Integrand size = 114, antiderivative size = 20 \[ \int \frac {e^{-\frac {2 x^2}{36+4 e^2+e (24-8 x)-24 x+4 x^2}} \left (1350+50 e^3+e^2 (450-150 x)-1350 x+300 x^2-50 x^3+e \left (1350-900 x+100 x^2\right )\right )}{27+e^3+e^2 (9-3 x)-27 x+9 x^2-x^3+e \left (27-18 x+3 x^2\right )} \, dx=50 e^{-\frac {x^2}{2 (3+e-x)^2}} x \] Output:
50/exp(1/2*x^2/(3-x+exp(1))/(6-2*x+2*exp(1)))^2*x
Time = 3.37 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {2 x^2}{36+4 e^2+e (24-8 x)-24 x+4 x^2}} \left (1350+50 e^3+e^2 (450-150 x)-1350 x+300 x^2-50 x^3+e \left (1350-900 x+100 x^2\right )\right )}{27+e^3+e^2 (9-3 x)-27 x+9 x^2-x^3+e \left (27-18 x+3 x^2\right )} \, dx=50 e^{-\frac {x^2}{2 (3+e-x)^2}} x \] Input:
Integrate[(1350 + 50*E^3 + E^2*(450 - 150*x) - 1350*x + 300*x^2 - 50*x^3 + E*(1350 - 900*x + 100*x^2))/(E^((2*x^2)/(36 + 4*E^2 + E*(24 - 8*x) - 24*x + 4*x^2))*(27 + E^3 + E^2*(9 - 3*x) - 27*x + 9*x^2 - x^3 + E*(27 - 18*x + 3*x^2))),x]
Output:
(50*x)/E^(x^2/(2*(3 + E - x)^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.
\(\displaystyle \int \frac {\left (-50 x^3+300 x^2+e \left (100 x^2-900 x+1350\right )-1350 x+e^2 (450-150 x)+50 e^3+1350\right ) \exp \left (-\frac {2 x^2}{4 x^2-24 x+e (24-8 x)+4 e^2+36}\right )}{-x^3+9 x^2+e \left (3 x^2-18 x+27\right )-27 x+e^2 (9-3 x)+e^3+27} \, dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {\left (-50 x^3+300 x^2+e \left (100 x^2-900 x+1350\right )-1350 x+e^2 (450-150 x)+50 e^3+1350\right ) \exp \left (-\frac {2 x^2}{4 x^2-24 x+e (24-8 x)+4 e^2+36}\right )}{(-x+e+3)^3}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (-50 x^3+100 (3+e) x^2-150 (3+e)^2 x+50 (3+e)^3\right ) \exp \left (-\frac {2 x^2}{4 x^2-8 (3+e) x+4 (3+e)^2}\right )}{(-x+e+3)^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (50 \exp \left (-\frac {2 x^2}{4 x^2-8 (3+e) x+4 (3+e)^2}\right )-\frac {50 (3+e) \exp \left (-\frac {2 x^2}{4 x^2-8 (3+e) x+4 (3+e)^2}\right )}{-x+e+3}+\frac {100 (3+e)^2 \exp \left (-\frac {2 x^2}{4 x^2-8 (3+e) x+4 (3+e)^2}\right )}{(-x+e+3)^2}-\frac {50 (3+e)^3 \exp \left (-\frac {2 x^2}{4 x^2-8 (3+e) x+4 (3+e)^2}\right )}{(-x+e+3)^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 50 \int \exp \left (-\frac {2 x^2}{4 x^2-8 (3+e) x+4 (3+e)^2}\right )dx-50 (3+e)^3 \int \frac {\exp \left (-\frac {2 x^2}{4 x^2-8 (3+e) x+4 (3+e)^2}\right )}{(-x+e+3)^3}dx+100 (3+e)^2 \int \frac {\exp \left (-\frac {2 x^2}{4 x^2-8 (3+e) x+4 (3+e)^2}\right )}{(-x+e+3)^2}dx-50 (3+e) \int \frac {\exp \left (-\frac {2 x^2}{4 x^2-8 (3+e) x+4 (3+e)^2}\right )}{-x+e+3}dx\) |
Input:
Int[(1350 + 50*E^3 + E^2*(450 - 150*x) - 1350*x + 300*x^2 - 50*x^3 + E*(13 50 - 900*x + 100*x^2))/(E^((2*x^2)/(36 + 4*E^2 + E*(24 - 8*x) - 24*x + 4*x ^2))*(27 + E^3 + E^2*(9 - 3*x) - 27*x + 9*x^2 - x^3 + E*(27 - 18*x + 3*x^2 ))),x]
Output:
$Aborted
Time = 0.86 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.75
method | result | size |
gosper | \(50 x \,{\mathrm e}^{-\frac {x^{2}}{2 \left (-2 x \,{\mathrm e}+x^{2}+6 \,{\mathrm e}+{\mathrm e}^{2}-6 x +9\right )}}\) | \(35\) |
risch | \(50 x \,{\mathrm e}^{\frac {x^{2}}{4 x \,{\mathrm e}-2 x^{2}-12 \,{\mathrm e}-2 \,{\mathrm e}^{2}+12 x -18}}\) | \(35\) |
norman | \(\frac {\left (\left (-100 \,{\mathrm e}-300\right ) x^{2}+\left (50 \,{\mathrm e}^{2}+300 \,{\mathrm e}+450\right ) x +50 x^{3}\right ) {\mathrm e}^{-\frac {2 x^{2}}{4 \,{\mathrm e}^{2}+\left (-8 x +24\right ) {\mathrm e}+4 x^{2}-24 x +36}}}{\left (3-x +{\mathrm e}\right )^{2}}\) | \(74\) |
parallelrisch | \(\frac {\left (200 \,{\mathrm e}^{2} x -400 x^{2} {\mathrm e}+200 x^{3}+1200 x \,{\mathrm e}-1200 x^{2}+1800 x \right ) {\mathrm e}^{-\frac {x^{2}}{2 \left (-2 x \,{\mathrm e}+x^{2}+6 \,{\mathrm e}+{\mathrm e}^{2}-6 x +9\right )}}}{4 \left (3-x +{\mathrm e}\right )^{2}}\) | \(76\) |
Input:
int((50*exp(1)^3+(-150*x+450)*exp(1)^2+(100*x^2-900*x+1350)*exp(1)-50*x^3+ 300*x^2-1350*x+1350)/(exp(1)^3+(-3*x+9)*exp(1)^2+(3*x^2-18*x+27)*exp(1)-x^ 3+9*x^2-27*x+27)/exp(x^2/(4*exp(1)^2+(-8*x+24)*exp(1)+4*x^2-24*x+36))^2,x, method=_RETURNVERBOSE)
Output:
50*x/exp(1/4*x^2/(exp(1)^2-2*x*exp(1)+x^2+6*exp(1)-6*x+9))^2
Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int \frac {e^{-\frac {2 x^2}{36+4 e^2+e (24-8 x)-24 x+4 x^2}} \left (1350+50 e^3+e^2 (450-150 x)-1350 x+300 x^2-50 x^3+e \left (1350-900 x+100 x^2\right )\right )}{27+e^3+e^2 (9-3 x)-27 x+9 x^2-x^3+e \left (27-18 x+3 x^2\right )} \, dx=50 \, x e^{\left (-\frac {x^{2}}{2 \, {\left (x^{2} - 2 \, {\left (x - 3\right )} e - 6 \, x + e^{2} + 9\right )}}\right )} \] Input:
integrate((50*exp(1)^3+(-150*x+450)*exp(1)^2+(100*x^2-900*x+1350)*exp(1)-5 0*x^3+300*x^2-1350*x+1350)/(exp(1)^3+(-3*x+9)*exp(1)^2+(3*x^2-18*x+27)*exp (1)-x^3+9*x^2-27*x+27)/exp(x^2/(4*exp(1)^2+(-8*x+24)*exp(1)+4*x^2-24*x+36) )^2,x, algorithm="fricas")
Output:
50*x*e^(-1/2*x^2/(x^2 - 2*(x - 3)*e - 6*x + e^2 + 9))
Time = 0.40 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60 \[ \int \frac {e^{-\frac {2 x^2}{36+4 e^2+e (24-8 x)-24 x+4 x^2}} \left (1350+50 e^3+e^2 (450-150 x)-1350 x+300 x^2-50 x^3+e \left (1350-900 x+100 x^2\right )\right )}{27+e^3+e^2 (9-3 x)-27 x+9 x^2-x^3+e \left (27-18 x+3 x^2\right )} \, dx=50 x e^{- \frac {2 x^{2}}{4 x^{2} - 24 x + e \left (24 - 8 x\right ) + 4 e^{2} + 36}} \] Input:
integrate((50*exp(1)**3+(-150*x+450)*exp(1)**2+(100*x**2-900*x+1350)*exp(1 )-50*x**3+300*x**2-1350*x+1350)/(exp(1)**3+(-3*x+9)*exp(1)**2+(3*x**2-18*x +27)*exp(1)-x**3+9*x**2-27*x+27)/exp(x**2/(4*exp(1)**2+(-8*x+24)*exp(1)+4* x**2-24*x+36))**2,x)
Output:
50*x*exp(-2*x**2/(4*x**2 - 24*x + E*(24 - 8*x) + 4*exp(2) + 36))
Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (18) = 36\).
Time = 0.17 (sec) , antiderivative size = 100, normalized size of antiderivative = 5.00 \[ \int \frac {e^{-\frac {2 x^2}{36+4 e^2+e (24-8 x)-24 x+4 x^2}} \left (1350+50 e^3+e^2 (450-150 x)-1350 x+300 x^2-50 x^3+e \left (1350-900 x+100 x^2\right )\right )}{27+e^3+e^2 (9-3 x)-27 x+9 x^2-x^3+e \left (27-18 x+3 x^2\right )} \, dx=50 \, x e^{\left (-\frac {e^{2}}{2 \, {\left (x^{2} - 2 \, x {\left (e + 3\right )} + e^{2} + 6 \, e + 9\right )}} - \frac {3 \, e}{x^{2} - 2 \, x {\left (e + 3\right )} + e^{2} + 6 \, e + 9} - \frac {e}{x - e - 3} - \frac {9}{2 \, {\left (x^{2} - 2 \, x {\left (e + 3\right )} + e^{2} + 6 \, e + 9\right )}} - \frac {3}{x - e - 3} - \frac {1}{2}\right )} \] Input:
integrate((50*exp(1)^3+(-150*x+450)*exp(1)^2+(100*x^2-900*x+1350)*exp(1)-5 0*x^3+300*x^2-1350*x+1350)/(exp(1)^3+(-3*x+9)*exp(1)^2+(3*x^2-18*x+27)*exp (1)-x^3+9*x^2-27*x+27)/exp(x^2/(4*exp(1)^2+(-8*x+24)*exp(1)+4*x^2-24*x+36) )^2,x, algorithm="maxima")
Output:
50*x*e^(-1/2*e^2/(x^2 - 2*x*(e + 3) + e^2 + 6*e + 9) - 3*e/(x^2 - 2*x*(e + 3) + e^2 + 6*e + 9) - e/(x - e - 3) - 9/2/(x^2 - 2*x*(e + 3) + e^2 + 6*e + 9) - 3/(x - e - 3) - 1/2)
Time = 0.16 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.50 \[ \int \frac {e^{-\frac {2 x^2}{36+4 e^2+e (24-8 x)-24 x+4 x^2}} \left (1350+50 e^3+e^2 (450-150 x)-1350 x+300 x^2-50 x^3+e \left (1350-900 x+100 x^2\right )\right )}{27+e^3+e^2 (9-3 x)-27 x+9 x^2-x^3+e \left (27-18 x+3 x^2\right )} \, dx=50 \, x e^{\left (-\frac {x^{2}}{2 \, {\left (x^{2} - 2 \, x e - 6 \, x + e^{2} + 6 \, e + 9\right )}}\right )} \] Input:
integrate((50*exp(1)^3+(-150*x+450)*exp(1)^2+(100*x^2-900*x+1350)*exp(1)-5 0*x^3+300*x^2-1350*x+1350)/(exp(1)^3+(-3*x+9)*exp(1)^2+(3*x^2-18*x+27)*exp (1)-x^3+9*x^2-27*x+27)/exp(x^2/(4*exp(1)^2+(-8*x+24)*exp(1)+4*x^2-24*x+36) )^2,x, algorithm="giac")
Output:
50*x*e^(-1/2*x^2/(x^2 - 2*x*e - 6*x + e^2 + 6*e + 9))
Time = 0.94 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.70 \[ \int \frac {e^{-\frac {2 x^2}{36+4 e^2+e (24-8 x)-24 x+4 x^2}} \left (1350+50 e^3+e^2 (450-150 x)-1350 x+300 x^2-50 x^3+e \left (1350-900 x+100 x^2\right )\right )}{27+e^3+e^2 (9-3 x)-27 x+9 x^2-x^3+e \left (27-18 x+3 x^2\right )} \, dx=50\,x\,{\mathrm {e}}^{-\frac {x^2}{12\,\mathrm {e}-12\,x+2\,{\mathrm {e}}^2-4\,x\,\mathrm {e}+2\,x^2+18}} \] Input:
int((exp(-(2*x^2)/(4*exp(2) - 24*x + 4*x^2 - exp(1)*(8*x - 24) + 36))*(50* exp(3) - 1350*x + exp(1)*(100*x^2 - 900*x + 1350) + 300*x^2 - 50*x^3 - exp (2)*(150*x - 450) + 1350))/(exp(3) - 27*x + exp(1)*(3*x^2 - 18*x + 27) + 9 *x^2 - x^3 - exp(2)*(3*x - 9) + 27),x)
Output:
50*x*exp(-x^2/(12*exp(1) - 12*x + 2*exp(2) - 4*x*exp(1) + 2*x^2 + 18))
Time = 0.24 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.75 \[ \int \frac {e^{-\frac {2 x^2}{36+4 e^2+e (24-8 x)-24 x+4 x^2}} \left (1350+50 e^3+e^2 (450-150 x)-1350 x+300 x^2-50 x^3+e \left (1350-900 x+100 x^2\right )\right )}{27+e^3+e^2 (9-3 x)-27 x+9 x^2-x^3+e \left (27-18 x+3 x^2\right )} \, dx=\frac {50 x}{e^{\frac {x^{2}}{2 e^{2}-4 e x +2 x^{2}+12 e -12 x +18}}} \] Input:
int((50*exp(1)^3+(-150*x+450)*exp(1)^2+(100*x^2-900*x+1350)*exp(1)-50*x^3+ 300*x^2-1350*x+1350)/(exp(1)^3+(-3*x+9)*exp(1)^2+(3*x^2-18*x+27)*exp(1)-x^ 3+9*x^2-27*x+27)/exp(x^2/(4*exp(1)^2+(-8*x+24)*exp(1)+4*x^2-24*x+36))^2,x)
Output:
(50*x)/e**(x**2/(2*e**2 - 4*e*x + 12*e + 2*x**2 - 12*x + 18))