Integrand size = 113, antiderivative size = 25 \[ \int \frac {20 x^4-10 x^5+e^{e^x} \left (e^{2 x} \left (-16 x^3+8 x^4-x^5\right )+e^x \left (48 x^2-40 x^3+11 x^4-x^5\right )\right )}{25 x^4+e^{2 e^x+2 x} \left (16-8 x+x^2\right )+e^{e^x+x} \left (40 x^2-10 x^3\right )} \, dx=-2+\frac {x}{-\frac {5}{-4+x}+\frac {e^{e^x+x}}{x^2}} \] Output:
x/(exp(x)/x^2*exp(exp(x))-5/(-4+x))-2
Time = 0.76 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {20 x^4-10 x^5+e^{e^x} \left (e^{2 x} \left (-16 x^3+8 x^4-x^5\right )+e^x \left (48 x^2-40 x^3+11 x^4-x^5\right )\right )}{25 x^4+e^{2 e^x+2 x} \left (16-8 x+x^2\right )+e^{e^x+x} \left (40 x^2-10 x^3\right )} \, dx=-\frac {(-4+x) x^3}{-e^{e^x+x} (-4+x)+5 x^2} \] Input:
Integrate[(20*x^4 - 10*x^5 + E^E^x*(E^(2*x)*(-16*x^3 + 8*x^4 - x^5) + E^x* (48*x^2 - 40*x^3 + 11*x^4 - x^5)))/(25*x^4 + E^(2*E^x + 2*x)*(16 - 8*x + x ^2) + E^(E^x + x)*(40*x^2 - 10*x^3)),x]
Output:
-(((-4 + x)*x^3)/(-(E^(E^x + x)*(-4 + x)) + 5*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.
\(\displaystyle \int \frac {-10 x^5+20 x^4+e^{e^x} \left (e^{2 x} \left (-x^5+8 x^4-16 x^3\right )+e^x \left (-x^5+11 x^4-40 x^3+48 x^2\right )\right )}{25 x^4+e^{2 x+2 e^x} \left (x^2-8 x+16\right )+e^{x+e^x} \left (40 x^2-10 x^3\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {x^2 \left (-10 (x-2) x^2-e^{x+e^x} (x-3) (x-4)^2-e^{2 x+e^x} x (x-4)^2\right )}{\left (e^{x+e^x} (x-4)-5 x^2\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-e^{-e^x} x^3-\frac {e^{-e^x} \left (10 x^3+e^{e^x} x^2-7 e^{e^x} x+12 e^{e^x}\right ) x^2}{-5 x^2+e^{x+e^x} x-4 e^{x+e^x}}-\frac {5 e^{-e^x} \left (5 x^3+e^{e^x} x^2-5 e^{e^x} x+8 e^{e^x}\right ) x^4}{\left (-5 x^2+e^{x+e^x} x-4 e^{x+e^x}\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int e^{-e^x} x^3dx-12 \int \frac {x^2}{-5 x^2+e^{x+e^x} x-4 e^{x+e^x}}dx-25 \int \frac {e^{-e^x} x^7}{\left (5 x^2-e^{x+e^x} x+4 e^{x+e^x}\right )^2}dx-5 \int \frac {x^6}{\left (-5 x^2+e^{x+e^x} x-4 e^{x+e^x}\right )^2}dx+25 \int \frac {x^5}{\left (-5 x^2+e^{x+e^x} x-4 e^{x+e^x}\right )^2}dx+10 \int \frac {e^{-e^x} x^5}{5 x^2-e^{x+e^x} x+4 e^{x+e^x}}dx-40 \int \frac {x^4}{\left (-5 x^2+e^{x+e^x} x-4 e^{x+e^x}\right )^2}dx-\int \frac {x^4}{-5 x^2+e^{x+e^x} x-4 e^{x+e^x}}dx+7 \int \frac {x^3}{-5 x^2+e^{x+e^x} x-4 e^{x+e^x}}dx\) |
Input:
Int[(20*x^4 - 10*x^5 + E^E^x*(E^(2*x)*(-16*x^3 + 8*x^4 - x^5) + E^x*(48*x^ 2 - 40*x^3 + 11*x^4 - x^5)))/(25*x^4 + E^(2*E^x + 2*x)*(16 - 8*x + x^2) + E^(E^x + x)*(40*x^2 - 10*x^3)),x]
Output:
$Aborted
Time = 0.66 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28
method | result | size |
risch | \(-\frac {\left (x -4\right ) x^{3}}{-x \,{\mathrm e}^{{\mathrm e}^{x}+x}+5 x^{2}+4 \,{\mathrm e}^{{\mathrm e}^{x}+x}}\) | \(32\) |
parallelrisch | \(-\frac {5 x^{4}-20 x^{3}}{5 \left (-x \,{\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{x}}+5 x^{2}+4 \,{\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{x}}\right )}\) | \(37\) |
Input:
int((((-x^5+8*x^4-16*x^3)*exp(x)^2+(-x^5+11*x^4-40*x^3+48*x^2)*exp(x))*exp (exp(x))-10*x^5+20*x^4)/((x^2-8*x+16)*exp(x)^2*exp(exp(x))^2+(-10*x^3+40*x ^2)*exp(x)*exp(exp(x))+25*x^4),x,method=_RETURNVERBOSE)
Output:
-(x-4)*x^3/(-x*exp(exp(x)+x)+5*x^2+4*exp(exp(x)+x))
Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {20 x^4-10 x^5+e^{e^x} \left (e^{2 x} \left (-16 x^3+8 x^4-x^5\right )+e^x \left (48 x^2-40 x^3+11 x^4-x^5\right )\right )}{25 x^4+e^{2 e^x+2 x} \left (16-8 x+x^2\right )+e^{e^x+x} \left (40 x^2-10 x^3\right )} \, dx=-\frac {x^{4} - 4 \, x^{3}}{5 \, x^{2} - {\left (x - 4\right )} e^{\left (x + e^{x}\right )}} \] Input:
integrate((((-x^5+8*x^4-16*x^3)*exp(x)^2+(-x^5+11*x^4-40*x^3+48*x^2)*exp(x ))*exp(exp(x))-10*x^5+20*x^4)/((x^2-8*x+16)*exp(x)^2*exp(exp(x))^2+(-10*x^ 3+40*x^2)*exp(x)*exp(exp(x))+25*x^4),x, algorithm="fricas")
Output:
-(x^4 - 4*x^3)/(5*x^2 - (x - 4)*e^(x + e^x))
Time = 0.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {20 x^4-10 x^5+e^{e^x} \left (e^{2 x} \left (-16 x^3+8 x^4-x^5\right )+e^x \left (48 x^2-40 x^3+11 x^4-x^5\right )\right )}{25 x^4+e^{2 e^x+2 x} \left (16-8 x+x^2\right )+e^{e^x+x} \left (40 x^2-10 x^3\right )} \, dx=\frac {x^{4} - 4 x^{3}}{- 5 x^{2} + \left (x e^{x} - 4 e^{x}\right ) e^{e^{x}}} \] Input:
integrate((((-x**5+8*x**4-16*x**3)*exp(x)**2+(-x**5+11*x**4-40*x**3+48*x** 2)*exp(x))*exp(exp(x))-10*x**5+20*x**4)/((x**2-8*x+16)*exp(x)**2*exp(exp(x ))**2+(-10*x**3+40*x**2)*exp(x)*exp(exp(x))+25*x**4),x)
Output:
(x**4 - 4*x**3)/(-5*x**2 + (x*exp(x) - 4*exp(x))*exp(exp(x)))
Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {20 x^4-10 x^5+e^{e^x} \left (e^{2 x} \left (-16 x^3+8 x^4-x^5\right )+e^x \left (48 x^2-40 x^3+11 x^4-x^5\right )\right )}{25 x^4+e^{2 e^x+2 x} \left (16-8 x+x^2\right )+e^{e^x+x} \left (40 x^2-10 x^3\right )} \, dx=-\frac {x^{4} - 4 \, x^{3}}{5 \, x^{2} - {\left (x - 4\right )} e^{\left (x + e^{x}\right )}} \] Input:
integrate((((-x^5+8*x^4-16*x^3)*exp(x)^2+(-x^5+11*x^4-40*x^3+48*x^2)*exp(x ))*exp(exp(x))-10*x^5+20*x^4)/((x^2-8*x+16)*exp(x)^2*exp(exp(x))^2+(-10*x^ 3+40*x^2)*exp(x)*exp(exp(x))+25*x^4),x, algorithm="maxima")
Output:
-(x^4 - 4*x^3)/(5*x^2 - (x - 4)*e^(x + e^x))
Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {20 x^4-10 x^5+e^{e^x} \left (e^{2 x} \left (-16 x^3+8 x^4-x^5\right )+e^x \left (48 x^2-40 x^3+11 x^4-x^5\right )\right )}{25 x^4+e^{2 e^x+2 x} \left (16-8 x+x^2\right )+e^{e^x+x} \left (40 x^2-10 x^3\right )} \, dx=-\frac {x^{4} - 4 \, x^{3}}{5 \, x^{2} - x e^{\left (x + e^{x}\right )} + 4 \, e^{\left (x + e^{x}\right )}} \] Input:
integrate((((-x^5+8*x^4-16*x^3)*exp(x)^2+(-x^5+11*x^4-40*x^3+48*x^2)*exp(x ))*exp(exp(x))-10*x^5+20*x^4)/((x^2-8*x+16)*exp(x)^2*exp(exp(x))^2+(-10*x^ 3+40*x^2)*exp(x)*exp(exp(x))+25*x^4),x, algorithm="giac")
Output:
-(x^4 - 4*x^3)/(5*x^2 - x*e^(x + e^x) + 4*e^(x + e^x))
Timed out. \[ \int \frac {20 x^4-10 x^5+e^{e^x} \left (e^{2 x} \left (-16 x^3+8 x^4-x^5\right )+e^x \left (48 x^2-40 x^3+11 x^4-x^5\right )\right )}{25 x^4+e^{2 e^x+2 x} \left (16-8 x+x^2\right )+e^{e^x+x} \left (40 x^2-10 x^3\right )} \, dx=\int -\frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,\left ({\mathrm {e}}^{2\,x}\,\left (x^5-8\,x^4+16\,x^3\right )-{\mathrm {e}}^x\,\left (-x^5+11\,x^4-40\,x^3+48\,x^2\right )\right )-20\,x^4+10\,x^5}{25\,x^4+{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,\left (x^2-8\,x+16\right )+{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^x\,\left (40\,x^2-10\,x^3\right )} \,d x \] Input:
int(-(exp(exp(x))*(exp(2*x)*(16*x^3 - 8*x^4 + x^5) - exp(x)*(48*x^2 - 40*x ^3 + 11*x^4 - x^5)) - 20*x^4 + 10*x^5)/(25*x^4 + exp(2*x)*exp(2*exp(x))*(x ^2 - 8*x + 16) + exp(exp(x))*exp(x)*(40*x^2 - 10*x^3)),x)
Output:
int(-(exp(exp(x))*(exp(2*x)*(16*x^3 - 8*x^4 + x^5) - exp(x)*(48*x^2 - 40*x ^3 + 11*x^4 - x^5)) - 20*x^4 + 10*x^5)/(25*x^4 + exp(2*x)*exp(2*exp(x))*(x ^2 - 8*x + 16) + exp(exp(x))*exp(x)*(40*x^2 - 10*x^3)), x)
Time = 0.21 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {20 x^4-10 x^5+e^{e^x} \left (e^{2 x} \left (-16 x^3+8 x^4-x^5\right )+e^x \left (48 x^2-40 x^3+11 x^4-x^5\right )\right )}{25 x^4+e^{2 e^x+2 x} \left (16-8 x+x^2\right )+e^{e^x+x} \left (40 x^2-10 x^3\right )} \, dx=\frac {x^{3} \left (x -4\right )}{e^{e^{x}+x} x -4 e^{e^{x}+x}-5 x^{2}} \] Input:
int((((-x^5+8*x^4-16*x^3)*exp(x)^2+(-x^5+11*x^4-40*x^3+48*x^2)*exp(x))*exp (exp(x))-10*x^5+20*x^4)/((x^2-8*x+16)*exp(x)^2*exp(exp(x))^2+(-10*x^3+40*x ^2)*exp(x)*exp(exp(x))+25*x^4),x)
Output:
(x**3*(x - 4))/(e**(e**x + x)*x - 4*e**(e**x + x) - 5*x**2)