Integrand size = 219, antiderivative size = 34 \[ \int \frac {2-4 e^9 x+2 e^{18} x^2+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 e^9 x-2 e^{18} x^2\right )}{1-2 x+x^2+e^{\frac {2 \left (e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)\right )}{-1+e^9 x}} \left (1-2 e^9 x+e^{18} x^2\right )+e^9 \left (-2 x+4 x^2-2 x^3\right )+e^{18} \left (x^2-2 x^3+x^4\right )+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 x+e^9 \left (4 x-4 x^2\right )+e^{18} \left (-2 x^2+2 x^3\right )\right )} \, dx=-5+\frac {2 x}{1-e^{-\frac {x}{\frac {1}{e^9}-x}+\log (2) \log (3)}-x} \]
Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {2-4 e^9 x+2 e^{18} x^2+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 e^9 x-2 e^{18} x^2\right )}{1-2 x+x^2+e^{\frac {2 \left (e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)\right )}{-1+e^9 x}} \left (1-2 e^9 x+e^{18} x^2\right )+e^9 \left (-2 x+4 x^2-2 x^3\right )+e^{18} \left (x^2-2 x^3+x^4\right )+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 x+e^9 \left (4 x-4 x^2\right )+e^{18} \left (-2 x^2+2 x^3\right )\right )} \, dx=-\frac {2 x}{-1+e^{1+\frac {1}{-1+e^9 x}+\log (2) \log (3)}+x} \]
Integrate[(2 - 4*E^9*x + 2*E^18*x^2 + E^((E^9*x + (-1 + E^9*x)*Log[2]*Log[ 3])/(-1 + E^9*x))*(-2 + 2*E^9*x - 2*E^18*x^2))/(1 - 2*x + x^2 + E^((2*(E^9 *x + (-1 + E^9*x)*Log[2]*Log[3]))/(-1 + E^9*x))*(1 - 2*E^9*x + E^18*x^2) + E^9*(-2*x + 4*x^2 - 2*x^3) + E^18*(x^2 - 2*x^3 + x^4) + E^((E^9*x + (-1 + E^9*x)*Log[2]*Log[3])/(-1 + E^9*x))*(-2 + 2*x + E^9*(4*x - 4*x^2) + E^18* (-2*x^2 + 2*x^3))),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.
\(\displaystyle \int \frac {\left (-2 e^{18} x^2+2 e^9 x-2\right ) \exp \left (\frac {e^9 x+\left (e^9 x-1\right ) \log (2) \log (3)}{e^9 x-1}\right )+2 e^{18} x^2-4 e^9 x+2}{\left (e^{18} x^2-2 e^9 x+1\right ) \exp \left (\frac {2 \left (e^9 x+\left (e^9 x-1\right ) \log (2) \log (3)\right )}{e^9 x-1}\right )+\left (e^9 \left (4 x-4 x^2\right )+e^{18} \left (2 x^3-2 x^2\right )+2 x-2\right ) \exp \left (\frac {e^9 x+\left (e^9 x-1\right ) \log (2) \log (3)}{e^9 x-1}\right )+x^2+e^9 \left (-2 x^3+4 x^2-2 x\right )+e^{18} \left (x^4-2 x^3+x^2\right )-2 x+1} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (-2 e^{18} x^2+2 e^9 x-2\right ) \exp \left (\frac {e^9 x+\left (e^9 x-1\right ) \log (2) \log (3)}{e^9 x-1}\right )+2 e^{18} x^2-4 e^9 x+2}{\left (1-e^9 x\right )^2 \left (-x-e^{\frac {e^9 x}{e^9 x-1}+\log (2) \log (3)}+1\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 x \left (e^{18} x^2-e^9 x-e^9+1\right )}{\left (e^9 x-1\right )^2 \left (x+e^{\frac {e^9 x}{e^9 x-1}+\log (2) \log (3)}-1\right )^2}-\frac {2 \left (e^{18} x^2-e^9 x+1\right )}{\left (e^9 x-1\right )^2 \left (x+e^{\frac {e^9 x}{e^9 x-1}+\log (2) \log (3)}-1\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \int \frac {1}{\left (x+e^{\frac {e^9 x}{e^9 x-1}+\log (2) \log (3)}-1\right )^2}dx}{e^9}+2 \int \frac {x}{\left (x+e^{\frac {e^9 x}{e^9 x-1}+\log (2) \log (3)}-1\right )^2}dx-2 \int \frac {1}{x+e^{\frac {e^9 x}{e^9 x-1}+\log (2) \log (3)}-1}dx-2 \left (1-\frac {1}{e^9}\right ) \int \frac {1}{\left (x+e^{\frac {e^9 x}{e^9 x-1}+\log (2) \log (3)}-1\right )^2 \left (e^9 x-1\right )^2}dx-2 \int \frac {1}{\left (x+e^{\frac {e^9 x}{e^9 x-1}+\log (2) \log (3)}-1\right ) \left (e^9 x-1\right )^2}dx+\frac {2 \left (2-e^9\right ) \int \frac {1}{\left (x+e^{\frac {e^9 x}{e^9 x-1}+\log (2) \log (3)}-1\right )^2 \left (e^9 x-1\right )}dx}{e^9}-2 \int \frac {1}{\left (x+e^{\frac {e^9 x}{e^9 x-1}+\log (2) \log (3)}-1\right ) \left (e^9 x-1\right )}dx\) |
Int[(2 - 4*E^9*x + 2*E^18*x^2 + E^((E^9*x + (-1 + E^9*x)*Log[2]*Log[3])/(- 1 + E^9*x))*(-2 + 2*E^9*x - 2*E^18*x^2))/(1 - 2*x + x^2 + E^((2*(E^9*x + ( -1 + E^9*x)*Log[2]*Log[3]))/(-1 + E^9*x))*(1 - 2*E^9*x + E^18*x^2) + E^9*( -2*x + 4*x^2 - 2*x^3) + E^18*(x^2 - 2*x^3 + x^4) + E^((E^9*x + (-1 + E^9*x )*Log[2]*Log[3])/(-1 + E^9*x))*(-2 + 2*x + E^9*(4*x - 4*x^2) + E^18*(-2*x^ 2 + 2*x^3))),x]
3.25.69.3.1 Defintions of rubi rules used
Time = 3.06 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12
method | result | size |
risch | \(-\frac {2 x}{{\mathrm e}^{\frac {\ln \left (3\right ) \ln \left (2\right ) {\mathrm e}^{9} x -\ln \left (2\right ) \ln \left (3\right )+x \,{\mathrm e}^{9}}{x \,{\mathrm e}^{9}-1}}+x -1}\) | \(38\) |
norman | \(\frac {2 x -2 x^{2} {\mathrm e}^{9}}{\left (x +{\mathrm e}^{\frac {\left (x \,{\mathrm e}^{9}-1\right ) \ln \left (2\right ) \ln \left (3\right )+x \,{\mathrm e}^{9}}{x \,{\mathrm e}^{9}-1}}-1\right ) \left (x \,{\mathrm e}^{9}-1\right )}\) | \(52\) |
parallelrisch | \(\frac {\left (2 \,{\mathrm e}^{36} {\mathrm e}^{\frac {\left (x \,{\mathrm e}^{9}-1\right ) \ln \left (2\right ) \ln \left (3\right )+x \,{\mathrm e}^{9}}{x \,{\mathrm e}^{9}-1}}-2 \,{\mathrm e}^{36}\right ) {\mathrm e}^{-36}}{x +{\mathrm e}^{\frac {\left (x \,{\mathrm e}^{9}-1\right ) \ln \left (2\right ) \ln \left (3\right )+x \,{\mathrm e}^{9}}{x \,{\mathrm e}^{9}-1}}-1}\) | \(76\) |
int(((-2*x^2*exp(9)^2+2*x*exp(9)-2)*exp(((x*exp(9)-1)*ln(2)*ln(3)+x*exp(9) )/(x*exp(9)-1))+2*x^2*exp(9)^2-4*x*exp(9)+2)/((x^2*exp(9)^2-2*x*exp(9)+1)* exp(((x*exp(9)-1)*ln(2)*ln(3)+x*exp(9))/(x*exp(9)-1))^2+((2*x^3-2*x^2)*exp (9)^2+(-4*x^2+4*x)*exp(9)+2*x-2)*exp(((x*exp(9)-1)*ln(2)*ln(3)+x*exp(9))/( x*exp(9)-1))+(x^4-2*x^3+x^2)*exp(9)^2+(-2*x^3+4*x^2-2*x)*exp(9)+x^2-2*x+1) ,x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {2-4 e^9 x+2 e^{18} x^2+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 e^9 x-2 e^{18} x^2\right )}{1-2 x+x^2+e^{\frac {2 \left (e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)\right )}{-1+e^9 x}} \left (1-2 e^9 x+e^{18} x^2\right )+e^9 \left (-2 x+4 x^2-2 x^3\right )+e^{18} \left (x^2-2 x^3+x^4\right )+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 x+e^9 \left (4 x-4 x^2\right )+e^{18} \left (-2 x^2+2 x^3\right )\right )} \, dx=-\frac {2 \, x}{x + e^{\left (\frac {{\left (x e^{9} - 1\right )} \log \left (3\right ) \log \left (2\right ) + x e^{9}}{x e^{9} - 1}\right )} - 1} \]
integrate(((-2*x^2*exp(9)^2+2*x*exp(9)-2)*exp(((x*exp(9)-1)*log(2)*log(3)+ x*exp(9))/(x*exp(9)-1))+2*x^2*exp(9)^2-4*x*exp(9)+2)/((x^2*exp(9)^2-2*x*ex p(9)+1)*exp(((x*exp(9)-1)*log(2)*log(3)+x*exp(9))/(x*exp(9)-1))^2+((2*x^3- 2*x^2)*exp(9)^2+(-4*x^2+4*x)*exp(9)+2*x-2)*exp(((x*exp(9)-1)*log(2)*log(3) +x*exp(9))/(x*exp(9)-1))+(x^4-2*x^3+x^2)*exp(9)^2+(-2*x^3+4*x^2-2*x)*exp(9 )+x^2-2*x+1),x, algorithm=\
Time = 0.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {2-4 e^9 x+2 e^{18} x^2+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 e^9 x-2 e^{18} x^2\right )}{1-2 x+x^2+e^{\frac {2 \left (e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)\right )}{-1+e^9 x}} \left (1-2 e^9 x+e^{18} x^2\right )+e^9 \left (-2 x+4 x^2-2 x^3\right )+e^{18} \left (x^2-2 x^3+x^4\right )+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 x+e^9 \left (4 x-4 x^2\right )+e^{18} \left (-2 x^2+2 x^3\right )\right )} \, dx=- \frac {2 x}{x + e^{\frac {x e^{9} + \left (x e^{9} - 1\right ) \log {\left (2 \right )} \log {\left (3 \right )}}{x e^{9} - 1}} - 1} \]
integrate(((-2*x**2*exp(9)**2+2*x*exp(9)-2)*exp(((x*exp(9)-1)*ln(2)*ln(3)+ x*exp(9))/(x*exp(9)-1))+2*x**2*exp(9)**2-4*x*exp(9)+2)/((x**2*exp(9)**2-2* x*exp(9)+1)*exp(((x*exp(9)-1)*ln(2)*ln(3)+x*exp(9))/(x*exp(9)-1))**2+((2*x **3-2*x**2)*exp(9)**2+(-4*x**2+4*x)*exp(9)+2*x-2)*exp(((x*exp(9)-1)*ln(2)* ln(3)+x*exp(9))/(x*exp(9)-1))+(x**4-2*x**3+x**2)*exp(9)**2+(-2*x**3+4*x**2 -2*x)*exp(9)+x**2-2*x+1),x)
Time = 0.43 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \frac {2-4 e^9 x+2 e^{18} x^2+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 e^9 x-2 e^{18} x^2\right )}{1-2 x+x^2+e^{\frac {2 \left (e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)\right )}{-1+e^9 x}} \left (1-2 e^9 x+e^{18} x^2\right )+e^9 \left (-2 x+4 x^2-2 x^3\right )+e^{18} \left (x^2-2 x^3+x^4\right )+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 x+e^9 \left (4 x-4 x^2\right )+e^{18} \left (-2 x^2+2 x^3\right )\right )} \, dx=-\frac {2 \, x}{2^{\log \left (3\right )} e^{\left (\frac {1}{x e^{9} - 1} + 1\right )} + x - 1} \]
integrate(((-2*x^2*exp(9)^2+2*x*exp(9)-2)*exp(((x*exp(9)-1)*log(2)*log(3)+ x*exp(9))/(x*exp(9)-1))+2*x^2*exp(9)^2-4*x*exp(9)+2)/((x^2*exp(9)^2-2*x*ex p(9)+1)*exp(((x*exp(9)-1)*log(2)*log(3)+x*exp(9))/(x*exp(9)-1))^2+((2*x^3- 2*x^2)*exp(9)^2+(-4*x^2+4*x)*exp(9)+2*x-2)*exp(((x*exp(9)-1)*log(2)*log(3) +x*exp(9))/(x*exp(9)-1))+(x^4-2*x^3+x^2)*exp(9)^2+(-2*x^3+4*x^2-2*x)*exp(9 )+x^2-2*x+1),x, algorithm=\
Time = 1.91 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int \frac {2-4 e^9 x+2 e^{18} x^2+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 e^9 x-2 e^{18} x^2\right )}{1-2 x+x^2+e^{\frac {2 \left (e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)\right )}{-1+e^9 x}} \left (1-2 e^9 x+e^{18} x^2\right )+e^9 \left (-2 x+4 x^2-2 x^3\right )+e^{18} \left (x^2-2 x^3+x^4\right )+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 x+e^9 \left (4 x-4 x^2\right )+e^{18} \left (-2 x^2+2 x^3\right )\right )} \, dx=-\frac {2 \, x}{x + e^{\left (\log \left (3\right ) \log \left (2\right ) + \frac {x e^{9}}{x e^{9} - 1}\right )} - 1} \]
integrate(((-2*x^2*exp(9)^2+2*x*exp(9)-2)*exp(((x*exp(9)-1)*log(2)*log(3)+ x*exp(9))/(x*exp(9)-1))+2*x^2*exp(9)^2-4*x*exp(9)+2)/((x^2*exp(9)^2-2*x*ex p(9)+1)*exp(((x*exp(9)-1)*log(2)*log(3)+x*exp(9))/(x*exp(9)-1))^2+((2*x^3- 2*x^2)*exp(9)^2+(-4*x^2+4*x)*exp(9)+2*x-2)*exp(((x*exp(9)-1)*log(2)*log(3) +x*exp(9))/(x*exp(9)-1))+(x^4-2*x^3+x^2)*exp(9)^2+(-2*x^3+4*x^2-2*x)*exp(9 )+x^2-2*x+1),x, algorithm=\
Time = 10.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {2-4 e^9 x+2 e^{18} x^2+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 e^9 x-2 e^{18} x^2\right )}{1-2 x+x^2+e^{\frac {2 \left (e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)\right )}{-1+e^9 x}} \left (1-2 e^9 x+e^{18} x^2\right )+e^9 \left (-2 x+4 x^2-2 x^3\right )+e^{18} \left (x^2-2 x^3+x^4\right )+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 x+e^9 \left (4 x-4 x^2\right )+e^{18} \left (-2 x^2+2 x^3\right )\right )} \, dx=-\frac {2\,x}{x+2^{\ln \left (3\right )}\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^9}{x\,{\mathrm {e}}^9-1}}-1} \]
int(-(4*x*exp(9) - 2*x^2*exp(18) + exp((x*exp(9) + log(2)*log(3)*(x*exp(9) - 1))/(x*exp(9) - 1))*(2*x^2*exp(18) - 2*x*exp(9) + 2) - 2)/(exp(18)*(x^2 - 2*x^3 + x^4) - 2*x - exp(9)*(2*x - 4*x^2 + 2*x^3) + exp((x*exp(9) + log (2)*log(3)*(x*exp(9) - 1))/(x*exp(9) - 1))*(2*x + exp(9)*(4*x - 4*x^2) - e xp(18)*(2*x^2 - 2*x^3) - 2) + exp((2*(x*exp(9) + log(2)*log(3)*(x*exp(9) - 1)))/(x*exp(9) - 1))*(x^2*exp(18) - 2*x*exp(9) + 1) + x^2 + 1),x)