Integrand size = 74, antiderivative size = 29 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=e^{7+x+\frac {1}{3} x \left (x-\left (\frac {3}{5}+x\right ) \log \left (e^{-e^x} x\right )\right )} \] Output:
exp(7+1/3*x*(x-ln(x/exp(exp(x)))*(x+3/5))+x)
Time = 1.98 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=e^{7+x+\frac {x^2}{3}} \left (e^{-e^x} x\right )^{-\frac {1}{15} x (3+5 x)} \] Input:
Integrate[(E^((105 + 15*x + 5*x^2 + (-3*x - 5*x^2)*Log[x/E^E^x])/15)*(12 + 5*x + E^x*(3*x + 5*x^2) + (-3 - 10*x)*Log[x/E^E^x]))/15,x]
Output:
E^(7 + x + x^2/3)/(x/E^E^x)^((x*(3 + 5*x))/15)
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.
| \(\displaystyle \int \frac {1}{15} \left (e^x \left (5 x^2+3 x\right )+5 x+(-10 x-3) \log \left (e^{-e^x} x\right )+12\right ) \exp \left (\frac {1}{15} \left (5 x^2+\left (-5 x^2-3 x\right ) \log \left (e^{-e^x} x\right )+15 x+105\right )\right ) \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{15} \int e^{\frac {1}{3} \left (x^2+3 x+21\right )} \left (e^{-e^x} x\right )^{\frac {1}{15} \left (-5 x^2-3 x\right )} \left (5 x+e^x \left (5 x^2+3 x\right )-(10 x+3) \log \left (e^{-e^x} x\right )+12\right )dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \frac {1}{15} \int e^{\frac {1}{3} \left (x^2+3 x+21\right )} \left (e^{-e^x} x\right )^{\frac {1}{15} (-5 x-3) x} \left (5 x+e^x \left (5 x^2+3 x\right )-(10 x+3) \log \left (e^{-e^x} x\right )+12\right )dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {1}{15} \int \left (12 e^{\frac {1}{3} \left (x^2+3 x+21\right )} \left (e^{-e^x} x\right )^{\frac {1}{15} (-5 x-3) x}+5 e^{\frac {1}{3} \left (x^2+3 x+21\right )} x \left (e^{-e^x} x\right )^{\frac {1}{15} (-5 x-3) x}+e^{x+\frac {1}{3} \left (x^2+3 x+21\right )} x (5 x+3) \left (e^{-e^x} x\right )^{\frac {1}{15} (-5 x-3) x}-e^{\frac {1}{3} \left (x^2+3 x+21\right )} (10 x+3) \log \left (e^{-e^x} x\right ) \left (e^{-e^x} x\right )^{\frac {1}{15} (-5 x-3) x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{15} \left (12 \int e^{\frac {1}{3} \left (x^2+3 x+21\right )} \left (e^{-e^x} x\right )^{\frac {1}{15} (-5 x-3) x}dx+3 \int e^{\frac {x^2}{3}+2 x+7} x \left (e^{-e^x} x\right )^{\frac {1}{15} (-5 x-3) x}dx+5 \int e^{\frac {1}{3} \left (x^2+3 x+21\right )} x \left (e^{-e^x} x\right )^{\frac {1}{15} (-5 x-3) x}dx+5 \int e^{\frac {x^2}{3}+2 x+7} x^2 \left (e^{-e^x} x\right )^{\frac {1}{15} (-5 x-3) x}dx-3 \int e^x \int e^{\frac {x^2}{3}+x+7} \left (e^{-e^x} x\right )^{-\frac {1}{15} x (5 x+3)}dxdx+3 \int \frac {\int e^{\frac {x^2}{3}+x+7} \left (e^{-e^x} x\right )^{-\frac {1}{15} x (5 x+3)}dx}{x}dx-10 \int e^x \int e^{\frac {x^2}{3}+x+7} x \left (e^{-e^x} x\right )^{-\frac {1}{15} x (5 x+3)}dxdx+10 \int \frac {\int e^{\frac {x^2}{3}+x+7} x \left (e^{-e^x} x\right )^{-\frac {1}{15} x (5 x+3)}dx}{x}dx-3 \log \left (e^{-e^x} x\right ) \int e^{\frac {1}{3} \left (x^2+3 x+21\right )} \left (e^{-e^x} x\right )^{\frac {1}{15} (-5 x-3) x}dx-10 \log \left (e^{-e^x} x\right ) \int e^{\frac {1}{3} \left (x^2+3 x+21\right )} x \left (e^{-e^x} x\right )^{\frac {1}{15} (-5 x-3) x}dx\right )\) |
Input:
Int[(E^((105 + 15*x + 5*x^2 + (-3*x - 5*x^2)*Log[x/E^E^x])/15)*(12 + 5*x + E^x*(3*x + 5*x^2) + (-3 - 10*x)*Log[x/E^E^x]))/15,x]
Output:
$Aborted
Time = 0.44 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00
| method | result | size |
| parallelrisch | \({\mathrm e}^{\frac {\left (-5 x^{2}-3 x \right ) \ln \left (x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )}{15}+\frac {x^{2}}{3}+x +7}\) | \(29\) |
| risch | \(x^{-\frac {x^{2}}{3}} x^{-\frac {x}{5}} \left ({\mathrm e}^{{\mathrm e}^{x}}\right )^{\frac {x^{2}}{3}} \left ({\mathrm e}^{{\mathrm e}^{x}}\right )^{\frac {x}{5}} {\mathrm e}^{7+\frac {i \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{3} \pi \,x^{2}}{6}+\frac {i \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{3} \pi x}{10}-\frac {i \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{2} \operatorname {csgn}\left (i x \right ) \pi \,x^{2}}{6}-\frac {i \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{2} \operatorname {csgn}\left (i x \right ) \pi x}{10}-\frac {i \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{2} \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{x}}\right ) x^{2}}{6}-\frac {i \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{2} \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{x}}\right ) x}{10}+\frac {i \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right ) \operatorname {csgn}\left (i x \right ) \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{x}}\right ) x^{2}}{6}+\frac {i \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right ) \operatorname {csgn}\left (i x \right ) \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{x}}\right ) x}{10}+\frac {x^{2}}{3}+x}\) | \(235\) |
Input:
int(1/15*((-10*x-3)*ln(x/exp(exp(x)))+(5*x^2+3*x)*exp(x)+5*x+12)*exp(1/15* (-5*x^2-3*x)*ln(x/exp(exp(x)))+1/3*x^2+x+7),x,method=_RETURNVERBOSE)
Output:
exp(1/15*(-5*x^2-3*x)*ln(x/exp(exp(x)))+1/3*x^2+x+7)
Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=e^{\left (\frac {1}{3} \, x^{2} - \frac {1}{15} \, {\left (5 \, x^{2} + 3 \, x\right )} \log \left (x e^{\left (-e^{x}\right )}\right ) + x + 7\right )} \] Input:
integrate(1/15*((-10*x-3)*log(x/exp(exp(x)))+(5*x^2+3*x)*exp(x)+5*x+12)*ex p(1/15*(-5*x^2-3*x)*log(x/exp(exp(x)))+1/3*x^2+x+7),x, algorithm="fricas")
Output:
e^(1/3*x^2 - 1/15*(5*x^2 + 3*x)*log(x*e^(-e^x)) + x + 7)
Timed out. \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=\text {Timed out} \] Input:
integrate(1/15*((-10*x-3)*ln(x/exp(exp(x)))+(5*x**2+3*x)*exp(x)+5*x+12)*ex p(1/15*(-5*x**2-3*x)*ln(x/exp(exp(x)))+1/3*x**2+x+7),x)
Output:
Timed out
Time = 0.19 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=e^{\left (\frac {1}{3} \, x^{2} e^{x} - \frac {1}{3} \, x^{2} \log \left (x\right ) + \frac {1}{3} \, x^{2} + \frac {1}{5} \, x e^{x} - \frac {1}{5} \, x \log \left (x\right ) + x + 7\right )} \] Input:
integrate(1/15*((-10*x-3)*log(x/exp(exp(x)))+(5*x^2+3*x)*exp(x)+5*x+12)*ex p(1/15*(-5*x^2-3*x)*log(x/exp(exp(x)))+1/3*x^2+x+7),x, algorithm="maxima")
Output:
e^(1/3*x^2*e^x - 1/3*x^2*log(x) + 1/3*x^2 + 1/5*x*e^x - 1/5*x*log(x) + x + 7)
Time = 0.15 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=e^{\left (-\frac {1}{3} \, x^{2} \log \left (x e^{\left (-e^{x}\right )}\right ) + \frac {1}{3} \, x^{2} - \frac {1}{5} \, x \log \left (x e^{\left (-e^{x}\right )}\right ) + x + 7\right )} \] Input:
integrate(1/15*((-10*x-3)*log(x/exp(exp(x)))+(5*x^2+3*x)*exp(x)+5*x+12)*ex p(1/15*(-5*x^2-3*x)*log(x/exp(exp(x)))+1/3*x^2+x+7),x, algorithm="giac")
Output:
e^(-1/3*x^2*log(x*e^(-e^x)) + 1/3*x^2 - 1/5*x*log(x*e^(-e^x)) + x + 7)
Time = 3.72 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=\frac {{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^x}{5}}\,{\mathrm {e}}^7\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^x}{3}}\,{\mathrm {e}}^{\frac {x^2}{3}}\,{\mathrm {e}}^x}{x^{\frac {x^2}{3}+\frac {x}{5}}} \] Input:
int((exp(x - (log(x*exp(-exp(x)))*(3*x + 5*x^2))/15 + x^2/3 + 7)*(5*x - lo g(x*exp(-exp(x)))*(10*x + 3) + exp(x)*(3*x + 5*x^2) + 12))/15,x)
Output:
(exp((x*exp(x))/5)*exp(7)*exp((x^2*exp(x))/3)*exp(x^2/3)*exp(x))/x^(x/5 + x^2/3)
\[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=\frac {e^{7} \left (12 \left (\int \frac {e^{\frac {e^{x} x^{2}}{3}+\frac {e^{x} x}{5}+\frac {x^{2}}{3}+x}}{x^{\frac {1}{3} x^{2}+\frac {1}{5} x}}d x \right )+5 \left (\int \frac {e^{\frac {e^{x} x^{2}}{3}+\frac {e^{x} x}{5}+\frac {x^{2}}{3}+2 x} x^{2}}{x^{\frac {1}{3} x^{2}+\frac {1}{5} x}}d x \right )+3 \left (\int \frac {e^{\frac {e^{x} x^{2}}{3}+\frac {e^{x} x}{5}+\frac {x^{2}}{3}+2 x} x}{x^{\frac {1}{3} x^{2}+\frac {1}{5} x}}d x \right )-10 \left (\int \frac {e^{\frac {e^{x} x^{2}}{3}+\frac {e^{x} x}{5}+\frac {x^{2}}{3}+x} \mathrm {log}\left (\frac {x}{e^{e^{x}}}\right ) x}{x^{\frac {1}{3} x^{2}+\frac {1}{5} x}}d x \right )-3 \left (\int \frac {e^{\frac {e^{x} x^{2}}{3}+\frac {e^{x} x}{5}+\frac {x^{2}}{3}+x} \mathrm {log}\left (\frac {x}{e^{e^{x}}}\right )}{x^{\frac {1}{3} x^{2}+\frac {1}{5} x}}d x \right )+5 \left (\int \frac {e^{\frac {e^{x} x^{2}}{3}+\frac {e^{x} x}{5}+\frac {x^{2}}{3}+x} x}{x^{\frac {1}{3} x^{2}+\frac {1}{5} x}}d x \right )\right )}{15} \] Input:
int(1/15*((-10*x-3)*log(x/exp(exp(x)))+(5*x^2+3*x)*exp(x)+5*x+12)*exp(1/15 *(-5*x^2-3*x)*log(x/exp(exp(x)))+1/3*x^2+x+7),x)
Output:
(e**7*(12*int(e**((5*e**x*x**2 + 3*e**x*x + 5*x**2 + 15*x)/15)/x**((5*x**2 + 3*x)/15),x) + 5*int((e**((5*e**x*x**2 + 3*e**x*x + 5*x**2 + 30*x)/15)*x **2)/x**((5*x**2 + 3*x)/15),x) + 3*int((e**((5*e**x*x**2 + 3*e**x*x + 5*x* *2 + 30*x)/15)*x)/x**((5*x**2 + 3*x)/15),x) - 10*int((e**((5*e**x*x**2 + 3 *e**x*x + 5*x**2 + 15*x)/15)*log(x/e**(e**x))*x)/x**((5*x**2 + 3*x)/15),x) - 3*int((e**((5*e**x*x**2 + 3*e**x*x + 5*x**2 + 15*x)/15)*log(x/e**(e**x) ))/x**((5*x**2 + 3*x)/15),x) + 5*int((e**((5*e**x*x**2 + 3*e**x*x + 5*x**2 + 15*x)/15)*x)/x**((5*x**2 + 3*x)/15),x)))/15