Integrand size = 144, antiderivative size = 30 \[ \int \frac {x^3+\left (2 x^2+x^3\right ) \log \left (\frac {1}{2} (6+3 x)\right )+e^{e^x} \left (-4-2 x-x^2+e^x \left (4 x+2 x^2\right )+\left (-4 x-2 x^2+e^x \left (2 x^2+x^3\right )\right ) \log \left (\frac {1}{2} (6+3 x)\right )\right )}{8 x^2+4 x^3+\left (8 x^3+4 x^4\right ) \log \left (\frac {1}{2} (6+3 x)\right )+\left (2 x^4+x^5\right ) \log ^2\left (\frac {1}{2} (6+3 x)\right )} \, dx=\frac {\log \left (e^{e^{e^x}-x}\right )}{x \left (2+x \log \left (3+\frac {3 x}{2}\right )\right )} \] Output:
ln(1/exp(x-exp(exp(x))))/(2+x*ln(3+3/2*x))/x
Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {x^3+\left (2 x^2+x^3\right ) \log \left (\frac {1}{2} (6+3 x)\right )+e^{e^x} \left (-4-2 x-x^2+e^x \left (4 x+2 x^2\right )+\left (-4 x-2 x^2+e^x \left (2 x^2+x^3\right )\right ) \log \left (\frac {1}{2} (6+3 x)\right )\right )}{8 x^2+4 x^3+\left (8 x^3+4 x^4\right ) \log \left (\frac {1}{2} (6+3 x)\right )+\left (2 x^4+x^5\right ) \log ^2\left (\frac {1}{2} (6+3 x)\right )} \, dx=\frac {e^{e^x}-x}{x \left (2+x \log \left (\frac {3 (2+x)}{2}\right )\right )} \] Input:
Integrate[(x^3 + (2*x^2 + x^3)*Log[(6 + 3*x)/2] + E^E^x*(-4 - 2*x - x^2 + E^x*(4*x + 2*x^2) + (-4*x - 2*x^2 + E^x*(2*x^2 + x^3))*Log[(6 + 3*x)/2]))/ (8*x^2 + 4*x^3 + (8*x^3 + 4*x^4)*Log[(6 + 3*x)/2] + (2*x^4 + x^5)*Log[(6 + 3*x)/2]^2),x]
Output:
(E^E^x - x)/(x*(2 + x*Log[(3*(2 + x))/2]))
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.
\(\displaystyle \int \frac {x^3+\left (x^3+2 x^2\right ) \log \left (\frac {1}{2} (3 x+6)\right )+e^{e^x} \left (-x^2+e^x \left (2 x^2+4 x\right )+\left (-2 x^2+e^x \left (x^3+2 x^2\right )-4 x\right ) \log \left (\frac {1}{2} (3 x+6)\right )-2 x-4\right )}{4 x^3+8 x^2+\left (x^5+2 x^4\right ) \log ^2\left (\frac {1}{2} (3 x+6)\right )+\left (4 x^4+8 x^3\right ) \log \left (\frac {1}{2} (3 x+6)\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {x^3-e^{e^x} \left (x^2+2 x+4\right )+2 e^{x+e^x} (x+2) x+(x+2) \left (e^{x+e^x} x+x-2 e^{e^x}\right ) x \log \left (\frac {3 (x+2)}{2}\right )}{x^2 (x+2) \left (x \log \left (\frac {3 (x+2)}{2}\right )+2\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {e^{e^x} \left (x^2+2 x+4\right )}{(x+2) x^2 \left (x \log \left (\frac {3 (x+2)}{2}\right )+2\right )^2}+\frac {x}{(x+2) \left (x \log \left (\frac {3 (x+2)}{2}\right )+2\right )^2}+\frac {\log \left (\frac {3 x}{2}+3\right )}{\left (x \log \left (\frac {3 (x+2)}{2}\right )+2\right )^2}-\frac {2 e^{e^x} \log \left (\frac {3 x}{2}+3\right )}{x \left (x \log \left (\frac {3 (x+2)}{2}\right )+2\right )^2}+\frac {e^{x+e^x}}{x \left (x \log \left (\frac {3 (x+2)}{2}\right )+2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \int \frac {e^{e^x}}{x^2 \left (x \log \left (\frac {3 (x+2)}{2}\right )+2\right )^2}dx+\int \frac {1}{\left (x \log \left (\frac {3 (x+2)}{2}\right )+2\right )^2}dx-2 \int \frac {1}{(x+2) \left (x \log \left (\frac {3 (x+2)}{2}\right )+2\right )^2}dx-\int \frac {e^{e^x}}{(x+2) \left (x \log \left (\frac {3 (x+2)}{2}\right )+2\right )^2}dx+\int \frac {e^{x+e^x}}{x \left (x \log \left (\frac {3 (x+2)}{2}\right )+2\right )}dx+\int \frac {\log \left (\frac {3 x}{2}+3\right )}{\left (x \log \left (\frac {3 (x+2)}{2}\right )+2\right )^2}dx-2 \int \frac {e^{e^x} \log \left (\frac {3 x}{2}+3\right )}{x \left (x \log \left (\frac {3 (x+2)}{2}\right )+2\right )^2}dx\) |
Input:
Int[(x^3 + (2*x^2 + x^3)*Log[(6 + 3*x)/2] + E^E^x*(-4 - 2*x - x^2 + E^x*(4 *x + 2*x^2) + (-4*x - 2*x^2 + E^x*(2*x^2 + x^3))*Log[(6 + 3*x)/2]))/(8*x^2 + 4*x^3 + (8*x^3 + 4*x^4)*Log[(6 + 3*x)/2] + (2*x^4 + x^5)*Log[(6 + 3*x)/ 2]^2),x]
Output:
$Aborted
Time = 9.42 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90
method | result | size |
parallelrisch | \(\frac {-4 x +4 \,{\mathrm e}^{{\mathrm e}^{x}}}{4 x \left (2+x \ln \left (3+\frac {3 x}{2}\right )\right )}\) | \(27\) |
risch | \(-\frac {1}{2+x \ln \left (3+\frac {3 x}{2}\right )}+\frac {{\mathrm e}^{{\mathrm e}^{x}}}{\left (2+x \ln \left (3+\frac {3 x}{2}\right )\right ) x}\) | \(35\) |
Input:
int(((((x^3+2*x^2)*exp(x)-2*x^2-4*x)*ln(3+3/2*x)+(2*x^2+4*x)*exp(x)-x^2-2* x-4)*exp(exp(x))+(x^3+2*x^2)*ln(3+3/2*x)+x^3)/((x^5+2*x^4)*ln(3+3/2*x)^2+( 4*x^4+8*x^3)*ln(3+3/2*x)+4*x^3+8*x^2),x,method=_RETURNVERBOSE)
Output:
1/4*(-4*x+4*exp(exp(x)))/x/(2+x*ln(3+3/2*x))
Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {x^3+\left (2 x^2+x^3\right ) \log \left (\frac {1}{2} (6+3 x)\right )+e^{e^x} \left (-4-2 x-x^2+e^x \left (4 x+2 x^2\right )+\left (-4 x-2 x^2+e^x \left (2 x^2+x^3\right )\right ) \log \left (\frac {1}{2} (6+3 x)\right )\right )}{8 x^2+4 x^3+\left (8 x^3+4 x^4\right ) \log \left (\frac {1}{2} (6+3 x)\right )+\left (2 x^4+x^5\right ) \log ^2\left (\frac {1}{2} (6+3 x)\right )} \, dx=-\frac {x - e^{\left (e^{x}\right )}}{x^{2} \log \left (\frac {3}{2} \, x + 3\right ) + 2 \, x} \] Input:
integrate(((((x^3+2*x^2)*exp(x)-2*x^2-4*x)*log(3+3/2*x)+(2*x^2+4*x)*exp(x) -x^2-2*x-4)*exp(exp(x))+(x^3+2*x^2)*log(3+3/2*x)+x^3)/((x^5+2*x^4)*log(3+3 /2*x)^2+(4*x^4+8*x^3)*log(3+3/2*x)+4*x^3+8*x^2),x, algorithm="fricas")
Output:
-(x - e^(e^x))/(x^2*log(3/2*x + 3) + 2*x)
Time = 0.19 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {x^3+\left (2 x^2+x^3\right ) \log \left (\frac {1}{2} (6+3 x)\right )+e^{e^x} \left (-4-2 x-x^2+e^x \left (4 x+2 x^2\right )+\left (-4 x-2 x^2+e^x \left (2 x^2+x^3\right )\right ) \log \left (\frac {1}{2} (6+3 x)\right )\right )}{8 x^2+4 x^3+\left (8 x^3+4 x^4\right ) \log \left (\frac {1}{2} (6+3 x)\right )+\left (2 x^4+x^5\right ) \log ^2\left (\frac {1}{2} (6+3 x)\right )} \, dx=\frac {e^{e^{x}}}{x^{2} \log {\left (\frac {3 x}{2} + 3 \right )} + 2 x} - \frac {1}{x \log {\left (\frac {3 x}{2} + 3 \right )} + 2} \] Input:
integrate(((((x**3+2*x**2)*exp(x)-2*x**2-4*x)*ln(3+3/2*x)+(2*x**2+4*x)*exp (x)-x**2-2*x-4)*exp(exp(x))+(x**3+2*x**2)*ln(3+3/2*x)+x**3)/((x**5+2*x**4) *ln(3+3/2*x)**2+(4*x**4+8*x**3)*ln(3+3/2*x)+4*x**3+8*x**2),x)
Output:
exp(exp(x))/(x**2*log(3*x/2 + 3) + 2*x) - 1/(x*log(3*x/2 + 3) + 2)
Time = 0.17 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {x^3+\left (2 x^2+x^3\right ) \log \left (\frac {1}{2} (6+3 x)\right )+e^{e^x} \left (-4-2 x-x^2+e^x \left (4 x+2 x^2\right )+\left (-4 x-2 x^2+e^x \left (2 x^2+x^3\right )\right ) \log \left (\frac {1}{2} (6+3 x)\right )\right )}{8 x^2+4 x^3+\left (8 x^3+4 x^4\right ) \log \left (\frac {1}{2} (6+3 x)\right )+\left (2 x^4+x^5\right ) \log ^2\left (\frac {1}{2} (6+3 x)\right )} \, dx=-\frac {x - e^{\left (e^{x}\right )}}{x^{2} {\left (\log \left (3\right ) - \log \left (2\right )\right )} + x^{2} \log \left (x + 2\right ) + 2 \, x} \] Input:
integrate(((((x^3+2*x^2)*exp(x)-2*x^2-4*x)*log(3+3/2*x)+(2*x^2+4*x)*exp(x) -x^2-2*x-4)*exp(exp(x))+(x^3+2*x^2)*log(3+3/2*x)+x^3)/((x^5+2*x^4)*log(3+3 /2*x)^2+(4*x^4+8*x^3)*log(3+3/2*x)+4*x^3+8*x^2),x, algorithm="maxima")
Output:
-(x - e^(e^x))/(x^2*(log(3) - log(2)) + x^2*log(x + 2) + 2*x)
Time = 0.17 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {x^3+\left (2 x^2+x^3\right ) \log \left (\frac {1}{2} (6+3 x)\right )+e^{e^x} \left (-4-2 x-x^2+e^x \left (4 x+2 x^2\right )+\left (-4 x-2 x^2+e^x \left (2 x^2+x^3\right )\right ) \log \left (\frac {1}{2} (6+3 x)\right )\right )}{8 x^2+4 x^3+\left (8 x^3+4 x^4\right ) \log \left (\frac {1}{2} (6+3 x)\right )+\left (2 x^4+x^5\right ) \log ^2\left (\frac {1}{2} (6+3 x)\right )} \, dx=-\frac {x - e^{\left (e^{x}\right )}}{x^{2} \log \left (\frac {3}{2} \, x + 3\right ) + 2 \, x} \] Input:
integrate(((((x^3+2*x^2)*exp(x)-2*x^2-4*x)*log(3+3/2*x)+(2*x^2+4*x)*exp(x) -x^2-2*x-4)*exp(exp(x))+(x^3+2*x^2)*log(3+3/2*x)+x^3)/((x^5+2*x^4)*log(3+3 /2*x)^2+(4*x^4+8*x^3)*log(3+3/2*x)+4*x^3+8*x^2),x, algorithm="giac")
Output:
-(x - e^(e^x))/(x^2*log(3/2*x + 3) + 2*x)
Time = 4.00 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {x^3+\left (2 x^2+x^3\right ) \log \left (\frac {1}{2} (6+3 x)\right )+e^{e^x} \left (-4-2 x-x^2+e^x \left (4 x+2 x^2\right )+\left (-4 x-2 x^2+e^x \left (2 x^2+x^3\right )\right ) \log \left (\frac {1}{2} (6+3 x)\right )\right )}{8 x^2+4 x^3+\left (8 x^3+4 x^4\right ) \log \left (\frac {1}{2} (6+3 x)\right )+\left (2 x^4+x^5\right ) \log ^2\left (\frac {1}{2} (6+3 x)\right )} \, dx=-\frac {x-{\mathrm {e}}^{{\mathrm {e}}^x}}{x\,\left (x\,\ln \left (\frac {3\,x}{2}+3\right )+2\right )} \] Input:
int((log((3*x)/2 + 3)*(2*x^2 + x^3) - exp(exp(x))*(2*x + log((3*x)/2 + 3)* (4*x - exp(x)*(2*x^2 + x^3) + 2*x^2) - exp(x)*(4*x + 2*x^2) + x^2 + 4) + x ^3)/(log((3*x)/2 + 3)*(8*x^3 + 4*x^4) + log((3*x)/2 + 3)^2*(2*x^4 + x^5) + 8*x^2 + 4*x^3),x)
Output:
-(x - exp(exp(x)))/(x*(x*log((3*x)/2 + 3) + 2))
Time = 0.18 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17 \[ \int \frac {x^3+\left (2 x^2+x^3\right ) \log \left (\frac {1}{2} (6+3 x)\right )+e^{e^x} \left (-4-2 x-x^2+e^x \left (4 x+2 x^2\right )+\left (-4 x-2 x^2+e^x \left (2 x^2+x^3\right )\right ) \log \left (\frac {1}{2} (6+3 x)\right )\right )}{8 x^2+4 x^3+\left (8 x^3+4 x^4\right ) \log \left (\frac {1}{2} (6+3 x)\right )+\left (2 x^4+x^5\right ) \log ^2\left (\frac {1}{2} (6+3 x)\right )} \, dx=\frac {2 e^{e^{x}}+\mathrm {log}\left (\frac {3 x}{2}+3\right ) x^{2}}{2 x \left (\mathrm {log}\left (\frac {3 x}{2}+3\right ) x +2\right )} \] Input:
int(((((x^3+2*x^2)*exp(x)-2*x^2-4*x)*log(3+3/2*x)+(2*x^2+4*x)*exp(x)-x^2-2 *x-4)*exp(exp(x))+(x^3+2*x^2)*log(3+3/2*x)+x^3)/((x^5+2*x^4)*log(3+3/2*x)^ 2+(4*x^4+8*x^3)*log(3+3/2*x)+4*x^3+8*x^2),x)
Output:
(2*e**(e**x) + log((3*x + 6)/2)*x**2)/(2*x*(log((3*x + 6)/2)*x + 2))