Integrand size = 56, antiderivative size = 28 \[ \int e^{-x+\frac {e^{-x} \left (-40 x-8 e x+e^x \left (e^5+20 x^2+4 e x^2\right )\right )}{5+e}} \left (-8+8 x+8 e^x x\right ) \, dx=e^{-\frac {e^5}{-5-e}-8 e^{-x} x+4 x^2} \] Output:
exp(4*x^2-exp(5)/(-exp(1)-5)-8*x/exp(x))
Time = 0.59 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int e^{-x+\frac {e^{-x} \left (-40 x-8 e x+e^x \left (e^5+20 x^2+4 e x^2\right )\right )}{5+e}} \left (-8+8 x+8 e^x x\right ) \, dx=e^{\frac {e^5}{5+e}-8 e^{-x} x+4 x^2} \] Input:
Integrate[E^(-x + (-40*x - 8*E*x + E^x*(E^5 + 20*x^2 + 4*E*x^2))/(E^x*(5 + E)))*(-8 + 8*x + 8*E^x*x),x]
Output:
E^(E^5/(5 + E) - (8*x)/E^x + 4*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.
\(\displaystyle \int \left (8 e^x x+8 x-8\right ) \exp \left (\frac {e^{-x} \left (e^x \left (4 e x^2+20 x^2+e^5\right )-8 e x-40 x\right )}{5+e}-x\right ) \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int 8 \left (e^x x+x-1\right ) \exp \left (\frac {e^{-x} \left (e^x \left (4 e x^2+20 x^2+e^5\right )-8 e x-40 x\right )}{5+e}-x\right )dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 8 \int -\exp \left (-x-\frac {e^{-x} \left (8 e x+40 x-e^x \left (4 e x^2+20 x^2+e^5\right )\right )}{5+e}\right ) \left (-e^x x-x+1\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -8 \int \exp \left (-x-\frac {e^{-x} \left (8 e x+40 x-e^x \left (4 e x^2+20 x^2+e^5\right )\right )}{5+e}\right ) \left (-e^x x-x+1\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -8 \int \left (-\exp \left (-\frac {e^{-x} \left (8 e x+40 x-e^x \left (4 e x^2+20 x^2+e^5\right )\right )}{5+e}\right ) x-\exp \left (-x-\frac {e^{-x} \left (8 e x+40 x-e^x \left (4 e x^2+20 x^2+e^5\right )\right )}{5+e}\right ) x+\exp \left (-x-\frac {e^{-x} \left (8 e x+40 x-e^x \left (4 e x^2+20 x^2+e^5\right )\right )}{5+e}\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -8 \left (\int \exp \left (-x-\frac {e^{-x} \left (8 e x+40 x-e^x \left (4 e x^2+20 x^2+e^5\right )\right )}{5+e}\right )dx-\int \exp \left (-\frac {e^{-x} \left (40 \left (1+\frac {e}{5}\right ) x-e^x \left (4 e x^2+20 x^2+e^5\right )\right )}{5+e}\right ) xdx-\int \exp \left (-x-\frac {e^{-x} \left (8 e x+40 x-e^x \left (4 e x^2+20 x^2+e^5\right )\right )}{5+e}\right ) xdx\right )\) |
Input:
Int[E^(-x + (-40*x - 8*E*x + E^x*(E^5 + 20*x^2 + 4*E*x^2))/(E^x*(5 + E)))* (-8 + 8*x + 8*E^x*x),x]
Output:
$Aborted
Time = 0.34 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43
method | result | size |
default | \({\mathrm e}^{\frac {\left (\left ({\mathrm e}^{5}+4 x^{2} {\mathrm e}+20 x^{2}\right ) {\mathrm e}^{x}-8 x \,{\mathrm e}-40 x \right ) {\mathrm e}^{-x}}{{\mathrm e}+5}}\) | \(40\) |
norman | \({\mathrm e}^{\frac {\left (\left ({\mathrm e}^{5}+4 x^{2} {\mathrm e}+20 x^{2}\right ) {\mathrm e}^{x}-8 x \,{\mathrm e}-40 x \right ) {\mathrm e}^{-x}}{{\mathrm e}+5}}\) | \(40\) |
parallelrisch | \({\mathrm e}^{\frac {\left (\left ({\mathrm e}^{5}+4 x^{2} {\mathrm e}+20 x^{2}\right ) {\mathrm e}^{x}-8 x \,{\mathrm e}-40 x \right ) {\mathrm e}^{-x}}{{\mathrm e}+5}}\) | \(40\) |
risch | \({\mathrm e}^{-\frac {\left (-4 x^{2} {\mathrm e}^{1+x}-20 \,{\mathrm e}^{x} x^{2}-{\mathrm e}^{5+x}+8 x \,{\mathrm e}+40 x \right ) {\mathrm e}^{-x}}{{\mathrm e}+5}}\) | \(45\) |
Input:
int((8*exp(x)*x+8*x-8)*exp(((exp(5)+4*x^2*exp(1)+20*x^2)*exp(x)-8*x*exp(1) -40*x)/(exp(1)+5)/exp(x))/exp(x),x,method=_RETURNVERBOSE)
Output:
exp(((exp(5)+4*x^2*exp(1)+20*x^2)*exp(x)-8*x*exp(1)-40*x)/(exp(1)+5)/exp(x ))
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (23) = 46\).
Time = 0.10 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.82 \[ \int e^{-x+\frac {e^{-x} \left (-40 x-8 e x+e^x \left (e^5+20 x^2+4 e x^2\right )\right )}{5+e}} \left (-8+8 x+8 e^x x\right ) \, dx=e^{\left (x - \frac {{\left (8 \, x e - {\left (20 \, x^{2} + {\left (4 \, x^{2} - x\right )} e - 5 \, x + e^{5}\right )} e^{x} + 40 \, x\right )} e^{\left (-x\right )}}{e + 5}\right )} \] Input:
integrate((8*exp(x)*x+8*x-8)*exp(((exp(5)+4*x^2*exp(1)+20*x^2)*exp(x)-8*ex p(1)*x-40*x)/(exp(1)+5)/exp(x))/exp(x),x, algorithm="fricas")
Output:
e^(x - (8*x*e - (20*x^2 + (4*x^2 - x)*e - 5*x + e^5)*e^x + 40*x)*e^(-x)/(e + 5))
Time = 0.12 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int e^{-x+\frac {e^{-x} \left (-40 x-8 e x+e^x \left (e^5+20 x^2+4 e x^2\right )\right )}{5+e}} \left (-8+8 x+8 e^x x\right ) \, dx=e^{\frac {\left (- 40 x - 8 e x + \left (4 e x^{2} + 20 x^{2} + e^{5}\right ) e^{x}\right ) e^{- x}}{e + 5}} \] Input:
integrate((8*exp(x)*x+8*x-8)*exp(((exp(5)+4*x**2*exp(1)+20*x**2)*exp(x)-8* exp(1)*x-40*x)/(exp(1)+5)/exp(x))/exp(x),x)
Output:
exp((-40*x - 8*E*x + (4*E*x**2 + 20*x**2 + exp(5))*exp(x))*exp(-x)/(E + 5) )
Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (23) = 46\).
Time = 0.29 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.25 \[ \int e^{-x+\frac {e^{-x} \left (-40 x-8 e x+e^x \left (e^5+20 x^2+4 e x^2\right )\right )}{5+e}} \left (-8+8 x+8 e^x x\right ) \, dx=e^{\left (\frac {4 \, x^{2} e}{e + 5} + \frac {20 \, x^{2}}{e + 5} - \frac {40 \, x e^{\left (-x\right )}}{e + 5} - \frac {8 \, x e^{\left (-x + 1\right )}}{e + 5} + \frac {e^{5}}{e + 5}\right )} \] Input:
integrate((8*exp(x)*x+8*x-8)*exp(((exp(5)+4*x^2*exp(1)+20*x^2)*exp(x)-8*ex p(1)*x-40*x)/(exp(1)+5)/exp(x))/exp(x),x, algorithm="maxima")
Output:
e^(4*x^2*e/(e + 5) + 20*x^2/(e + 5) - 40*x*e^(-x)/(e + 5) - 8*x*e^(-x + 1) /(e + 5) + e^5/(e + 5))
\[ \int e^{-x+\frac {e^{-x} \left (-40 x-8 e x+e^x \left (e^5+20 x^2+4 e x^2\right )\right )}{5+e}} \left (-8+8 x+8 e^x x\right ) \, dx=\int { 8 \, {\left (x e^{x} + x - 1\right )} e^{\left (-x - \frac {{\left (8 \, x e - {\left (4 \, x^{2} e + 20 \, x^{2} + e^{5}\right )} e^{x} + 40 \, x\right )} e^{\left (-x\right )}}{e + 5}\right )} \,d x } \] Input:
integrate((8*exp(x)*x+8*x-8)*exp(((exp(5)+4*x^2*exp(1)+20*x^2)*exp(x)-8*ex p(1)*x-40*x)/(exp(1)+5)/exp(x))/exp(x),x, algorithm="giac")
Output:
integrate(8*(x*e^x + x - 1)*e^(-x - (8*x*e - (4*x^2*e + 20*x^2 + e^5)*e^x + 40*x)*e^(-x)/(e + 5)), x)
Time = 2.98 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.39 \[ \int e^{-x+\frac {e^{-x} \left (-40 x-8 e x+e^x \left (e^5+20 x^2+4 e x^2\right )\right )}{5+e}} \left (-8+8 x+8 e^x x\right ) \, dx={\mathrm {e}}^{-\frac {8\,x\,{\mathrm {e}}^{-x}\,\mathrm {e}}{\mathrm {e}+5}}\,{\mathrm {e}}^{\frac {4\,x^2\,\mathrm {e}}{\mathrm {e}+5}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^5}{\mathrm {e}+5}}\,{\mathrm {e}}^{-\frac {40\,x\,{\mathrm {e}}^{-x}}{\mathrm {e}+5}}\,{\mathrm {e}}^{\frac {20\,x^2}{\mathrm {e}+5}} \] Input:
int(exp(-x)*exp(-(exp(-x)*(40*x + 8*x*exp(1) - exp(x)*(exp(5) + 4*x^2*exp( 1) + 20*x^2)))/(exp(1) + 5))*(8*x + 8*x*exp(x) - 8),x)
Output:
exp(-(8*x*exp(-x)*exp(1))/(exp(1) + 5))*exp((4*x^2*exp(1))/(exp(1) + 5))*e xp(exp(5)/(exp(1) + 5))*exp(-(40*x*exp(-x))/(exp(1) + 5))*exp((20*x^2)/(ex p(1) + 5))
Time = 0.19 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.89 \[ \int e^{-x+\frac {e^{-x} \left (-40 x-8 e x+e^x \left (e^5+20 x^2+4 e x^2\right )\right )}{5+e}} \left (-8+8 x+8 e^x x\right ) \, dx=\frac {e^{e^{4}+25 e^{2}+4 x^{2}} e^{625}}{e^{\frac {5 e^{x} e^{4}+25 e^{x} e^{3}+125 e^{x} e^{2}+625 e^{x} e +3125 e^{x}+8 e x +40 x}{e^{x} e +5 e^{x}}}} \] Input:
int((8*exp(x)*x+8*x-8)*exp(((exp(5)+4*x^2*exp(1)+20*x^2)*exp(x)-8*exp(1)*x -40*x)/(exp(1)+5)/exp(x))/exp(x),x)
Output:
(e**(e**4 + 25*e**2 + 4*x**2)*e**625)/e**((5*e**x*e**4 + 25*e**x*e**3 + 12 5*e**x*e**2 + 625*e**x*e + 3125*e**x + 8*e*x + 40*x)/(e**x*e + 5*e**x))