Integrand size = 11, antiderivative size = 85 \[ \int x \log \left (e^x \log (x) \sin (x)\right ) \, dx=\left (-\frac {1}{6}+\frac {i}{6}\right ) x^3-\frac {1}{2} \operatorname {ExpIntegralEi}(2 \log (x))-\frac {1}{2} x^2 \log \left (1-e^{2 i x}\right )+\frac {1}{2} x^2 \log \left (e^x \log (x) \sin (x)\right )+\frac {1}{2} i x \operatorname {PolyLog}\left (2,e^{2 i x}\right )-\frac {1}{4} \operatorname {PolyLog}\left (3,e^{2 i x}\right ) \]
(-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,exp(2*I*x))-1/4*polylog(3,exp(2*I*x))
Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.96 \[ \int x \log \left (e^x \log (x) \sin (x)\right ) \, dx=\frac {1}{48} \left (i \pi ^3-(8+8 i) x^3-24 \operatorname {ExpIntegralEi}(2 \log (x))-24 x^2 \log \left (1-e^{-2 i x}\right )+24 x^2 \log \left (e^x \log (x) \sin (x)\right )-24 i x \operatorname {PolyLog}\left (2,e^{-2 i x}\right )-12 \operatorname {PolyLog}\left (3,e^{-2 i x}\right )\right ) \]
(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]*Sin[x]] - (24*I)*x*PolyLog[2, E^((-2*I)* x)] - 12*PolyLog[3, E^((-2*I)*x)])/48
Time = 0.32 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, 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.
\(\displaystyle \int x \log \left (e^x \log (x) \sin (x)\right ) \, dx\) |
\(\Big \downarrow \) 3035 |
\(\displaystyle \frac {1}{2} x^2 \log \left (e^x \log (x) \sin (x)\right )-\int \frac {1}{2} x^2 \left (\cot (x)+\frac {1}{x \log (x)}+1\right )dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} x^2 \log \left (e^x \log (x) \sin (x)\right )-\frac {1}{2} \int x^2 \left (\cot (x)+\frac {1}{x \log (x)}+1\right )dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \frac {1}{2} x^2 \log \left (e^x \log (x) \sin (x)\right )-\frac {1}{2} \int \left ((\cot (x)+1) x^2+\frac {x}{\log (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} x^2 \log \left (e^x \log (x) \sin (x)\right )+\frac {1}{2} \left (-\operatorname {ExpIntegralEi}(2 \log (x))+i x \operatorname {PolyLog}\left (2,e^{2 i x}\right )-\frac {1}{2} \operatorname {PolyLog}\left (3,e^{2 i x}\right )+\left (-\frac {1}{3}+\frac {i}{3}\right ) x^3-x^2 \log \left (1-e^{2 i x}\right )\right )\) |
(x^2*Log[E^x*Log[x]*Sin[x]])/2 + ((-1/3 + I/3)*x^3 - ExpIntegralEi[2*Log[x ]] - x^2*Log[1 - E^((2*I)*x)] + I*x*PolyLog[2, E^((2*I)*x)] - PolyLog[3, E ^((2*I)*x)]/2)/2
3.4.11.3.1 Defintions of rubi rules used
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 = 1.65 (sec) , antiderivative size = 615, normalized size of antiderivative = 7.24
-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*polyl og(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)*(ex p((1+I)*x)-exp((1-I)*x)))*csgn((exp((1+I)*x)-exp((1-I)*x))*ln(x))^2+I*Pi*c sgn(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))*csg n(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*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(l n(x)*sin(x))^2-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))-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
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 \[ \int x \log \left (e^x \log (x) \sin (x)\right ) \, dx=-\frac {1}{6} \, x^{3} + \frac {1}{2} \, x^{2} \log \left (e^{x} \log \left (x\right ) \sin \left (x\right )\right ) - \frac {1}{4} \, x^{2} \log \left (\cos \left (x\right ) + i \, \sin \left (x\right ) + 1\right ) - \frac {1}{4} \, x^{2} \log \left (\cos \left (x\right ) - i \, \sin \left (x\right ) + 1\right ) - \frac {1}{4} \, x^{2} \log \left (-\cos \left (x\right ) + i \, \sin \left (x\right ) + 1\right ) - \frac {1}{4} \, x^{2} \log \left (-\cos \left (x\right ) - i \, \sin \left (x\right ) + 1\right ) + \frac {1}{2} i \, x {\rm Li}_2\left (\cos \left (x\right ) + i \, \sin \left (x\right )\right ) - \frac {1}{2} i \, x {\rm Li}_2\left (\cos \left (x\right ) - i \, \sin \left (x\right )\right ) - \frac {1}{2} i \, x {\rm Li}_2\left (-\cos \left (x\right ) + i \, \sin \left (x\right )\right ) + \frac {1}{2} i \, x {\rm Li}_2\left (-\cos \left (x\right ) - i \, \sin \left (x\right )\right ) - \frac {1}{2} \, \operatorname {log\_integral}\left (x^{2}\right ) - \frac {1}{2} \, {\rm polylog}\left (3, \cos \left (x\right ) + i \, \sin \left (x\right )\right ) - \frac {1}{2} \, {\rm polylog}\left (3, \cos \left (x\right ) - i \, \sin \left (x\right )\right ) - \frac {1}{2} \, {\rm polylog}\left (3, -\cos \left (x\right ) + i \, \sin \left (x\right )\right ) - \frac {1}{2} \, {\rm polylog}\left (3, -\cos \left (x\right ) - i \, \sin \left (x\right )\right ) \]
-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*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_integral(x^2) - 1/2*polylo g(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))
\[ \int x \log \left (e^x \log (x) \sin (x)\right ) \, dx=\int x \log {\left (e^{x} \log {\left (x \right )} \sin {\left (x \right )} \right )}\, dx \]
Time = 0.36 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.82 \[ \int x \log \left (e^x \log (x) \sin (x)\right ) \, dx=-\frac {1}{4} \, {\left (-i \, \pi + 2 \, \log \left (2\right )\right )} x^{2} - \left (\frac {1}{3} i - \frac {1}{3}\right ) \, x^{3} + \frac {1}{2} \, x^{2} \log \left (\log \left (x\right )\right ) + i \, x {\rm Li}_2\left (-e^{\left (i \, x\right )}\right ) + i \, x {\rm Li}_2\left (e^{\left (i \, x\right )}\right ) - \frac {1}{2} \, {\rm Ei}\left (2 \, \log \left (x\right )\right ) - {\rm Li}_{3}(-e^{\left (i \, x\right )}) - {\rm Li}_{3}(e^{\left (i \, x\right )}) \]
-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))
\[ \int x \log \left (e^x \log (x) \sin (x)\right ) \, dx=\int { x \log \left (e^{x} \log \left (x\right ) \sin \left (x\right )\right ) \,d x } \]
Timed out. \[ \int x \log \left (e^x \log (x) \sin (x)\right ) \, dx=\int x\,\ln \left ({\mathrm {e}}^x\,\ln \left (x\right )\,\sin \left (x\right )\right ) \,d x \]