Integrand size = 72, antiderivative size = 25 \[ \int \frac {9-12 e^x+e^{x^2 \log ^2(2)} \left (-3-6 x^2 \log ^2(2)+e^x \left (3+6 x \log ^2(2)\right )\right )}{-4 e^x+e^{x^2 \log ^2(2)} \left (e^x-x\right )+3 x} \, dx=3 \log \left (x+\left (-4+e^{x^2 \log ^2(2)}\right ) \left (-e^x+x\right )\right ) \]
\[ \int \frac {9-12 e^x+e^{x^2 \log ^2(2)} \left (-3-6 x^2 \log ^2(2)+e^x \left (3+6 x \log ^2(2)\right )\right )}{-4 e^x+e^{x^2 \log ^2(2)} \left (e^x-x\right )+3 x} \, dx=\int \frac {9-12 e^x+e^{x^2 \log ^2(2)} \left (-3-6 x^2 \log ^2(2)+e^x \left (3+6 x \log ^2(2)\right )\right )}{-4 e^x+e^{x^2 \log ^2(2)} \left (e^x-x\right )+3 x} \, dx \]
Integrate[(9 - 12*E^x + E^(x^2*Log[2]^2)*(-3 - 6*x^2*Log[2]^2 + E^x*(3 + 6 *x*Log[2]^2)))/(-4*E^x + E^(x^2*Log[2]^2)*(E^x - x) + 3*x),x]
Integrate[(9 - 12*E^x + E^(x^2*Log[2]^2)*(-3 - 6*x^2*Log[2]^2 + E^x*(3 + 6 *x*Log[2]^2)))/(-4*E^x + E^(x^2*Log[2]^2)*(E^x - x) + 3*x), 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.
\(\displaystyle \int \frac {e^{x^2 \log ^2(2)} \left (-6 x^2 \log ^2(2)+e^x \left (6 x \log ^2(2)+3\right )-3\right )-12 e^x+9}{\left (e^x-x\right ) e^{x^2 \log ^2(2)}-4 e^x+3 x} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {3 \left (-2 x^2 \log ^2(2)+e^x+2 e^x x \log ^2(2)-1\right )}{e^x-x}+\frac {3 \left (6 x^3 \log ^2(2)-14 e^x x^2 \log ^2(2)+e^x x-e^x+8 e^{2 x} x \log ^2(2)\right )}{\left (e^x-x\right ) \left (x \left (-e^{x^2 \log ^2(2)}\right )+e^{x^2 \log ^2(2)+x}+3 x-4 e^x\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -3 \int \frac {e^x}{\left (e^x-x\right ) \left (-e^{x^2 \log ^2(2)} x+3 x-4 e^x+e^{\log ^2(2) x^2+x}\right )}dx+3 \int \frac {e^x x}{\left (e^x-x\right ) \left (-e^{x^2 \log ^2(2)} x+3 x-4 e^x+e^{\log ^2(2) x^2+x}\right )}dx+24 \log ^2(2) \int \frac {e^{2 x} x}{\left (e^x-x\right ) \left (-e^{x^2 \log ^2(2)} x+3 x-4 e^x+e^{\log ^2(2) x^2+x}\right )}dx-42 \log ^2(2) \int \frac {e^x x^2}{\left (e^x-x\right ) \left (-e^{x^2 \log ^2(2)} x+3 x-4 e^x+e^{\log ^2(2) x^2+x}\right )}dx+18 \log ^2(2) \int \frac {x^3}{\left (e^x-x\right ) \left (-e^{x^2 \log ^2(2)} x+3 x-4 e^x+e^{\log ^2(2) x^2+x}\right )}dx-3 \int \frac {1}{e^x-x}dx+3 \int \frac {x}{e^x-x}dx+3 x^2 \log ^2(2)+3 x\) |
Int[(9 - 12*E^x + E^(x^2*Log[2]^2)*(-3 - 6*x^2*Log[2]^2 + E^x*(3 + 6*x*Log [2]^2)))/(-4*E^x + E^(x^2*Log[2]^2)*(E^x - x) + 3*x),x]
3.23.19.3.1 Defintions of rubi rules used
Time = 0.21 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44
method | result | size |
norman | \(3 \ln \left ({\mathrm e}^{x^{2} \ln \left (2\right )^{2}} x -{\mathrm e}^{x^{2} \ln \left (2\right )^{2}} {\mathrm e}^{x}-3 x +4 \,{\mathrm e}^{x}\right )\) | \(36\) |
parallelrisch | \(3 \ln \left ({\mathrm e}^{x^{2} \ln \left (2\right )^{2}} x -{\mathrm e}^{x^{2} \ln \left (2\right )^{2}} {\mathrm e}^{x}-3 x +4 \,{\mathrm e}^{x}\right )\) | \(36\) |
risch | \(3 \ln \left ({\mathrm e}^{x}-x \right )+3 \ln \left ({\mathrm e}^{x^{2} \ln \left (2\right )^{2}}-\frac {3 x -4 \,{\mathrm e}^{x}}{x -{\mathrm e}^{x}}\right )\) | \(42\) |
int((((6*x*ln(2)^2+3)*exp(x)-6*x^2*ln(2)^2-3)*exp(x^2*ln(2)^2)-12*exp(x)+9 )/((exp(x)-x)*exp(x^2*ln(2)^2)-4*exp(x)+3*x),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.84 \[ \int \frac {9-12 e^x+e^{x^2 \log ^2(2)} \left (-3-6 x^2 \log ^2(2)+e^x \left (3+6 x \log ^2(2)\right )\right )}{-4 e^x+e^{x^2 \log ^2(2)} \left (e^x-x\right )+3 x} \, dx=3 \, \log \left (-x + e^{x}\right ) + 3 \, \log \left (\frac {{\left (x - e^{x}\right )} e^{\left (x^{2} \log \left (2\right )^{2}\right )} - 3 \, x + 4 \, e^{x}}{x - e^{x}}\right ) \]
integrate((((6*x*log(2)^2+3)*exp(x)-6*x^2*log(2)^2-3)*exp(x^2*log(2)^2)-12 *exp(x)+9)/((exp(x)-x)*exp(x^2*log(2)^2)-4*exp(x)+3*x),x, algorithm=\
Time = 0.31 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {9-12 e^x+e^{x^2 \log ^2(2)} \left (-3-6 x^2 \log ^2(2)+e^x \left (3+6 x \log ^2(2)\right )\right )}{-4 e^x+e^{x^2 \log ^2(2)} \left (e^x-x\right )+3 x} \, dx=3 \log {\left (- x + e^{x} \right )} + 3 \log {\left (\frac {- 3 x + 4 e^{x}}{x - e^{x}} + e^{x^{2} \log {\left (2 \right )}^{2}} \right )} \]
integrate((((6*x*ln(2)**2+3)*exp(x)-6*x**2*ln(2)**2-3)*exp(x**2*ln(2)**2)- 12*exp(x)+9)/((exp(x)-x)*exp(x**2*ln(2)**2)-4*exp(x)+3*x),x)
Time = 0.30 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.84 \[ \int \frac {9-12 e^x+e^{x^2 \log ^2(2)} \left (-3-6 x^2 \log ^2(2)+e^x \left (3+6 x \log ^2(2)\right )\right )}{-4 e^x+e^{x^2 \log ^2(2)} \left (e^x-x\right )+3 x} \, dx=3 \, \log \left (-x + e^{x}\right ) + 3 \, \log \left (\frac {{\left (x - e^{x}\right )} e^{\left (x^{2} \log \left (2\right )^{2}\right )} - 3 \, x + 4 \, e^{x}}{x - e^{x}}\right ) \]
integrate((((6*x*log(2)^2+3)*exp(x)-6*x^2*log(2)^2-3)*exp(x^2*log(2)^2)-12 *exp(x)+9)/((exp(x)-x)*exp(x^2*log(2)^2)-4*exp(x)+3*x),x, algorithm=\
Time = 0.31 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int \frac {9-12 e^x+e^{x^2 \log ^2(2)} \left (-3-6 x^2 \log ^2(2)+e^x \left (3+6 x \log ^2(2)\right )\right )}{-4 e^x+e^{x^2 \log ^2(2)} \left (e^x-x\right )+3 x} \, dx=3 \, \log \left (x e^{\left (x^{2} \log \left (2\right )^{2}\right )} - 3 \, x - e^{\left (x^{2} \log \left (2\right )^{2} + x\right )} + 4 \, e^{x}\right ) \]
integrate((((6*x*log(2)^2+3)*exp(x)-6*x^2*log(2)^2-3)*exp(x^2*log(2)^2)-12 *exp(x)+9)/((exp(x)-x)*exp(x^2*log(2)^2)-4*exp(x)+3*x),x, algorithm=\
Time = 12.65 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {9-12 e^x+e^{x^2 \log ^2(2)} \left (-3-6 x^2 \log ^2(2)+e^x \left (3+6 x \log ^2(2)\right )\right )}{-4 e^x+e^{x^2 \log ^2(2)} \left (e^x-x\right )+3 x} \, dx=3\,\ln \left (3\,x+{\mathrm {e}}^{{\ln \left (2\right )}^2\,x^2+x}-4\,{\mathrm {e}}^x-x\,{\mathrm {e}}^{x^2\,{\ln \left (2\right )}^2}\right ) \]