Integrand size = 135, antiderivative size = 27 \[ \int \frac {e^{\frac {-50 x-2 x^2}{-43-53 x-2 x^2+e^x \left (50 x+2 x^2\right )}} \left (2150+172 x+6 x^2+e^x \left (2500 x^2+200 x^3+4 x^4\right )\right )}{1849+4558 x+2981 x^2+212 x^3+4 x^4+e^x \left (-4300 x-5472 x^2-412 x^3-8 x^4\right )+e^{2 x} \left (2500 x^2+200 x^3+4 x^4\right )} \, dx=e^{\frac {x}{1+x-e^x x+\frac {-7+x}{2 (25+x)}}} \] Output:
exp(x/(1-exp(x)*x+x+(-7+x)/(2*x+50)))
Time = 0.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {e^{\frac {-50 x-2 x^2}{-43-53 x-2 x^2+e^x \left (50 x+2 x^2\right )}} \left (2150+172 x+6 x^2+e^x \left (2500 x^2+200 x^3+4 x^4\right )\right )}{1849+4558 x+2981 x^2+212 x^3+4 x^4+e^x \left (-4300 x-5472 x^2-412 x^3-8 x^4\right )+e^{2 x} \left (2500 x^2+200 x^3+4 x^4\right )} \, dx=e^{-\frac {2 x (25+x)}{-43+\left (-53+50 e^x\right ) x+2 \left (-1+e^x\right ) x^2}} \] Input:
Integrate[(E^((-50*x - 2*x^2)/(-43 - 53*x - 2*x^2 + E^x*(50*x + 2*x^2)))*( 2150 + 172*x + 6*x^2 + E^x*(2500*x^2 + 200*x^3 + 4*x^4)))/(1849 + 4558*x + 2981*x^2 + 212*x^3 + 4*x^4 + E^x*(-4300*x - 5472*x^2 - 412*x^3 - 8*x^4) + E^(2*x)*(2500*x^2 + 200*x^3 + 4*x^4)),x]
Output:
E^((-2*x*(25 + x))/(-43 + (-53 + 50*E^x)*x + 2*(-1 + E^x)*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.
\(\displaystyle \int \frac {\left (6 x^2+e^x \left (4 x^4+200 x^3+2500 x^2\right )+172 x+2150\right ) \exp \left (\frac {-2 x^2-50 x}{-2 x^2+e^x \left (2 x^2+50 x\right )-53 x-43}\right )}{4 x^4+212 x^3+2981 x^2+e^x \left (-8 x^4-412 x^3-5472 x^2-4300 x\right )+e^{2 x} \left (4 x^4+200 x^3+2500 x^2\right )+4558 x+1849} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (6 x^2+e^x \left (4 x^4+200 x^3+2500 x^2\right )+172 x+2150\right ) \exp \left (\frac {(-2 x-50) x}{-2 x^2+e^x \left (2 x^2+50 x\right )-53 x-43}\right )}{\left (-2 e^x x^2+2 x^2-50 e^x x+53 x+43\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 x (x+25) \exp \left (\frac {(-2 x-50) x}{-2 x^2+e^x \left (2 x^2+50 x\right )-53 x-43}\right )}{2 e^x x^2-2 x^2+50 e^x x-53 x-43}+\frac {2 \left (2 x^4+103 x^3+1371 x^2+1161 x+1075\right ) \exp \left (\frac {(-2 x-50) x}{-2 x^2+e^x \left (2 x^2+50 x\right )-53 x-43}\right )}{\left (2 e^x x^2-2 x^2+50 e^x x-53 x-43\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2150 \int \frac {\exp \left (\frac {(-2 x-50) x}{-2 x^2-53 x+e^x \left (2 x^2+50 x\right )-43}\right )}{\left (2 e^x x^2-2 x^2+50 e^x x-53 x-43\right )^2}dx+2322 \int \frac {\exp \left (\frac {(-2 x-50) x}{-2 x^2-53 x+e^x \left (2 x^2+50 x\right )-43}\right ) x}{\left (2 e^x x^2-2 x^2+50 e^x x-53 x-43\right )^2}dx+2742 \int \frac {\exp \left (\frac {(-2 x-50) x}{-2 x^2-53 x+e^x \left (2 x^2+50 x\right )-43}\right ) x^2}{\left (2 e^x x^2-2 x^2+50 e^x x-53 x-43\right )^2}dx+50 \int \frac {\exp \left (\frac {(-2 x-50) x}{-2 x^2-53 x+e^x \left (2 x^2+50 x\right )-43}\right ) x}{2 e^x x^2-2 x^2+50 e^x x-53 x-43}dx+2 \int \frac {\exp \left (\frac {(-2 x-50) x}{-2 x^2-53 x+e^x \left (2 x^2+50 x\right )-43}\right ) x^2}{2 e^x x^2-2 x^2+50 e^x x-53 x-43}dx+4 \int \frac {\exp \left (\frac {(-2 x-50) x}{-2 x^2-53 x+e^x \left (2 x^2+50 x\right )-43}\right ) x^4}{\left (2 e^x x^2-2 x^2+50 e^x x-53 x-43\right )^2}dx+206 \int \frac {\exp \left (\frac {(-2 x-50) x}{-2 x^2-53 x+e^x \left (2 x^2+50 x\right )-43}\right ) x^3}{\left (2 e^x x^2-2 x^2+50 e^x x-53 x-43\right )^2}dx\) |
Input:
Int[(E^((-50*x - 2*x^2)/(-43 - 53*x - 2*x^2 + E^x*(50*x + 2*x^2)))*(2150 + 172*x + 6*x^2 + E^x*(2500*x^2 + 200*x^3 + 4*x^4)))/(1849 + 4558*x + 2981* x^2 + 212*x^3 + 4*x^4 + E^x*(-4300*x - 5472*x^2 - 412*x^3 - 8*x^4) + E^(2* x)*(2500*x^2 + 200*x^3 + 4*x^4)),x]
Output:
$Aborted
Time = 2.48 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19
method | result | size |
risch | \({\mathrm e}^{-\frac {2 x \left (x +25\right )}{2 \,{\mathrm e}^{x} x^{2}+50 \,{\mathrm e}^{x} x -2 x^{2}-53 x -43}}\) | \(32\) |
parallelrisch | \({\mathrm e}^{\frac {-2 x^{2}-50 x}{2 \,{\mathrm e}^{x} x^{2}+50 \,{\mathrm e}^{x} x -2 x^{2}-53 x -43}}\) | \(36\) |
norman | \(\frac {-53 x \,{\mathrm e}^{\frac {-2 x^{2}-50 x}{\left (2 x^{2}+50 x \right ) {\mathrm e}^{x}-2 x^{2}-53 x -43}}-2 x^{2} {\mathrm e}^{\frac {-2 x^{2}-50 x}{\left (2 x^{2}+50 x \right ) {\mathrm e}^{x}-2 x^{2}-53 x -43}}+50 \,{\mathrm e}^{x} x \,{\mathrm e}^{\frac {-2 x^{2}-50 x}{\left (2 x^{2}+50 x \right ) {\mathrm e}^{x}-2 x^{2}-53 x -43}}+2 \,{\mathrm e}^{x} x^{2} {\mathrm e}^{\frac {-2 x^{2}-50 x}{\left (2 x^{2}+50 x \right ) {\mathrm e}^{x}-2 x^{2}-53 x -43}}-43 \,{\mathrm e}^{\frac {-2 x^{2}-50 x}{\left (2 x^{2}+50 x \right ) {\mathrm e}^{x}-2 x^{2}-53 x -43}}}{2 \,{\mathrm e}^{x} x^{2}+50 \,{\mathrm e}^{x} x -2 x^{2}-53 x -43}\) | \(224\) |
Input:
int(((4*x^4+200*x^3+2500*x^2)*exp(x)+6*x^2+172*x+2150)*exp((-2*x^2-50*x)/( (2*x^2+50*x)*exp(x)-2*x^2-53*x-43))/((4*x^4+200*x^3+2500*x^2)*exp(x)^2+(-8 *x^4-412*x^3-5472*x^2-4300*x)*exp(x)+4*x^4+212*x^3+2981*x^2+4558*x+1849),x ,method=_RETURNVERBOSE)
Output:
exp(-2*x*(x+25)/(2*exp(x)*x^2+50*exp(x)*x-2*x^2-53*x-43))
Time = 0.10 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {e^{\frac {-50 x-2 x^2}{-43-53 x-2 x^2+e^x \left (50 x+2 x^2\right )}} \left (2150+172 x+6 x^2+e^x \left (2500 x^2+200 x^3+4 x^4\right )\right )}{1849+4558 x+2981 x^2+212 x^3+4 x^4+e^x \left (-4300 x-5472 x^2-412 x^3-8 x^4\right )+e^{2 x} \left (2500 x^2+200 x^3+4 x^4\right )} \, dx=e^{\left (\frac {2 \, {\left (x^{2} + 25 \, x\right )}}{2 \, x^{2} - 2 \, {\left (x^{2} + 25 \, x\right )} e^{x} + 53 \, x + 43}\right )} \] Input:
integrate(((4*x^4+200*x^3+2500*x^2)*exp(x)+6*x^2+172*x+2150)*exp((-2*x^2-5 0*x)/((2*x^2+50*x)*exp(x)-2*x^2-53*x-43))/((4*x^4+200*x^3+2500*x^2)*exp(x) ^2+(-8*x^4-412*x^3-5472*x^2-4300*x)*exp(x)+4*x^4+212*x^3+2981*x^2+4558*x+1 849),x, algorithm="fricas")
Output:
e^(2*(x^2 + 25*x)/(2*x^2 - 2*(x^2 + 25*x)*e^x + 53*x + 43))
Time = 0.48 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {e^{\frac {-50 x-2 x^2}{-43-53 x-2 x^2+e^x \left (50 x+2 x^2\right )}} \left (2150+172 x+6 x^2+e^x \left (2500 x^2+200 x^3+4 x^4\right )\right )}{1849+4558 x+2981 x^2+212 x^3+4 x^4+e^x \left (-4300 x-5472 x^2-412 x^3-8 x^4\right )+e^{2 x} \left (2500 x^2+200 x^3+4 x^4\right )} \, dx=e^{\frac {- 2 x^{2} - 50 x}{- 2 x^{2} - 53 x + \left (2 x^{2} + 50 x\right ) e^{x} - 43}} \] Input:
integrate(((4*x**4+200*x**3+2500*x**2)*exp(x)+6*x**2+172*x+2150)*exp((-2*x **2-50*x)/((2*x**2+50*x)*exp(x)-2*x**2-53*x-43))/((4*x**4+200*x**3+2500*x* *2)*exp(x)**2+(-8*x**4-412*x**3-5472*x**2-4300*x)*exp(x)+4*x**4+212*x**3+2 981*x**2+4558*x+1849),x)
Output:
exp((-2*x**2 - 50*x)/(-2*x**2 - 53*x + (2*x**2 + 50*x)*exp(x) - 43))
Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (26) = 52\).
Time = 0.33 (sec) , antiderivative size = 163, normalized size of antiderivative = 6.04 \[ \int \frac {e^{\frac {-50 x-2 x^2}{-43-53 x-2 x^2+e^x \left (50 x+2 x^2\right )}} \left (2150+172 x+6 x^2+e^x \left (2500 x^2+200 x^3+4 x^4\right )\right )}{1849+4558 x+2981 x^2+212 x^3+4 x^4+e^x \left (-4300 x-5472 x^2-412 x^3-8 x^4\right )+e^{2 x} \left (2500 x^2+200 x^3+4 x^4\right )} \, dx=e^{\left (\frac {50 \, x e^{x}}{2 \, x^{2} + 2 \, {\left (x^{2} + 25 \, x\right )} e^{\left (2 \, x\right )} - {\left (4 \, x^{2} + 103 \, x + 43\right )} e^{x} + 53 \, x + 43} - \frac {53 \, x}{2 \, x^{2} + 2 \, {\left (x^{2} + 25 \, x\right )} e^{\left (2 \, x\right )} - {\left (4 \, x^{2} + 103 \, x + 43\right )} e^{x} + 53 \, x + 43} + \frac {50 \, x}{2 \, x^{2} - 2 \, {\left (x^{2} + 25 \, x\right )} e^{x} + 53 \, x + 43} - \frac {43}{2 \, x^{2} + 2 \, {\left (x^{2} + 25 \, x\right )} e^{\left (2 \, x\right )} - {\left (4 \, x^{2} + 103 \, x + 43\right )} e^{x} + 53 \, x + 43} - \frac {1}{e^{x} - 1}\right )} \] Input:
integrate(((4*x^4+200*x^3+2500*x^2)*exp(x)+6*x^2+172*x+2150)*exp((-2*x^2-5 0*x)/((2*x^2+50*x)*exp(x)-2*x^2-53*x-43))/((4*x^4+200*x^3+2500*x^2)*exp(x) ^2+(-8*x^4-412*x^3-5472*x^2-4300*x)*exp(x)+4*x^4+212*x^3+2981*x^2+4558*x+1 849),x, algorithm="maxima")
Output:
e^(50*x*e^x/(2*x^2 + 2*(x^2 + 25*x)*e^(2*x) - (4*x^2 + 103*x + 43)*e^x + 5 3*x + 43) - 53*x/(2*x^2 + 2*(x^2 + 25*x)*e^(2*x) - (4*x^2 + 103*x + 43)*e^ x + 53*x + 43) + 50*x/(2*x^2 - 2*(x^2 + 25*x)*e^x + 53*x + 43) - 43/(2*x^2 + 2*(x^2 + 25*x)*e^(2*x) - (4*x^2 + 103*x + 43)*e^x + 53*x + 43) - 1/(e^x - 1))
Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (26) = 52\).
Time = 0.15 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.15 \[ \int \frac {e^{\frac {-50 x-2 x^2}{-43-53 x-2 x^2+e^x \left (50 x+2 x^2\right )}} \left (2150+172 x+6 x^2+e^x \left (2500 x^2+200 x^3+4 x^4\right )\right )}{1849+4558 x+2981 x^2+212 x^3+4 x^4+e^x \left (-4300 x-5472 x^2-412 x^3-8 x^4\right )+e^{2 x} \left (2500 x^2+200 x^3+4 x^4\right )} \, dx=e^{\left (-\frac {2 \, x^{2}}{2 \, x^{2} e^{x} - 2 \, x^{2} + 50 \, x e^{x} - 53 \, x - 43} - \frac {50 \, x}{2 \, x^{2} e^{x} - 2 \, x^{2} + 50 \, x e^{x} - 53 \, x - 43}\right )} \] Input:
integrate(((4*x^4+200*x^3+2500*x^2)*exp(x)+6*x^2+172*x+2150)*exp((-2*x^2-5 0*x)/((2*x^2+50*x)*exp(x)-2*x^2-53*x-43))/((4*x^4+200*x^3+2500*x^2)*exp(x) ^2+(-8*x^4-412*x^3-5472*x^2-4300*x)*exp(x)+4*x^4+212*x^3+2981*x^2+4558*x+1 849),x, algorithm="giac")
Output:
e^(-2*x^2/(2*x^2*e^x - 2*x^2 + 50*x*e^x - 53*x - 43) - 50*x/(2*x^2*e^x - 2 *x^2 + 50*x*e^x - 53*x - 43))
Time = 0.40 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {e^{\frac {-50 x-2 x^2}{-43-53 x-2 x^2+e^x \left (50 x+2 x^2\right )}} \left (2150+172 x+6 x^2+e^x \left (2500 x^2+200 x^3+4 x^4\right )\right )}{1849+4558 x+2981 x^2+212 x^3+4 x^4+e^x \left (-4300 x-5472 x^2-412 x^3-8 x^4\right )+e^{2 x} \left (2500 x^2+200 x^3+4 x^4\right )} \, dx={\mathrm {e}}^{\frac {2\,x^2+50\,x}{53\,x-2\,x^2\,{\mathrm {e}}^x-50\,x\,{\mathrm {e}}^x+2\,x^2+43}} \] Input:
int((exp((50*x + 2*x^2)/(53*x - exp(x)*(50*x + 2*x^2) + 2*x^2 + 43))*(172* x + exp(x)*(2500*x^2 + 200*x^3 + 4*x^4) + 6*x^2 + 2150))/(4558*x - exp(x)* (4300*x + 5472*x^2 + 412*x^3 + 8*x^4) + exp(2*x)*(2500*x^2 + 200*x^3 + 4*x ^4) + 2981*x^2 + 212*x^3 + 4*x^4 + 1849),x)
Output:
exp((50*x + 2*x^2)/(53*x - 2*x^2*exp(x) - 50*x*exp(x) + 2*x^2 + 43))
Time = 0.18 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48 \[ \int \frac {e^{\frac {-50 x-2 x^2}{-43-53 x-2 x^2+e^x \left (50 x+2 x^2\right )}} \left (2150+172 x+6 x^2+e^x \left (2500 x^2+200 x^3+4 x^4\right )\right )}{1849+4558 x+2981 x^2+212 x^3+4 x^4+e^x \left (-4300 x-5472 x^2-412 x^3-8 x^4\right )+e^{2 x} \left (2500 x^2+200 x^3+4 x^4\right )} \, dx=\frac {1}{e^{\frac {2 x^{2}+50 x}{2 e^{x} x^{2}+50 e^{x} x -2 x^{2}-53 x -43}}} \] Input:
int(((4*x^4+200*x^3+2500*x^2)*exp(x)+6*x^2+172*x+2150)*exp((-2*x^2-50*x)/( (2*x^2+50*x)*exp(x)-2*x^2-53*x-43))/((4*x^4+200*x^3+2500*x^2)*exp(x)^2+(-8 *x^4-412*x^3-5472*x^2-4300*x)*exp(x)+4*x^4+212*x^3+2981*x^2+4558*x+1849),x )
Output:
1/e**((2*x**2 + 50*x)/(2*e**x*x**2 + 50*e**x*x - 2*x**2 - 53*x - 43))