (b_.)*Sin[v_])^(n_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*(a + b*Sin[ExpandToSum[v, x
]])^n, x] /; FreeQ[{a, b, m, n}, x] && LinearQ[{u, v}, x] &&  !LinearMatchQ[{u, v}, x]

Rubi steps

\begin {align*} \int \cos ^2(a+b x) \log (x) \, dx &=\frac {1}{2} x \log (x)+\frac {\cos (a+b x) \log (x) \sin (a+b x)}{2 b}-\int \frac {1}{4} \left (2+\frac {\sin (2 (a+b x))}{b x}\right ) \, dx\\ &=\frac {1}{2} x \log (x)+\frac {\cos (a+b x) \log (x) \sin (a+b x)}{2 b}-\frac {1}{4} \int \left (2+\frac {\sin (2 (a+b x))}{b x}\right ) \, dx\\ &=-\frac {x}{2}+\frac {1}{2} x \log (x)+\frac {\cos (a+b x) \log (x) \sin (a+b x)}{2 b}-\frac {\int \frac {\sin (2 (a+b x))}{x} \, dx}{4 b}\\ &=-\frac {x}{2}+\frac {1}{2} x \log (x)+\frac {\cos (a+b x) \log (x) \sin (a+b x)}{2 b}-\frac {\int \frac {\sin (2 a+2 b x)}{x} \, dx}{4 b}\\ &=-\frac {x}{2}+\frac {1}{2} x \log (x)+\frac {\cos (a+b x) \log (x) \sin (a+b x)}{2 b}-\frac {\cos (2 a) \int \frac {\sin (2 b x)}{x} \, dx}{4 b}-\frac {\sin (2 a) \int \frac {\cos (2 b x)}{x} \, dx}{4 b}\\ &=-\frac {x}{2}+\frac {1}{2} x \log (x)-\frac {\text {Ci}(2 b x) \sin (2 a)}{4 b}+\frac {\cos (a+b x) \log (x) \sin (a+b x)}{2 b}-\frac {\cos (2 a) \text {Si}(2 b x)}{4 b}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 50, normalized size = 0.76 \[ -\frac {\sin (2 a) \text {Ci}(2 b x)+\cos (2 a) \text {Si}(2 b x)-\log (x) \sin (2 (a+b x))+2 b x-2 b x \log (x)}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^2*Log[x],x]

[Out]

-1/4*(2*b*x - 2*b*x*Log[x] + CosIntegral[2*b*x]*Sin[2*a] - Log[x]*Sin[2*(a + b*x)] + Cos[2*a]*SinIntegral[2*b*
x])/b

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fricas [A]  time = 0.55, size = 60, normalized size = 0.91 \[ \frac {4 \, b x \log \relax (x) + 4 \, \cos \left (b x + a\right ) \log \relax (x) \sin \left (b x + a\right ) - 4 \, b x - {\left (\operatorname {Ci}\left (2 \, b x\right ) + \operatorname {Ci}\left (-2 \, b x\right )\right )} \sin \left (2 \, a\right ) - 2 \, \cos \left (2 \, a\right ) \operatorname {Si}\left (2 \, b x\right )}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2*log(x),x, algorithm="fricas")

[Out]

1/8*(4*b*x*log(x) + 4*cos(b*x + a)*log(x)*sin(b*x + a) - 4*b*x - (cos_integral(2*b*x) + cos_integral(-2*b*x))*
sin(2*a) - 2*cos(2*a)*sin_integral(2*b*x))/b

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giac [C]  time = 0.21, size = 122, normalized size = 1.85 \[ \frac {1}{4} \, {\left (2 \, x + \frac {\sin \left (2 \, b x + 2 \, a\right )}{b}\right )} \log \relax (x) - \frac {4 \, b x \tan \relax (a)^{2} - \Im \left (\operatorname {Ci}\left (2 \, b x\right ) \right ) \tan \relax (a)^{2} + \Im \left (\operatorname {Ci}\left (-2 \, b x\right ) \right ) \tan \relax (a)^{2} - 2 \, \operatorname {Si}\left (2 \, b x\right ) \tan \relax (a)^{2} + 4 \, b x + 2 \, \Re \left (\operatorname {Ci}\left (2 \, b x\right ) \right ) \tan \relax (a) + 2 \, \Re \left (\operatorname {Ci}\left (-2 \, b x\right ) \right ) \tan \relax (a) + \Im \left (\operatorname {Ci}\left (2 \, b x\right ) \right ) - \Im \left (\operatorname {Ci}\left (-2 \, b x\right ) \right ) + 2 \, \operatorname {Si}\left (2 \, b x\right )}{8 \, {\left (b \tan \relax (a)^{2} + b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2*log(x),x, algorithm="giac")

[Out]

1/4*(2*x + sin(2*b*x + 2*a)/b)*log(x) - 1/8*(4*b*x*tan(a)^2 - imag_part(cos_integral(2*b*x))*tan(a)^2 + imag_p
art(cos_integral(-2*b*x))*tan(a)^2 - 2*sin_integral(2*b*x)*tan(a)^2 + 4*b*x + 2*real_part(cos_integral(2*b*x))
*tan(a) + 2*real_part(cos_integral(-2*b*x))*tan(a) + imag_part(cos_integral(2*b*x)) - imag_part(cos_integral(-
2*b*x)) + 2*sin_integral(2*b*x))/(b*tan(a)^2 + b)

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maple [C]  time = 1.33, size = 132, normalized size = 2.00 \[ \frac {x \ln \relax (x )}{2}+\frac {a \ln \left (i b x \right )}{2 b}-\frac {a \ln \left (a +i \left (i b x +i a \right )\right )}{2 b}+\frac {i \Ei \left (1, -2 i b x \right ) {\mathrm e}^{-2 i a}}{8 b}-\frac {i \Ei \left (1, -2 i b x \right ) {\mathrm e}^{2 i a}}{8 b}-\frac {\Si \left (2 b x \right ) {\mathrm e}^{-2 i a}}{4 b}+\frac {\pi \,\mathrm {csgn}\left (b x \right ) {\mathrm e}^{-2 i a}}{8 b}+\frac {\ln \relax (x ) \sin \left (2 b x +2 a \right )}{4 b}-\frac {x}{2}-\frac {a}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^2*ln(x),x)

[Out]

1/2*x*ln(x)+1/4/b*ln(x)*sin(2*b*x+2*a)+1/8*Pi/b*csgn(b*x)*exp(-2*I*a)-1/4/b*exp(-2*I*a)*Si(2*b*x)+1/8*I/b*exp(
-2*I*a)*Ei(1,-2*I*b*x)+1/2*a/b*ln(I*b*x)-1/2*x-1/2*a/b-1/2*a/b*ln(a+I*(I*b*x+I*a))-1/8*I/b*exp(2*I*a)*Ei(1,-2*
I*b*x)

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maxima [C]  time = 1.12, size = 76, normalized size = 1.15 \[ \frac {{\left (2 \, b x + 2 \, a + \sin \left (2 \, b x + 2 \, a\right )\right )} \log \relax (x)}{4 \, b} - \frac {4 \, b x + {\left (-i \, {\rm Ei}\left (2 i \, b x\right ) + i \, {\rm Ei}\left (-2 i \, b x\right )\right )} \cos \left (2 \, a\right ) + 4 \, a \log \relax (x) + {\left ({\rm Ei}\left (2 i \, b x\right ) + {\rm Ei}\left (-2 i \, b x\right )\right )} \sin \left (2 \, a\right )}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2*log(x),x, algorithm="maxima")

[Out]

1/4*(2*b*x + 2*a + sin(2*b*x + 2*a))*log(x)/b - 1/8*(4*b*x + (-I*Ei(2*I*b*x) + I*Ei(-2*I*b*x))*cos(2*a) + 4*a*
log(x) + (Ei(2*I*b*x) + Ei(-2*I*b*x))*sin(2*a))/b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\cos \left (a+b\,x\right )}^2\,\ln \relax (x) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^2*log(x),x)

[Out]

int(cos(a + b*x)^2*log(x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \log {\relax (x )} \cos ^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**2*ln(x),x)

[Out]

Integral(log(x)*cos(a + b*x)**2, x)

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