Integrand size = 71, antiderivative size = 28 \[ \int e^{-x+e^{-x} \left (-5 x+e^x \left (1+x^3\right )\right )+\frac {3-e^{e^{-x} \left (-5 x+e^x \left (1+x^3\right )\right )} \log (3)}{\log (3)}} \left (5-5 x-3 e^x x^2\right ) \, dx=e^{-e^{1+x \left (-5 e^{-x}+x^2\right )}+\frac {3}{\log (3)}} \] Output:
exp(3/ln(3)-exp(1+x*(x^2-5/exp(x))))
Time = 1.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int e^{-x+e^{-x} \left (-5 x+e^x \left (1+x^3\right )\right )+\frac {3-e^{e^{-x} \left (-5 x+e^x \left (1+x^3\right )\right )} \log (3)}{\log (3)}} \left (5-5 x-3 e^x x^2\right ) \, dx=e^{-e^{1-5 e^{-x} x+x^3}+\frac {3}{\log (3)}} \] Input:
Integrate[E^(-x + (-5*x + E^x*(1 + x^3))/E^x + (3 - E^((-5*x + E^x*(1 + x^ 3))/E^x)*Log[3])/Log[3])*(5 - 5*x - 3*E^x*x^2),x]
Output:
E^(-E^(1 - (5*x)/E^x + x^3) + 3/Log[3])
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.
| \(\displaystyle \int \left (-3 e^x x^2-5 x+5\right ) \exp \left (e^{-x} \left (e^x \left (x^3+1\right )-5 x\right )+\frac {3-e^{e^{-x} \left (e^x \left (x^3+1\right )-5 x\right )} \log (3)}{\log (3)}-x\right ) \, dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-5 x \exp \left (e^{-x} \left (e^x \left (x^3+1\right )-5 x\right )+\frac {3-e^{e^{-x} \left (e^x \left (x^3+1\right )-5 x\right )} \log (3)}{\log (3)}-x\right )+5 \exp \left (e^{-x} \left (e^x \left (x^3+1\right )-5 x\right )+\frac {3-e^{e^{-x} \left (e^x \left (x^3+1\right )-5 x\right )} \log (3)}{\log (3)}-x\right )-3 x^2 \exp \left (e^{-x} \left (e^x \left (x^3+1\right )-5 x\right )+\frac {3-e^{e^{-x} \left (e^x \left (x^3+1\right )-5 x\right )} \log (3)}{\log (3)}\right )\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 5 \int \exp \left (-x+e^{-x} \left (e^x \left (x^3+1\right )-5 x\right )+\frac {3-e^{e^{-x} \left (e^x \left (x^3+1\right )-5 x\right )} \log (3)}{\log (3)}\right )dx-5 \int \exp \left (-x+e^{-x} \left (e^x \left (x^3+1\right )-5 x\right )+\frac {3-e^{e^{-x} \left (e^x \left (x^3+1\right )-5 x\right )} \log (3)}{\log (3)}\right ) xdx-3 \int \exp \left (e^{-x} \left (e^x \left (x^3+1\right )-5 x\right )+\frac {3-e^{e^{-x} \left (e^x \left (x^3+1\right )-5 x\right )} \log (3)}{\log (3)}\right ) x^2dx\) |
Input:
Int[E^(-x + (-5*x + E^x*(1 + x^3))/E^x + (3 - E^((-5*x + E^x*(1 + x^3))/E^ x)*Log[3])/Log[3])*(5 - 5*x - 3*E^x*x^2),x]
Output:
$Aborted
Time = 0.33 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11
| method | result | size |
| norman | \({\mathrm e}^{\frac {-\ln \left (3\right ) {\mathrm e}^{\left (\left (x^{3}+1\right ) {\mathrm e}^{x}-5 x \right ) {\mathrm e}^{-x}}+3}{\ln \left (3\right )}}\) | \(31\) |
| risch | \({\mathrm e}^{-\frac {\ln \left (3\right ) {\mathrm e}^{\left ({\mathrm e}^{x} x^{3}+{\mathrm e}^{x}-5 x \right ) {\mathrm e}^{-x}}-3}{\ln \left (3\right )}}\) | \(31\) |
| parallelrisch | \({\mathrm e}^{-\frac {\ln \left (3\right ) {\mathrm e}^{\left ({\mathrm e}^{x} x^{3}+{\mathrm e}^{x}-5 x \right ) {\mathrm e}^{-x}}-3}{\ln \left (3\right )}}\) | \(31\) |
Input:
int((-3*exp(x)*x^2-5*x+5)*exp(((x^3+1)*exp(x)-5*x)/exp(x))*exp((-ln(3)*exp (((x^3+1)*exp(x)-5*x)/exp(x))+3)/ln(3))/exp(x),x,method=_RETURNVERBOSE)
Output:
exp((-ln(3)*exp(((x^3+1)*exp(x)-5*x)/exp(x))+3)/ln(3))
Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (25) = 50\).
Time = 0.10 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.71 \[ \int e^{-x+e^{-x} \left (-5 x+e^x \left (1+x^3\right )\right )+\frac {3-e^{e^{-x} \left (-5 x+e^x \left (1+x^3\right )\right )} \log (3)}{\log (3)}} \left (5-5 x-3 e^x x^2\right ) \, dx=e^{\left (-{\left ({\left (x^{3} + 1\right )} e^{x} - 5 \, x\right )} e^{\left (-x\right )} + x + \frac {{\left ({\left ({\left (x^{3} - x + 1\right )} \log \left (3\right ) + 3\right )} e^{x} - 5 \, x \log \left (3\right ) - e^{\left ({\left ({\left (x^{3} + 1\right )} e^{x} - 5 \, x\right )} e^{\left (-x\right )} + x\right )} \log \left (3\right )\right )} e^{\left (-x\right )}}{\log \left (3\right )}\right )} \] Input:
integrate((-3*exp(x)*x^2-5*x+5)*exp(((x^3+1)*exp(x)-5*x)/exp(x))*exp((-log (3)*exp(((x^3+1)*exp(x)-5*x)/exp(x))+3)/log(3))/exp(x),x, algorithm="frica s")
Output:
e^(-((x^3 + 1)*e^x - 5*x)*e^(-x) + x + (((x^3 - x + 1)*log(3) + 3)*e^x - 5 *x*log(3) - e^(((x^3 + 1)*e^x - 5*x)*e^(-x) + x)*log(3))*e^(-x)/log(3))
Time = 0.40 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int e^{-x+e^{-x} \left (-5 x+e^x \left (1+x^3\right )\right )+\frac {3-e^{e^{-x} \left (-5 x+e^x \left (1+x^3\right )\right )} \log (3)}{\log (3)}} \left (5-5 x-3 e^x x^2\right ) \, dx=e^{\frac {- e^{\left (- 5 x + \left (x^{3} + 1\right ) e^{x}\right ) e^{- x}} \log {\left (3 \right )} + 3}{\log {\left (3 \right )}}} \] Input:
integrate((-3*exp(x)*x**2-5*x+5)*exp(((x**3+1)*exp(x)-5*x)/exp(x))*exp((-l n(3)*exp(((x**3+1)*exp(x)-5*x)/exp(x))+3)/ln(3))/exp(x),x)
Output:
exp((-exp((-5*x + (x**3 + 1)*exp(x))*exp(-x))*log(3) + 3)/log(3))
Time = 0.41 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int e^{-x+e^{-x} \left (-5 x+e^x \left (1+x^3\right )\right )+\frac {3-e^{e^{-x} \left (-5 x+e^x \left (1+x^3\right )\right )} \log (3)}{\log (3)}} \left (5-5 x-3 e^x x^2\right ) \, dx=e^{\left (\frac {3}{\log \left (3\right )} - e^{\left (x^{3} - 5 \, x e^{\left (-x\right )} + 1\right )}\right )} \] Input:
integrate((-3*exp(x)*x^2-5*x+5)*exp(((x^3+1)*exp(x)-5*x)/exp(x))*exp((-log (3)*exp(((x^3+1)*exp(x)-5*x)/exp(x))+3)/log(3))/exp(x),x, algorithm="maxim a")
Output:
e^(3/log(3) - e^(x^3 - 5*x*e^(-x) + 1))
\[ \int e^{-x+e^{-x} \left (-5 x+e^x \left (1+x^3\right )\right )+\frac {3-e^{e^{-x} \left (-5 x+e^x \left (1+x^3\right )\right )} \log (3)}{\log (3)}} \left (5-5 x-3 e^x x^2\right ) \, dx=\int { -{\left (3 \, x^{2} e^{x} + 5 \, x - 5\right )} e^{\left ({\left ({\left (x^{3} + 1\right )} e^{x} - 5 \, x\right )} e^{\left (-x\right )} - x - \frac {e^{\left ({\left ({\left (x^{3} + 1\right )} e^{x} - 5 \, x\right )} e^{\left (-x\right )}\right )} \log \left (3\right ) - 3}{\log \left (3\right )}\right )} \,d x } \] Input:
integrate((-3*exp(x)*x^2-5*x+5)*exp(((x^3+1)*exp(x)-5*x)/exp(x))*exp((-log (3)*exp(((x^3+1)*exp(x)-5*x)/exp(x))+3)/log(3))/exp(x),x, algorithm="giac" )
Output:
integrate(-(3*x^2*e^x + 5*x - 5)*e^(((x^3 + 1)*e^x - 5*x)*e^(-x) - x - (e^ (((x^3 + 1)*e^x - 5*x)*e^(-x))*log(3) - 3)/log(3)), x)
Time = 2.69 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int e^{-x+e^{-x} \left (-5 x+e^x \left (1+x^3\right )\right )+\frac {3-e^{e^{-x} \left (-5 x+e^x \left (1+x^3\right )\right )} \log (3)}{\log (3)}} \left (5-5 x-3 e^x x^2\right ) \, dx={\mathrm {e}}^{\frac {3}{\ln \left (3\right )}}\,{\mathrm {e}}^{-{\mathrm {e}}^{x^3}\,\mathrm {e}\,{\mathrm {e}}^{-5\,x\,{\mathrm {e}}^{-x}}} \] Input:
int(-exp(-x)*exp(-(exp(-exp(-x)*(5*x - exp(x)*(x^3 + 1)))*log(3) - 3)/log( 3))*exp(-exp(-x)*(5*x - exp(x)*(x^3 + 1)))*(5*x + 3*x^2*exp(x) - 5),x)
Output:
exp(3/log(3))*exp(-exp(x^3)*exp(1)*exp(-5*x*exp(-x)))
\[ \int e^{-x+e^{-x} \left (-5 x+e^x \left (1+x^3\right )\right )+\frac {3-e^{e^{-x} \left (-5 x+e^x \left (1+x^3\right )\right )} \log (3)}{\log (3)}} \left (5-5 x-3 e^x x^2\right ) \, dx=e \left (5 \left (\int \frac {e^{\frac {\mathrm {log}\left (3\right ) x^{3}+3}{\mathrm {log}\left (3\right )}}}{e^{\frac {e^{x^{3}+x} e +e^{\frac {e^{x} x +5 x}{e^{x}}} x +5 e^{\frac {5 x}{e^{x}}} x}{e^{\frac {e^{x} x +5 x}{e^{x}}}}}}d x \right )-3 \left (\int \frac {e^{\frac {\mathrm {log}\left (3\right ) x^{3}+3}{\mathrm {log}\left (3\right )}} x^{2}}{e^{\frac {e^{x^{3}+x} e +5 e^{\frac {5 x}{e^{x}}} x}{e^{\frac {e^{x} x +5 x}{e^{x}}}}}}d x \right )-5 \left (\int \frac {e^{\frac {\mathrm {log}\left (3\right ) x^{3}+3}{\mathrm {log}\left (3\right )}} x}{e^{\frac {e^{x^{3}+x} e +e^{\frac {e^{x} x +5 x}{e^{x}}} x +5 e^{\frac {5 x}{e^{x}}} x}{e^{\frac {e^{x} x +5 x}{e^{x}}}}}}d x \right )\right ) \] Input:
int((-3*exp(x)*x^2-5*x+5)*exp(((x^3+1)*exp(x)-5*x)/exp(x))*exp((-log(3)*ex p(((x^3+1)*exp(x)-5*x)/exp(x))+3)/log(3))/exp(x),x)
Output:
e*(5*int(e**((log(3)*x**3 + 3)/log(3))/e**((e**(x**3 + x)*e + e**((e**x*x + 5*x)/e**x)*x + 5*e**((5*x)/e**x)*x)/e**((e**x*x + 5*x)/e**x)),x) - 3*int ((e**((log(3)*x**3 + 3)/log(3))*x**2)/e**((e**(x**3 + x)*e + 5*e**((5*x)/e **x)*x)/e**((e**x*x + 5*x)/e**x)),x) - 5*int((e**((log(3)*x**3 + 3)/log(3) )*x)/e**((e**(x**3 + x)*e + e**((e**x*x + 5*x)/e**x)*x + 5*e**((5*x)/e**x) *x)/e**((e**x*x + 5*x)/e**x)),x))