3.306 \(\int x \log (\log (x) \sin (x)) \, dx\)

Optimal. Leaf size=80 \[ -\frac {1}{2} \text {Ei}(2 \log (x))+\frac {1}{2} i x \text {Li}_2\left (e^{2 i x}\right )-\frac {1}{4} \text {Li}_3\left (e^{2 i x}\right )+\frac {i x^3}{6}-\frac {1}{2} x^2 \log \left (1-e^{2 i x}\right )+\frac {1}{2} x^2 \log (\log (x) \sin (x)) \]

[Out]

1/6*I*x^3-1/2*Ei(2*ln(x))-1/2*x^2*ln(1-exp(2*I*x))+1/2*x^2*ln(ln(x)*sin(x))+1/2*I*x*polylog(2,exp(2*I*x))-1/4*
polylog(3,exp(2*I*x))

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Rubi [A]  time = 0.18, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.500, Rules used = {30, 2555, 12, 6688, 14, 3717, 2190, 2531, 2282, 6589, 2309, 2178} \[ \frac {1}{2} i x \text {PolyLog}\left (2,e^{2 i x}\right )-\frac {1}{4} \text {PolyLog}\left (3,e^{2 i x}\right )-\frac {1}{2} \text {Ei}(2 \log (x))+\frac {i x^3}{6}-\frac {1}{2} x^2 \log \left (1-e^{2 i x}\right )+\frac {1}{2} x^2 \log (\log (x) \sin (x)) \]

Antiderivative was successfully verified.

[In]

Int[x*Log[Log[x]*Sin[x]],x]

[Out]

(I/6)*x^3 - ExpIntegralEi[2*Log[x]]/2 - (x^2*Log[1 - E^((2*I)*x)])/2 + (x^2*Log[Log[x]*Sin[x]])/2 + (I/2)*x*Po
lyLog[2, E^((2*I)*x)] - PolyLog[3, E^((2*I)*x)]/4

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2555

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[w*Simplify
[D[u, x]/u], x], x] /; InverseFunctionFreeQ[w, x]] /; ProductQ[u]

Rule 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*
(m + 1)), x] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {align*} \int x \log (\log (x) \sin (x)) \, dx &=\frac {1}{2} x^2 \log (\log (x) \sin (x))-\int \frac {x (1+x \cot (x) \log (x))}{2 \log (x)} \, dx\\ &=\frac {1}{2} x^2 \log (\log (x) \sin (x))-\frac {1}{2} \int \frac {x (1+x \cot (x) \log (x))}{\log (x)} \, dx\\ &=\frac {1}{2} x^2 \log (\log (x) \sin (x))-\frac {1}{2} \int x \left (x \cot (x)+\frac {1}{\log (x)}\right ) \, dx\\ &=\frac {1}{2} x^2 \log (\log (x) \sin (x))-\frac {1}{2} \int \left (x^2 \cot (x)+\frac {x}{\log (x)}\right ) \, dx\\ &=\frac {1}{2} x^2 \log (\log (x) \sin (x))-\frac {1}{2} \int x^2 \cot (x) \, dx-\frac {1}{2} \int \frac {x}{\log (x)} \, dx\\ &=\frac {i x^3}{6}+\frac {1}{2} x^2 \log (\log (x) \sin (x))+i \int \frac {e^{2 i x} x^2}{1-e^{2 i x}} \, dx-\frac {1}{2} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=\frac {i x^3}{6}-\frac {1}{2} \text {Ei}(2 \log (x))-\frac {1}{2} x^2 \log \left (1-e^{2 i x}\right )+\frac {1}{2} x^2 \log (\log (x) \sin (x))+\int x \log \left (1-e^{2 i x}\right ) \, dx\\ &=\frac {i x^3}{6}-\frac {1}{2} \text {Ei}(2 \log (x))-\frac {1}{2} x^2 \log \left (1-e^{2 i x}\right )+\frac {1}{2} x^2 \log (\log (x) \sin (x))+\frac {1}{2} i x \text {Li}_2\left (e^{2 i x}\right )-\frac {1}{2} i \int \text {Li}_2\left (e^{2 i x}\right ) \, dx\\ &=\frac {i x^3}{6}-\frac {1}{2} \text {Ei}(2 \log (x))-\frac {1}{2} x^2 \log \left (1-e^{2 i x}\right )+\frac {1}{2} x^2 \log (\log (x) \sin (x))+\frac {1}{2} i x \text {Li}_2\left (e^{2 i x}\right )-\frac {1}{4} \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{2 i x}\right )\\ &=\frac {i x^3}{6}-\frac {1}{2} \text {Ei}(2 \log (x))-\frac {1}{2} x^2 \log \left (1-e^{2 i x}\right )+\frac {1}{2} x^2 \log (\log (x) \sin (x))+\frac {1}{2} i x \text {Li}_2\left (e^{2 i x}\right )-\frac {1}{4} \text {Li}_3\left (e^{2 i x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 79, normalized size = 0.99 \[ \frac {1}{48} \left (-24 \text {Ei}(2 \log (x))-24 i x \text {Li}_2\left (e^{-2 i x}\right )-12 \text {Li}_3\left (e^{-2 i x}\right )-8 i x^3-24 x^2 \log \left (1-e^{-2 i x}\right )+24 x^2 \log (\log (x) \sin (x))+i \pi ^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*Log[Log[x]*Sin[x]],x]

[Out]

(I*Pi^3 - (8*I)*x^3 - 24*ExpIntegralEi[2*Log[x]] - 24*x^2*Log[1 - E^((-2*I)*x)] + 24*x^2*Log[Log[x]*Sin[x]] -
(24*I)*x*PolyLog[2, E^((-2*I)*x)] - 12*PolyLog[3, E^((-2*I)*x)])/48

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fricas [C]  time = 0.47, size = 174, normalized size = 2.18 \[ \frac {1}{2} \, x^{2} \log \left (\log \relax (x) \sin \relax (x)\right ) - \frac {1}{4} \, x^{2} \log \left (\cos \relax (x) + i \, \sin \relax (x) + 1\right ) - \frac {1}{4} \, x^{2} \log \left (\cos \relax (x) - i \, \sin \relax (x) + 1\right ) - \frac {1}{4} \, x^{2} \log \left (-\cos \relax (x) + i \, \sin \relax (x) + 1\right ) - \frac {1}{4} \, x^{2} \log \left (-\cos \relax (x) - i \, \sin \relax (x) + 1\right ) + \frac {1}{2} i \, x {\rm Li}_2\left (\cos \relax (x) + i \, \sin \relax (x)\right ) - \frac {1}{2} i \, x {\rm Li}_2\left (\cos \relax (x) - i \, \sin \relax (x)\right ) - \frac {1}{2} i \, x {\rm Li}_2\left (-\cos \relax (x) + i \, \sin \relax (x)\right ) + \frac {1}{2} i \, x {\rm Li}_2\left (-\cos \relax (x) - i \, \sin \relax (x)\right ) - \frac {1}{2} \, \operatorname {log\_integral}\left (x^{2}\right ) - \frac {1}{2} \, {\rm polylog}\left (3, \cos \relax (x) + i \, \sin \relax (x)\right ) - \frac {1}{2} \, {\rm polylog}\left (3, \cos \relax (x) - i \, \sin \relax (x)\right ) - \frac {1}{2} \, {\rm polylog}\left (3, -\cos \relax (x) + i \, \sin \relax (x)\right ) - \frac {1}{2} \, {\rm polylog}\left (3, -\cos \relax (x) - i \, \sin \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*log(log(x)*sin(x)) - 1/4*x^2*log(cos(x) + I*sin(x) + 1) - 1/4*x^2*log(cos(x) - I*sin(x) + 1) - 1/4*x^2
*log(-cos(x) + I*sin(x) + 1) - 1/4*x^2*log(-cos(x) - I*sin(x) + 1) + 1/2*I*x*dilog(cos(x) + I*sin(x)) - 1/2*I*
x*dilog(cos(x) - I*sin(x)) - 1/2*I*x*dilog(-cos(x) + I*sin(x)) + 1/2*I*x*dilog(-cos(x) - I*sin(x)) - 1/2*log_i
ntegral(x^2) - 1/2*polylog(3, cos(x) + I*sin(x)) - 1/2*polylog(3, cos(x) - I*sin(x)) - 1/2*polylog(3, -cos(x)
+ I*sin(x)) - 1/2*polylog(3, -cos(x) - I*sin(x))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \log \left (\log \relax (x) \sin \relax (x)\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*log(log(x)*sin(x)), x)

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maple [F]  time = 1.33, size = 0, normalized size = 0.00 \[ \int x \ln \left (\ln \relax (x ) \sin \relax (x )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*ln(ln(x)*sin(x)),x)

[Out]

int(x*ln(ln(x)*sin(x)),x)

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maxima [A]  time = 1.92, size = 70, normalized size = 0.88 \[ \frac {1}{12} \, {\left (3 i \, \pi - 6 \, \log \relax (2)\right )} x^{2} - \frac {1}{3} i \, x^{3} + \frac {1}{2} \, x^{2} \log \left (\log \relax (x)\right ) + i \, x {\rm Li}_2\left (-e^{\left (i \, x\right )}\right ) + i \, x {\rm Li}_2\left (e^{\left (i \, x\right )}\right ) - \frac {1}{2} \, {\rm Ei}\left (2 \, \log \relax (x)\right ) - {\rm Li}_{3}(-e^{\left (i \, x\right )}) - {\rm Li}_{3}(e^{\left (i \, x\right )}) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*I*pi - 6*log(2))*x^2 - 1/3*I*x^3 + 1/2*x^2*log(log(x)) + I*x*dilog(-e^(I*x)) + I*x*dilog(e^(I*x)) - 1/
2*Ei(2*log(x)) - polylog(3, -e^(I*x)) - polylog(3, e^(I*x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\ln \left (\ln \relax (x)\,\sin \relax (x)\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*log(log(x)*sin(x)),x)

[Out]

int(x*log(log(x)*sin(x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*ln(ln(x)*sin(x)),x)

[Out]

Integral(x*log(log(x)*sin(x)), x)

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