Integrand size = 100, antiderivative size = 33 \[ \int \frac {-6 x^2-4 x^3+e^x \left (6 x^2+4 x^3-x^4\right )+\left (-18 x-18 x^2+e^x \left (18 x+18 x^2-6 x^3\right )\right ) \log (x)+\left (-18 x+e^x \left (18 x-9 x^2\right )\right ) \log ^2(x)}{9-18 e^x+9 e^{2 x}} \, dx=\frac {e^{-x} x \left (\frac {x^2}{3}+x \log (x)\right )^2}{x-e^{-x} x} \] Output:
(1/3*x^2+x*ln(x))^2/(x-x/exp(x))/exp(x)*x
Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.67 \[ \int \frac {-6 x^2-4 x^3+e^x \left (6 x^2+4 x^3-x^4\right )+\left (-18 x-18 x^2+e^x \left (18 x+18 x^2-6 x^3\right )\right ) \log (x)+\left (-18 x+e^x \left (18 x-9 x^2\right )\right ) \log ^2(x)}{9-18 e^x+9 e^{2 x}} \, dx=\frac {x^2 (x+3 \log (x))^2}{9 \left (-1+e^x\right )} \] Input:
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]
Output:
(x^2*(x + 3*Log[x])^2)/(9*(-1 + E^x))
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.
\(\displaystyle \int \frac {-4 x^3-6 x^2+\left (e^x \left (18 x-9 x^2\right )-18 x\right ) \log ^2(x)+\left (-18 x^2+e^x \left (-6 x^3+18 x^2+18 x\right )-18 x\right ) \log (x)+e^x \left (-x^4+4 x^3+6 x^2\right )}{-18 e^x+9 e^{2 x}+9} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {x (x+3 \log (x)) \left (-e^x \left (x^2-4 x-6\right )-4 x-3 \left (e^x (x-2)+2\right ) \log (x)-6\right )}{9 \left (1-e^x\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{9} \int -\frac {x (x+3 \log (x)) \left (4 x-e^x \left (-x^2+4 x+6\right )+3 \left (2-e^x (2-x)\right ) \log (x)+6\right )}{\left (1-e^x\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{9} \int \frac {x (x+3 \log (x)) \left (4 x-e^x \left (-x^2+4 x+6\right )+3 \left (2-e^x (2-x)\right ) \log (x)+6\right )}{\left (1-e^x\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{9} \int \left (\frac {x^2 (x+3 \log (x))^2}{\left (-1+e^x\right )^2}+\frac {x \left (x^3+6 \log (x) x^2-4 x^2+9 \log ^2(x) x-18 \log (x) x-6 x-18 \log ^2(x)-18 \log (x)\right )}{-1+e^x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{9} \left (-6 \int \frac {x^3 \log (x)}{\left (-1+e^x\right )^2}dx-6 \int \frac {x^3 \log (x)}{-1+e^x}dx-9 \int \frac {x^2 \log ^2(x)}{\left (-1+e^x\right )^2}dx-9 \int \frac {x^2 \log ^2(x)}{-1+e^x}dx+18 \int \frac {x^2 \log (x)}{-1+e^x}dx+18 \int \frac {x \log ^2(x)}{-1+e^x}dx+18 \int \frac {x \log (x)}{-1+e^x}dx+12 x \operatorname {PolyLog}\left (2,e^x\right )-12 \operatorname {PolyLog}\left (3,e^x\right )-\frac {x^4}{1-e^x}-2 x^3+6 x^2 \log \left (1-e^x\right )\right )\) |
Input:
Int[(-6*x^2 - 4*x^3 + E^x*(6*x^2 + 4*x^3 - x^4) + (-18*x - 18*x^2 + E^x*(1 8*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]
Output:
$Aborted
Time = 211.52 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94
method | result | size |
parallelrisch | \(-\frac {-x^{4}-6 x^{3} \ln \left (x \right )-9 x^{2} \ln \left (x \right )^{2}}{9 \left ({\mathrm e}^{x}-1\right )}\) | \(31\) |
risch | \(\frac {x^{2} \ln \left (x \right )^{2}}{{\mathrm e}^{x}-1}+\frac {2 x^{3} \ln \left (x \right )}{3 \left ({\mathrm e}^{x}-1\right )}+\frac {x^{4}}{9 \,{\mathrm e}^{x}-9}\) | \(40\) |
Input:
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)*exp(x)-4*x^3-6*x^2)/(9*exp(x)^2-18*exp(x )+9),x,method=_RETURNVERBOSE)
Output:
-1/9*(-x^4-6*x^3*ln(x)-9*x^2*ln(x)^2)/(exp(x)-1)
Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {-6 x^2-4 x^3+e^x \left (6 x^2+4 x^3-x^4\right )+\left (-18 x-18 x^2+e^x \left (18 x+18 x^2-6 x^3\right )\right ) \log (x)+\left (-18 x+e^x \left (18 x-9 x^2\right )\right ) \log ^2(x)}{9-18 e^x+9 e^{2 x}} \, dx=\frac {x^{4} + 6 \, x^{3} \log \left (x\right ) + 9 \, x^{2} \log \left (x\right )^{2}}{9 \, {\left (e^{x} - 1\right )}} \] Input:
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")
Output:
1/9*(x^4 + 6*x^3*log(x) + 9*x^2*log(x)^2)/(e^x - 1)
Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {-6 x^2-4 x^3+e^x \left (6 x^2+4 x^3-x^4\right )+\left (-18 x-18 x^2+e^x \left (18 x+18 x^2-6 x^3\right )\right ) \log (x)+\left (-18 x+e^x \left (18 x-9 x^2\right )\right ) \log ^2(x)}{9-18 e^x+9 e^{2 x}} \, dx=\frac {x^{4} + 6 x^{3} \log {\left (x \right )} + 9 x^{2} \log {\left (x \right )}^{2}}{9 e^{x} - 9} \] Input:
integrate((((-9*x**2+18*x)*exp(x)-18*x)*ln(x)**2+((-6*x**3+18*x**2+18*x)*e xp(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*e xp(x)**2-18*exp(x)+9),x)
Output:
(x**4 + 6*x**3*log(x) + 9*x**2*log(x)**2)/(9*exp(x) - 9)
Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {-6 x^2-4 x^3+e^x \left (6 x^2+4 x^3-x^4\right )+\left (-18 x-18 x^2+e^x \left (18 x+18 x^2-6 x^3\right )\right ) \log (x)+\left (-18 x+e^x \left (18 x-9 x^2\right )\right ) \log ^2(x)}{9-18 e^x+9 e^{2 x}} \, dx=\frac {x^{4} + 6 \, x^{3} \log \left (x\right ) + 9 \, x^{2} \log \left (x\right )^{2}}{9 \, {\left (e^{x} - 1\right )}} \] Input:
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")
Output:
1/9*(x^4 + 6*x^3*log(x) + 9*x^2*log(x)^2)/(e^x - 1)
Time = 0.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {-6 x^2-4 x^3+e^x \left (6 x^2+4 x^3-x^4\right )+\left (-18 x-18 x^2+e^x \left (18 x+18 x^2-6 x^3\right )\right ) \log (x)+\left (-18 x+e^x \left (18 x-9 x^2\right )\right ) \log ^2(x)}{9-18 e^x+9 e^{2 x}} \, dx=\frac {x^{4} + 6 \, x^{3} \log \left (x\right ) + 9 \, x^{2} \log \left (x\right )^{2}}{9 \, {\left (e^{x} - 1\right )}} \] Input:
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")
Output:
1/9*(x^4 + 6*x^3*log(x) + 9*x^2*log(x)^2)/(e^x - 1)
Time = 3.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.61 \[ \int \frac {-6 x^2-4 x^3+e^x \left (6 x^2+4 x^3-x^4\right )+\left (-18 x-18 x^2+e^x \left (18 x+18 x^2-6 x^3\right )\right ) \log (x)+\left (-18 x+e^x \left (18 x-9 x^2\right )\right ) \log ^2(x)}{9-18 e^x+9 e^{2 x}} \, dx=\frac {x^2\,{\left (x+3\,\ln \left (x\right )\right )}^2}{9\,\left ({\mathrm {e}}^x-1\right )} \] Input:
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)
Output:
(x^2*(x + 3*log(x))^2)/(9*(exp(x) - 1))
Time = 0.17 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {-6 x^2-4 x^3+e^x \left (6 x^2+4 x^3-x^4\right )+\left (-18 x-18 x^2+e^x \left (18 x+18 x^2-6 x^3\right )\right ) \log (x)+\left (-18 x+e^x \left (18 x-9 x^2\right )\right ) \log ^2(x)}{9-18 e^x+9 e^{2 x}} \, dx=\frac {x^{2} \left (9 \mathrm {log}\left (x \right )^{2}+6 \,\mathrm {log}\left (x \right ) x +x^{2}\right )}{9 e^{x}-9} \] Input:
int((((-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)
Output:
(x**2*(9*log(x)**2 + 6*log(x)*x + x**2))/(9*(e**x - 1))