Integrand size = 97, antiderivative size = 29 \[ \int \frac {1055 x^2+2 x^3+e^{2 x} (1055+2 x)+e^x \left (-2110 x-4 x^2\right )+e^{-\frac {1}{e^x-x}} \left (-24+8 e^{2 x}+e^x (24-8 x)-8 x+8 x^2\right )}{4 e^{2 x}-8 e^x x+4 x^2} \, dx=-2+x+\left (256+2 \left (3+e^{\frac {1}{-e^x+x}}\right )+\frac {x}{4}\right ) (3+x) \] Output:
(3+x)*(2*exp(1/(x-exp(x)))+262+1/4*x)-2+x
Time = 5.10 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {1055 x^2+2 x^3+e^{2 x} (1055+2 x)+e^x \left (-2110 x-4 x^2\right )+e^{-\frac {1}{e^x-x}} \left (-24+8 e^{2 x}+e^x (24-8 x)-8 x+8 x^2\right )}{4 e^{2 x}-8 e^x x+4 x^2} \, dx=\frac {1}{4} \left (1055 x+x^2+e^{-\frac {1}{e^x-x}} (24+8 x)\right ) \] Input:
Integrate[(1055*x^2 + 2*x^3 + E^(2*x)*(1055 + 2*x) + E^x*(-2110*x - 4*x^2) + (-24 + 8*E^(2*x) + E^x*(24 - 8*x) - 8*x + 8*x^2)/E^(E^x - x)^(-1))/(4*E ^(2*x) - 8*E^x*x + 4*x^2),x]
Output:
(1055*x + x^2 + (24 + 8*x)/E^(E^x - x)^(-1))/4
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.
\(\displaystyle \int \frac {2 x^3+1055 x^2+e^x \left (-4 x^2-2110 x\right )+e^{-\frac {1}{e^x-x}} \left (8 x^2-8 x+8 e^{2 x}+e^x (24-8 x)-24\right )+e^{2 x} (2 x+1055)}{4 x^2-8 e^x x+4 e^{2 x}} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2 x^3+1055 x^2+e^x \left (-4 x^2-2110 x\right )+e^{-\frac {1}{e^x-x}} \left (8 x^2-8 x+8 e^{2 x}+e^x (24-8 x)-24\right )+e^{2 x} (2 x+1055)}{4 \left (e^x-x\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int \frac {2 x^3+1055 x^2+e^{2 x} (2 x+1055)-8 e^{-\frac {1}{e^x-x}} \left (-x^2+x-e^{2 x}-e^x (3-x)+3\right )-2 e^x \left (2 x^2+1055 x\right )}{\left (e^x-x\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{4} \int \left (\frac {8 e^{-\frac {1}{e^x-x}} (x+3)}{e^x-x}+e^{-\frac {1}{e^x-x}} \left (2 e^{\frac {1}{e^x-x}} x+1055 e^{\frac {1}{e^x-x}}+8\right )+\frac {8 e^{-\frac {1}{e^x-x}} \left (x^2+2 x-3\right )}{\left (e^x-x\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (8 \int \frac {e^{-\frac {1}{e^x-x}} x^2}{\left (e^x-x\right )^2}dx+8 \int e^{-\frac {1}{e^x-x}}dx-24 \int \frac {e^{-\frac {1}{e^x-x}}}{\left (e^x-x\right )^2}dx+24 \int \frac {e^{-\frac {1}{e^x-x}}}{e^x-x}dx+16 \int \frac {e^{-\frac {1}{e^x-x}} x}{\left (e^x-x\right )^2}dx+8 \int \frac {e^{-\frac {1}{e^x-x}} x}{e^x-x}dx+x^2+1055 x\right )\) |
Input:
Int[(1055*x^2 + 2*x^3 + E^(2*x)*(1055 + 2*x) + E^x*(-2110*x - 4*x^2) + (-2 4 + 8*E^(2*x) + E^x*(24 - 8*x) - 8*x + 8*x^2)/E^(E^x - x)^(-1))/(4*E^(2*x) - 8*E^x*x + 4*x^2),x]
Output:
$Aborted
Time = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93
method | result | size |
risch | \(\frac {x^{2}}{4}+\frac {1055 x}{4}+\left (2 x +6\right ) {\mathrm e}^{-\frac {1}{{\mathrm e}^{x}-x}}\) | \(27\) |
parallelrisch | \(\frac {x^{2}}{4}+\frac {1055 x}{4}+2 \,{\mathrm e}^{-\frac {1}{{\mathrm e}^{x}-x}} x +6 \,{\mathrm e}^{-\frac {1}{{\mathrm e}^{x}-x}}\) | \(37\) |
norman | \(\frac {\frac {1055 x^{2}}{4}+\frac {x^{3}}{4}-\frac {1055 \,{\mathrm e}^{x} x}{4}-\frac {{\mathrm e}^{x} x^{2}}{4}-6 \,{\mathrm e}^{x} {\mathrm e}^{-\frac {1}{{\mathrm e}^{x}-x}}+6 \,{\mathrm e}^{-\frac {1}{{\mathrm e}^{x}-x}} x +2 \,{\mathrm e}^{-\frac {1}{{\mathrm e}^{x}-x}} x^{2}-2 \,{\mathrm e}^{x} {\mathrm e}^{-\frac {1}{{\mathrm e}^{x}-x}} x}{x -{\mathrm e}^{x}}\) | \(94\) |
Input:
int(((8*exp(x)^2+(-8*x+24)*exp(x)+8*x^2-8*x-24)*exp(-1/(exp(x)-x))+(2*x+10 55)*exp(x)^2+(-4*x^2-2110*x)*exp(x)+2*x^3+1055*x^2)/(4*exp(x)^2-8*exp(x)*x +4*x^2),x,method=_RETURNVERBOSE)
Output:
1/4*x^2+1055/4*x+(2*x+6)*exp(-1/(exp(x)-x))
Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {1055 x^2+2 x^3+e^{2 x} (1055+2 x)+e^x \left (-2110 x-4 x^2\right )+e^{-\frac {1}{e^x-x}} \left (-24+8 e^{2 x}+e^x (24-8 x)-8 x+8 x^2\right )}{4 e^{2 x}-8 e^x x+4 x^2} \, dx=\frac {1}{4} \, x^{2} + 2 \, {\left (x + 3\right )} e^{\left (\frac {1}{x - e^{x}}\right )} + \frac {1055}{4} \, x \] Input:
integrate(((8*exp(x)^2+(-8*x+24)*exp(x)+8*x^2-8*x-24)*exp(-1/(exp(x)-x))+( 2*x+1055)*exp(x)^2+(-4*x^2-2110*x)*exp(x)+2*x^3+1055*x^2)/(4*exp(x)^2-8*ex p(x)*x+4*x^2),x, algorithm="fricas")
Output:
1/4*x^2 + 2*(x + 3)*e^(1/(x - e^x)) + 1055/4*x
Time = 0.79 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {1055 x^2+2 x^3+e^{2 x} (1055+2 x)+e^x \left (-2110 x-4 x^2\right )+e^{-\frac {1}{e^x-x}} \left (-24+8 e^{2 x}+e^x (24-8 x)-8 x+8 x^2\right )}{4 e^{2 x}-8 e^x x+4 x^2} \, dx=\frac {x^{2}}{4} + \frac {1055 x}{4} + \left (2 x + 6\right ) e^{- \frac {1}{- x + e^{x}}} \] Input:
integrate(((8*exp(x)**2+(-8*x+24)*exp(x)+8*x**2-8*x-24)*exp(-1/(exp(x)-x)) +(2*x+1055)*exp(x)**2+(-4*x**2-2110*x)*exp(x)+2*x**3+1055*x**2)/(4*exp(x)* *2-8*exp(x)*x+4*x**2),x)
Output:
x**2/4 + 1055*x/4 + (2*x + 6)*exp(-1/(-x + exp(x)))
\[ \int \frac {1055 x^2+2 x^3+e^{2 x} (1055+2 x)+e^x \left (-2110 x-4 x^2\right )+e^{-\frac {1}{e^x-x}} \left (-24+8 e^{2 x}+e^x (24-8 x)-8 x+8 x^2\right )}{4 e^{2 x}-8 e^x x+4 x^2} \, dx=\int { \frac {2 \, x^{3} + 1055 \, x^{2} + {\left (2 \, x + 1055\right )} e^{\left (2 \, x\right )} - 2 \, {\left (2 \, x^{2} + 1055 \, x\right )} e^{x} + 8 \, {\left (x^{2} - {\left (x - 3\right )} e^{x} - x + e^{\left (2 \, x\right )} - 3\right )} e^{\left (\frac {1}{x - e^{x}}\right )}}{4 \, {\left (x^{2} - 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )}} \,d x } \] Input:
integrate(((8*exp(x)^2+(-8*x+24)*exp(x)+8*x^2-8*x-24)*exp(-1/(exp(x)-x))+( 2*x+1055)*exp(x)^2+(-4*x^2-2110*x)*exp(x)+2*x^3+1055*x^2)/(4*exp(x)^2-8*ex p(x)*x+4*x^2),x, algorithm="maxima")
Output:
1/4*x^2 + 1055/4*x + 1/4*integrate(8*(x^2 - (x - 3)*e^x - x + e^(2*x) - 3) *e^(1/(x - e^x))/(x^2 - 2*x*e^x + e^(2*x)), x)
Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (22) = 44\).
Time = 0.12 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.17 \[ \int \frac {1055 x^2+2 x^3+e^{2 x} (1055+2 x)+e^x \left (-2110 x-4 x^2\right )+e^{-\frac {1}{e^x-x}} \left (-24+8 e^{2 x}+e^x (24-8 x)-8 x+8 x^2\right )}{4 e^{2 x}-8 e^x x+4 x^2} \, dx=\frac {1}{4} \, {\left (x^{2} e^{x} + 1055 \, x e^{x} + 8 \, x e^{\left (\frac {x^{2} - x e^{x} + 1}{x - e^{x}}\right )} + 24 \, e^{\left (\frac {x^{2} - x e^{x} + 1}{x - e^{x}}\right )}\right )} e^{\left (-x\right )} \] Input:
integrate(((8*exp(x)^2+(-8*x+24)*exp(x)+8*x^2-8*x-24)*exp(-1/(exp(x)-x))+( 2*x+1055)*exp(x)^2+(-4*x^2-2110*x)*exp(x)+2*x^3+1055*x^2)/(4*exp(x)^2-8*ex p(x)*x+4*x^2),x, algorithm="giac")
Output:
1/4*(x^2*e^x + 1055*x*e^x + 8*x*e^((x^2 - x*e^x + 1)/(x - e^x)) + 24*e^((x ^2 - x*e^x + 1)/(x - e^x)))*e^(-x)
Time = 2.70 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {1055 x^2+2 x^3+e^{2 x} (1055+2 x)+e^x \left (-2110 x-4 x^2\right )+e^{-\frac {1}{e^x-x}} \left (-24+8 e^{2 x}+e^x (24-8 x)-8 x+8 x^2\right )}{4 e^{2 x}-8 e^x x+4 x^2} \, dx=\frac {1055\,x}{4}+{\mathrm {e}}^{\frac {1}{x-{\mathrm {e}}^x}}\,\left (2\,x+6\right )+\frac {x^2}{4} \] Input:
int((exp(2*x)*(2*x + 1055) - exp(1/(x - exp(x)))*(8*x - 8*exp(2*x) + exp(x )*(8*x - 24) - 8*x^2 + 24) - exp(x)*(2110*x + 4*x^2) + 1055*x^2 + 2*x^3)/( 4*exp(2*x) - 8*x*exp(x) + 4*x^2),x)
Output:
(1055*x)/4 + exp(1/(x - exp(x)))*(2*x + 6) + x^2/4
Time = 0.18 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.69 \[ \int \frac {1055 x^2+2 x^3+e^{2 x} (1055+2 x)+e^x \left (-2110 x-4 x^2\right )+e^{-\frac {1}{e^x-x}} \left (-24+8 e^{2 x}+e^x (24-8 x)-8 x+8 x^2\right )}{4 e^{2 x}-8 e^x x+4 x^2} \, dx=\frac {e^{\frac {1}{e^{x}-x}} x^{2}+1055 e^{\frac {1}{e^{x}-x}} x +8 x +24}{4 e^{\frac {1}{e^{x}-x}}} \] Input:
int(((8*exp(x)^2+(-8*x+24)*exp(x)+8*x^2-8*x-24)*exp(-1/(exp(x)-x))+(2*x+10 55)*exp(x)^2+(-4*x^2-2110*x)*exp(x)+2*x^3+1055*x^2)/(4*exp(x)^2-8*exp(x)*x +4*x^2),x)
Output:
(e**(1/(e**x - x))*x**2 + 1055*e**(1/(e**x - x))*x + 8*x + 24)/(4*e**(1/(e **x - x)))