### 3.312 $$\int \log (e^x \log (x) \sin (x)) \, dx$$

Optimal. Leaf size=57 $\frac{1}{2} i \text{PolyLog}\left (2,e^{2 i x}\right )-\text{li}(x)+\left (-\frac{1}{2}+\frac{i}{2}\right ) x^2-x \log \left (1-e^{2 i x}\right )+x \log \left (e^x \log (x) \sin (x)\right )$

[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)]

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Rubi [A]  time = 0.0642922, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 9, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.667, Rules used = {2549, 3717, 2190, 2279, 2391, 2298} $\frac{1}{2} i \text{PolyLog}\left (2,e^{2 i x}\right )-\text{li}(x)+\left (-\frac{1}{2}+\frac{i}{2}\right ) x^2-x \log \left (1-e^{2 i x}\right )+x \log \left (e^x \log (x) \sin (x)\right )$

Antiderivative was successfully veriﬁed.

[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 2549

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

Rule 3717

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 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2279

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 2391

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 2298

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

Rubi steps

\begin{align*} \int \log \left (e^x \log (x) \sin (x)\right ) \, dx &=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 \operatorname{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]  time = 0.0297676, size = 56, normalized size = 0.98 $\frac{1}{2} \left (i \text{PolyLog}\left (2,e^{2 i x}\right )-2 \text{li}(x)+(-1+i) x^2-2 x \log \left (1-e^{2 i x}\right )+2 x \log \left (e^x \log (x) \sin (x)\right )\right )$

Antiderivative was successfully veriﬁed.

[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

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Maple [C]  time = 0.244, size = 583, normalized size = 10.2 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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(x))*csgn(I*ln(x)*(exp((1+I)*x)-exp((1-I)*x)))^2*x-ln(2)*x-1/2*I*Pi*csgn(I*(exp(2*I*x)-1))*csgn(I*ln(x))*cs
gn(I*ln(x)*(exp(2*I*x)-1))*x-1/2*I*Pi*csgn(I*ln(x)*(exp(2*I*x)-1))^3*x+x*ln(ln(x))-I*dilog(exp(I*x))+1/2*I*x^2
-1/2*I*Pi*x+I*ln(exp(I*x))*ln(exp(I*x)+1)+1/2*I*Pi*csgn(ln(x)*sin(x))^3*x+1/2*I*Pi*csgn(I*exp(-I*x))*csgn(ln(x
)*sin(x))^2*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*csgn(ln(x)*sin(x))*csgn(I*
ln(x)*(exp((1+I)*x)-exp((1-I)*x)))^2*x-I*ln(exp(I*x))*ln(exp(2*I*x)-1)+1/2*ln(exp(x))^2+1/2*I*Pi*csgn(I*exp(-I
*x))*csgn(I*ln(x)*(exp(2*I*x)-1))*csgn(ln(x)*sin(x))*x+I*dilog(exp(I*x)+1)+1/2*I*Pi*csgn((exp((1+I)*x)-exp((1-
I)*x))*ln(x))^2*x-1/2*I*Pi*csgn(I*ln(x)*(exp((1+I)*x)-exp((1-I)*x)))^3*x-x*ln(exp(I*x))+Ei(1,-ln(x))+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))^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(I*(exp(2*I*x)-1))*csgn(I*ln(x)*(e
xp(2*I*x)-1))^2*x+1/2*I*Pi*csgn(I*ln(x))*csgn(I*ln(x)*(exp(2*I*x)-1))^2*x-1/2*I*Pi*csgn((exp((1+I)*x)-exp((1-I
)*x))*ln(x))^3*x

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Maxima [A]  time = 1.80966, size = 58, normalized size = 1.02 \begin{align*} \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 ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)
)

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Fricas [B]  time = 2.42055, size = 447, normalized size = 7.84 \begin{align*} -\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 ) - \logintegral \left (x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError