Integrand size = 81, antiderivative size = 26 \[ \int e^{-x+\frac {1}{5} e^{-x} \left (1125 x+150 x^2+5 x^3+e^x \left (225 x^3+30 x^4+x^5\right )\right )} \left (225-165 x-27 x^2-x^3+e^x \left (135 x^2+24 x^3+x^4\right )\right ) \, dx=e^{\left (\frac {e^{-x}}{x}+\frac {x}{5}\right ) x^2 (15+x)^2} \] Output:
exp(x^2*(x+15)^2*(1/exp(x)/x+1/5*x))
Time = 0.85 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int e^{-x+\frac {1}{5} e^{-x} \left (1125 x+150 x^2+5 x^3+e^x \left (225 x^3+30 x^4+x^5\right )\right )} \left (225-165 x-27 x^2-x^3+e^x \left (135 x^2+24 x^3+x^4\right )\right ) \, dx=e^{\frac {1}{5} e^{-x} x (15+x)^2 \left (5+e^x x^2\right )} \] Input:
Integrate[E^(-x + (1125*x + 150*x^2 + 5*x^3 + E^x*(225*x^3 + 30*x^4 + x^5) )/(5*E^x))*(225 - 165*x - 27*x^2 - x^3 + E^x*(135*x^2 + 24*x^3 + x^4)),x]
Output:
E^((x*(15 + x)^2*(5 + E^x*x^2))/(5*E^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.
\(\displaystyle \int \left (-x^3-27 x^2+e^x \left (x^4+24 x^3+135 x^2\right )-165 x+225\right ) \exp \left (\frac {1}{5} e^{-x} \left (5 x^3+150 x^2+e^x \left (x^5+30 x^4+225 x^3\right )+1125 x\right )-x\right ) \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int (x+15) \left (e^x x^3+9 e^x x^2-x^2-12 x+15\right ) \exp \left (\frac {1}{5} e^{-x} \left (5 x^3+150 x^2+e^x \left (x^5+30 x^4+225 x^3\right )+1125 x\right )-x\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (x^3 \left (-\exp \left (\frac {1}{5} e^{-x} \left (5 x^3+150 x^2+e^x \left (x^5+30 x^4+225 x^3\right )+1125 x\right )-x\right )\right )-27 x^2 \exp \left (\frac {1}{5} e^{-x} \left (5 x^3+150 x^2+e^x \left (x^5+30 x^4+225 x^3\right )+1125 x\right )-x\right )+\left (x^2+24 x+135\right ) x^2 \exp \left (\frac {1}{5} e^{-x} \left (5 x^3+150 x^2+e^x \left (x^5+30 x^4+225 x^3\right )+1125 x\right )\right )-165 x \exp \left (\frac {1}{5} e^{-x} \left (5 x^3+150 x^2+e^x \left (x^5+30 x^4+225 x^3\right )+1125 x\right )-x\right )+225 \exp \left (\frac {1}{5} e^{-x} \left (5 x^3+150 x^2+e^x \left (x^5+30 x^4+225 x^3\right )+1125 x\right )-x\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 225 \int \exp \left (\frac {1}{5} e^{-x} \left (5 x^3+150 x^2+1125 x+e^x \left (x^5+30 x^4+225 x^3\right )\right )-x\right )dx-165 \int \exp \left (\frac {1}{5} e^{-x} \left (5 x^3+150 x^2+1125 x+e^x \left (x^5+30 x^4+225 x^3\right )\right )-x\right ) xdx-27 \int \exp \left (\frac {1}{5} e^{-x} \left (5 x^3+150 x^2+1125 x+e^x \left (x^5+30 x^4+225 x^3\right )\right )-x\right ) x^2dx-\int \exp \left (\frac {1}{5} e^{-x} \left (5 x^3+150 x^2+1125 x+e^x \left (x^5+30 x^4+225 x^3\right )\right )-x\right ) x^3dx+135 \int e^{\frac {1}{5} e^{-x} x (x+15)^2 \left (e^x x^2+5\right )} x^2dx+\int e^{\frac {1}{5} e^{-x} x (x+15)^2 \left (e^x x^2+5\right )} x^4dx+24 \int e^{\frac {1}{5} e^{-x} x (x+15)^2 \left (e^x x^2+5\right )} x^3dx\) |
Input:
Int[E^(-x + (1125*x + 150*x^2 + 5*x^3 + E^x*(225*x^3 + 30*x^4 + x^5))/(5*E ^x))*(225 - 165*x - 27*x^2 - x^3 + E^x*(135*x^2 + 24*x^3 + x^4)),x]
Output:
$Aborted
Time = 0.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85
method | result | size |
risch | \({\mathrm e}^{\frac {x \left (x +15\right )^{2} \left ({\mathrm e}^{x} x^{2}+5\right ) {\mathrm e}^{-x}}{5}}\) | \(22\) |
norman | \({\mathrm e}^{\frac {\left (\left (x^{5}+30 x^{4}+225 x^{3}\right ) {\mathrm e}^{x}+5 x^{3}+150 x^{2}+1125 x \right ) {\mathrm e}^{-x}}{5}}\) | \(39\) |
parallelrisch | \({\mathrm e}^{\frac {\left (\left (x^{5}+30 x^{4}+225 x^{3}\right ) {\mathrm e}^{x}+5 x^{3}+150 x^{2}+1125 x \right ) {\mathrm e}^{-x}}{5}}\) | \(39\) |
Input:
int(((x^4+24*x^3+135*x^2)*exp(x)-x^3-27*x^2-165*x+225)*exp(1/5*((x^5+30*x^ 4+225*x^3)*exp(x)+5*x^3+150*x^2+1125*x)/exp(x))/exp(x),x,method=_RETURNVER BOSE)
Output:
exp(1/5*x*(x+15)^2*(exp(x)*x^2+5)*exp(-x))
Time = 0.09 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65 \[ \int e^{-x+\frac {1}{5} e^{-x} \left (1125 x+150 x^2+5 x^3+e^x \left (225 x^3+30 x^4+x^5\right )\right )} \left (225-165 x-27 x^2-x^3+e^x \left (135 x^2+24 x^3+x^4\right )\right ) \, dx=e^{\left (\frac {1}{5} \, {\left (5 \, x^{3} + 150 \, x^{2} + {\left (x^{5} + 30 \, x^{4} + 225 \, x^{3} - 5 \, x\right )} e^{x} + 1125 \, x\right )} e^{\left (-x\right )} + x\right )} \] Input:
integrate(((x^4+24*x^3+135*x^2)*exp(x)-x^3-27*x^2-165*x+225)*exp(1/5*((x^5 +30*x^4+225*x^3)*exp(x)+5*x^3+150*x^2+1125*x)/exp(x))/exp(x),x, algorithm= "fricas")
Output:
e^(1/5*(5*x^3 + 150*x^2 + (x^5 + 30*x^4 + 225*x^3 - 5*x)*e^x + 1125*x)*e^( -x) + x)
Time = 0.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int e^{-x+\frac {1}{5} e^{-x} \left (1125 x+150 x^2+5 x^3+e^x \left (225 x^3+30 x^4+x^5\right )\right )} \left (225-165 x-27 x^2-x^3+e^x \left (135 x^2+24 x^3+x^4\right )\right ) \, dx=e^{\left (x^{3} + 30 x^{2} + 225 x + \frac {\left (x^{5} + 30 x^{4} + 225 x^{3}\right ) e^{x}}{5}\right ) e^{- x}} \] Input:
integrate(((x**4+24*x**3+135*x**2)*exp(x)-x**3-27*x**2-165*x+225)*exp(1/5* ((x**5+30*x**4+225*x**3)*exp(x)+5*x**3+150*x**2+1125*x)/exp(x))/exp(x),x)
Output:
exp((x**3 + 30*x**2 + 225*x + (x**5 + 30*x**4 + 225*x**3)*exp(x)/5)*exp(-x ))
Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int e^{-x+\frac {1}{5} e^{-x} \left (1125 x+150 x^2+5 x^3+e^x \left (225 x^3+30 x^4+x^5\right )\right )} \left (225-165 x-27 x^2-x^3+e^x \left (135 x^2+24 x^3+x^4\right )\right ) \, dx=e^{\left (\frac {1}{5} \, x^{5} + 6 \, x^{4} + x^{3} e^{\left (-x\right )} + 45 \, x^{3} + 30 \, x^{2} e^{\left (-x\right )} + 225 \, x e^{\left (-x\right )}\right )} \] Input:
integrate(((x^4+24*x^3+135*x^2)*exp(x)-x^3-27*x^2-165*x+225)*exp(1/5*((x^5 +30*x^4+225*x^3)*exp(x)+5*x^3+150*x^2+1125*x)/exp(x))/exp(x),x, algorithm= "maxima")
Output:
e^(1/5*x^5 + 6*x^4 + x^3*e^(-x) + 45*x^3 + 30*x^2*e^(-x) + 225*x*e^(-x))
\[ \int e^{-x+\frac {1}{5} e^{-x} \left (1125 x+150 x^2+5 x^3+e^x \left (225 x^3+30 x^4+x^5\right )\right )} \left (225-165 x-27 x^2-x^3+e^x \left (135 x^2+24 x^3+x^4\right )\right ) \, dx=\int { -{\left (x^{3} + 27 \, x^{2} - {\left (x^{4} + 24 \, x^{3} + 135 \, x^{2}\right )} e^{x} + 165 \, x - 225\right )} e^{\left (\frac {1}{5} \, {\left (5 \, x^{3} + 150 \, x^{2} + {\left (x^{5} + 30 \, x^{4} + 225 \, x^{3}\right )} e^{x} + 1125 \, x\right )} e^{\left (-x\right )} - x\right )} \,d x } \] Input:
integrate(((x^4+24*x^3+135*x^2)*exp(x)-x^3-27*x^2-165*x+225)*exp(1/5*((x^5 +30*x^4+225*x^3)*exp(x)+5*x^3+150*x^2+1125*x)/exp(x))/exp(x),x, algorithm= "giac")
Output:
integrate(-(x^3 + 27*x^2 - (x^4 + 24*x^3 + 135*x^2)*e^x + 165*x - 225)*e^( 1/5*(5*x^3 + 150*x^2 + (x^5 + 30*x^4 + 225*x^3)*e^x + 1125*x)*e^(-x) - x), x)
Time = 2.73 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.77 \[ \int e^{-x+\frac {1}{5} e^{-x} \left (1125 x+150 x^2+5 x^3+e^x \left (225 x^3+30 x^4+x^5\right )\right )} \left (225-165 x-27 x^2-x^3+e^x \left (135 x^2+24 x^3+x^4\right )\right ) \, dx={\mathrm {e}}^{225\,x\,{\mathrm {e}}^{-x}}\,{\mathrm {e}}^{6\,x^4}\,{\mathrm {e}}^{\frac {x^5}{5}}\,{\mathrm {e}}^{45\,x^3}\,{\mathrm {e}}^{x^3\,{\mathrm {e}}^{-x}}\,{\mathrm {e}}^{30\,x^2\,{\mathrm {e}}^{-x}} \] Input:
int(-exp(exp(-x)*(225*x + 30*x^2 + x^3 + (exp(x)*(225*x^3 + 30*x^4 + x^5)) /5))*exp(-x)*(165*x + 27*x^2 + x^3 - exp(x)*(135*x^2 + 24*x^3 + x^4) - 225 ),x)
Output:
exp(225*x*exp(-x))*exp(6*x^4)*exp(x^5/5)*exp(45*x^3)*exp(x^3*exp(-x))*exp( 30*x^2*exp(-x))
Time = 5.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.77 \[ \int e^{-x+\frac {1}{5} e^{-x} \left (1125 x+150 x^2+5 x^3+e^x \left (225 x^3+30 x^4+x^5\right )\right )} \left (225-165 x-27 x^2-x^3+e^x \left (135 x^2+24 x^3+x^4\right )\right ) \, dx=e^{\frac {e^{x} x^{5}+30 e^{x} x^{4}+225 e^{x} x^{3}+5 x^{3}+150 x^{2}+1125 x}{5 e^{x}}} \] Input:
int(((x^4+24*x^3+135*x^2)*exp(x)-x^3-27*x^2-165*x+225)*exp(1/5*((x^5+30*x^ 4+225*x^3)*exp(x)+5*x^3+150*x^2+1125*x)/exp(x))/exp(x),x)
Output:
e**((e**x*x**5 + 30*e**x*x**4 + 225*e**x*x**3 + 5*x**3 + 150*x**2 + 1125*x )/(5*e**x))