\(\int x^2 \log (e^x \log (x) \sin (x)) \, dx\) [310]

Optimal result

Integrand size = 13, antiderivative size = 103 \[ \int x^2 \log \left (e^x \log (x) \sin (x)\right ) \, dx=\left (-\frac {1}{12}+\frac {i}{12}\right ) x^4-\frac {1}{3} \operatorname {ExpIntegralEi}(3 \log (x))-\frac {1}{3} x^3 \log \left (1-e^{2 i x}\right )+\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )+\frac {1}{2} i x^2 \operatorname {PolyLog}\left (2,e^{2 i x}\right )-\frac {1}{2} x \operatorname {PolyLog}\left (3,e^{2 i x}\right )-\frac {1}{4} i \operatorname {PolyLog}\left (4,e^{2 i x}\right ) \]

[Out]

(-1/12+1/12*I)*x^4-1/3*Ei(3*ln(x))-1/3*x^3*ln(1-exp(2*I*x))+1/3*x^3*ln(exp(x)*ln(x)*sin(x))+1/2*I*x^2*polylog(
2,exp(2*I*x))-1/2*x*polylog(3,exp(2*I*x))-1/4*I*polylog(4,exp(2*I*x))

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.923, Rules used = {30, 2635, 12, 14, 3798, 2221, 2611, 6744, 2320, 6724, 2346, 2209} \[ \int x^2 \log \left (e^x \log (x) \sin (x)\right ) \, dx=-\frac {1}{3} \operatorname {ExpIntegralEi}(3 \log (x))+\frac {1}{2} i x^2 \operatorname {PolyLog}\left (2,e^{2 i x}\right )-\frac {1}{2} x \operatorname {PolyLog}\left (3,e^{2 i x}\right )-\frac {1}{4} i \operatorname {PolyLog}\left (4,e^{2 i x}\right )+\left (-\frac {1}{12}+\frac {i}{12}\right ) x^4-\frac {1}{3} x^3 \log \left (1-e^{2 i x}\right )+\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right ) \]

[In]

Int[x^2*Log[E^x*Log[x]*Sin[x]],x]

[Out]

(-1/12 + I/12)*x^4 - ExpIntegralEi[3*Log[x]]/3 - (x^3*Log[1 - E^((2*I)*x)])/3 + (x^3*Log[E^x*Log[x]*Sin[x]])/3
+ (I/2)*x^2*PolyLog[2, E^((2*I)*x)] - (x*PolyLog[3, E^((2*I)*x)])/2 - (I/4)*PolyLog[4, E^((2*I)*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 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]

Rule 6744

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )-\int \frac {1}{3} x^3 \left (1+\cot (x)+\frac {1}{x \log (x)}\right ) \, dx \\ & = \frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )-\frac {1}{3} \int x^3 \left (1+\cot (x)+\frac {1}{x \log (x)}\right ) \, dx \\ & = \frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )-\frac {1}{3} \int \left (x^3 (1+\cot (x))+\frac {x^2}{\log (x)}\right ) \, dx \\ & = \frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )-\frac {1}{3} \int x^3 (1+\cot (x)) \, dx-\frac {1}{3} \int \frac {x^2}{\log (x)} \, dx \\ & = \frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )-\frac {1}{3} \int \left (x^3+x^3 \cot (x)\right ) \, dx-\frac {1}{3} \text {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right ) \\ & = -\frac {x^4}{12}-\frac {1}{3} \text {Ei}(3 \log (x))+\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )-\frac {1}{3} \int x^3 \cot (x) \, dx \\ & = \left (-\frac {1}{12}+\frac {i}{12}\right ) x^4-\frac {1}{3} \text {Ei}(3 \log (x))+\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )+\frac {2}{3} i \int \frac {e^{2 i x} x^3}{1-e^{2 i x}} \, dx \\ & = \left (-\frac {1}{12}+\frac {i}{12}\right ) x^4-\frac {1}{3} \text {Ei}(3 \log (x))-\frac {1}{3} x^3 \log \left (1-e^{2 i x}\right )+\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )+\int x^2 \log \left (1-e^{2 i x}\right ) \, dx \\ & = \left (-\frac {1}{12}+\frac {i}{12}\right ) x^4-\frac {1}{3} \text {Ei}(3 \log (x))-\frac {1}{3} x^3 \log \left (1-e^{2 i x}\right )+\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )+\frac {1}{2} i x^2 \text {Li}_2\left (e^{2 i x}\right )-i \int x \text {Li}_2\left (e^{2 i x}\right ) \, dx \\ & = \left (-\frac {1}{12}+\frac {i}{12}\right ) x^4-\frac {1}{3} \text {Ei}(3 \log (x))-\frac {1}{3} x^3 \log \left (1-e^{2 i x}\right )+\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )+\frac {1}{2} i x^2 \text {Li}_2\left (e^{2 i x}\right )-\frac {1}{2} x \text {Li}_3\left (e^{2 i x}\right )+\frac {1}{2} \int \text {Li}_3\left (e^{2 i x}\right ) \, dx \\ & = \left (-\frac {1}{12}+\frac {i}{12}\right ) x^4-\frac {1}{3} \text {Ei}(3 \log (x))-\frac {1}{3} x^3 \log \left (1-e^{2 i x}\right )+\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )+\frac {1}{2} i x^2 \text {Li}_2\left (e^{2 i x}\right )-\frac {1}{2} x \text {Li}_3\left (e^{2 i x}\right )-\frac {1}{4} i \text {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{2 i x}\right ) \\ & = \left (-\frac {1}{12}+\frac {i}{12}\right ) x^4-\frac {1}{3} \text {Ei}(3 \log (x))-\frac {1}{3} x^3 \log \left (1-e^{2 i x}\right )+\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )+\frac {1}{2} i x^2 \text {Li}_2\left (e^{2 i x}\right )-\frac {1}{2} x \text {Li}_3\left (e^{2 i x}\right )-\frac {1}{4} i \text {Li}_4\left (e^{2 i x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.97 \[ \int x^2 \log \left (e^x \log (x) \sin (x)\right ) \, dx=\frac {1}{192} i \left (\pi ^4-(16-16 i) x^4+64 i \operatorname {ExpIntegralEi}(3 \log (x))+64 i x^3 \log \left (1-e^{-2 i x}\right )-64 i x^3 \log \left (e^x \log (x) \sin (x)\right )-96 x^2 \operatorname {PolyLog}\left (2,e^{-2 i x}\right )+96 i x \operatorname {PolyLog}\left (3,e^{-2 i x}\right )+48 \operatorname {PolyLog}\left (4,e^{-2 i x}\right )\right ) \]

[In]

Integrate[x^2*Log[E^x*Log[x]*Sin[x]],x]

[Out]

(I/192)*(Pi^4 - (16 - 16*I)*x^4 + (64*I)*ExpIntegralEi[3*Log[x]] + (64*I)*x^3*Log[1 - E^((-2*I)*x)] - (64*I)*x
^3*Log[E^x*Log[x]*Sin[x]] - 96*x^2*PolyLog[2, E^((-2*I)*x)] + (96*I)*x*PolyLog[3, E^((-2*I)*x)] + 48*PolyLog[4
, E^((-2*I)*x)])

Maple [C] (warning: unable to verify)

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

Time = 2.03 (sec) , antiderivative size = 643, normalized size of antiderivative = 6.24

method	result	size
risch	\(\text {Expression too large to display}\)	\(643\)

[In]

int(x^2*ln(exp(x)*ln(x)*sin(x)),x,method=_RETURNVERBOSE)

[Out]

-1/3*x^3*ln(exp(I*x))+1/6*(-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*cs
gn(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*Pi*csg
n(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)))*csgn((e
xp((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^3+
1/3*x^3*ln(exp(2*I*x)-1)-1/3*x^3*ln(exp(I*x)+1)+1/12*I*x^4-2*x*polylog(3,-exp(I*x))-2*I*polylog(4,exp(I*x))-1/
3*x^3*ln(1-exp(I*x))+I*x^2*polylog(2,-exp(I*x))-2*x*polylog(3,exp(I*x))-2*I*polylog(4,-exp(I*x))+1/3*x^3*ln(ex
p(x))-1/12*x^4+1/3*x^3*ln(ln(x))+1/3*Ei(1,-3*ln(x))+I*x^2*polylog(2,exp(I*x))

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. 241 vs. \(2 (67) = 134\).

Time = 0.32 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.34 \[ \int x^2 \log \left (e^x \log (x) \sin (x)\right ) \, dx=-\frac {1}{12} \, x^{4} + \frac {1}{3} \, x^{3} \log \left (e^{x} \log \left (x\right ) \sin \left (x\right )\right ) - \frac {1}{6} \, x^{3} \log \left (\cos \left (x\right ) + i \, \sin \left (x\right ) + 1\right ) - \frac {1}{6} \, x^{3} \log \left (\cos \left (x\right ) - i \, \sin \left (x\right ) + 1\right ) - \frac {1}{6} \, x^{3} \log \left (-\cos \left (x\right ) + i \, \sin \left (x\right ) + 1\right ) - \frac {1}{6} \, x^{3} \log \left (-\cos \left (x\right ) - i \, \sin \left (x\right ) + 1\right ) + \frac {1}{2} i \, x^{2} {\rm Li}_2\left (\cos \left (x\right ) + i \, \sin \left (x\right )\right ) - \frac {1}{2} i \, x^{2} {\rm Li}_2\left (\cos \left (x\right ) - i \, \sin \left (x\right )\right ) - \frac {1}{2} i \, x^{2} {\rm Li}_2\left (-\cos \left (x\right ) + i \, \sin \left (x\right )\right ) + \frac {1}{2} i \, x^{2} {\rm Li}_2\left (-\cos \left (x\right ) - i \, \sin \left (x\right )\right ) - x {\rm polylog}\left (3, \cos \left (x\right ) + i \, \sin \left (x\right )\right ) - x {\rm polylog}\left (3, \cos \left (x\right ) - i \, \sin \left (x\right )\right ) - x {\rm polylog}\left (3, -\cos \left (x\right ) + i \, \sin \left (x\right )\right ) - x {\rm polylog}\left (3, -\cos \left (x\right ) - i \, \sin \left (x\right )\right ) - \frac {1}{3} \, \operatorname {log\_integral}\left (x^{3}\right ) - i \, {\rm polylog}\left (4, \cos \left (x\right ) + i \, \sin \left (x\right )\right ) + i \, {\rm polylog}\left (4, \cos \left (x\right ) - i \, \sin \left (x\right )\right ) + i \, {\rm polylog}\left (4, -\cos \left (x\right ) + i \, \sin \left (x\right )\right ) - i \, {\rm polylog}\left (4, -\cos \left (x\right ) - i \, \sin \left (x\right )\right ) \]

[In]

integrate(x^2*log(exp(x)*log(x)*sin(x)),x, algorithm="fricas")

[Out]

-1/12*x^4 + 1/3*x^3*log(e^x*log(x)*sin(x)) - 1/6*x^3*log(cos(x) + I*sin(x) + 1) - 1/6*x^3*log(cos(x) - I*sin(x
) + 1) - 1/6*x^3*log(-cos(x) + I*sin(x) + 1) - 1/6*x^3*log(-cos(x) - I*sin(x) + 1) + 1/2*I*x^2*dilog(cos(x) +
I*sin(x)) - 1/2*I*x^2*dilog(cos(x) - I*sin(x)) - 1/2*I*x^2*dilog(-cos(x) + I*sin(x)) + 1/2*I*x^2*dilog(-cos(x)
- I*sin(x)) - x*polylog(3, cos(x) + I*sin(x)) - x*polylog(3, cos(x) - I*sin(x)) - x*polylog(3, -cos(x) + I*si
n(x)) - x*polylog(3, -cos(x) - I*sin(x)) - 1/3*log_integral(x^3) - I*polylog(4, cos(x) + I*sin(x)) + I*polylog
(4, cos(x) - I*sin(x)) + I*polylog(4, -cos(x) + I*sin(x)) - I*polylog(4, -cos(x) - I*sin(x))

Sympy [F]

\[ \int x^2 \log \left (e^x \log (x) \sin (x)\right ) \, dx=\int x^{2} \log {\left (e^{x} \log {\left (x \right )} \sin {\left (x \right )} \right )}\, dx \]

[In]

integrate(x**2*ln(exp(x)*ln(x)*sin(x)),x)

[Out]

Integral(x**2*log(exp(x)*log(x)*sin(x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.91 \[ \int x^2 \log \left (e^x \log (x) \sin (x)\right ) \, dx=-\frac {1}{6} \, {\left (-i \, \pi + 2 \, \log \left (2\right )\right )} x^{3} - \left (\frac {1}{4} i - \frac {1}{4}\right ) \, x^{4} + \frac {1}{3} \, x^{3} \log \left (\log \left (x\right )\right ) + i \, x^{2} {\rm Li}_2\left (-e^{\left (i \, x\right )}\right ) + i \, x^{2} {\rm Li}_2\left (e^{\left (i \, x\right )}\right ) - 2 \, x {\rm Li}_{3}(-e^{\left (i \, x\right )}) - 2 \, x {\rm Li}_{3}(e^{\left (i \, x\right )}) - \frac {1}{3} \, {\rm Ei}\left (3 \, \log \left (x\right )\right ) - 2 i \, {\rm Li}_{4}(-e^{\left (i \, x\right )}) - 2 i \, {\rm Li}_{4}(e^{\left (i \, x\right )}) \]

[In]

integrate(x^2*log(exp(x)*log(x)*sin(x)),x, algorithm="maxima")

[Out]

-1/6*(-I*pi + 2*log(2))*x^3 - (1/4*I - 1/4)*x^4 + 1/3*x^3*log(log(x)) + I*x^2*dilog(-e^(I*x)) + I*x^2*dilog(e^
(I*x)) - 2*x*polylog(3, -e^(I*x)) - 2*x*polylog(3, e^(I*x)) - 1/3*Ei(3*log(x)) - 2*I*polylog(4, -e^(I*x)) - 2*
I*polylog(4, e^(I*x))

Giac [F]

\[ \int x^2 \log \left (e^x \log (x) \sin (x)\right ) \, dx=\int { x^{2} \log \left (e^{x} \log \left (x\right ) \sin \left (x\right )\right ) \,d x } \]

[In]

integrate(x^2*log(exp(x)*log(x)*sin(x)),x, algorithm="giac")

[Out]

integrate(x^2*log(e^x*log(x)*sin(x)), x)

Mupad [F(-1)]

Timed out. \[ \int x^2 \log \left (e^x \log (x) \sin (x)\right ) \, dx=\int x^2\,\ln \left ({\mathrm {e}}^x\,\ln \left (x\right )\,\sin \left (x\right )\right ) \,d x \]

[In]

int(x^2*log(exp(x)*log(x)*sin(x)),x)

[Out]

int(x^2*log(exp(x)*log(x)*sin(x)), x)