Integrand size = 123, antiderivative size = 25 \[ \int \frac {e^{\frac {1}{2} (2 x+3 \log (x-\log (x)))} \left (11 x-x^2-2 x^3+\left (4-19 x-9 x^2-2 x^3\right ) \log (x)+\left (8+8 x+2 x^2\right ) \log ^2(x)\right )}{-2 x^3+\left (-14 x^2-4 x^3\right ) \log (x)+\left (-16 x-12 x^2-2 x^3\right ) \log ^2(x)+\left (32+16 x+2 x^2\right ) \log ^3(x)} \, dx=\frac {e^x x (x-\log (x))^{3/2}}{x+(4+x) \log (x)} \] Output:
exp(3/2*ln(x-ln(x))+x)/((4+x)*ln(x)+x)*x
Time = 5.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1}{2} (2 x+3 \log (x-\log (x)))} \left (11 x-x^2-2 x^3+\left (4-19 x-9 x^2-2 x^3\right ) \log (x)+\left (8+8 x+2 x^2\right ) \log ^2(x)\right )}{-2 x^3+\left (-14 x^2-4 x^3\right ) \log (x)+\left (-16 x-12 x^2-2 x^3\right ) \log ^2(x)+\left (32+16 x+2 x^2\right ) \log ^3(x)} \, dx=\frac {e^x x (x-\log (x))^{3/2}}{x+(4+x) \log (x)} \] Input:
Integrate[(E^((2*x + 3*Log[x - Log[x]])/2)*(11*x - x^2 - 2*x^3 + (4 - 19*x - 9*x^2 - 2*x^3)*Log[x] + (8 + 8*x + 2*x^2)*Log[x]^2))/(-2*x^3 + (-14*x^2 - 4*x^3)*Log[x] + (-16*x - 12*x^2 - 2*x^3)*Log[x]^2 + (32 + 16*x + 2*x^2) *Log[x]^3),x]
Output:
(E^x*x*(x - Log[x])^(3/2))/(x + (4 + x)*Log[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.
| \(\displaystyle \int \frac {e^{\frac {1}{2} (2 x+3 \log (x-\log (x)))} \left (-2 x^3-x^2+\left (2 x^2+8 x+8\right ) \log ^2(x)+\left (-2 x^3-9 x^2-19 x+4\right ) \log (x)+11 x\right )}{-2 x^3+\left (2 x^2+16 x+32\right ) \log ^3(x)+\left (-2 x^3-12 x^2-16 x\right ) \log ^2(x)+\left (-4 x^3-14 x^2\right ) \log (x)} \, dx\) |
| \(\Big \downarrow \) 2704 |
| \(\displaystyle \int \frac {e^x (x-\log (x))^{3/2} \left (-2 x^3-x^2+\left (2 x^2+8 x+8\right ) \log ^2(x)+\left (-2 x^3-9 x^2-19 x+4\right ) \log (x)+11 x\right )}{-2 x^3+\left (2 x^2+16 x+32\right ) \log ^3(x)+\left (-2 x^3-12 x^2-16 x\right ) \log ^2(x)+\left (-4 x^3-14 x^2\right ) \log (x)}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {e^x \sqrt {x-\log (x)} \left (x \left (2 x^2+x-11\right )+\left (2 x^3+9 x^2+19 x-4\right ) \log (x)-2 (x+2)^2 \log ^2(x)\right )}{2 (x+(x+4) \log (x))^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int -\frac {e^x \sqrt {x-\log (x)} \left (2 (x+2)^2 \log ^2(x)+\left (-2 x^3-9 x^2-19 x+4\right ) \log (x)+x \left (-2 x^2-x+11\right )\right )}{(x+(x+4) \log (x))^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{2} \int \frac {e^x \sqrt {x-\log (x)} \left (2 (x+2)^2 \log ^2(x)+\left (-2 x^3-9 x^2-19 x+4\right ) \log (x)+x \left (-2 x^2-x+11\right )\right )}{(x+(x+4) \log (x))^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {1}{2} \int \left (\frac {2 e^x \sqrt {x-\log (x)} (x+2)^2}{(x+4)^2}+\frac {e^x \left (-2 x^4-21 x^3-71 x^2-88 x+16\right ) \sqrt {x-\log (x)}}{(x+4)^2 (\log (x) x+x+4 \log (x))}+\frac {2 e^x x \left (x^3+17 x^2+76 x+80\right ) \sqrt {x-\log (x)}}{(x+4)^2 (\log (x) x+x+4 \log (x))^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (-2 \int \frac {e^x x^2 \sqrt {x-\log (x)}}{(\log (x) x+x+4 \log (x))^2}dx+2 \int \frac {e^x x^2 \sqrt {x-\log (x)}}{\log (x) x+x+4 \log (x)}dx-2 \int e^x \sqrt {x-\log (x)}dx-8 \int \frac {e^x \sqrt {x-\log (x)}}{(x+4)^2}dx+8 \int \frac {e^x \sqrt {x-\log (x)}}{x+4}dx+24 \int \frac {e^x \sqrt {x-\log (x)}}{(\log (x) x+x+4 \log (x))^2}dx-18 \int \frac {e^x x \sqrt {x-\log (x)}}{(\log (x) x+x+4 \log (x))^2}dx-128 \int \frac {e^x \sqrt {x-\log (x)}}{(x+4)^2 (\log (x) x+x+4 \log (x))^2}dx-64 \int \frac {e^x \sqrt {x-\log (x)}}{(x+4) (\log (x) x+x+4 \log (x))^2}dx-\int \frac {e^x \sqrt {x-\log (x)}}{\log (x) x+x+4 \log (x)}dx+5 \int \frac {e^x x \sqrt {x-\log (x)}}{\log (x) x+x+4 \log (x)}dx-64 \int \frac {e^x \sqrt {x-\log (x)}}{(x+4)^2 (\log (x) x+x+4 \log (x))}dx+16 \int \frac {e^x \sqrt {x-\log (x)}}{(x+4) (\log (x) x+x+4 \log (x))}dx\right )\) |
Input:
Int[(E^((2*x + 3*Log[x - Log[x]])/2)*(11*x - x^2 - 2*x^3 + (4 - 19*x - 9*x ^2 - 2*x^3)*Log[x] + (8 + 8*x + 2*x^2)*Log[x]^2))/(-2*x^3 + (-14*x^2 - 4*x ^3)*Log[x] + (-16*x - 12*x^2 - 2*x^3)*Log[x]^2 + (32 + 16*x + 2*x^2)*Log[x ]^3),x]
Output:
$Aborted
Time = 4.60 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00
| method | result | size |
| risch | \(\frac {x \left (x -\ln \left (x \right )\right )^{\frac {3}{2}} {\mathrm e}^{x}}{x \ln \left (x \right )+4 \ln \left (x \right )+x}\) | \(25\) |
| parallelrisch | \(\frac {x \,{\mathrm e}^{\frac {3 \ln \left (x -\ln \left (x \right )\right )}{2}+x}}{x \ln \left (x \right )+4 \ln \left (x \right )+x}\) | \(27\) |
Input:
int(((2*x^2+8*x+8)*ln(x)^2+(-2*x^3-9*x^2-19*x+4)*ln(x)-2*x^3-x^2+11*x)*exp (3/2*ln(x-ln(x))+x)/((2*x^2+16*x+32)*ln(x)^3+(-2*x^3-12*x^2-16*x)*ln(x)^2+ (-4*x^3-14*x^2)*ln(x)-2*x^3),x,method=_RETURNVERBOSE)
Output:
x/(x*ln(x)+4*ln(x)+x)*(x-ln(x))^(3/2)*exp(x)
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {1}{2} (2 x+3 \log (x-\log (x)))} \left (11 x-x^2-2 x^3+\left (4-19 x-9 x^2-2 x^3\right ) \log (x)+\left (8+8 x+2 x^2\right ) \log ^2(x)\right )}{-2 x^3+\left (-14 x^2-4 x^3\right ) \log (x)+\left (-16 x-12 x^2-2 x^3\right ) \log ^2(x)+\left (32+16 x+2 x^2\right ) \log ^3(x)} \, dx=\frac {x e^{\left (x + \frac {3}{2} \, \log \left (x - \log \left (x\right )\right )\right )}}{{\left (x + 4\right )} \log \left (x\right ) + x} \] Input:
integrate(((2*x^2+8*x+8)*log(x)^2+(-2*x^3-9*x^2-19*x+4)*log(x)-2*x^3-x^2+1 1*x)*exp(3/2*log(x-log(x))+x)/((2*x^2+16*x+32)*log(x)^3+(-2*x^3-12*x^2-16* x)*log(x)^2+(-4*x^3-14*x^2)*log(x)-2*x^3),x, algorithm="fricas")
Output:
x*e^(x + 3/2*log(x - log(x)))/((x + 4)*log(x) + x)
Timed out. \[ \int \frac {e^{\frac {1}{2} (2 x+3 \log (x-\log (x)))} \left (11 x-x^2-2 x^3+\left (4-19 x-9 x^2-2 x^3\right ) \log (x)+\left (8+8 x+2 x^2\right ) \log ^2(x)\right )}{-2 x^3+\left (-14 x^2-4 x^3\right ) \log (x)+\left (-16 x-12 x^2-2 x^3\right ) \log ^2(x)+\left (32+16 x+2 x^2\right ) \log ^3(x)} \, dx=\text {Timed out} \] Input:
integrate(((2*x**2+8*x+8)*ln(x)**2+(-2*x**3-9*x**2-19*x+4)*ln(x)-2*x**3-x* *2+11*x)*exp(3/2*ln(x-ln(x))+x)/((2*x**2+16*x+32)*ln(x)**3+(-2*x**3-12*x** 2-16*x)*ln(x)**2+(-4*x**3-14*x**2)*ln(x)-2*x**3),x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {e^{\frac {1}{2} (2 x+3 \log (x-\log (x)))} \left (11 x-x^2-2 x^3+\left (4-19 x-9 x^2-2 x^3\right ) \log (x)+\left (8+8 x+2 x^2\right ) \log ^2(x)\right )}{-2 x^3+\left (-14 x^2-4 x^3\right ) \log (x)+\left (-16 x-12 x^2-2 x^3\right ) \log ^2(x)+\left (32+16 x+2 x^2\right ) \log ^3(x)} \, dx=\frac {{\left (x^{2} - x \log \left (x\right )\right )} \sqrt {x - \log \left (x\right )} e^{x}}{{\left (x + 4\right )} \log \left (x\right ) + x} \] Input:
integrate(((2*x^2+8*x+8)*log(x)^2+(-2*x^3-9*x^2-19*x+4)*log(x)-2*x^3-x^2+1 1*x)*exp(3/2*log(x-log(x))+x)/((2*x^2+16*x+32)*log(x)^3+(-2*x^3-12*x^2-16* x)*log(x)^2+(-4*x^3-14*x^2)*log(x)-2*x^3),x, algorithm="maxima")
Output:
(x^2 - x*log(x))*sqrt(x - log(x))*e^x/((x + 4)*log(x) + x)
\[ \int \frac {e^{\frac {1}{2} (2 x+3 \log (x-\log (x)))} \left (11 x-x^2-2 x^3+\left (4-19 x-9 x^2-2 x^3\right ) \log (x)+\left (8+8 x+2 x^2\right ) \log ^2(x)\right )}{-2 x^3+\left (-14 x^2-4 x^3\right ) \log (x)+\left (-16 x-12 x^2-2 x^3\right ) \log ^2(x)+\left (32+16 x+2 x^2\right ) \log ^3(x)} \, dx=\int { -\frac {{\left (2 \, x^{3} - 2 \, {\left (x^{2} + 4 \, x + 4\right )} \log \left (x\right )^{2} + x^{2} + {\left (2 \, x^{3} + 9 \, x^{2} + 19 \, x - 4\right )} \log \left (x\right ) - 11 \, x\right )} e^{\left (x + \frac {3}{2} \, \log \left (x - \log \left (x\right )\right )\right )}}{2 \, {\left ({\left (x^{2} + 8 \, x + 16\right )} \log \left (x\right )^{3} - x^{3} - {\left (x^{3} + 6 \, x^{2} + 8 \, x\right )} \log \left (x\right )^{2} - {\left (2 \, x^{3} + 7 \, x^{2}\right )} \log \left (x\right )\right )}} \,d x } \] Input:
integrate(((2*x^2+8*x+8)*log(x)^2+(-2*x^3-9*x^2-19*x+4)*log(x)-2*x^3-x^2+1 1*x)*exp(3/2*log(x-log(x))+x)/((2*x^2+16*x+32)*log(x)^3+(-2*x^3-12*x^2-16* x)*log(x)^2+(-4*x^3-14*x^2)*log(x)-2*x^3),x, algorithm="giac")
Output:
integrate(-1/2*(2*x^3 - 2*(x^2 + 4*x + 4)*log(x)^2 + x^2 + (2*x^3 + 9*x^2 + 19*x - 4)*log(x) - 11*x)*e^(x + 3/2*log(x - log(x)))/((x^2 + 8*x + 16)*l og(x)^3 - x^3 - (x^3 + 6*x^2 + 8*x)*log(x)^2 - (2*x^3 + 7*x^2)*log(x)), x)
Time = 2.91 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {1}{2} (2 x+3 \log (x-\log (x)))} \left (11 x-x^2-2 x^3+\left (4-19 x-9 x^2-2 x^3\right ) \log (x)+\left (8+8 x+2 x^2\right ) \log ^2(x)\right )}{-2 x^3+\left (-14 x^2-4 x^3\right ) \log (x)+\left (-16 x-12 x^2-2 x^3\right ) \log ^2(x)+\left (32+16 x+2 x^2\right ) \log ^3(x)} \, dx=\frac {x\,{\mathrm {e}}^x\,{\left (x-\ln \left (x\right )\right )}^{3/2}}{x+4\,\ln \left (x\right )+x\,\ln \left (x\right )} \] Input:
int((exp(x + (3*log(x - log(x)))/2)*(x^2 - log(x)^2*(8*x + 2*x^2 + 8) - 11 *x + 2*x^3 + log(x)*(19*x + 9*x^2 + 2*x^3 - 4)))/(log(x)*(14*x^2 + 4*x^3) - log(x)^3*(16*x + 2*x^2 + 32) + log(x)^2*(16*x + 12*x^2 + 2*x^3) + 2*x^3) ,x)
Output:
(x*exp(x)*(x - log(x))^(3/2))/(x + 4*log(x) + x*log(x))
Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {e^{\frac {1}{2} (2 x+3 \log (x-\log (x)))} \left (11 x-x^2-2 x^3+\left (4-19 x-9 x^2-2 x^3\right ) \log (x)+\left (8+8 x+2 x^2\right ) \log ^2(x)\right )}{-2 x^3+\left (-14 x^2-4 x^3\right ) \log (x)+\left (-16 x-12 x^2-2 x^3\right ) \log ^2(x)+\left (32+16 x+2 x^2\right ) \log ^3(x)} \, dx=\frac {e^{x} \sqrt {x -\mathrm {log}\left (x \right )}\, x \left (x -\mathrm {log}\left (x \right )\right )}{\mathrm {log}\left (x \right ) x +4 \,\mathrm {log}\left (x \right )+x} \] Input:
int(((2*x^2+8*x+8)*log(x)^2+(-2*x^3-9*x^2-19*x+4)*log(x)-2*x^3-x^2+11*x)*e xp(3/2*log(x-log(x))+x)/((2*x^2+16*x+32)*log(x)^3+(-2*x^3-12*x^2-16*x)*log (x)^2+(-4*x^3-14*x^2)*log(x)-2*x^3),x)
Output:
(e**x*sqrt( - log(x) + x)*x*( - log(x) + x))/(log(x)*x + 4*log(x) + x)