\(\int \log (e^x \log (x) \sin (x)) \, dx\) [312]

Optimal result

Integrand size = 9, antiderivative size = 57 \[ \int \log \left (e^x \log (x) \sin (x)\right ) \, dx=\left (-\frac {1}{2}+\frac {i}{2}\right ) x^2-x \log \left (1-e^{2 i x}\right )+x \log \left (e^x \log (x) \sin (x)\right )-\operatorname {LogIntegral}(x)+\frac {1}{2} i \operatorname {PolyLog}\left (2,e^{2 i x}\right ) \]

[Out]

(-1/2+1/2*I)*x^2-Li(x)-x*ln(1-exp(2*I*x))+x*ln(exp(x)*ln(x)*sin(x))+1/2*I*polylog(2,exp(2*I*x))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {2629, 3798, 2221, 2317, 2438, 2335} \[ \int \log \left (e^x \log (x) \sin (x)\right ) \, dx=-\operatorname {LogIntegral}(x)+\frac {1}{2} i \operatorname {PolyLog}\left (2,e^{2 i x}\right )+\left (-\frac {1}{2}+\frac {i}{2}\right ) x^2-x \log \left (1-e^{2 i x}\right )+x \log \left (e^x \log (x) \sin (x)\right ) \]

[In]

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

[Out]

(-1/2 + I/2)*x^2 - x*Log[1 - E^((2*I)*x)] + x*Log[E^x*Log[x]*Sin[x]] - LogIntegral[x] + (I/2)*PolyLog[2, E^((2
*I)*x)]

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 2317

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

Rule 2335

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2629

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[x*Simplify[D[u, x]/u], x], 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]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.98 \[ \int \log \left (e^x \log (x) \sin (x)\right ) \, dx=\frac {1}{2} \left ((-1+i) x^2-2 x \log \left (1-e^{2 i x}\right )+2 x \log \left (e^x \log (x) \sin (x)\right )-2 \operatorname {LogIntegral}(x)+i \operatorname {PolyLog}\left (2,e^{2 i x}\right )\right ) \]

[In]

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

[Out]

((-1 + I)*x^2 - 2*x*Log[1 - E^((2*I)*x)] + 2*x*Log[E^x*Log[x]*Sin[x]] - 2*LogIntegral[x] + I*PolyLog[2, E^((2*
I)*x)])/2

Maple [C] (warning: unable to verify)

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

Time = 1.02 (sec) , antiderivative size = 583, normalized size of antiderivative = 10.23

method	result	size
risch	\(\text {Expression too large to display}\)	\(583\)

[In]

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

[Out]

1/2*I*Pi*csgn(I*ln(x))*csgn(I*ln(x)*(exp(2*I*x)-1))^2*x+1/2*ln(exp(x))^2-1/2*I*Pi*csgn((exp((1+I)*x)-exp((1-I)
*x))*ln(x))^3*x-1/2*I*Pi*csgn(I*ln(x)*(exp((1+I)*x)-exp((1-I)*x)))^3*x+1/2*I*Pi*csgn(I*(exp(2*I*x)-1))*csgn(I*
ln(x)*(exp(2*I*x)-1))^2*x+1/2*I*Pi*csgn(I*exp(x))*csgn(ln(x)*sin(x))*csgn(I*ln(x)*(exp((1+I)*x)-exp((1-I)*x)))
*x+1/2*I*Pi*csgn((exp((1+I)*x)-exp((1-I)*x))*ln(x))^2*x-x*ln(2)-1/2*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))*x+1/2*I*Pi*csgn(I*exp(-I*x))*csgn(ln(x)*sin(x))^2*x+1/2*I*Pi*csgn
(I*exp(x))*csgn(I*ln(x)*(exp((1+I)*x)-exp((1-I)*x)))^2*x+1/2*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))^2*x+1/2*I*Pi*csgn(ln(x)*sin(x))^3*x+I*ln(exp(I*x))*ln(exp(I*x)+1)+Ei(1,-
ln(x))+1/2*I*Pi*csgn(I*exp(-I*x))*csgn(I*ln(x)*(exp(2*I*x)-1))*csgn(ln(x)*sin(x))*x-1/2*I*Pi*csgn(I*ln(x)*(exp
(2*I*x)-1))^3*x+1/2*I*Pi*csgn(I*ln(x)*(exp(2*I*x)-1))*csgn(ln(x)*sin(x))^2*x-1/2*I*Pi*x-I*ln(exp(I*x))*ln(exp(
2*I*x)-1)-1/2*I*Pi*csgn(ln(x)*sin(x))*csgn(I*ln(x)*(exp((1+I)*x)-exp((1-I)*x)))^2*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))*x+1/2*I*x^2+ln(ln(x))*x+I*dilog(exp(I*x)+1)-I*dilog(exp(I*x))
-x*ln(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. 116 vs. \(2 (39) = 78\).

Time = 0.31 (sec) , antiderivative size = 116, normalized size of antiderivative = 2.04 \[ \int \log \left (e^x \log (x) \sin (x)\right ) \, dx=-\frac {1}{2} \, x^{2} + x \log \left (e^{x} \log \left (x\right ) \sin \left (x\right )\right ) - \frac {1}{2} \, x \log \left (\cos \left (x\right ) + i \, \sin \left (x\right ) + 1\right ) - \frac {1}{2} \, x \log \left (\cos \left (x\right ) - i \, \sin \left (x\right ) + 1\right ) - \frac {1}{2} \, x \log \left (-\cos \left (x\right ) + i \, \sin \left (x\right ) + 1\right ) - \frac {1}{2} \, x \log \left (-\cos \left (x\right ) - i \, \sin \left (x\right ) + 1\right ) + \frac {1}{2} i \, {\rm Li}_2\left (\cos \left (x\right ) + i \, \sin \left (x\right )\right ) - \frac {1}{2} i \, {\rm Li}_2\left (\cos \left (x\right ) - i \, \sin \left (x\right )\right ) - \frac {1}{2} i \, {\rm Li}_2\left (-\cos \left (x\right ) + i \, \sin \left (x\right )\right ) + \frac {1}{2} i \, {\rm Li}_2\left (-\cos \left (x\right ) - i \, \sin \left (x\right )\right ) - \operatorname {log\_integral}\left (x\right ) \]

[In]

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

[Out]

-1/2*x^2 + x*log(e^x*log(x)*sin(x)) - 1/2*x*log(cos(x) + I*sin(x) + 1) - 1/2*x*log(cos(x) - I*sin(x) + 1) - 1/
2*x*log(-cos(x) + I*sin(x) + 1) - 1/2*x*log(-cos(x) - I*sin(x) + 1) + 1/2*I*dilog(cos(x) + I*sin(x)) - 1/2*I*d
ilog(cos(x) - I*sin(x)) - 1/2*I*dilog(-cos(x) + I*sin(x)) + 1/2*I*dilog(-cos(x) - I*sin(x)) - log_integral(x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.75 \[ \int \log \left (e^x \log (x) \sin (x)\right ) \, dx=\frac {1}{2} \, {\left (i \, \pi - 2 \, \log \left (2\right )\right )} x - \left (\frac {1}{2} i - \frac {1}{2}\right ) \, x^{2} + x \log \left (\log \left (x\right )\right ) - {\rm Ei}\left (\log \left (x\right )\right ) + i \, {\rm Li}_2\left (-e^{\left (i \, x\right )}\right ) + i \, {\rm Li}_2\left (e^{\left (i \, x\right )}\right ) \]

[In]

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

[Out]

1/2*(I*pi - 2*log(2))*x - (1/2*I - 1/2)*x^2 + x*log(log(x)) - Ei(log(x)) + I*dilog(-e^(I*x)) + I*dilog(e^(I*x)
)

Giac [F]

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

[In]

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

[Out]

integrate(log(e^x*log(x)*sin(x)), x)

Mupad [F(-1)]

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

[In]

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

[Out]

int(log(exp(x)*log(x)*sin(x)), x)