Integrand size = 94, antiderivative size = 34 \[ \int \frac {2+e^{2+e^{e^{e^x}-x} (-4+x)-x} \left (-16+8 x-x^2+e^{e^{e^x}-x} \left (80-56 x+13 x^2-x^3+e^{e^x+x} \left (-64+48 x-12 x^2+x^3\right )\right )\right )}{16-8 x+x^2} \, dx=e^{2+e^{e^{e^x}-x} (-4+x)-x}+\frac {2-x}{-4+x} \] Output:
(2-x)/(-4+x)+exp((-4+x)*exp(exp(exp(x))-x)+2-x)
Time = 1.39 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \frac {2+e^{2+e^{e^{e^x}-x} (-4+x)-x} \left (-16+8 x-x^2+e^{e^{e^x}-x} \left (80-56 x+13 x^2-x^3+e^{e^x+x} \left (-64+48 x-12 x^2+x^3\right )\right )\right )}{16-8 x+x^2} \, dx=e^{2+e^{e^{e^x}-x} (-4+x)-x}-\frac {2}{-4+x} \] Input:
Integrate[(2 + E^(2 + E^(E^E^x - x)*(-4 + x) - x)*(-16 + 8*x - x^2 + E^(E^ E^x - x)*(80 - 56*x + 13*x^2 - x^3 + E^(E^x + x)*(-64 + 48*x - 12*x^2 + x^ 3))))/(16 - 8*x + x^2),x]
Output:
E^(2 + E^(E^E^x - x)*(-4 + x) - x) - 2/(-4 + 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.
\(\displaystyle \int \frac {e^{e^{e^{e^x}-x} (x-4)-x+2} \left (-x^2+e^{e^{e^x}-x} \left (-x^3+13 x^2+e^{x+e^x} \left (x^3-12 x^2+48 x-64\right )-56 x+80\right )+8 x-16\right )+2}{x^2-8 x+16} \, dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 4 \int \frac {2-e^{-e^{e^{e^x}-x} (4-x)-x+2} \left (x^2-8 x-e^{e^{e^x}-x} \left (-x^3+13 x^2-56 x-e^{x+e^x} \left (-x^3+12 x^2-48 x+64\right )+80\right )+16\right )}{4 (4-x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {2-e^{-e^{e^{e^x}-x} (4-x)-x+2} \left (x^2-e^{e^{e^x}-x} \left (-x^3+13 x^2-e^{x+e^x} \left (-x^3+12 x^2-48 x+64\right )-56 x+80\right )-8 x+16\right )}{(4-x)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-e^{e^{e^{e^x}-x} (x-4)+e^{e^x}-2 x+2} (x-5)+e^{e^{e^{e^x}-x} (x-4)-x+2} \left (e^{e^{e^x}+e^x} x-4 e^{e^{e^x}+e^x}-1\right )+\frac {2}{(x-4)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 5 \int e^{e^{e^{e^x}-x} (x-4)+e^{e^x}-2 x+2}dx-\int e^{e^{e^{e^x}-x} (x-4)-x+2}dx-4 \int e^{e^{e^{e^x}-x} (x-4)+e^{e^x}+e^x-x+2}dx-\int e^{e^{e^{e^x}-x} (x-4)+e^{e^x}-2 x+2} xdx+\int e^{e^{e^{e^x}-x} (x-4)+e^{e^x}+e^x-x+2} xdx+\frac {2}{4-x}\) |
Input:
Int[(2 + E^(2 + E^(E^E^x - x)*(-4 + x) - x)*(-16 + 8*x - x^2 + E^(E^E^x - x)*(80 - 56*x + 13*x^2 - x^3 + E^(E^x + x)*(-64 + 48*x - 12*x^2 + x^3))))/ (16 - 8*x + x^2),x]
Output:
$Aborted
Time = 3.63 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03
method | result | size |
risch | \(-\frac {2}{x -4}+{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{x}}-x} x -4 \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{x}}-x}-x +2}\) | \(35\) |
parallelrisch | \(\frac {{\mathrm e}^{\left (x -4\right ) {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{x}}-x}+2-x} x -2-4 \,{\mathrm e}^{\left (x -4\right ) {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{x}}-x}+2-x}}{x -4}\) | \(49\) |
Input:
int(((((x^3-12*x^2+48*x-64)*exp(x)*exp(exp(x))-x^3+13*x^2-56*x+80)*exp(exp (exp(x))-x)-x^2+8*x-16)*exp((x-4)*exp(exp(exp(x))-x)+2-x)+2)/(x^2-8*x+16), x,method=_RETURNVERBOSE)
Output:
-2/(x-4)+exp(exp(exp(exp(x))-x)*x-4*exp(exp(exp(x))-x)-x+2)
Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {2+e^{2+e^{e^{e^x}-x} (-4+x)-x} \left (-16+8 x-x^2+e^{e^{e^x}-x} \left (80-56 x+13 x^2-x^3+e^{e^x+x} \left (-64+48 x-12 x^2+x^3\right )\right )\right )}{16-8 x+x^2} \, dx=\frac {{\left (x - 4\right )} e^{\left ({\left (x - 4\right )} e^{\left (-{\left (x e^{x} - e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )}\right )} - x + 2\right )} - 2}{x - 4} \] Input:
integrate(((((x^3-12*x^2+48*x-64)*exp(x)*exp(exp(x))-x^3+13*x^2-56*x+80)*e xp(exp(exp(x))-x)-x^2+8*x-16)*exp((-4+x)*exp(exp(exp(x))-x)+2-x)+2)/(x^2-8 *x+16),x, algorithm="fricas")
Output:
((x - 4)*e^((x - 4)*e^(-(x*e^x - e^(x + e^x))*e^(-x)) - x + 2) - 2)/(x - 4 )
Time = 1.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.59 \[ \int \frac {2+e^{2+e^{e^{e^x}-x} (-4+x)-x} \left (-16+8 x-x^2+e^{e^{e^x}-x} \left (80-56 x+13 x^2-x^3+e^{e^x+x} \left (-64+48 x-12 x^2+x^3\right )\right )\right )}{16-8 x+x^2} \, dx=e^{- x + \left (x - 4\right ) e^{- x + e^{e^{x}}} + 2} - \frac {2}{x - 4} \] Input:
integrate(((((x**3-12*x**2+48*x-64)*exp(x)*exp(exp(x))-x**3+13*x**2-56*x+8 0)*exp(exp(exp(x))-x)-x**2+8*x-16)*exp((-4+x)*exp(exp(exp(x))-x)+2-x)+2)/( x**2-8*x+16),x)
Output:
exp(-x + (x - 4)*exp(-x + exp(exp(x))) + 2) - 2/(x - 4)
Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {2+e^{2+e^{e^{e^x}-x} (-4+x)-x} \left (-16+8 x-x^2+e^{e^{e^x}-x} \left (80-56 x+13 x^2-x^3+e^{e^x+x} \left (-64+48 x-12 x^2+x^3\right )\right )\right )}{16-8 x+x^2} \, dx=-\frac {2}{x - 4} + e^{\left (x e^{\left (-x + e^{\left (e^{x}\right )}\right )} - x - 4 \, e^{\left (-x + e^{\left (e^{x}\right )}\right )} + 2\right )} \] Input:
integrate(((((x^3-12*x^2+48*x-64)*exp(x)*exp(exp(x))-x^3+13*x^2-56*x+80)*e xp(exp(exp(x))-x)-x^2+8*x-16)*exp((-4+x)*exp(exp(exp(x))-x)+2-x)+2)/(x^2-8 *x+16),x, algorithm="maxima")
Output:
-2/(x - 4) + e^(x*e^(-x + e^(e^x)) - x - 4*e^(-x + e^(e^x)) + 2)
\[ \int \frac {2+e^{2+e^{e^{e^x}-x} (-4+x)-x} \left (-16+8 x-x^2+e^{e^{e^x}-x} \left (80-56 x+13 x^2-x^3+e^{e^x+x} \left (-64+48 x-12 x^2+x^3\right )\right )\right )}{16-8 x+x^2} \, dx=\int { -\frac {{\left (x^{2} + {\left (x^{3} - 13 \, x^{2} - {\left (x^{3} - 12 \, x^{2} + 48 \, x - 64\right )} e^{\left (x + e^{x}\right )} + 56 \, x - 80\right )} e^{\left (-x + e^{\left (e^{x}\right )}\right )} - 8 \, x + 16\right )} e^{\left ({\left (x - 4\right )} e^{\left (-x + e^{\left (e^{x}\right )}\right )} - x + 2\right )} - 2}{x^{2} - 8 \, x + 16} \,d x } \] Input:
integrate(((((x^3-12*x^2+48*x-64)*exp(x)*exp(exp(x))-x^3+13*x^2-56*x+80)*e xp(exp(exp(x))-x)-x^2+8*x-16)*exp((-4+x)*exp(exp(exp(x))-x)+2-x)+2)/(x^2-8 *x+16),x, algorithm="giac")
Output:
integrate(-((x^2 + (x^3 - 13*x^2 - (x^3 - 12*x^2 + 48*x - 64)*e^(x + e^x) + 56*x - 80)*e^(-x + e^(e^x)) - 8*x + 16)*e^((x - 4)*e^(-x + e^(e^x)) - x + 2) - 2)/(x^2 - 8*x + 16), x)
Time = 3.61 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {2+e^{2+e^{e^{e^x}-x} (-4+x)-x} \left (-16+8 x-x^2+e^{e^{e^x}-x} \left (80-56 x+13 x^2-x^3+e^{e^x+x} \left (-64+48 x-12 x^2+x^3\right )\right )\right )}{16-8 x+x^2} \, dx={\mathrm {e}}^{x\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^x}}}\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^2\,{\mathrm {e}}^{-4\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^x}}}-\frac {2}{x-4} \] Input:
int((exp(exp(exp(exp(x)) - x)*(x - 4) - x + 2)*(8*x + exp(exp(exp(x)) - x) *(13*x^2 - 56*x - x^3 + exp(exp(x))*exp(x)*(48*x - 12*x^2 + x^3 - 64) + 80 ) - x^2 - 16) + 2)/(x^2 - 8*x + 16),x)
Output:
exp(x*exp(-x)*exp(exp(exp(x))))*exp(-x)*exp(2)*exp(-4*exp(-x)*exp(exp(exp( x)))) - 2/(x - 4)
Time = 0.18 (sec) , antiderivative size = 102, normalized size of antiderivative = 3.00 \[ \int \frac {2+e^{2+e^{e^{e^x}-x} (-4+x)-x} \left (-16+8 x-x^2+e^{e^{e^x}-x} \left (80-56 x+13 x^2-x^3+e^{e^x+x} \left (-64+48 x-12 x^2+x^3\right )\right )\right )}{16-8 x+x^2} \, dx=\frac {-e^{\frac {4 e^{e^{e^{x}}}+e^{x} x}{e^{x}}} x +2 e^{\frac {e^{e^{e^{x}}} x}{e^{x}}} e^{2} x -8 e^{\frac {e^{e^{e^{x}}} x}{e^{x}}} e^{2}}{2 e^{\frac {4 e^{e^{e^{x}}}+e^{x} x}{e^{x}}} \left (x -4\right )} \] Input:
int(((((x^3-12*x^2+48*x-64)*exp(x)*exp(exp(x))-x^3+13*x^2-56*x+80)*exp(exp (exp(x))-x)-x^2+8*x-16)*exp((-4+x)*exp(exp(exp(x))-x)+2-x)+2)/(x^2-8*x+16) ,x)
Output:
( - e**((4*e**(e**(e**x)) + e**x*x)/e**x)*x + 2*e**((e**(e**(e**x))*x)/e** x)*e**2*x - 8*e**((e**(e**(e**x))*x)/e**x)*e**2)/(2*e**((4*e**(e**(e**x)) + e**x*x)/e**x)*(x - 4))