Integrand size = 111, antiderivative size = 24 \[ \int \frac {e^{-3+e^{\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}} x+\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}} \left (-e^5 x+e^3 \left (9+18 x^2+27 x^3\right ) \log (2)\right )}{9 \log (2)} \, dx=e^{e^{x \left (x+x^2-\frac {e^2}{9 \log (2)}\right )} x} \] Output:
exp(x*exp(x*(x^2+x-1/9*exp(5)/exp(3)/ln(2))))
\[ \int \frac {e^{-3+e^{\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}} x+\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}} \left (-e^5 x+e^3 \left (9+18 x^2+27 x^3\right ) \log (2)\right )}{9 \log (2)} \, dx=\int \frac {e^{-3+e^{\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}} x+\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}} \left (-e^5 x+e^3 \left (9+18 x^2+27 x^3\right ) \log (2)\right )}{9 \log (2)} \, dx \] Input:
Integrate[(E^(-3 + E^((-(E^5*x) + E^3*(9*x^2 + 9*x^3)*Log[2])/(9*E^3*Log[2 ]))*x + (-(E^5*x) + E^3*(9*x^2 + 9*x^3)*Log[2])/(9*E^3*Log[2]))*(-(E^5*x) + E^3*(9 + 18*x^2 + 27*x^3)*Log[2]))/(9*Log[2]),x]
Output:
Integrate[E^(-3 + E^((-(E^5*x) + E^3*(9*x^2 + 9*x^3)*Log[2])/(9*E^3*Log[2] ))*x + (-(E^5*x) + E^3*(9*x^2 + 9*x^3)*Log[2])/(9*E^3*Log[2]))*(-(E^5*x) + E^3*(9 + 18*x^2 + 27*x^3)*Log[2]), x]/(9*Log[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.
| \(\displaystyle \int \frac {\left (e^3 \left (27 x^3+18 x^2+9\right ) \log (2)-e^5 x\right ) \exp \left (x \exp \left (\frac {e^3 \left (9 x^3+9 x^2\right ) \log (2)-e^5 x}{9 e^3 \log (2)}\right )+\frac {e^3 \left (9 x^3+9 x^2\right ) \log (2)-e^5 x}{9 e^3 \log (2)}-3\right )}{9 \log (2)} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int -\exp \left (2^{\frac {9 x^3+9 x^2}{9 \log (2)}} e^{-\frac {e^2 x}{9 \log (2)}} x-\frac {e^2 x-9 \left (x^3+x^2\right ) \log (2)}{9 \log (2)}-3\right ) \left (e^5 x-9 e^3 \left (3 x^3+2 x^2+1\right ) \log (2)\right )dx}{9 \log (2)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \exp \left (2^{\frac {x^3+x^2}{\log (2)}} e^{-\frac {e^2 x}{9 \log (2)}} x-\frac {e^2 x-9 \left (x^3+x^2\right ) \log (2)}{9 \log (2)}-3\right ) \left (e^5 x-9 e^3 \left (3 x^3+2 x^2+1\right ) \log (2)\right )dx}{9 \log (2)}\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle -\frac {\int \exp \left (\frac {\log (512) x^3+\log (512) x^2+e^{x \left (x^2+x-\frac {e^2}{\log (512)}\right )} \log (512) x-e^2 x-3 \log (512)}{\log (512)}\right ) \left (-27 e^3 \log (2) x^3-18 e^3 \log (2) x^2+e^5 x-9 e^3 \log (2)\right )dx}{9 \log (2)}\) |
| \(\Big \downarrow \) 2704 |
| \(\displaystyle -\frac {\int 512^{-\frac {3}{\log (512)}} \exp \left (\frac {\log (512) x^3+\log (512) x^2+e^{x \left (x^2+x-\frac {e^2}{\log (512)}\right )} \log (512) x-e^2 x}{\log (512)}\right ) \left (-27 e^3 \log (2) x^3-18 e^3 \log (2) x^2+e^5 x-9 e^3 \log (2)\right )dx}{9 \log (2)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int 512^{\frac {x^3}{\log (512)}+\frac {x^2}{\log (512)}+\frac {e^{x \left (x^2+x-\frac {e^2}{\log (512)}\right )} x}{\log (512)}} e^{-\frac {e^2 x}{\log (512)}} \left (-27 e^3 \log (2) x^3-18 e^3 \log (2) x^2+e^5 x-9 e^3 \log (2)\right )dx}{9 e^3 \log (2)}\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle -\frac {\int 512^{\frac {x \left (x^2+x+e^{x \left (x^2+x-\frac {e^2}{\log (512)}\right )}\right )}{\log (512)}} e^{-\frac {e^2 x}{\log (512)}} \left (-27 e^3 \log (2) x^3-18 e^3 \log (2) x^2+e^5 x-9 e^3 \log (2)\right )dx}{9 e^3 \log (2)}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {\int \left (-27 512^{\frac {x \left (x^2+x+e^{x \left (x^2+x-\frac {e^2}{\log (512)}\right )}\right )}{\log (512)}} e^{3-\frac {e^2 x}{\log (512)}} \log (2) x^3-9\ 2^{\frac {9 x \left (x^2+x+e^{x \left (x^2+x-\frac {e^2}{\log (512)}\right )}\right )}{\log (512)}+1} e^{3-\frac {e^2 x}{\log (512)}} \log (2) x^2+512^{\frac {x \left (x^2+x+e^{x \left (x^2+x-\frac {e^2}{\log (512)}\right )}\right )}{\log (512)}} e^{5-\frac {e^2 x}{\log (512)}} x-9\ 512^{\frac {x \left (x^2+x+e^{x \left (x^2+x-\frac {e^2}{\log (512)}\right )}\right )}{\log (512)}} e^{3-\frac {e^2 x}{\log (512)}} \log (2)\right )dx}{9 e^3 \log (2)}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {-9 \log (2) \int 512^{\frac {x \left (x^2+x+e^{x \left (x^2+x-\frac {e^2}{\log (512)}\right )}\right )}{\log (512)}} e^{3-\frac {e^2 x}{\log (512)}}dx+\int 512^{\frac {x \left (x^2+x+e^{x \left (x^2+x-\frac {e^2}{\log (512)}\right )}\right )}{\log (512)}} e^{5-\frac {e^2 x}{\log (512)}} xdx-9 \log (2) \int 2^{\frac {9 x \left (x^2+x+e^{x \left (x^2+x-\frac {e^2}{\log (512)}\right )}\right )}{\log (512)}+1} e^{3-\frac {e^2 x}{\log (512)}} x^2dx-27 \log (2) \int 512^{\frac {x \left (x^2+x+e^{x \left (x^2+x-\frac {e^2}{\log (512)}\right )}\right )}{\log (512)}} e^{3-\frac {e^2 x}{\log (512)}} x^3dx}{9 e^3 \log (2)}\) |
Input:
Int[(E^(-3 + E^((-(E^5*x) + E^3*(9*x^2 + 9*x^3)*Log[2])/(9*E^3*Log[2]))*x + (-(E^5*x) + E^3*(9*x^2 + 9*x^3)*Log[2])/(9*E^3*Log[2]))*(-(E^5*x) + E^3* (9 + 18*x^2 + 27*x^3)*Log[2]))/(9*Log[2]),x]
Output:
$Aborted
Time = 1.12 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46
| method | result | size |
| risch | \({\mathrm e}^{x \,{\mathrm e}^{\frac {x \left (9 \ln \left (2\right ) {\mathrm e}^{3} x^{2}+9 x \,{\mathrm e}^{3} \ln \left (2\right )-{\mathrm e}^{5}\right ) {\mathrm e}^{-3}}{9 \ln \left (2\right )}}}\) | \(35\) |
| norman | \({\mathrm e}^{x \,{\mathrm e}^{\frac {\left (\left (9 x^{3}+9 x^{2}\right ) {\mathrm e}^{3} \ln \left (2\right )-x \,{\mathrm e}^{5}\right ) {\mathrm e}^{-3}}{9 \ln \left (2\right )}}}\) | \(37\) |
| parallelrisch | \({\mathrm e}^{x \,{\mathrm e}^{\frac {\left (\left (9 x^{3}+9 x^{2}\right ) {\mathrm e}^{3} \ln \left (2\right )-x \,{\mathrm e}^{5}\right ) {\mathrm e}^{-3}}{9 \ln \left (2\right )}}}\) | \(37\) |
Input:
int(1/9*((27*x^3+18*x^2+9)*exp(3)*ln(2)-x*exp(5))*exp(1/9*((9*x^3+9*x^2)*e xp(3)*ln(2)-x*exp(5))/exp(3)/ln(2))*exp(x*exp(1/9*((9*x^3+9*x^2)*exp(3)*ln (2)-x*exp(5))/exp(3)/ln(2)))/exp(3)/ln(2),x,method=_RETURNVERBOSE)
Output:
exp(x*exp(1/9*x*(9*ln(2)*exp(3)*x^2+9*x*exp(3)*ln(2)-exp(5))*exp(-3)/ln(2) ))
Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (24) = 48\).
Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 3.21 \[ \int \frac {e^{-3+e^{\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}} x+\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}} \left (-e^5 x+e^3 \left (9+18 x^2+27 x^3\right ) \log (2)\right )}{9 \log (2)} \, dx=e^{\left (\frac {9 \, x e^{\left (-\frac {x e^{2} - 9 \, {\left (x^{3} + x^{2}\right )} \log \left (2\right )}{9 \, \log \left (2\right )}\right )} \log \left (2\right ) - x e^{2} + 9 \, {\left (x^{3} + x^{2} - 3\right )} \log \left (2\right )}{9 \, \log \left (2\right )} + \frac {x e^{2} - 9 \, {\left (x^{3} + x^{2}\right )} \log \left (2\right )}{9 \, \log \left (2\right )} + 3\right )} \] Input:
integrate(1/9*((27*x^3+18*x^2+9)*exp(3)*log(2)-x*exp(5))*exp(1/9*((9*x^3+9 *x^2)*exp(3)*log(2)-x*exp(5))/exp(3)/log(2))*exp(x*exp(1/9*((9*x^3+9*x^2)* exp(3)*log(2)-x*exp(5))/exp(3)/log(2)))/exp(3)/log(2),x, algorithm="fricas ")
Output:
e^(1/9*(9*x*e^(-1/9*(x*e^2 - 9*(x^3 + x^2)*log(2))/log(2))*log(2) - x*e^2 + 9*(x^3 + x^2 - 3)*log(2))/log(2) + 1/9*(x*e^2 - 9*(x^3 + x^2)*log(2))/lo g(2) + 3)
Time = 0.21 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {e^{-3+e^{\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}} x+\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}} \left (-e^5 x+e^3 \left (9+18 x^2+27 x^3\right ) \log (2)\right )}{9 \log (2)} \, dx=e^{x e^{\frac {- \frac {x e^{5}}{9} + \frac {\left (9 x^{3} + 9 x^{2}\right ) e^{3} \log {\left (2 \right )}}{9}}{e^{3} \log {\left (2 \right )}}}} \] Input:
integrate(1/9*((27*x**3+18*x**2+9)*exp(3)*ln(2)-x*exp(5))*exp(1/9*((9*x**3 +9*x**2)*exp(3)*ln(2)-x*exp(5))/exp(3)/ln(2))*exp(x*exp(1/9*((9*x**3+9*x** 2)*exp(3)*ln(2)-x*exp(5))/exp(3)/ln(2)))/exp(3)/ln(2),x)
Output:
exp(x*exp((-x*exp(5)/9 + (9*x**3 + 9*x**2)*exp(3)*log(2)/9)*exp(-3)/log(2) ))
Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-3+e^{\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}} x+\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}} \left (-e^5 x+e^3 \left (9+18 x^2+27 x^3\right ) \log (2)\right )}{9 \log (2)} \, dx=e^{\left (x e^{\left (x^{3} + x^{2} - \frac {x e^{2}}{9 \, \log \left (2\right )}\right )}\right )} \] Input:
integrate(1/9*((27*x^3+18*x^2+9)*exp(3)*log(2)-x*exp(5))*exp(1/9*((9*x^3+9 *x^2)*exp(3)*log(2)-x*exp(5))/exp(3)/log(2))*exp(x*exp(1/9*((9*x^3+9*x^2)* exp(3)*log(2)-x*exp(5))/exp(3)/log(2)))/exp(3)/log(2),x, algorithm="maxima ")
Output:
e^(x*e^(x^3 + x^2 - 1/9*x*e^2/log(2)))
\[ \int \frac {e^{-3+e^{\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}} x+\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}} \left (-e^5 x+e^3 \left (9+18 x^2+27 x^3\right ) \log (2)\right )}{9 \log (2)} \, dx=\int { \frac {{\left (9 \, {\left (3 \, x^{3} + 2 \, x^{2} + 1\right )} e^{3} \log \left (2\right ) - x e^{5}\right )} e^{\left (x e^{\left (\frac {{\left (9 \, {\left (x^{3} + x^{2}\right )} e^{3} \log \left (2\right ) - x e^{5}\right )} e^{\left (-3\right )}}{9 \, \log \left (2\right )}\right )} + \frac {{\left (9 \, {\left (x^{3} + x^{2}\right )} e^{3} \log \left (2\right ) - x e^{5}\right )} e^{\left (-3\right )}}{9 \, \log \left (2\right )} - 3\right )}}{9 \, \log \left (2\right )} \,d x } \] Input:
integrate(1/9*((27*x^3+18*x^2+9)*exp(3)*log(2)-x*exp(5))*exp(1/9*((9*x^3+9 *x^2)*exp(3)*log(2)-x*exp(5))/exp(3)/log(2))*exp(x*exp(1/9*((9*x^3+9*x^2)* exp(3)*log(2)-x*exp(5))/exp(3)/log(2)))/exp(3)/log(2),x, algorithm="giac")
Output:
integrate(1/9*(9*(3*x^3 + 2*x^2 + 1)*e^3*log(2) - x*e^5)*e^(x*e^(1/9*(9*(x ^3 + x^2)*e^3*log(2) - x*e^5)*e^(-3)/log(2)) + 1/9*(9*(x^3 + x^2)*e^3*log( 2) - x*e^5)*e^(-3)/log(2) - 3)/log(2), x)
Time = 2.81 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-3+e^{\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}} x+\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}} \left (-e^5 x+e^3 \left (9+18 x^2+27 x^3\right ) \log (2)\right )}{9 \log (2)} \, dx={\mathrm {e}}^{x\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{x^3}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^2}{9\,\ln \left (2\right )}}} \] Input:
int(-(exp(-3)*exp(-(exp(-3)*((x*exp(5))/9 - (exp(3)*log(2)*(9*x^2 + 9*x^3) )/9))/log(2))*exp(x*exp(-(exp(-3)*((x*exp(5))/9 - (exp(3)*log(2)*(9*x^2 + 9*x^3))/9))/log(2)))*(x*exp(5) - exp(3)*log(2)*(18*x^2 + 27*x^3 + 9)))/(9* log(2)),x)
Output:
exp(x*exp(x^2)*exp(x^3)*exp(-(x*exp(2))/(9*log(2))))
Time = 0.78 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-3+e^{\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}} x+\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}} \left (-e^5 x+e^3 \left (9+18 x^2+27 x^3\right ) \log (2)\right )}{9 \log (2)} \, dx=e^{\frac {e^{x^{3}+x^{2}} x}{e^{\frac {e^{2} x}{9 \,\mathrm {log}\left (2\right )}}}} \] Input:
int(1/9*((27*x^3+18*x^2+9)*exp(3)*log(2)-x*exp(5))*exp(1/9*((9*x^3+9*x^2)* exp(3)*log(2)-x*exp(5))/exp(3)/log(2))*exp(x*exp(1/9*((9*x^3+9*x^2)*exp(3) *log(2)-x*exp(5))/exp(3)/log(2)))/exp(3)/log(2),x)
Output:
e**((e**(x**3 + x**2)*x)/e**((e**2*x)/(9*log(2))))