Integrand size = 125, antiderivative size = 32 \[ \int e^{e^{-\frac {2+x+2 x^2-\log (4)}{x}} x-\frac {2+x+2 x^2-\log (4)}{x}} \left (2+3 x-x^2-2 x^3+e^{\frac {2+x+2 x^2-\log (4)}{x}} (1+2 x)+e^{e^4} \left (2+e^{\frac {2+x+2 x^2-\log (4)}{x}}+x-2 x^2-\log (4)\right )+(-1-x) \log (4)\right ) \, dx=e^{e^{-2 x-\frac {2+x-\log (4)}{x}} x} x \left (1+e^{e^4}+x\right ) \]
Time = 0.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int e^{e^{-\frac {2+x+2 x^2-\log (4)}{x}} x-\frac {2+x+2 x^2-\log (4)}{x}} \left (2+3 x-x^2-2 x^3+e^{\frac {2+x+2 x^2-\log (4)}{x}} (1+2 x)+e^{e^4} \left (2+e^{\frac {2+x+2 x^2-\log (4)}{x}}+x-2 x^2-\log (4)\right )+(-1-x) \log (4)\right ) \, dx=e^{4^{\frac {1}{x}} e^{-1-\frac {2}{x}-2 x} x} x \left (1+e^{e^4}+x\right ) \]
Integrate[E^(x/E^((2 + x + 2*x^2 - Log[4])/x) - (2 + x + 2*x^2 - Log[4])/x )*(2 + 3*x - x^2 - 2*x^3 + E^((2 + x + 2*x^2 - Log[4])/x)*(1 + 2*x) + E^E^ 4*(2 + E^((2 + x + 2*x^2 - Log[4])/x) + x - 2*x^2 - Log[4]) + (-1 - x)*Log [4]),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.
| \(\displaystyle \int \left (-2 x^3-x^2+(2 x+1) e^{\frac {2 x^2+x+2-\log (4)}{x}}+e^{e^4} \left (-2 x^2+e^{\frac {2 x^2+x+2-\log (4)}{x}}+x+2-\log (4)\right )+3 x+(-x-1) \log (4)+2\right ) \exp \left (x e^{-\frac {2 x^2+x+2-\log (4)}{x}}-\frac {2 x^2+x+2-\log (4)}{x}\right ) \, dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-x^2 \exp \left (x e^{-\frac {2 x^2+x+2-\log (4)}{x}}-\frac {2 x^2+x+2-\log (4)}{x}\right )+3 x \exp \left (x e^{-\frac {2 x^2+x+2-\log (4)}{x}}-\frac {2 x^2+x+2-\log (4)}{x}\right )+2 \exp \left (x e^{-\frac {2 x^2+x+2-\log (4)}{x}}-\frac {2 x^2+x+2-\log (4)}{x}\right )+4^{-1/x} (2 x+1) \exp \left (x e^{-\frac {2 x^2+x+2-\log (4)}{x}}-\frac {2 x^2+x+2-\log (4)}{x}+2 x+\frac {2}{x}+1\right )+2^{-2/x} \left (-2^{\frac {2}{x}+1} x^2+2^{2/x} x+e^{2 x+\frac {2}{x}+1}+2^{\frac {2}{x}+1} (1-\log (2))\right ) \exp \left (x e^{-\frac {2 x^2+x+2-\log (4)}{x}}-\frac {2 x^2+x+2-\log (4)}{x}+e^4\right )-(x+1) \log (4) \exp \left (x e^{-\frac {2 x^2+x+2-\log (4)}{x}}-\frac {2 x^2+x+2-\log (4)}{x}\right )-2 x^3 \exp \left (x e^{-\frac {2 x^2+x+2-\log (4)}{x}}-\frac {2 x^2+x+2-\log (4)}{x}\right )\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\log (4) \int \exp \left (e^{-\frac {2 x^2+x-\log (4)+2}{x}} x-\frac {2 x^2+x-\log (4)+2}{x}\right )dx+2 \int \exp \left (e^{-\frac {2 x^2+x-\log (4)+2}{x}} x-\frac {2 x^2+x-\log (4)+2}{x}\right )dx+2 (1-\log (2)) \int \exp \left (e^{-\frac {2 x^2+x-\log (4)+2}{x}} x+e^4-\frac {2 x^2+x-\log (4)+2}{x}\right )dx-\log (4) \int \exp \left (e^{-\frac {2 x^2+x-\log (4)+2}{x}} x-\frac {2 x^2+x-\log (4)+2}{x}\right ) xdx+3 \int \exp \left (e^{-\frac {2 x^2+x-\log (4)+2}{x}} x-\frac {2 x^2+x-\log (4)+2}{x}\right ) xdx+\int \exp \left (e^{-\frac {2 x^2+x-\log (4)+2}{x}} x+e^4-\frac {2 x^2+x-\log (4)+2}{x}\right ) xdx-\int \exp \left (e^{-\frac {2 x^2+x-\log (4)+2}{x}} x-\frac {2 x^2+x-\log (4)+2}{x}\right ) x^2dx-2 \int \exp \left (e^{-\frac {2 x^2+x-\log (4)+2}{x}} x+e^4-\frac {2 x^2+x-\log (4)+2}{x}\right ) x^2dx-2 \int \exp \left (e^{-\frac {2 x^2+x-\log (4)+2}{x}} x-\frac {2 x^2+x-\log (4)+2}{x}\right ) x^3dx+\int e^{4^{\frac {1}{x}} e^{-2 x-1-\frac {2}{x}} x}dx+\int e^{4^{\frac {1}{x}} e^{-2 x-1-\frac {2}{x}} x+e^4}dx+2 \int e^{4^{\frac {1}{x}} e^{-2 x-1-\frac {2}{x}} x} xdx\) |
Int[E^(x/E^((2 + x + 2*x^2 - Log[4])/x) - (2 + x + 2*x^2 - Log[4])/x)*(2 + 3*x - x^2 - 2*x^3 + E^((2 + x + 2*x^2 - Log[4])/x)*(1 + 2*x) + E^E^4*(2 + E^((2 + x + 2*x^2 - Log[4])/x) + x - 2*x^2 - Log[4]) + (-1 - x)*Log[4]),x ]
3.17.93.3.1 Defintions of rubi rules used
Time = 0.83 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97
| method | result | size |
| risch | \(x \left ({\mathrm e}^{{\mathrm e}^{4}}+x +1\right ) {\mathrm e}^{x \,{\mathrm e}^{\frac {-2 x^{2}+2 \ln \left (2\right )-x -2}{x}}}\) | \(31\) |
| parallelrisch | \(\frac {x^{2} {\mathrm e}^{x \,{\mathrm e}^{\frac {-2 x^{2}+2 \ln \left (2\right )-x -2}{x}}} {\mathrm e}^{{\mathrm e}^{4}}+{\mathrm e}^{x \,{\mathrm e}^{\frac {-2 x^{2}+2 \ln \left (2\right )-x -2}{x}}} x^{3}+x^{2} {\mathrm e}^{x \,{\mathrm e}^{\frac {-2 x^{2}+2 \ln \left (2\right )-x -2}{x}}}}{x}\) | \(96\) |
| norman | \(\left (x^{2} {\mathrm e}^{\frac {-2 \ln \left (2\right )+2 x^{2}+x +2}{x}} {\mathrm e}^{x \,{\mathrm e}^{-\frac {-2 \ln \left (2\right )+2 x^{2}+x +2}{x}}}+\left ({\mathrm e}^{{\mathrm e}^{4}}+1\right ) x \,{\mathrm e}^{\frac {-2 \ln \left (2\right )+2 x^{2}+x +2}{x}} {\mathrm e}^{x \,{\mathrm e}^{-\frac {-2 \ln \left (2\right )+2 x^{2}+x +2}{x}}}\right ) {\mathrm e}^{-\frac {-2 \ln \left (2\right )+2 x^{2}+x +2}{x}}\) | \(111\) |
int(((exp((-2*ln(2)+2*x^2+x+2)/x)-2*ln(2)-2*x^2+x+2)*exp(exp(4))+(1+2*x)*e xp((-2*ln(2)+2*x^2+x+2)/x)+2*(-1-x)*ln(2)-2*x^3-x^2+3*x+2)*exp(x/exp((-2*l n(2)+2*x^2+x+2)/x))/exp((-2*ln(2)+2*x^2+x+2)/x),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (28) = 56\).
Time = 0.26 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.16 \[ \int e^{e^{-\frac {2+x+2 x^2-\log (4)}{x}} x-\frac {2+x+2 x^2-\log (4)}{x}} \left (2+3 x-x^2-2 x^3+e^{\frac {2+x+2 x^2-\log (4)}{x}} (1+2 x)+e^{e^4} \left (2+e^{\frac {2+x+2 x^2-\log (4)}{x}}+x-2 x^2-\log (4)\right )+(-1-x) \log (4)\right ) \, dx={\left (x^{2} + x e^{\left (e^{4}\right )} + x\right )} e^{\left (\frac {x^{2} e^{\left (-\frac {2 \, x^{2} + x - 2 \, \log \left (2\right ) + 2}{x}\right )} - 2 \, x^{2} - x + 2 \, \log \left (2\right ) - 2}{x} + \frac {2 \, x^{2} + x - 2 \, \log \left (2\right ) + 2}{x}\right )} \]
integrate(((exp((-2*log(2)+2*x^2+x+2)/x)-2*log(2)-2*x^2+x+2)*exp(exp(4))+( 1+2*x)*exp((-2*log(2)+2*x^2+x+2)/x)+2*(-1-x)*log(2)-2*x^3-x^2+3*x+2)*exp(x /exp((-2*log(2)+2*x^2+x+2)/x))/exp((-2*log(2)+2*x^2+x+2)/x),x, algorithm=\
(x^2 + x*e^(e^4) + x)*e^((x^2*e^(-(2*x^2 + x - 2*log(2) + 2)/x) - 2*x^2 - x + 2*log(2) - 2)/x + (2*x^2 + x - 2*log(2) + 2)/x)
Time = 1.43 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int e^{e^{-\frac {2+x+2 x^2-\log (4)}{x}} x-\frac {2+x+2 x^2-\log (4)}{x}} \left (2+3 x-x^2-2 x^3+e^{\frac {2+x+2 x^2-\log (4)}{x}} (1+2 x)+e^{e^4} \left (2+e^{\frac {2+x+2 x^2-\log (4)}{x}}+x-2 x^2-\log (4)\right )+(-1-x) \log (4)\right ) \, dx=\left (x^{2} + x + x e^{e^{4}}\right ) e^{x e^{- \frac {2 x^{2} + x - 2 \log {\left (2 \right )} + 2}{x}}} \]
integrate(((exp((-2*ln(2)+2*x**2+x+2)/x)-2*ln(2)-2*x**2+x+2)*exp(exp(4))+( 1+2*x)*exp((-2*ln(2)+2*x**2+x+2)/x)+2*(-1-x)*ln(2)-2*x**3-x**2+3*x+2)*exp( x/exp((-2*ln(2)+2*x**2+x+2)/x))/exp((-2*ln(2)+2*x**2+x+2)/x),x)
Time = 0.32 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int e^{e^{-\frac {2+x+2 x^2-\log (4)}{x}} x-\frac {2+x+2 x^2-\log (4)}{x}} \left (2+3 x-x^2-2 x^3+e^{\frac {2+x+2 x^2-\log (4)}{x}} (1+2 x)+e^{e^4} \left (2+e^{\frac {2+x+2 x^2-\log (4)}{x}}+x-2 x^2-\log (4)\right )+(-1-x) \log (4)\right ) \, dx={\left (x^{2} + x {\left (e^{\left (e^{4}\right )} + 1\right )}\right )} e^{\left (x e^{\left (-2 \, x + \frac {2 \, \log \left (2\right )}{x} - \frac {2}{x} - 1\right )}\right )} \]
integrate(((exp((-2*log(2)+2*x^2+x+2)/x)-2*log(2)-2*x^2+x+2)*exp(exp(4))+( 1+2*x)*exp((-2*log(2)+2*x^2+x+2)/x)+2*(-1-x)*log(2)-2*x^3-x^2+3*x+2)*exp(x /exp((-2*log(2)+2*x^2+x+2)/x))/exp((-2*log(2)+2*x^2+x+2)/x),x, algorithm=\
\[ \int e^{e^{-\frac {2+x+2 x^2-\log (4)}{x}} x-\frac {2+x+2 x^2-\log (4)}{x}} \left (2+3 x-x^2-2 x^3+e^{\frac {2+x+2 x^2-\log (4)}{x}} (1+2 x)+e^{e^4} \left (2+e^{\frac {2+x+2 x^2-\log (4)}{x}}+x-2 x^2-\log (4)\right )+(-1-x) \log (4)\right ) \, dx=\int { -{\left (2 \, x^{3} + x^{2} - {\left (2 \, x + 1\right )} e^{\left (\frac {2 \, x^{2} + x - 2 \, \log \left (2\right ) + 2}{x}\right )} + {\left (2 \, x^{2} - x - e^{\left (\frac {2 \, x^{2} + x - 2 \, \log \left (2\right ) + 2}{x}\right )} + 2 \, \log \left (2\right ) - 2\right )} e^{\left (e^{4}\right )} + 2 \, {\left (x + 1\right )} \log \left (2\right ) - 3 \, x - 2\right )} e^{\left (x e^{\left (-\frac {2 \, x^{2} + x - 2 \, \log \left (2\right ) + 2}{x}\right )} - \frac {2 \, x^{2} + x - 2 \, \log \left (2\right ) + 2}{x}\right )} \,d x } \]
integrate(((exp((-2*log(2)+2*x^2+x+2)/x)-2*log(2)-2*x^2+x+2)*exp(exp(4))+( 1+2*x)*exp((-2*log(2)+2*x^2+x+2)/x)+2*(-1-x)*log(2)-2*x^3-x^2+3*x+2)*exp(x /exp((-2*log(2)+2*x^2+x+2)/x))/exp((-2*log(2)+2*x^2+x+2)/x),x, algorithm=\
integrate(-(2*x^3 + x^2 - (2*x + 1)*e^((2*x^2 + x - 2*log(2) + 2)/x) + (2* x^2 - x - e^((2*x^2 + x - 2*log(2) + 2)/x) + 2*log(2) - 2)*e^(e^4) + 2*(x + 1)*log(2) - 3*x - 2)*e^(x*e^(-(2*x^2 + x - 2*log(2) + 2)/x) - (2*x^2 + x - 2*log(2) + 2)/x), x)
Timed out. \[ \int e^{e^{-\frac {2+x+2 x^2-\log (4)}{x}} x-\frac {2+x+2 x^2-\log (4)}{x}} \left (2+3 x-x^2-2 x^3+e^{\frac {2+x+2 x^2-\log (4)}{x}} (1+2 x)+e^{e^4} \left (2+e^{\frac {2+x+2 x^2-\log (4)}{x}}+x-2 x^2-\log (4)\right )+(-1-x) \log (4)\right ) \, dx=\int {\mathrm {e}}^{-\frac {2\,x^2+x-2\,\ln \left (2\right )+2}{x}}\,{\mathrm {e}}^{x\,{\mathrm {e}}^{-\frac {2\,x^2+x-2\,\ln \left (2\right )+2}{x}}}\,\left (3\,x+{\mathrm {e}}^{\frac {2\,x^2+x-2\,\ln \left (2\right )+2}{x}}\,\left (2\,x+1\right )+{\mathrm {e}}^{{\mathrm {e}}^4}\,\left (x-2\,\ln \left (2\right )+{\mathrm {e}}^{\frac {2\,x^2+x-2\,\ln \left (2\right )+2}{x}}-2\,x^2+2\right )-2\,\ln \left (2\right )\,\left (x+1\right )-x^2-2\,x^3+2\right ) \,d x \]
int(exp(-(x - 2*log(2) + 2*x^2 + 2)/x)*exp(x*exp(-(x - 2*log(2) + 2*x^2 + 2)/x))*(3*x + exp((x - 2*log(2) + 2*x^2 + 2)/x)*(2*x + 1) + exp(exp(4))*(x - 2*log(2) + exp((x - 2*log(2) + 2*x^2 + 2)/x) - 2*x^2 + 2) - 2*log(2)*(x + 1) - x^2 - 2*x^3 + 2),x)