\(\int x \log (e^x \log (x) \sin (x)) \, dx\) [311]

Optimal result

Integrand size = 11, antiderivative size = 85 \[ \int x \log \left (e^x \log (x) \sin (x)\right ) \, dx=\left (-\frac {1}{6}+\frac {i}{6}\right ) x^3-\frac {1}{2} \operatorname {ExpIntegralEi}(2 \log (x))-\frac {1}{2} x^2 \log \left (1-e^{2 i x}\right )+\frac {1}{2} x^2 \log \left (e^x \log (x) \sin (x)\right )+\frac {1}{2} i x \operatorname {PolyLog}\left (2,e^{2 i x}\right )-\frac {1}{4} \operatorname {PolyLog}\left (3,e^{2 i x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {30, 2635, 12, 14, 3798, 2221, 2611, 2320, 6724, 2346, 2209} \[ \int x \log \left (e^x \log (x) \sin (x)\right ) \, dx=-\frac {1}{2} \operatorname {ExpIntegralEi}(2 \log (x))+\frac {1}{2} i x \operatorname {PolyLog}\left (2,e^{2 i x}\right )-\frac {1}{4} \operatorname {PolyLog}\left (3,e^{2 i x}\right )+\left (-\frac {1}{6}+\frac {i}{6}\right ) x^3-\frac {1}{2} x^2 \log \left (1-e^{2 i x}\right )+\frac {1}{2} x^2 \log \left (e^x \log (x) \sin (x)\right ) \]

[In]

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

[Out]

(-1/6 + I/6)*x^3 - ExpIntegralEi[2*Log[x]]/2 - (x^2*Log[1 - E^((2*I)*x)])/2 + (x^2*Log[E^x*Log[x]*Sin[x]])/2 +
(I/2)*x*PolyLog[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 2209

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

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2320

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 2346

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 2611

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 2635

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 3798

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 6724

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]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.96 \[ \int x \log \left (e^x \log (x) \sin (x)\right ) \, dx=\frac {1}{48} \left (i \pi ^3-(8+8 i) x^3-24 \operatorname {ExpIntegralEi}(2 \log (x))-24 x^2 \log \left (1-e^{-2 i x}\right )+24 x^2 \log \left (e^x \log (x) \sin (x)\right )-24 i x \operatorname {PolyLog}\left (2,e^{-2 i x}\right )-12 \operatorname {PolyLog}\left (3,e^{-2 i x}\right )\right ) \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 1.65 (sec) , antiderivative size = 615, normalized size of antiderivative = 7.24

method	result	size
risch	\(\text {Expression too large to display}\)	\(615\)

[In]

int(x*ln(exp(x)*ln(x)*sin(x)),x,method=_RETURNVERBOSE)

[Out]

-1/2*x^2*ln(exp(I*x))+1/2*x^2*ln(exp(2*I*x)-1)-1/2*x^2*ln(exp(I*x)+1)+I*x*polylog(2,-exp(I*x))-polylog(3,-exp(
I*x))-1/2*x^2*ln(1-exp(I*x))+I*x*polylog(2,exp(I*x))-polylog(3,exp(I*x))+1/2*ln(exp(x))*x^2-1/6*x^3+1/2*ln(ln(
x))*x^2+1/2*Ei(1,-2*ln(x))+1/4*(-I*Pi*csgn(I*ln(x)*(exp((1+I)*x)-exp((1-I)*x)))^3-I*Pi*csgn((exp((1+I)*x)-exp(
(1-I)*x))*ln(x))^3+I*Pi*csgn(I*ln(x)*(exp((1+I)*x)-exp((1-I)*x)))*csgn((exp((1+I)*x)-exp((1-I)*x))*ln(x))^2+I*
Pi*csgn(I*exp(x))*csgn(ln(x)*sin(x))*csgn(I*ln(x)*(exp((1+I)*x)-exp((1-I)*x)))+I*Pi*csgn((exp((1+I)*x)-exp((1-
I)*x))*ln(x))^2+I*Pi*csgn(I*exp(-I*x))*csgn(I*ln(x)*(exp(2*I*x)-1))*csgn(ln(x)*sin(x))+I*Pi*csgn(I*ln(x))*csgn
(I*ln(x)*(exp(2*I*x)-1))^2-I*Pi*csgn(I*(exp(2*I*x)-1))*csgn(I*ln(x))*csgn(I*ln(x)*(exp(2*I*x)-1))+I*Pi*csgn(I*
ln(x)*(exp(2*I*x)-1))*csgn(ln(x)*sin(x))^2+I*Pi*csgn(I*exp(x))*csgn(I*ln(x)*(exp((1+I)*x)-exp((1-I)*x)))^2-I*P
i*csgn(I*ln(x)*(exp(2*I*x)-1))^3-I*Pi*csgn(ln(x)*sin(x))*csgn(I*ln(x)*(exp((1+I)*x)-exp((1-I)*x)))^2+I*Pi*csgn
(ln(x)*sin(x))^3+I*Pi*csgn(I*exp(-I*x))*csgn(ln(x)*sin(x))^2-I*Pi*csgn(I*ln(x)*(exp((1+I)*x)-exp((1-I)*x)))*cs
gn((exp((1+I)*x)-exp((1-I)*x))*ln(x))-I*Pi+I*Pi*csgn(I*(exp(2*I*x)-1))*csgn(I*ln(x)*(exp(2*I*x)-1))^2-2*ln(2))
*x^2+1/6*I*x^3

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 181 vs. \(2 (56) = 112\).

Time = 0.32 (sec) , antiderivative size = 181, normalized size of antiderivative = 2.13 \[ \int x \log \left (e^x \log (x) \sin (x)\right ) \, dx=-\frac {1}{6} \, x^{3} + \frac {1}{2} \, x^{2} \log \left (e^{x} \log \left (x\right ) \sin \left (x\right )\right ) - \frac {1}{4} \, x^{2} \log \left (\cos \left (x\right ) + i \, \sin \left (x\right ) + 1\right ) - \frac {1}{4} \, x^{2} \log \left (\cos \left (x\right ) - i \, \sin \left (x\right ) + 1\right ) - \frac {1}{4} \, x^{2} \log \left (-\cos \left (x\right ) + i \, \sin \left (x\right ) + 1\right ) - \frac {1}{4} \, x^{2} \log \left (-\cos \left (x\right ) - i \, \sin \left (x\right ) + 1\right ) + \frac {1}{2} i \, x {\rm Li}_2\left (\cos \left (x\right ) + i \, \sin \left (x\right )\right ) - \frac {1}{2} i \, x {\rm Li}_2\left (\cos \left (x\right ) - i \, \sin \left (x\right )\right ) - \frac {1}{2} i \, x {\rm Li}_2\left (-\cos \left (x\right ) + i \, \sin \left (x\right )\right ) + \frac {1}{2} i \, x {\rm Li}_2\left (-\cos \left (x\right ) - i \, \sin \left (x\right )\right ) - \frac {1}{2} \, \operatorname {log\_integral}\left (x^{2}\right ) - \frac {1}{2} \, {\rm polylog}\left (3, \cos \left (x\right ) + i \, \sin \left (x\right )\right ) - \frac {1}{2} \, {\rm polylog}\left (3, \cos \left (x\right ) - i \, \sin \left (x\right )\right ) - \frac {1}{2} \, {\rm polylog}\left (3, -\cos \left (x\right ) + i \, \sin \left (x\right )\right ) - \frac {1}{2} \, {\rm polylog}\left (3, -\cos \left (x\right ) - i \, \sin \left (x\right )\right ) \]

[In]

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

[Out]

-1/6*x^3 + 1/2*x^2*log(e^x*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*s
in(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_integral(x^2) - 1/2*polylog(3, cos(x) + I*sin(x)) - 1/2*polylog(3, cos(x) - I*sin(x)) - 1/2*poly
log(3, -cos(x) + I*sin(x)) - 1/2*polylog(3, -cos(x) - I*sin(x))

Sympy [F]

\[ \int x \log \left (e^x \log (x) \sin (x)\right ) \, dx=\int x \log {\left (e^{x} \log {\left (x \right )} \sin {\left (x \right )} \right )}\, dx \]

[In]

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

[Out]

Integral(x*log(exp(x)*log(x)*sin(x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.82 \[ \int x \log \left (e^x \log (x) \sin (x)\right ) \, dx=-\frac {1}{4} \, {\left (-i \, \pi + 2 \, \log \left (2\right )\right )} x^{2} - \left (\frac {1}{3} i - \frac {1}{3}\right ) \, x^{3} + \frac {1}{2} \, x^{2} \log \left (\log \left (x\right )\right ) + i \, x {\rm Li}_2\left (-e^{\left (i \, x\right )}\right ) + i \, x {\rm Li}_2\left (e^{\left (i \, x\right )}\right ) - \frac {1}{2} \, {\rm Ei}\left (2 \, \log \left (x\right )\right ) - {\rm Li}_{3}(-e^{\left (i \, x\right )}) - {\rm Li}_{3}(e^{\left (i \, x\right )}) \]

[In]

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

[Out]

-1/4*(-I*pi + 2*log(2))*x^2 - (1/3*I - 1/3)*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))

Giac [F]

\[ \int x \log \left (e^x \log (x) \sin (x)\right ) \, dx=\int { x \log \left (e^{x} \log \left (x\right ) \sin \left (x\right )\right ) \,d x } \]

[In]

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

[Out]

integrate(x*log(e^x*log(x)*sin(x)), x)

Mupad [F(-1)]

Timed out. \[ \int x \log \left (e^x \log (x) \sin (x)\right ) \, dx=\int x\,\ln \left ({\mathrm {e}}^x\,\ln \left (x\right )\,\sin \left (x\right )\right ) \,d x \]

[In]

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

[Out]

int(x*log(exp(x)*log(x)*sin(x)), x)