Integrand size = 80, antiderivative size = 28 \[ \int \frac {e^{e^x-2 x+x^2+2 x^3+x^4+\left (-8 x-8 x^2\right ) \log (x)+16 \log ^2(x)} \left (-20 x+2 e^x x-12 x^2+12 x^3+8 x^4+\left (64-16 x-32 x^2\right ) \log (x)\right )}{x} \, dx=3+2 e^{e^x-2 x+\left (-x-x^2+4 \log (x)\right )^2} \] Output:
3+2*exp(exp(x)-2*x+(4*ln(x)-x^2-x)^2)
Time = 5.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {e^{e^x-2 x+x^2+2 x^3+x^4+\left (-8 x-8 x^2\right ) \log (x)+16 \log ^2(x)} \left (-20 x+2 e^x x-12 x^2+12 x^3+8 x^4+\left (64-16 x-32 x^2\right ) \log (x)\right )}{x} \, dx=2 e^{e^x-2 x+x^2+2 x^3+x^4+16 \log ^2(x)} x^{-8 x (1+x)} \] Input:
Integrate[(E^(E^x - 2*x + x^2 + 2*x^3 + x^4 + (-8*x - 8*x^2)*Log[x] + 16*L og[x]^2)*(-20*x + 2*E^x*x - 12*x^2 + 12*x^3 + 8*x^4 + (64 - 16*x - 32*x^2) *Log[x]))/x,x]
Output:
(2*E^(E^x - 2*x + x^2 + 2*x^3 + x^4 + 16*Log[x]^2))/x^(8*x*(1 + 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.
\(\displaystyle \int \frac {\left (8 x^4+12 x^3-12 x^2+\left (-32 x^2-16 x+64\right ) \log (x)+2 e^x x-20 x\right ) \exp \left (x^4+2 x^3+x^2+\left (-8 x^2-8 x\right ) \log (x)-2 x+e^x+16 \log ^2(x)\right )}{x} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {4 \left (2 x^4+3 x^3-3 x^2-8 x^2 \log (x)-5 x-4 x \log (x)+16 \log (x)\right ) \exp \left (x^4+2 x^3+x^2+\left (-8 x^2-8 x\right ) \log (x)-2 x+e^x+16 \log ^2(x)\right )}{x}+2 \exp \left (x^4+2 x^3+x^2+\left (-8 x^2-8 x\right ) \log (x)-x+e^x+16 \log ^2(x)\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -20 \int \exp \left (x^4+2 x^3+x^2-2 x+e^x+16 \log ^2(x)+\left (-8 x^2-8 x\right ) \log (x)\right )dx+2 \int \exp \left (x^4+2 x^3+x^2-x+e^x+16 \log ^2(x)+\left (-8 x^2-8 x\right ) \log (x)\right )dx-12 \int \exp \left (x^4+2 x^3+x^2-2 x+e^x+16 \log ^2(x)+\left (-8 x^2-8 x\right ) \log (x)\right ) xdx+12 \int \exp \left (x^4+2 x^3+x^2-2 x+e^x+16 \log ^2(x)+\left (-8 x^2-8 x\right ) \log (x)\right ) x^2dx+8 \int \exp \left (x^4+2 x^3+x^2-2 x+e^x+16 \log ^2(x)+\left (-8 x^2-8 x\right ) \log (x)\right ) x^3dx-16 \int \exp \left (x^4+2 x^3+x^2-2 x+e^x+16 \log ^2(x)+\left (-8 x^2-8 x\right ) \log (x)\right ) \log (x)dx+64 \int \frac {\exp \left (x^4+2 x^3+x^2-2 x+e^x+16 \log ^2(x)+\left (-8 x^2-8 x\right ) \log (x)\right ) \log (x)}{x}dx-32 \int \exp \left (x^4+2 x^3+x^2-2 x+e^x+16 \log ^2(x)+\left (-8 x^2-8 x\right ) \log (x)\right ) x \log (x)dx\) |
Input:
Int[(E^(E^x - 2*x + x^2 + 2*x^3 + x^4 + (-8*x - 8*x^2)*Log[x] + 16*Log[x]^ 2)*(-20*x + 2*E^x*x - 12*x^2 + 12*x^3 + 8*x^4 + (64 - 16*x - 32*x^2)*Log[x ]))/x,x]
Output:
$Aborted
Time = 0.42 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25
method | result | size |
risch | \(2 x^{-8 \left (1+x \right ) x} {\mathrm e}^{16 \ln \left (x \right )^{2}+{\mathrm e}^{x}+x^{4}+2 x^{3}+x^{2}-2 x}\) | \(35\) |
parallelrisch | \(2 \,{\mathrm e}^{16 \ln \left (x \right )^{2}+\left (-8 x^{2}-8 x \right ) \ln \left (x \right )+{\mathrm e}^{x}+x^{4}+2 x^{3}+x^{2}-2 x}\) | \(39\) |
Input:
int(((-32*x^2-16*x+64)*ln(x)+2*exp(x)*x+8*x^4+12*x^3-12*x^2-20*x)*exp(16*l n(x)^2+(-8*x^2-8*x)*ln(x)+exp(x)+x^4+2*x^3+x^2-2*x)/x,x,method=_RETURNVERB OSE)
Output:
2*x^(-8*(1+x)*x)*exp(16*ln(x)^2+exp(x)+x^4+2*x^3+x^2-2*x)
Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {e^{e^x-2 x+x^2+2 x^3+x^4+\left (-8 x-8 x^2\right ) \log (x)+16 \log ^2(x)} \left (-20 x+2 e^x x-12 x^2+12 x^3+8 x^4+\left (64-16 x-32 x^2\right ) \log (x)\right )}{x} \, dx=2 \, e^{\left (x^{4} + 2 \, x^{3} + x^{2} - 8 \, {\left (x^{2} + x\right )} \log \left (x\right ) + 16 \, \log \left (x\right )^{2} - 2 \, x + e^{x}\right )} \] Input:
integrate(((-32*x^2-16*x+64)*log(x)+2*exp(x)*x+8*x^4+12*x^3-12*x^2-20*x)*e xp(16*log(x)^2+(-8*x^2-8*x)*log(x)+exp(x)+x^4+2*x^3+x^2-2*x)/x,x, algorith m="fricas")
Output:
2*e^(x^4 + 2*x^3 + x^2 - 8*(x^2 + x)*log(x) + 16*log(x)^2 - 2*x + e^x)
Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \frac {e^{e^x-2 x+x^2+2 x^3+x^4+\left (-8 x-8 x^2\right ) \log (x)+16 \log ^2(x)} \left (-20 x+2 e^x x-12 x^2+12 x^3+8 x^4+\left (64-16 x-32 x^2\right ) \log (x)\right )}{x} \, dx=2 e^{x^{4} + 2 x^{3} + x^{2} - 2 x + \left (- 8 x^{2} - 8 x\right ) \log {\left (x \right )} + e^{x} + 16 \log {\left (x \right )}^{2}} \] Input:
integrate(((-32*x**2-16*x+64)*ln(x)+2*exp(x)*x+8*x**4+12*x**3-12*x**2-20*x )*exp(16*ln(x)**2+(-8*x**2-8*x)*ln(x)+exp(x)+x**4+2*x**3+x**2-2*x)/x,x)
Output:
2*exp(x**4 + 2*x**3 + x**2 - 2*x + (-8*x**2 - 8*x)*log(x) + exp(x) + 16*lo g(x)**2)
Time = 0.18 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {e^{e^x-2 x+x^2+2 x^3+x^4+\left (-8 x-8 x^2\right ) \log (x)+16 \log ^2(x)} \left (-20 x+2 e^x x-12 x^2+12 x^3+8 x^4+\left (64-16 x-32 x^2\right ) \log (x)\right )}{x} \, dx=2 \, e^{\left (x^{4} + 2 \, x^{3} - 8 \, x^{2} \log \left (x\right ) + x^{2} - 8 \, x \log \left (x\right ) + 16 \, \log \left (x\right )^{2} - 2 \, x + e^{x}\right )} \] Input:
integrate(((-32*x^2-16*x+64)*log(x)+2*exp(x)*x+8*x^4+12*x^3-12*x^2-20*x)*e xp(16*log(x)^2+(-8*x^2-8*x)*log(x)+exp(x)+x^4+2*x^3+x^2-2*x)/x,x, algorith m="maxima")
Output:
2*e^(x^4 + 2*x^3 - 8*x^2*log(x) + x^2 - 8*x*log(x) + 16*log(x)^2 - 2*x + e ^x)
Time = 0.13 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {e^{e^x-2 x+x^2+2 x^3+x^4+\left (-8 x-8 x^2\right ) \log (x)+16 \log ^2(x)} \left (-20 x+2 e^x x-12 x^2+12 x^3+8 x^4+\left (64-16 x-32 x^2\right ) \log (x)\right )}{x} \, dx=2 \, e^{\left (x^{4} + 2 \, x^{3} - 8 \, x^{2} \log \left (x\right ) + x^{2} - 8 \, x \log \left (x\right ) + 16 \, \log \left (x\right )^{2} - 2 \, x + e^{x}\right )} \] Input:
integrate(((-32*x^2-16*x+64)*log(x)+2*exp(x)*x+8*x^4+12*x^3-12*x^2-20*x)*e xp(16*log(x)^2+(-8*x^2-8*x)*log(x)+exp(x)+x^4+2*x^3+x^2-2*x)/x,x, algorith m="giac")
Output:
2*e^(x^4 + 2*x^3 - 8*x^2*log(x) + x^2 - 8*x*log(x) + 16*log(x)^2 - 2*x + e ^x)
Time = 3.07 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.54 \[ \int \frac {e^{e^x-2 x+x^2+2 x^3+x^4+\left (-8 x-8 x^2\right ) \log (x)+16 \log ^2(x)} \left (-20 x+2 e^x x-12 x^2+12 x^3+8 x^4+\left (64-16 x-32 x^2\right ) \log (x)\right )}{x} \, dx=\frac {2\,{\mathrm {e}}^{16\,{\ln \left (x\right )}^2}\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{x^4}\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{2\,x^3}}{x^{8\,x^2+8\,x}} \] Input:
int(-(exp(exp(x) - 2*x + 16*log(x)^2 - log(x)*(8*x + 8*x^2) + x^2 + 2*x^3 + x^4)*(20*x + log(x)*(16*x + 32*x^2 - 64) - 2*x*exp(x) + 12*x^2 - 12*x^3 - 8*x^4))/x,x)
Output:
(2*exp(16*log(x)^2)*exp(-2*x)*exp(x^2)*exp(x^4)*exp(exp(x))*exp(2*x^3))/x^ (8*x + 8*x^2)
\[ \int \frac {e^{e^x-2 x+x^2+2 x^3+x^4+\left (-8 x-8 x^2\right ) \log (x)+16 \log ^2(x)} \left (-20 x+2 e^x x-12 x^2+12 x^3+8 x^4+\left (64-16 x-32 x^2\right ) \log (x)\right )}{x} \, dx=\int \frac {\left (\left (-32 x^{2}-16 x +64\right ) \mathrm {log}\left (x \right )+2 \,{\mathrm e}^{x} x +8 x^{4}+12 x^{3}-12 x^{2}-20 x \right ) {\mathrm e}^{16 \mathrm {log}\left (x \right )^{2}+\left (-8 x^{2}-8 x \right ) \mathrm {log}\left (x \right )+{\mathrm e}^{x}+x^{4}+2 x^{3}+x^{2}-2 x}}{x}d x \] Input:
int(((-32*x^2-16*x+64)*log(x)+2*exp(x)*x+8*x^4+12*x^3-12*x^2-20*x)*exp(16* log(x)^2+(-8*x^2-8*x)*log(x)+exp(x)+x^4+2*x^3+x^2-2*x)/x,x)
Output:
int(((-32*x^2-16*x+64)*log(x)+2*exp(x)*x+8*x^4+12*x^3-12*x^2-20*x)*exp(16* log(x)^2+(-8*x^2-8*x)*log(x)+exp(x)+x^4+2*x^3+x^2-2*x)/x,x)