Integrand size = 104, antiderivative size = 34 \[ \int \frac {e^{\frac {x-e^{2 e^{2 x}} x+16 x^2-x^3+4 x^2 \log (4)}{4+x}} \left (4+128 x+4 x^2-2 x^3+e^{2 e^{2 x}} \left (-4+e^{2 x} \left (-16 x-4 x^2\right )\right )+\left (32 x+4 x^2\right ) \log (4)\right )}{16+8 x+x^2} \, dx=e^{x \left (-x+\frac {1-e^{2 e^{2 x}}+4 x (5+\log (4))}{4+x}\right )} \] Output:
exp(x*((1-exp(exp(x)^2)^2+4*x*(2*ln(2)+5))/(4+x)-x))
Time = 0.39 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\frac {x-e^{2 e^{2 x}} x+16 x^2-x^3+4 x^2 \log (4)}{4+x}} \left (4+128 x+4 x^2-2 x^3+e^{2 e^{2 x}} \left (-4+e^{2 x} \left (-16 x-4 x^2\right )\right )+\left (32 x+4 x^2\right ) \log (4)\right )}{16+8 x+x^2} \, dx=2^{\frac {8 x^2}{4+x}} e^{-\frac {x \left (-1+e^{2 e^{2 x}}-16 x+x^2\right )}{4+x}} \] Input:
Integrate[(E^((x - E^(2*E^(2*x))*x + 16*x^2 - x^3 + 4*x^2*Log[4])/(4 + x)) *(4 + 128*x + 4*x^2 - 2*x^3 + E^(2*E^(2*x))*(-4 + E^(2*x)*(-16*x - 4*x^2)) + (32*x + 4*x^2)*Log[4]))/(16 + 8*x + x^2),x]
Output:
2^((8*x^2)/(4 + x))/E^((x*(-1 + E^(2*E^(2*x)) - 16*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.
| \(\displaystyle \int \frac {\left (-2 x^3+4 x^2+e^{2 e^{2 x}} \left (e^{2 x} \left (-4 x^2-16 x\right )-4\right )+\left (4 x^2+32 x\right ) \log (4)+128 x+4\right ) \exp \left (\frac {-x^3+16 x^2+4 x^2 \log (4)-e^{2 e^{2 x}} x+x}{x+4}\right )}{x^2+8 x+16} \, dx\) |
| \(\Big \downarrow \) 2007 |
| \(\displaystyle \int \frac {\left (-2 x^3+4 x^2+e^{2 e^{2 x}} \left (e^{2 x} \left (-4 x^2-16 x\right )-4\right )+\left (4 x^2+32 x\right ) \log (4)+128 x+4\right ) \exp \left (\frac {-x^3+16 x^2+4 x^2 \log (4)-e^{2 e^{2 x}} x+x}{x+4}\right )}{(x+4)^2}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {\left (-2 x^3+4 x^2+e^{2 e^{2 x}} \left (e^{2 x} \left (-4 x^2-16 x\right )-4\right )+\left (4 x^2+32 x\right ) \log (4)+128 x+4\right ) \exp \left (\frac {-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )-e^{2 e^{2 x}} x+x}{x+4}\right )}{(x+4)^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {2 x^3 \exp \left (\frac {-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )-e^{2 e^{2 x}} x+x}{x+4}\right )}{(x+4)^2}+\frac {4 x^2 \exp \left (\frac {-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )-e^{2 e^{2 x}} x+x}{x+4}\right )}{(x+4)^2}+\frac {4 x \exp \left (\frac {-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )-e^{2 e^{2 x}} x+x}{x+4}+2 \left (x+e^{2 x}\right )\right )}{-x-4}+\frac {128 x \exp \left (\frac {-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )-e^{2 e^{2 x}} x+x}{x+4}\right )}{(x+4)^2}+\frac {4 (x+8) x \log (4) \exp \left (\frac {-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )-e^{2 e^{2 x}} x+x}{x+4}\right )}{(x+4)^2}-\frac {4 \exp \left (\frac {-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )-e^{2 e^{2 x}} x+x}{x+4}+2 e^{2 x}\right )}{(x+4)^2}+\frac {4 \exp \left (\frac {-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )-e^{2 e^{2 x}} x+x}{x+4}\right )}{(x+4)^2}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {2 \left (-x^3+2 x^2 (1+\log (4))-2 e^{2 \left (x+e^{2 x}\right )} (x+4) x-2 e^{2 e^{2 x}}+16 x (4+\log (4))+2\right ) \exp \left (-\frac {x \left (x^2+e^{2 e^{2 x}}-8 x (2+\log (2))-1\right )}{x+4}\right )}{(x+4)^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 \int \frac {\exp \left (\frac {x \left (-x^2+8 (2+\log (2)) x-e^{2 e^{2 x}}+1\right )}{x+4}\right ) \left (-x^3+2 (1+\log (4)) x^2-2 e^{2 \left (x+e^{2 x}\right )} (x+4) x+16 (4+\log (4)) x-2 e^{2 e^{2 x}}+2\right )}{(x+4)^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 2 \int \left (-\frac {\exp \left (\frac {x \left (-x^2+8 (2+\log (2)) x-e^{2 e^{2 x}}+1\right )}{x+4}\right ) x^3}{(x+4)^2}+\frac {2 \exp \left (\frac {x \left (-x^2+8 (2+\log (2)) x-e^{2 e^{2 x}}+1\right )}{x+4}\right ) (1+\log (4)) x^2}{(x+4)^2}+\frac {2 \exp \left (2 \left (x+e^{2 x}\right )+\frac {x \left (-x^2+8 (2+\log (2)) x-e^{2 e^{2 x}}+1\right )}{x+4}\right ) x}{-x-4}+\frac {16 \exp \left (\frac {x \left (-x^2+8 (2+\log (2)) x-e^{2 e^{2 x}}+1\right )}{x+4}\right ) (4+\log (4)) x}{(x+4)^2}+\frac {2 \exp \left (\frac {x \left (-x^2+8 (2+\log (2)) x-e^{2 e^{2 x}}+1\right )}{x+4}\right )}{(x+4)^2}-\frac {2 \exp \left (\frac {x \left (-x^2+8 (2+\log (2)) x-e^{2 e^{2 x}}+1\right )}{x+4}+2 e^{2 x}\right )}{(x+4)^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 \left (2 (1+\log (4)) \int \exp \left (\frac {x \left (-x^2+8 (2+\log (2)) x-e^{2 e^{2 x}}+1\right )}{x+4}\right )dx+8 \int \exp \left (\frac {x \left (-x^2+8 (2+\log (2)) x-e^{2 e^{2 x}}+1\right )}{x+4}\right )dx-2 \int \exp \left (2 \left (x+e^{2 x}\right )+\frac {x \left (-x^2+8 (2+\log (2)) x-e^{2 e^{2 x}}+1\right )}{x+4}\right )dx-\int \exp \left (\frac {x \left (-x^2+8 (2+\log (2)) x-e^{2 e^{2 x}}+1\right )}{x+4}\right ) xdx-64 (4+\log (4)) \int \frac {\exp \left (\frac {x \left (-x^2+8 (2+\log (2)) x-e^{2 e^{2 x}}+1\right )}{x+4}\right )}{(x+4)^2}dx+32 (1+\log (4)) \int \frac {\exp \left (\frac {x \left (-x^2+8 (2+\log (2)) x-e^{2 e^{2 x}}+1\right )}{x+4}\right )}{(x+4)^2}dx+66 \int \frac {\exp \left (\frac {x \left (-x^2+8 (2+\log (2)) x-e^{2 e^{2 x}}+1\right )}{x+4}\right )}{(x+4)^2}dx-2 \int \frac {\exp \left (\frac {x \left (-x^2+8 (2+\log (2)) x-e^{2 e^{2 x}}+1\right )}{x+4}+2 e^{2 x}\right )}{(x+4)^2}dx+16 (4+\log (4)) \int \frac {\exp \left (\frac {x \left (-x^2+8 (2+\log (2)) x-e^{2 e^{2 x}}+1\right )}{x+4}\right )}{x+4}dx-16 (1+\log (4)) \int \frac {\exp \left (\frac {x \left (-x^2+8 (2+\log (2)) x-e^{2 e^{2 x}}+1\right )}{x+4}\right )}{x+4}dx-48 \int \frac {\exp \left (\frac {x \left (-x^2+8 (2+\log (2)) x-e^{2 e^{2 x}}+1\right )}{x+4}\right )}{x+4}dx+8 \int \frac {\exp \left (2 \left (x+e^{2 x}\right )+\frac {x \left (-x^2+8 (2+\log (2)) x-e^{2 e^{2 x}}+1\right )}{x+4}\right )}{x+4}dx\right )\) |
Input:
Int[(E^((x - E^(2*E^(2*x))*x + 16*x^2 - x^3 + 4*x^2*Log[4])/(4 + x))*(4 + 128*x + 4*x^2 - 2*x^3 + E^(2*E^(2*x))*(-4 + E^(2*x)*(-16*x - 4*x^2)) + (32 *x + 4*x^2)*Log[4]))/(16 + 8*x + x^2),x]
Output:
$Aborted
Time = 45.69 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97
| method | result | size |
| risch | \({\mathrm e}^{\frac {x \left (8 x \ln \left (2\right )-x^{2}-{\mathrm e}^{2 \,{\mathrm e}^{2 x}}+16 x +1\right )}{4+x}}\) | \(33\) |
| parallelrisch | \({\mathrm e}^{\frac {-{\mathrm e}^{2 \,{\mathrm e}^{2 x}} x +8 x^{2} \ln \left (2\right )-x^{3}+16 x^{2}+x}{4+x}}\) | \(37\) |
Input:
int((((-4*x^2-16*x)*exp(x)^2-4)*exp(exp(x)^2)^2+2*(4*x^2+32*x)*ln(2)-2*x^3 +4*x^2+128*x+4)*exp((-x*exp(exp(x)^2)^2+8*x^2*ln(2)-x^3+16*x^2+x)/(4+x))/( x^2+8*x+16),x,method=_RETURNVERBOSE)
Output:
exp(x*(8*x*ln(2)-x^2-exp(2*exp(2*x))+16*x+1)/(4+x))
Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {e^{\frac {x-e^{2 e^{2 x}} x+16 x^2-x^3+4 x^2 \log (4)}{4+x}} \left (4+128 x+4 x^2-2 x^3+e^{2 e^{2 x}} \left (-4+e^{2 x} \left (-16 x-4 x^2\right )\right )+\left (32 x+4 x^2\right ) \log (4)\right )}{16+8 x+x^2} \, dx=e^{\left (-\frac {x^{3} - 8 \, x^{2} \log \left (2\right ) - 16 \, x^{2} + x e^{\left (2 \, e^{\left (2 \, x\right )}\right )} - x}{x + 4}\right )} \] Input:
integrate((((-4*x^2-16*x)*exp(x)^2-4)*exp(exp(x)^2)^2+2*(4*x^2+32*x)*log(2 )-2*x^3+4*x^2+128*x+4)*exp((-x*exp(exp(x)^2)^2+8*x^2*log(2)-x^3+16*x^2+x)/ (4+x))/(x^2+8*x+16),x, algorithm="fricas")
Output:
e^(-(x^3 - 8*x^2*log(2) - 16*x^2 + x*e^(2*e^(2*x)) - x)/(x + 4))
Time = 0.52 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {x-e^{2 e^{2 x}} x+16 x^2-x^3+4 x^2 \log (4)}{4+x}} \left (4+128 x+4 x^2-2 x^3+e^{2 e^{2 x}} \left (-4+e^{2 x} \left (-16 x-4 x^2\right )\right )+\left (32 x+4 x^2\right ) \log (4)\right )}{16+8 x+x^2} \, dx=e^{\frac {- x^{3} + 8 x^{2} \log {\left (2 \right )} + 16 x^{2} - x e^{2 e^{2 x}} + x}{x + 4}} \] Input:
integrate((((-4*x**2-16*x)*exp(x)**2-4)*exp(exp(x)**2)**2+2*(4*x**2+32*x)* ln(2)-2*x**3+4*x**2+128*x+4)*exp((-x*exp(exp(x)**2)**2+8*x**2*ln(2)-x**3+1 6*x**2+x)/(4+x))/(x**2+8*x+16),x)
Output:
exp((-x**3 + 8*x**2*log(2) + 16*x**2 - x*exp(2*exp(2*x)) + x)/(x + 4))
Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.68 \[ \int \frac {e^{\frac {x-e^{2 e^{2 x}} x+16 x^2-x^3+4 x^2 \log (4)}{4+x}} \left (4+128 x+4 x^2-2 x^3+e^{2 e^{2 x}} \left (-4+e^{2 x} \left (-16 x-4 x^2\right )\right )+\left (32 x+4 x^2\right ) \log (4)\right )}{16+8 x+x^2} \, dx=\frac {1}{4294967296} \, e^{\left (-x^{2} + 8 \, x \log \left (2\right ) + 20 \, x + \frac {4 \, e^{\left (2 \, e^{\left (2 \, x\right )}\right )}}{x + 4} + \frac {128 \, \log \left (2\right )}{x + 4} + \frac {316}{x + 4} - e^{\left (2 \, e^{\left (2 \, x\right )}\right )} - 79\right )} \] Input:
integrate((((-4*x^2-16*x)*exp(x)^2-4)*exp(exp(x)^2)^2+2*(4*x^2+32*x)*log(2 )-2*x^3+4*x^2+128*x+4)*exp((-x*exp(exp(x)^2)^2+8*x^2*log(2)-x^3+16*x^2+x)/ (4+x))/(x^2+8*x+16),x, algorithm="maxima")
Output:
1/4294967296*e^(-x^2 + 8*x*log(2) + 20*x + 4*e^(2*e^(2*x))/(x + 4) + 128*l og(2)/(x + 4) + 316/(x + 4) - e^(2*e^(2*x)) - 79)
Time = 0.26 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.65 \[ \int \frac {e^{\frac {x-e^{2 e^{2 x}} x+16 x^2-x^3+4 x^2 \log (4)}{4+x}} \left (4+128 x+4 x^2-2 x^3+e^{2 e^{2 x}} \left (-4+e^{2 x} \left (-16 x-4 x^2\right )\right )+\left (32 x+4 x^2\right ) \log (4)\right )}{16+8 x+x^2} \, dx=e^{\left (-\frac {x^{3}}{x + 4} + \frac {8 \, x^{2} \log \left (2\right )}{x + 4} + \frac {16 \, x^{2}}{x + 4} - \frac {x e^{\left (2 \, e^{\left (2 \, x\right )}\right )}}{x + 4} + \frac {x}{x + 4}\right )} \] Input:
integrate((((-4*x^2-16*x)*exp(x)^2-4)*exp(exp(x)^2)^2+2*(4*x^2+32*x)*log(2 )-2*x^3+4*x^2+128*x+4)*exp((-x*exp(exp(x)^2)^2+8*x^2*log(2)-x^3+16*x^2+x)/ (4+x))/(x^2+8*x+16),x, algorithm="giac")
Output:
e^(-x^3/(x + 4) + 8*x^2*log(2)/(x + 4) + 16*x^2/(x + 4) - x*e^(2*e^(2*x))/ (x + 4) + x/(x + 4))
Time = 3.29 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.74 \[ \int \frac {e^{\frac {x-e^{2 e^{2 x}} x+16 x^2-x^3+4 x^2 \log (4)}{4+x}} \left (4+128 x+4 x^2-2 x^3+e^{2 e^{2 x}} \left (-4+e^{2 x} \left (-16 x-4 x^2\right )\right )+\left (32 x+4 x^2\right ) \log (4)\right )}{16+8 x+x^2} \, dx=2^{\frac {8\,x^2}{x+4}}\,{\mathrm {e}}^{\frac {x}{x+4}}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,x}}}{x+4}}\,{\mathrm {e}}^{-\frac {x^3}{x+4}}\,{\mathrm {e}}^{\frac {16\,x^2}{x+4}} \] Input:
int((exp((x + 8*x^2*log(2) + 16*x^2 - x^3 - x*exp(2*exp(2*x)))/(x + 4))*(1 28*x + 2*log(2)*(32*x + 4*x^2) - exp(2*exp(2*x))*(exp(2*x)*(16*x + 4*x^2) + 4) + 4*x^2 - 2*x^3 + 4))/(8*x + x^2 + 16),x)
Output:
2^((8*x^2)/(x + 4))*exp(x/(x + 4))*exp(-(x*exp(2*exp(2*x)))/(x + 4))*exp(- x^3/(x + 4))*exp((16*x^2)/(x + 4))
Time = 0.52 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.76 \[ \int \frac {e^{\frac {x-e^{2 e^{2 x}} x+16 x^2-x^3+4 x^2 \log (4)}{4+x}} \left (4+128 x+4 x^2-2 x^3+e^{2 e^{2 x}} \left (-4+e^{2 x} \left (-16 x-4 x^2\right )\right )+\left (32 x+4 x^2\right ) \log (4)\right )}{16+8 x+x^2} \, dx=\frac {e^{\frac {4 e^{2 e^{2 x}}+128 \,\mathrm {log}\left (2\right )+20 x^{2}+80 x +316}{x +4}} 2^{8 x}}{4294967296 e^{e^{2 e^{2 x}}+x^{2}} e^{79}} \] Input:
int((((-4*x^2-16*x)*exp(x)^2-4)*exp(exp(x)^2)^2+2*(4*x^2+32*x)*log(2)-2*x^ 3+4*x^2+128*x+4)*exp((-x*exp(exp(x)^2)^2+8*x^2*log(2)-x^3+16*x^2+x)/(4+x)) /(x^2+8*x+16),x)
Output:
(e**((4*e**(2*e**(2*x)) + 128*log(2) + 20*x**2 + 80*x + 316)/(x + 4))*2**( 8*x))/(4294967296*e**(e**(2*e**(2*x)) + x**2)*e**79)