Integrand size = 13, antiderivative size = 103 \[ \int x^2 \log \left (e^x \log (x) \sin (x)\right ) \, dx=\left (-\frac {1}{12}+\frac {i}{12}\right ) x^4-\frac {1}{3} \operatorname {ExpIntegralEi}(3 \log (x))-\frac {1}{3} x^3 \log \left (1-e^{2 i x}\right )+\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )+\frac {1}{2} i x^2 \operatorname {PolyLog}\left (2,e^{2 i x}\right )-\frac {1}{2} x \operatorname {PolyLog}\left (3,e^{2 i x}\right )-\frac {1}{4} i \operatorname {PolyLog}\left (4,e^{2 i x}\right ) \] Output:
(-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))
Time = 0.09 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.97 \[ \int x^2 \log \left (e^x \log (x) \sin (x)\right ) \, dx=\frac {1}{192} i \left (\pi ^4-(16-16 i) x^4+64 i \operatorname {ExpIntegralEi}(3 \log (x))+64 i x^3 \log \left (1-e^{-2 i x}\right )-64 i x^3 \log \left (e^x \log (x) \sin (x)\right )-96 x^2 \operatorname {PolyLog}\left (2,e^{-2 i x}\right )+96 i x \operatorname {PolyLog}\left (3,e^{-2 i x}\right )+48 \operatorname {PolyLog}\left (4,e^{-2 i x}\right )\right ) \] Input:
Integrate[x^2*Log[E^x*Log[x]*Sin[x]],x]
Output:
(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*Pol yLog[2, E^((-2*I)*x)] + (96*I)*x*PolyLog[3, E^((-2*I)*x)] + 48*PolyLog[4, E^((-2*I)*x)])
Time = 0.44 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.01, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3035, 27, 2010, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^2 \log \left (e^x \log (x) \sin (x)\right ) \, dx\) |
\(\Big \downarrow \) 3035 |
\(\displaystyle \frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )-\int \frac {1}{3} x^3 \left (\cot (x)+\frac {1}{x \log (x)}+1\right )dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )-\frac {1}{3} \int x^3 \left (\cot (x)+\frac {1}{x \log (x)}+1\right )dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )-\frac {1}{3} \int \left ((\cot (x)+1) x^3+\frac {x^2}{\log (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )+\frac {1}{3} \left (-\operatorname {ExpIntegralEi}(3 \log (x))+\frac {3}{2} i x^2 \operatorname {PolyLog}\left (2,e^{2 i x}\right )-\frac {3}{2} x \operatorname {PolyLog}\left (3,e^{2 i x}\right )-\frac {3}{4} i \operatorname {PolyLog}\left (4,e^{2 i x}\right )+\left (-\frac {1}{4}+\frac {i}{4}\right ) x^4-x^3 \log \left (1-e^{2 i x}\right )\right )\) |
Input:
Int[x^2*Log[E^x*Log[x]*Sin[x]],x]
Output:
(x^3*Log[E^x*Log[x]*Sin[x]])/3 + ((-1/4 + I/4)*x^4 - ExpIntegralEi[3*Log[x ]] - x^3*Log[1 - E^((2*I)*x)] + ((3*I)/2)*x^2*PolyLog[2, E^((2*I)*x)] - (3 *x*PolyLog[3, E^((2*I)*x)])/2 - ((3*I)/4)*PolyLog[4, E^((2*I)*x)])/3
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]]
Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Simp[Log[u] w, x ] - Int[SimplifyIntegrand[w*Simplify[D[u, x]/u], x], x] /; InverseFunctionF reeQ[w, x]] /; ProductQ[u]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.28 (sec) , antiderivative size = 643, normalized size of antiderivative = 6.24
Input:
int(x^2*ln(exp(x)*ln(x)*sin(x)),x,method=_RETURNVERBOSE)
Output:
-1/3*x^3*ln(exp(I*x))+1/6*(I*Pi*csgn(I*exp(x))*csgn(I*ln(x)*(exp((1+I)*x)- exp((1-I)*x)))^2-I*Pi*csgn(I*ln(x)*(exp((1+I)*x)-exp((1-I)*x)))*csgn(ln(x) *(exp((1+I)*x)-exp((1-I)*x)))-I*Pi*csgn(ln(x)*(exp((1+I)*x)-exp((1-I)*x))) ^3+I*Pi*csgn(I*exp(-I*x))*csgn(I*ln(x)*(exp(2*I*x)-1))*csgn(ln(x)*sin(x))+ I*Pi*csgn(ln(x)*sin(x))^3-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))*csgn(I*ln(x)*(exp(2*I*x)-1))^2+I *Pi*csgn(I*(exp(2*I*x)-1))*csgn(I*ln(x)*(exp(2*I*x)-1))^2-I*Pi*csgn(I*ln(x )*(exp((1+I)*x)-exp((1-I)*x)))^3+I*Pi*csgn(I*ln(x)*(exp(2*I*x)-1))*csgn(ln (x)*sin(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(ln(x)*sin(x))*csgn(I*ln(x)*(exp((1+I)*x)-ex p((1-I)*x)))^2-I*Pi*csgn(I*ln(x)*(exp(2*I*x)-1))^3-I*Pi+I*Pi*csgn(ln(x)*(e xp((1+I)*x)-exp((1-I)*x)))^2+I*Pi*csgn(I*ln(x)*(exp((1+I)*x)-exp((1-I)*x)) )*csgn(ln(x)*(exp((1+I)*x)-exp((1-I)*x)))^2+I*Pi*csgn(I*exp(-I*x))*csgn(ln (x)*sin(x))^2-2*ln(2))*x^3+1/3*x^3*ln(exp(2*I*x)-1)-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(exp(I*x)+1)-2*I*polylog(4,-exp(I*x))-2*x*polylog(3,-exp(I*x))+ 1/12*I*x^4+1/3*x^3*ln(exp(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))
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 = 0.11 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.34 \[ \int x^2 \log \left (e^x \log (x) \sin (x)\right ) \, dx=-\frac {1}{12} \, x^{4} + \frac {1}{3} \, x^{3} \log \left (e^{x} \log \left (x\right ) \sin \left (x\right )\right ) - \frac {1}{6} \, x^{3} \log \left (\cos \left (x\right ) + i \, \sin \left (x\right ) + 1\right ) - \frac {1}{6} \, x^{3} \log \left (\cos \left (x\right ) - i \, \sin \left (x\right ) + 1\right ) - \frac {1}{6} \, x^{3} \log \left (-\cos \left (x\right ) + i \, \sin \left (x\right ) + 1\right ) - \frac {1}{6} \, x^{3} \log \left (-\cos \left (x\right ) - i \, \sin \left (x\right ) + 1\right ) + \frac {1}{2} i \, x^{2} {\rm Li}_2\left (\cos \left (x\right ) + i \, \sin \left (x\right )\right ) - \frac {1}{2} i \, x^{2} {\rm Li}_2\left (\cos \left (x\right ) - i \, \sin \left (x\right )\right ) - \frac {1}{2} i \, x^{2} {\rm Li}_2\left (-\cos \left (x\right ) + i \, \sin \left (x\right )\right ) + \frac {1}{2} i \, x^{2} {\rm Li}_2\left (-\cos \left (x\right ) - i \, \sin \left (x\right )\right ) - x {\rm polylog}\left (3, \cos \left (x\right ) + i \, \sin \left (x\right )\right ) - x {\rm polylog}\left (3, \cos \left (x\right ) - i \, \sin \left (x\right )\right ) - x {\rm polylog}\left (3, -\cos \left (x\right ) + i \, \sin \left (x\right )\right ) - x {\rm polylog}\left (3, -\cos \left (x\right ) - i \, \sin \left (x\right )\right ) - \frac {1}{3} \, \operatorname {log\_integral}\left (x^{3}\right ) - i \, {\rm polylog}\left (4, \cos \left (x\right ) + i \, \sin \left (x\right )\right ) + i \, {\rm polylog}\left (4, \cos \left (x\right ) - i \, \sin \left (x\right )\right ) + i \, {\rm polylog}\left (4, -\cos \left (x\right ) + i \, \sin \left (x\right )\right ) - i \, {\rm polylog}\left (4, -\cos \left (x\right ) - i \, \sin \left (x\right )\right ) \] Input:
integrate(x^2*log(exp(x)*log(x)*sin(x)),x, algorithm="fricas")
Output:
-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*s in(x)) - x*polylog(3, cos(x) - I*sin(x)) - x*polylog(3, -cos(x) + I*sin(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))
\[ \int x^2 \log \left (e^x \log (x) \sin (x)\right ) \, dx=\int x^{2} \log {\left (e^{x} \log {\left (x \right )} \sin {\left (x \right )} \right )}\, dx \] Input:
integrate(x**2*ln(exp(x)*ln(x)*sin(x)),x)
Output:
Integral(x**2*log(exp(x)*log(x)*sin(x)), x)
Time = 0.21 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.91 \[ \int x^2 \log \left (e^x \log (x) \sin (x)\right ) \, dx=-\frac {1}{6} \, {\left (-i \, \pi + 2 \, \log \left (2\right )\right )} x^{3} - \left (\frac {1}{4} i - \frac {1}{4}\right ) \, x^{4} + \frac {1}{3} \, x^{3} \log \left (\log \left (x\right )\right ) + i \, x^{2} {\rm Li}_2\left (-e^{\left (i \, x\right )}\right ) + i \, x^{2} {\rm Li}_2\left (e^{\left (i \, x\right )}\right ) - 2 \, x {\rm Li}_{3}(-e^{\left (i \, x\right )}) - 2 \, x {\rm Li}_{3}(e^{\left (i \, x\right )}) - \frac {1}{3} \, {\rm Ei}\left (3 \, \log \left (x\right )\right ) - 2 i \, {\rm Li}_{4}(-e^{\left (i \, x\right )}) - 2 i \, {\rm Li}_{4}(e^{\left (i \, x\right )}) \] Input:
integrate(x^2*log(exp(x)*log(x)*sin(x)),x, algorithm="maxima")
Output:
-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))
\[ \int x^2 \log \left (e^x \log (x) \sin (x)\right ) \, dx=\int { x^{2} \log \left (e^{x} \log \left (x\right ) \sin \left (x\right )\right ) \,d x } \] Input:
integrate(x^2*log(exp(x)*log(x)*sin(x)),x, algorithm="giac")
Output:
integrate(x^2*log(e^x*log(x)*sin(x)), x)
Timed out. \[ \int x^2 \log \left (e^x \log (x) \sin (x)\right ) \, dx=\int x^2\,\ln \left ({\mathrm {e}}^x\,\ln \left (x\right )\,\sin \left (x\right )\right ) \,d x \] Input:
int(x^2*log(exp(x)*log(x)*sin(x)),x)
Output:
int(x^2*log(exp(x)*log(x)*sin(x)), x)
\[ \int x^2 \log \left (e^x \log (x) \sin (x)\right ) \, dx=\int x^{2} \mathrm {log}\left ({\mathrm e}^{x} \mathrm {log}\left (x \right ) \sin \left (x \right )\right )d x \] Input:
int(x^2*log(exp(x)*log(x)*sin(x)),x)
Output:
int(x^2*log(exp(x)*log(x)*sin(x)),x)