Integrand size = 87, antiderivative size = 23 \[ \int \frac {6+2 e^x+2 x+e^x (-2+2 x) \log (x)+e^{3+e^e} \left (-3-e^x-x+e^x (1-x) \log (x)\right )}{9+\left (-6 e^x-6 x\right ) \log (x)+\left (e^{2 x}+2 e^x x+x^2\right ) \log ^2(x)} \, dx=\frac {\left (-2+e^{3+e^e}\right ) x}{-3+\left (e^x+x\right ) \log (x)} \] Output:
x*(exp(exp(exp(1))+3)-2)/((exp(x)+x)*ln(x)-3)
Time = 0.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {6+2 e^x+2 x+e^x (-2+2 x) \log (x)+e^{3+e^e} \left (-3-e^x-x+e^x (1-x) \log (x)\right )}{9+\left (-6 e^x-6 x\right ) \log (x)+\left (e^{2 x}+2 e^x x+x^2\right ) \log ^2(x)} \, dx=\frac {\left (-2+e^{3+e^e}\right ) x}{-3+e^x \log (x)+x \log (x)} \] Input:
Integrate[(6 + 2*E^x + 2*x + E^x*(-2 + 2*x)*Log[x] + E^(3 + E^E)*(-3 - E^x - x + E^x*(1 - x)*Log[x]))/(9 + (-6*E^x - 6*x)*Log[x] + (E^(2*x) + 2*E^x* x + x^2)*Log[x]^2),x]
Output:
((-2 + E^(3 + E^E))*x)/(-3 + E^x*Log[x] + 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.
\(\displaystyle \int \frac {2 x+2 e^x+e^x (2 x-2) \log (x)+e^{3+e^e} \left (-x-e^x+e^x (1-x) \log (x)-3\right )+6}{\left (x^2+2 e^x x+e^{2 x}\right ) \log ^2(x)+\left (-6 x-6 e^x\right ) \log (x)+9} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {\left (2-e^{3+e^e}\right ) \left (x+e^x+e^x (x-1) \log (x)+3\right )}{\left (3-\left (x+e^x\right ) \log (x)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \left (2-e^{3+e^e}\right ) \int \frac {x+e^x-e^x (1-x) \log (x)+3}{\left (3-\left (x+e^x\right ) \log (x)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \left (2-e^{3+e^e}\right ) \int \left (\frac {x \log (x)-\log (x)+1}{\log (x) \left (e^x \log (x)+x \log (x)-3\right )}-\frac {x^2 \log ^2(x)-x \log ^2(x)-3 x \log (x)-3}{\log (x) \left (e^x \log (x)+x \log (x)-3\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \left (2-e^{3+e^e}\right ) \left (-\int \frac {x^2 \log (x)}{\left (e^x \log (x)+x \log (x)-3\right )^2}dx+3 \int \frac {x}{\left (e^x \log (x)+x \log (x)-3\right )^2}dx+3 \int \frac {1}{\log (x) \left (e^x \log (x)+x \log (x)-3\right )^2}dx+\int \frac {x \log (x)}{\left (e^x \log (x)+x \log (x)-3\right )^2}dx-\int \frac {1}{e^x \log (x)+x \log (x)-3}dx+\int \frac {x}{e^x \log (x)+x \log (x)-3}dx+\int \frac {1}{\log (x) \left (e^x \log (x)+x \log (x)-3\right )}dx\right )\) |
Input:
Int[(6 + 2*E^x + 2*x + E^x*(-2 + 2*x)*Log[x] + E^(3 + E^E)*(-3 - E^x - x + E^x*(1 - x)*Log[x]))/(9 + (-6*E^x - 6*x)*Log[x] + (E^(2*x) + 2*E^x*x + x^ 2)*Log[x]^2),x]
Output:
$Aborted
Time = 0.53 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04
method | result | size |
risch | \(\frac {x \left ({\mathrm e}^{{\mathrm e}^{{\mathrm e}}+3}-2\right )}{{\mathrm e}^{x} \ln \left (x \right )+x \ln \left (x \right )-3}\) | \(24\) |
parallelrisch | \(\frac {{\mathrm e}^{{\mathrm e}^{{\mathrm e}}+3} x -2 x}{{\mathrm e}^{x} \ln \left (x \right )+x \ln \left (x \right )-3}\) | \(27\) |
Input:
int((((1-x)*exp(x)*ln(x)-exp(x)-3-x)*exp(exp(exp(1))+3)+(-2+2*x)*exp(x)*ln (x)+2*exp(x)+2*x+6)/((exp(x)^2+2*exp(x)*x+x^2)*ln(x)^2+(-6*exp(x)-6*x)*ln( x)+9),x,method=_RETURNVERBOSE)
Output:
x*(exp(exp(exp(1))+3)-2)/(exp(x)*ln(x)+x*ln(x)-3)
Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {6+2 e^x+2 x+e^x (-2+2 x) \log (x)+e^{3+e^e} \left (-3-e^x-x+e^x (1-x) \log (x)\right )}{9+\left (-6 e^x-6 x\right ) \log (x)+\left (e^{2 x}+2 e^x x+x^2\right ) \log ^2(x)} \, dx=\frac {x e^{\left (e^{e} + 3\right )} - 2 \, x}{{\left (x + e^{x}\right )} \log \left (x\right ) - 3} \] Input:
integrate((((1-x)*exp(x)*log(x)-exp(x)-3-x)*exp(exp(exp(1))+3)+(2*x-2)*exp (x)*log(x)+2*exp(x)+2*x+6)/((exp(x)^2+2*exp(x)*x+x^2)*log(x)^2+(-6*exp(x)- 6*x)*log(x)+9),x, algorithm="fricas")
Output:
(x*e^(e^e + 3) - 2*x)/((x + e^x)*log(x) - 3)
Time = 0.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {6+2 e^x+2 x+e^x (-2+2 x) \log (x)+e^{3+e^e} \left (-3-e^x-x+e^x (1-x) \log (x)\right )}{9+\left (-6 e^x-6 x\right ) \log (x)+\left (e^{2 x}+2 e^x x+x^2\right ) \log ^2(x)} \, dx=\frac {- 2 x + x e^{3} e^{e^{e}}}{x \log {\left (x \right )} + e^{x} \log {\left (x \right )} - 3} \] Input:
integrate((((1-x)*exp(x)*ln(x)-exp(x)-3-x)*exp(exp(exp(1))+3)+(2*x-2)*exp( x)*ln(x)+2*exp(x)+2*x+6)/((exp(x)**2+2*exp(x)*x+x**2)*ln(x)**2+(-6*exp(x)- 6*x)*ln(x)+9),x)
Output:
(-2*x + x*exp(3)*exp(exp(E)))/(x*log(x) + exp(x)*log(x) - 3)
Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {6+2 e^x+2 x+e^x (-2+2 x) \log (x)+e^{3+e^e} \left (-3-e^x-x+e^x (1-x) \log (x)\right )}{9+\left (-6 e^x-6 x\right ) \log (x)+\left (e^{2 x}+2 e^x x+x^2\right ) \log ^2(x)} \, dx=\frac {x {\left (e^{\left (e^{e} + 3\right )} - 2\right )}}{x \log \left (x\right ) + e^{x} \log \left (x\right ) - 3} \] Input:
integrate((((1-x)*exp(x)*log(x)-exp(x)-3-x)*exp(exp(exp(1))+3)+(2*x-2)*exp (x)*log(x)+2*exp(x)+2*x+6)/((exp(x)^2+2*exp(x)*x+x^2)*log(x)^2+(-6*exp(x)- 6*x)*log(x)+9),x, algorithm="maxima")
Output:
x*(e^(e^e + 3) - 2)/(x*log(x) + e^x*log(x) - 3)
Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {6+2 e^x+2 x+e^x (-2+2 x) \log (x)+e^{3+e^e} \left (-3-e^x-x+e^x (1-x) \log (x)\right )}{9+\left (-6 e^x-6 x\right ) \log (x)+\left (e^{2 x}+2 e^x x+x^2\right ) \log ^2(x)} \, dx=\frac {x e^{\left (e^{e} + 3\right )} - 2 \, x}{x \log \left (x\right ) + e^{x} \log \left (x\right ) - 3} \] Input:
integrate((((1-x)*exp(x)*log(x)-exp(x)-3-x)*exp(exp(exp(1))+3)+(2*x-2)*exp (x)*log(x)+2*exp(x)+2*x+6)/((exp(x)^2+2*exp(x)*x+x^2)*log(x)^2+(-6*exp(x)- 6*x)*log(x)+9),x, algorithm="giac")
Output:
(x*e^(e^e + 3) - 2*x)/(x*log(x) + e^x*log(x) - 3)
Timed out. \[ \int \frac {6+2 e^x+2 x+e^x (-2+2 x) \log (x)+e^{3+e^e} \left (-3-e^x-x+e^x (1-x) \log (x)\right )}{9+\left (-6 e^x-6 x\right ) \log (x)+\left (e^{2 x}+2 e^x x+x^2\right ) \log ^2(x)} \, dx=\int \frac {2\,x+2\,{\mathrm {e}}^x-{\mathrm {e}}^{{\mathrm {e}}^{\mathrm {e}}+3}\,\left (x+{\mathrm {e}}^x+{\mathrm {e}}^x\,\ln \left (x\right )\,\left (x-1\right )+3\right )+{\mathrm {e}}^x\,\ln \left (x\right )\,\left (2\,x-2\right )+6}{\left ({\mathrm {e}}^{2\,x}+2\,x\,{\mathrm {e}}^x+x^2\right )\,{\ln \left (x\right )}^2+\left (-6\,x-6\,{\mathrm {e}}^x\right )\,\ln \left (x\right )+9} \,d x \] Input:
int((2*x + 2*exp(x) - exp(exp(exp(1)) + 3)*(x + exp(x) + exp(x)*log(x)*(x - 1) + 3) + exp(x)*log(x)*(2*x - 2) + 6)/(log(x)^2*(exp(2*x) + 2*x*exp(x) + x^2) - log(x)*(6*x + 6*exp(x)) + 9),x)
Output:
int((2*x + 2*exp(x) - exp(exp(exp(1)) + 3)*(x + exp(x) + exp(x)*log(x)*(x - 1) + 3) + exp(x)*log(x)*(2*x - 2) + 6)/(log(x)^2*(exp(2*x) + 2*x*exp(x) + x^2) - log(x)*(6*x + 6*exp(x)) + 9), x)
Time = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {6+2 e^x+2 x+e^x (-2+2 x) \log (x)+e^{3+e^e} \left (-3-e^x-x+e^x (1-x) \log (x)\right )}{9+\left (-6 e^x-6 x\right ) \log (x)+\left (e^{2 x}+2 e^x x+x^2\right ) \log ^2(x)} \, dx=\frac {x \left (e^{e^{e}} e^{3}-2\right )}{e^{x} \mathrm {log}\left (x \right )+\mathrm {log}\left (x \right ) x -3} \] Input:
int((((1-x)*exp(x)*log(x)-exp(x)-3-x)*exp(exp(exp(1))+3)+(2*x-2)*exp(x)*lo g(x)+2*exp(x)+2*x+6)/((exp(x)^2+2*exp(x)*x+x^2)*log(x)^2+(-6*exp(x)-6*x)*l og(x)+9),x)
Output:
(x*(e**(e**e)*e**3 - 2))/(e**x*log(x) + log(x)*x - 3)