3.79.89 \(\int \frac {-6 x^2-4 x^3+e^x (6 x^2+4 x^3-x^4)+(-18 x-18 x^2+e^x (18 x+18 x^2-6 x^3)) \log (x)+(-18 x+e^x (18 x-9 x^2)) \log ^2(x)}{9-18 e^x+9 e^{2 x}} \, dx\)

Optimal. Leaf size=33 \[ \frac {e^{-x} x \left (\frac {x^2}{3}+x \log (x)\right )^2}{x-e^{-x} x} \]

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Rubi [F]  time = 2.83, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-6 x^2-4 x^3+e^x \left (6 x^2+4 x^3-x^4\right )+\left (-18 x-18 x^2+e^x \left (18 x+18 x^2-6 x^3\right )\right ) \log (x)+\left (-18 x+e^x \left (18 x-9 x^2\right )\right ) \log ^2(x)}{9-18 e^x+9 e^{2 x}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-6*x^2 - 4*x^3 + E^x*(6*x^2 + 4*x^3 - x^4) + (-18*x - 18*x^2 + E^x*(18*x + 18*x^2 - 6*x^3))*Log[x] + (-18
*x + E^x*(18*x - 9*x^2))*Log[x]^2)/(9 - 18*E^x + 9*E^(2*x)),x]

[Out]

(-2*x^3)/9 - x^4/(9*(1 - E^x)) + (2*x^2*Log[1 - E^x])/3 + (4*x*PolyLog[2, E^x])/3 - (4*PolyLog[3, E^x])/3 + 2*
Defer[Int][(x*Log[x])/(-1 + E^x), x] + 2*Defer[Int][(x^2*Log[x])/(-1 + E^x), x] - (2*Defer[Int][(x^3*Log[x])/(
-1 + E^x)^2, x])/3 - (2*Defer[Int][(x^3*Log[x])/(-1 + E^x), x])/3 + 2*Defer[Int][(x*Log[x]^2)/(-1 + E^x), x] -
Defer[Int][(x^2*Log[x]^2)/(-1 + E^x)^2, x] - Defer[Int][(x^2*Log[x]^2)/(-1 + E^x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x (x+3 \log (x)) \left (-6-4 x-e^x \left (-6-4 x+x^2\right )-3 \left (2+e^x (-2+x)\right ) \log (x)\right )}{9 \left (1-e^x\right )^2} \, dx\\ &=\frac {1}{9} \int \frac {x (x+3 \log (x)) \left (-6-4 x-e^x \left (-6-4 x+x^2\right )-3 \left (2+e^x (-2+x)\right ) \log (x)\right )}{\left (1-e^x\right )^2} \, dx\\ &=\frac {1}{9} \int \left (-\frac {x^2 (x+3 \log (x))^2}{\left (-1+e^x\right )^2}-\frac {x \left (-6 x-4 x^2+x^3-18 \log (x)-18 x \log (x)+6 x^2 \log (x)-18 \log ^2(x)+9 x \log ^2(x)\right )}{-1+e^x}\right ) \, dx\\ &=-\left (\frac {1}{9} \int \frac {x^2 (x+3 \log (x))^2}{\left (-1+e^x\right )^2} \, dx\right )-\frac {1}{9} \int \frac {x \left (-6 x-4 x^2+x^3-18 \log (x)-18 x \log (x)+6 x^2 \log (x)-18 \log ^2(x)+9 x \log ^2(x)\right )}{-1+e^x} \, dx\\ &=-\left (\frac {1}{9} \int \left (\frac {x^4}{\left (-1+e^x\right )^2}+\frac {6 x^3 \log (x)}{\left (-1+e^x\right )^2}+\frac {9 x^2 \log ^2(x)}{\left (-1+e^x\right )^2}\right ) \, dx\right )-\frac {1}{9} \int \left (-\frac {6 x^2}{-1+e^x}-\frac {4 x^3}{-1+e^x}+\frac {x^4}{-1+e^x}-\frac {18 x \log (x)}{-1+e^x}-\frac {18 x^2 \log (x)}{-1+e^x}+\frac {6 x^3 \log (x)}{-1+e^x}-\frac {18 x \log ^2(x)}{-1+e^x}+\frac {9 x^2 \log ^2(x)}{-1+e^x}\right ) \, dx\\ &=-\left (\frac {1}{9} \int \frac {x^4}{\left (-1+e^x\right )^2} \, dx\right )-\frac {1}{9} \int \frac {x^4}{-1+e^x} \, dx+\frac {4}{9} \int \frac {x^3}{-1+e^x} \, dx+\frac {2}{3} \int \frac {x^2}{-1+e^x} \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{\left (-1+e^x\right )^2} \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log (x)}{-1+e^x} \, dx+2 \int \frac {x^2 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log ^2(x)}{-1+e^x} \, dx-\int \frac {x^2 \log ^2(x)}{\left (-1+e^x\right )^2} \, dx-\int \frac {x^2 \log ^2(x)}{-1+e^x} \, dx\\ &=-\frac {2 x^3}{9}-\frac {x^4}{9}+\frac {x^5}{45}-\frac {1}{9} \int \frac {e^x x^4}{\left (-1+e^x\right )^2} \, dx+\frac {1}{9} \int \frac {x^4}{-1+e^x} \, dx-\frac {1}{9} \int \frac {e^x x^4}{-1+e^x} \, dx+\frac {4}{9} \int \frac {e^x x^3}{-1+e^x} \, dx+\frac {2}{3} \int \frac {e^x x^2}{-1+e^x} \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{\left (-1+e^x\right )^2} \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log (x)}{-1+e^x} \, dx+2 \int \frac {x^2 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log ^2(x)}{-1+e^x} \, dx-\int \frac {x^2 \log ^2(x)}{\left (-1+e^x\right )^2} \, dx-\int \frac {x^2 \log ^2(x)}{-1+e^x} \, dx\\ &=-\frac {2 x^3}{9}-\frac {x^4}{9}-\frac {x^4}{9 \left (1-e^x\right )}+\frac {2}{3} x^2 \log \left (1-e^x\right )+\frac {4}{9} x^3 \log \left (1-e^x\right )-\frac {1}{9} x^4 \log \left (1-e^x\right )+\frac {1}{9} \int \frac {e^x x^4}{-1+e^x} \, dx-\frac {4}{9} \int \frac {x^3}{-1+e^x} \, dx+\frac {4}{9} \int x^3 \log \left (1-e^x\right ) \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{\left (-1+e^x\right )^2} \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{-1+e^x} \, dx-\frac {4}{3} \int x \log \left (1-e^x\right ) \, dx-\frac {4}{3} \int x^2 \log \left (1-e^x\right ) \, dx+2 \int \frac {x \log (x)}{-1+e^x} \, dx+2 \int \frac {x^2 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log ^2(x)}{-1+e^x} \, dx-\int \frac {x^2 \log ^2(x)}{\left (-1+e^x\right )^2} \, dx-\int \frac {x^2 \log ^2(x)}{-1+e^x} \, dx\\ &=-\frac {2 x^3}{9}-\frac {x^4}{9 \left (1-e^x\right )}+\frac {2}{3} x^2 \log \left (1-e^x\right )+\frac {4}{9} x^3 \log \left (1-e^x\right )+\frac {4 x \text {Li}_2\left (e^x\right )}{3}+\frac {4}{3} x^2 \text {Li}_2\left (e^x\right )-\frac {4}{9} x^3 \text {Li}_2\left (e^x\right )-\frac {4}{9} \int \frac {e^x x^3}{-1+e^x} \, dx-\frac {4}{9} \int x^3 \log \left (1-e^x\right ) \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{\left (-1+e^x\right )^2} \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{-1+e^x} \, dx-\frac {4}{3} \int \text {Li}_2\left (e^x\right ) \, dx+\frac {4}{3} \int x^2 \text {Li}_2\left (e^x\right ) \, dx+2 \int \frac {x \log (x)}{-1+e^x} \, dx+2 \int \frac {x^2 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log ^2(x)}{-1+e^x} \, dx-\frac {8}{3} \int x \text {Li}_2\left (e^x\right ) \, dx-\int \frac {x^2 \log ^2(x)}{\left (-1+e^x\right )^2} \, dx-\int \frac {x^2 \log ^2(x)}{-1+e^x} \, dx\\ &=-\frac {2 x^3}{9}-\frac {x^4}{9 \left (1-e^x\right )}+\frac {2}{3} x^2 \log \left (1-e^x\right )+\frac {4 x \text {Li}_2\left (e^x\right )}{3}+\frac {4}{3} x^2 \text {Li}_2\left (e^x\right )-\frac {8 x \text {Li}_3\left (e^x\right )}{3}+\frac {4}{3} x^2 \text {Li}_3\left (e^x\right )-\frac {2}{3} \int \frac {x^3 \log (x)}{\left (-1+e^x\right )^2} \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{-1+e^x} \, dx+\frac {4}{3} \int x^2 \log \left (1-e^x\right ) \, dx-\frac {4}{3} \int x^2 \text {Li}_2\left (e^x\right ) \, dx-\frac {4}{3} \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^x\right )+2 \int \frac {x \log (x)}{-1+e^x} \, dx+2 \int \frac {x^2 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log ^2(x)}{-1+e^x} \, dx+\frac {8}{3} \int \text {Li}_3\left (e^x\right ) \, dx-\frac {8}{3} \int x \text {Li}_3\left (e^x\right ) \, dx-\int \frac {x^2 \log ^2(x)}{\left (-1+e^x\right )^2} \, dx-\int \frac {x^2 \log ^2(x)}{-1+e^x} \, dx\\ &=-\frac {2 x^3}{9}-\frac {x^4}{9 \left (1-e^x\right )}+\frac {2}{3} x^2 \log \left (1-e^x\right )+\frac {4 x \text {Li}_2\left (e^x\right )}{3}-\frac {4 \text {Li}_3\left (e^x\right )}{3}-\frac {8 x \text {Li}_3\left (e^x\right )}{3}-\frac {8 x \text {Li}_4\left (e^x\right )}{3}-\frac {2}{3} \int \frac {x^3 \log (x)}{\left (-1+e^x\right )^2} \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log (x)}{-1+e^x} \, dx+2 \int \frac {x^2 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log ^2(x)}{-1+e^x} \, dx+\frac {8}{3} \int x \text {Li}_2\left (e^x\right ) \, dx+\frac {8}{3} \int x \text {Li}_3\left (e^x\right ) \, dx+\frac {8}{3} \int \text {Li}_4\left (e^x\right ) \, dx+\frac {8}{3} \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^x\right )-\int \frac {x^2 \log ^2(x)}{\left (-1+e^x\right )^2} \, dx-\int \frac {x^2 \log ^2(x)}{-1+e^x} \, dx\\ &=-\frac {2 x^3}{9}-\frac {x^4}{9 \left (1-e^x\right )}+\frac {2}{3} x^2 \log \left (1-e^x\right )+\frac {4 x \text {Li}_2\left (e^x\right )}{3}-\frac {4 \text {Li}_3\left (e^x\right )}{3}+\frac {8 \text {Li}_4\left (e^x\right )}{3}-\frac {2}{3} \int \frac {x^3 \log (x)}{\left (-1+e^x\right )^2} \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log (x)}{-1+e^x} \, dx+2 \int \frac {x^2 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log ^2(x)}{-1+e^x} \, dx-\frac {8}{3} \int \text {Li}_3\left (e^x\right ) \, dx-\frac {8}{3} \int \text {Li}_4\left (e^x\right ) \, dx+\frac {8}{3} \operatorname {Subst}\left (\int \frac {\text {Li}_4(x)}{x} \, dx,x,e^x\right )-\int \frac {x^2 \log ^2(x)}{\left (-1+e^x\right )^2} \, dx-\int \frac {x^2 \log ^2(x)}{-1+e^x} \, dx\\ &=-\frac {2 x^3}{9}-\frac {x^4}{9 \left (1-e^x\right )}+\frac {2}{3} x^2 \log \left (1-e^x\right )+\frac {4 x \text {Li}_2\left (e^x\right )}{3}-\frac {4 \text {Li}_3\left (e^x\right )}{3}+\frac {8 \text {Li}_4\left (e^x\right )}{3}+\frac {8 \text {Li}_5\left (e^x\right )}{3}-\frac {2}{3} \int \frac {x^3 \log (x)}{\left (-1+e^x\right )^2} \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log (x)}{-1+e^x} \, dx+2 \int \frac {x^2 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log ^2(x)}{-1+e^x} \, dx-\frac {8}{3} \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^x\right )-\frac {8}{3} \operatorname {Subst}\left (\int \frac {\text {Li}_4(x)}{x} \, dx,x,e^x\right )-\int \frac {x^2 \log ^2(x)}{\left (-1+e^x\right )^2} \, dx-\int \frac {x^2 \log ^2(x)}{-1+e^x} \, dx\\ &=-\frac {2 x^3}{9}-\frac {x^4}{9 \left (1-e^x\right )}+\frac {2}{3} x^2 \log \left (1-e^x\right )+\frac {4 x \text {Li}_2\left (e^x\right )}{3}-\frac {4 \text {Li}_3\left (e^x\right )}{3}-\frac {2}{3} \int \frac {x^3 \log (x)}{\left (-1+e^x\right )^2} \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log (x)}{-1+e^x} \, dx+2 \int \frac {x^2 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log ^2(x)}{-1+e^x} \, dx-\int \frac {x^2 \log ^2(x)}{\left (-1+e^x\right )^2} \, dx-\int \frac {x^2 \log ^2(x)}{-1+e^x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.38, size = 22, normalized size = 0.67 \begin {gather*} \frac {x^2 (x+3 \log (x))^2}{9 \left (-1+e^x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6*x^2 - 4*x^3 + E^x*(6*x^2 + 4*x^3 - x^4) + (-18*x - 18*x^2 + E^x*(18*x + 18*x^2 - 6*x^3))*Log[x]
+ (-18*x + E^x*(18*x - 9*x^2))*Log[x]^2)/(9 - 18*E^x + 9*E^(2*x)),x]

[Out]

(x^2*(x + 3*Log[x])^2)/(9*(-1 + E^x))

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fricas [A]  time = 0.90, size = 28, normalized size = 0.85 \begin {gather*} \frac {x^{4} + 6 \, x^{3} \log \relax (x) + 9 \, x^{2} \log \relax (x)^{2}}{9 \, {\left (e^{x} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-9*x^2+18*x)*exp(x)-18*x)*log(x)^2+((-6*x^3+18*x^2+18*x)*exp(x)-18*x^2-18*x)*log(x)+(-x^4+4*x^3+6
*x^2)*exp(x)-4*x^3-6*x^2)/(9*exp(x)^2-18*exp(x)+9),x, algorithm="fricas")

[Out]

1/9*(x^4 + 6*x^3*log(x) + 9*x^2*log(x)^2)/(e^x - 1)

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giac [A]  time = 0.17, size = 28, normalized size = 0.85 \begin {gather*} \frac {x^{4} + 6 \, x^{3} \log \relax (x) + 9 \, x^{2} \log \relax (x)^{2}}{9 \, {\left (e^{x} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-9*x^2+18*x)*exp(x)-18*x)*log(x)^2+((-6*x^3+18*x^2+18*x)*exp(x)-18*x^2-18*x)*log(x)+(-x^4+4*x^3+6
*x^2)*exp(x)-4*x^3-6*x^2)/(9*exp(x)^2-18*exp(x)+9),x, algorithm="giac")

[Out]

1/9*(x^4 + 6*x^3*log(x) + 9*x^2*log(x)^2)/(e^x - 1)

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maple [A]  time = 0.04, size = 40, normalized size = 1.21

method	result	size
risch	\(\frac {x^{2} \ln \relax (x )^{2}}{{\mathrm e}^{x}-1}+\frac {2 x^{3} \ln \relax (x )}{3 \left ({\mathrm e}^{x}-1\right )}+\frac {x^{4}}{9 \,{\mathrm e}^{x}-9}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-9*x^2+18*x)*exp(x)-18*x)*ln(x)^2+((-6*x^3+18*x^2+18*x)*exp(x)-18*x^2-18*x)*ln(x)+(-x^4+4*x^3+6*x^2)*ex
p(x)-4*x^3-6*x^2)/(9*exp(x)^2-18*exp(x)+9),x,method=_RETURNVERBOSE)

[Out]

x^2/(exp(x)-1)*ln(x)^2+2/3*x^3/(exp(x)-1)*ln(x)+1/9*x^4/(exp(x)-1)

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maxima [A]  time = 0.40, size = 28, normalized size = 0.85 \begin {gather*} \frac {x^{4} + 6 \, x^{3} \log \relax (x) + 9 \, x^{2} \log \relax (x)^{2}}{9 \, {\left (e^{x} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-9*x^2+18*x)*exp(x)-18*x)*log(x)^2+((-6*x^3+18*x^2+18*x)*exp(x)-18*x^2-18*x)*log(x)+(-x^4+4*x^3+6
*x^2)*exp(x)-4*x^3-6*x^2)/(9*exp(x)^2-18*exp(x)+9),x, algorithm="maxima")

[Out]

1/9*(x^4 + 6*x^3*log(x) + 9*x^2*log(x)^2)/(e^x - 1)

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mupad [B]  time = 5.28, size = 20, normalized size = 0.61 \begin {gather*} \frac {x^2\,{\left (x+3\,\ln \relax (x)\right )}^2}{9\,\left ({\mathrm {e}}^x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)^2*(18*x - exp(x)*(18*x - 9*x^2)) - exp(x)*(6*x^2 + 4*x^3 - x^4) + log(x)*(18*x + 18*x^2 - exp(x)*
(18*x + 18*x^2 - 6*x^3)) + 6*x^2 + 4*x^3)/(9*exp(2*x) - 18*exp(x) + 9),x)

[Out]

(x^2*(x + 3*log(x))^2)/(9*(exp(x) - 1))

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sympy [A]  time = 0.28, size = 27, normalized size = 0.82 \begin {gather*} \frac {x^{4} + 6 x^{3} \log {\relax (x )} + 9 x^{2} \log {\relax (x )}^{2}}{9 e^{x} - 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-9*x**2+18*x)*exp(x)-18*x)*ln(x)**2+((-6*x**3+18*x**2+18*x)*exp(x)-18*x**2-18*x)*ln(x)+(-x**4+4*x
**3+6*x**2)*exp(x)-4*x**3-6*x**2)/(9*exp(x)**2-18*exp(x)+9),x)

[Out]

(x**4 + 6*x**3*log(x) + 9*x**2*log(x)**2)/(9*exp(x) - 9)

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