Integrand size = 109, antiderivative size = 28 \[ \int \frac {1}{2} e^{e^{\frac {2-x}{2}}-e^{2 x^3+2 e^x x^3+2 x^3 \log (x)}} \left (-e^{\frac {2-x}{2}}+e^{2 x^3+2 e^x x^3+2 x^3 \log (x)} \left (-16 x^2+e^x \left (-12 x^2-4 x^3\right )-12 x^2 \log (x)\right )\right ) \, dx=e^{e^{1-\frac {x}{2}}-e^{2 x^3 \left (1+e^x+\log (x)\right )}} \] Output:
exp(exp(1-1/2*x)-exp(2*x^3*(exp(x)+ln(x)+1)))
Time = 0.22 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {1}{2} e^{e^{\frac {2-x}{2}}-e^{2 x^3+2 e^x x^3+2 x^3 \log (x)}} \left (-e^{\frac {2-x}{2}}+e^{2 x^3+2 e^x x^3+2 x^3 \log (x)} \left (-16 x^2+e^x \left (-12 x^2-4 x^3\right )-12 x^2 \log (x)\right )\right ) \, dx=e^{e^{1-\frac {x}{2}}-e^{2 \left (1+e^x\right ) x^3} x^{2 x^3}} \] Input:
Integrate[(E^(E^((2 - x)/2) - E^(2*x^3 + 2*E^x*x^3 + 2*x^3*Log[x]))*(-E^(( 2 - x)/2) + E^(2*x^3 + 2*E^x*x^3 + 2*x^3*Log[x])*(-16*x^2 + E^x*(-12*x^2 - 4*x^3) - 12*x^2*Log[x])))/2,x]
Output:
E^(E^(1 - x/2) - E^(2*(1 + E^x)*x^3)*x^(2*x^3))
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.
| \(\displaystyle \int \frac {1}{2} e^{e^{\frac {2-x}{2}}-e^{2 e^x x^3+2 x^3+2 x^3 \log (x)}} \left (e^{2 e^x x^3+2 x^3+2 x^3 \log (x)} \left (-16 x^2-12 x^2 \log (x)+e^x \left (-4 x^3-12 x^2\right )\right )-e^{\frac {2-x}{2}}\right ) \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int -e^{e^{\frac {2-x}{2}}-e^{2 e^x x^3+2 x^3} x^{2 x^3}} \left (4 e^{2 e^x x^3+2 x^3} \left (3 \log (x) x^2+4 x^2+e^x \left (x^3+3 x^2\right )\right ) x^{2 x^3}+e^{\frac {2-x}{2}}\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{2} \int e^{e^{\frac {2-x}{2}}-e^{2 e^x x^3+2 x^3} x^{2 x^3}} \left (4 e^{2 e^x x^3+2 x^3} \left (3 \log (x) x^2+4 x^2+e^x \left (x^3+3 x^2\right )\right ) x^{2 x^3}+e^{\frac {2-x}{2}}\right )dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle -\frac {1}{2} \int \exp \left (-e^{-x/2} \left (e^{2 \left (1+e^x\right ) x^3+\frac {x}{2}} x^{2 x^3}-e\right )\right ) \left (4 e^{2 e^x x^3+2 x^3} \left (3 \log (x) x^2+4 x^2+e^x \left (x^3+3 x^2\right )\right ) x^{2 x^3}+e^{\frac {2-x}{2}}\right )dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {1}{2} \int \left (4 \exp \left (2 \left (1+e^x\right ) x^3-e^{-x/2} \left (e^{2 \left (1+e^x\right ) x^3+\frac {x}{2}} x^{2 x^3}-e\right )\right ) \left (e^x x+3 e^x+3 \log (x)+4\right ) x^{2 x^3+2}+\exp \left (-\frac {x}{2}-e^{-x/2} \left (e^{2 \left (1+e^x\right ) x^3+\frac {x}{2}} x^{2 x^3}-e\right )+1\right )\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (-\int \exp \left (-\frac {x}{2}-e^{-x/2} \left (e^{2 \left (1+e^x\right ) x^3+\frac {x}{2}} x^{2 x^3}-e\right )+1\right )dx-16 \int \exp \left (2 \left (1+e^x\right ) x^3-e^{-x/2} \left (e^{2 \left (1+e^x\right ) x^3+\frac {x}{2}} x^{2 x^3}-e\right )\right ) x^{2 x^3+2}dx-12 \int \exp \left (2 \left (1+e^x\right ) x^3+x-e^{-x/2} \left (e^{2 \left (1+e^x\right ) x^3+\frac {x}{2}} x^{2 x^3}-e\right )\right ) x^{2 x^3+2}dx-4 \int \exp \left (2 \left (1+e^x\right ) x^3+x-e^{-x/2} \left (e^{2 \left (1+e^x\right ) x^3+\frac {x}{2}} x^{2 x^3}-e\right )\right ) x^{2 x^3+3}dx+12 \int \frac {\int \exp \left (-e^{2 \left (1+e^x\right ) x^3} x^{2 x^3}+2 \left (1+e^x\right ) x^3+e^{1-\frac {x}{2}}\right ) x^{2 x^3+2}dx}{x}dx-12 \log (x) \int \exp \left (2 \left (1+e^x\right ) x^3-e^{-x/2} \left (e^{2 \left (1+e^x\right ) x^3+\frac {x}{2}} x^{2 x^3}-e\right )\right ) x^{2 x^3+2}dx\right )\) |
Input:
Int[(E^(E^((2 - x)/2) - E^(2*x^3 + 2*E^x*x^3 + 2*x^3*Log[x]))*(-E^((2 - x) /2) + E^(2*x^3 + 2*E^x*x^3 + 2*x^3*Log[x])*(-16*x^2 + E^x*(-12*x^2 - 4*x^3 ) - 12*x^2*Log[x])))/2,x]
Output:
$Aborted
Time = 3.31 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82
| method | result | size |
| parallelrisch | \({\mathrm e}^{{\mathrm e}^{1-\frac {x}{2}}-{\mathrm e}^{2 x^{3} \left ({\mathrm e}^{x}+\ln \left (x \right )+1\right )}}\) | \(23\) |
| risch | \({\mathrm e}^{-x^{2 x^{3}} {\mathrm e}^{2 x^{3} \left ({\mathrm e}^{x}+1\right )}+{\mathrm e}^{1-\frac {x}{2}}}\) | \(28\) |
Input:
int(1/2*((-12*x^2*ln(x)+(-4*x^3-12*x^2)*exp(x)-16*x^2)*exp(2*x^3*ln(x)+2*e xp(x)*x^3+2*x^3)-exp(1-1/2*x))*exp(-exp(2*x^3*ln(x)+2*exp(x)*x^3+2*x^3)+ex p(1-1/2*x)),x,method=_RETURNVERBOSE)
Output:
exp(exp(1-1/2*x)-exp(2*x^3*(exp(x)+ln(x)+1)))
Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (22) = 44\).
Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {1}{2} e^{e^{\frac {2-x}{2}}-e^{2 x^3+2 e^x x^3+2 x^3 \log (x)}} \left (-e^{\frac {2-x}{2}}+e^{2 x^3+2 e^x x^3+2 x^3 \log (x)} \left (-16 x^2+e^x \left (-12 x^2-4 x^3\right )-12 x^2 \log (x)\right )\right ) \, dx=e^{\left (-e^{\left (2 \, {\left (x^{3} e^{\left (-x + 2\right )} \log \left (x\right ) + x^{3} e^{2} + x^{3} e^{\left (-x + 2\right )}\right )} e^{\left (x - 2\right )}\right )} + e^{\left (-\frac {1}{2} \, x + 1\right )}\right )} \] Input:
integrate(1/2*((-12*x^2*log(x)+(-4*x^3-12*x^2)*exp(x)-16*x^2)*exp(2*x^3*lo g(x)+2*exp(x)*x^3+2*x^3)-exp(1-1/2*x))*exp(-exp(2*x^3*log(x)+2*exp(x)*x^3+ 2*x^3)+exp(1-1/2*x)),x, algorithm="fricas")
Output:
e^(-e^(2*(x^3*e^(-x + 2)*log(x) + x^3*e^2 + x^3*e^(-x + 2))*e^(x - 2)) + e ^(-1/2*x + 1))
Time = 6.37 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {1}{2} e^{e^{\frac {2-x}{2}}-e^{2 x^3+2 e^x x^3+2 x^3 \log (x)}} \left (-e^{\frac {2-x}{2}}+e^{2 x^3+2 e^x x^3+2 x^3 \log (x)} \left (-16 x^2+e^x \left (-12 x^2-4 x^3\right )-12 x^2 \log (x)\right )\right ) \, dx=e^{- e^{2 x^{3} e^{x} + 2 x^{3} \log {\left (x \right )} + 2 x^{3}} + \frac {e}{\sqrt {e^{x}}}} \] Input:
integrate(1/2*((-12*x**2*ln(x)+(-4*x**3-12*x**2)*exp(x)-16*x**2)*exp(2*x** 3*ln(x)+2*exp(x)*x**3+2*x**3)-exp(1-1/2*x))*exp(-exp(2*x**3*ln(x)+2*exp(x) *x**3+2*x**3)+exp(1-1/2*x)),x)
Output:
exp(-exp(2*x**3*exp(x) + 2*x**3*log(x) + 2*x**3) + E/sqrt(exp(x)))
\[ \int \frac {1}{2} e^{e^{\frac {2-x}{2}}-e^{2 x^3+2 e^x x^3+2 x^3 \log (x)}} \left (-e^{\frac {2-x}{2}}+e^{2 x^3+2 e^x x^3+2 x^3 \log (x)} \left (-16 x^2+e^x \left (-12 x^2-4 x^3\right )-12 x^2 \log (x)\right )\right ) \, dx=\int { -\frac {1}{2} \, {\left (4 \, {\left (3 \, x^{2} \log \left (x\right ) + 4 \, x^{2} + {\left (x^{3} + 3 \, x^{2}\right )} e^{x}\right )} e^{\left (2 \, x^{3} e^{x} + 2 \, x^{3} \log \left (x\right ) + 2 \, x^{3}\right )} + e^{\left (-\frac {1}{2} \, x + 1\right )}\right )} e^{\left (-e^{\left (2 \, x^{3} e^{x} + 2 \, x^{3} \log \left (x\right ) + 2 \, x^{3}\right )} + e^{\left (-\frac {1}{2} \, x + 1\right )}\right )} \,d x } \] Input:
integrate(1/2*((-12*x^2*log(x)+(-4*x^3-12*x^2)*exp(x)-16*x^2)*exp(2*x^3*lo g(x)+2*exp(x)*x^3+2*x^3)-exp(1-1/2*x))*exp(-exp(2*x^3*log(x)+2*exp(x)*x^3+ 2*x^3)+exp(1-1/2*x)),x, algorithm="maxima")
Output:
-1/2*integrate((4*(3*x^2*log(x) + 4*x^2 + (x^3 + 3*x^2)*e^x)*e^(2*x^3*e^x + 2*x^3*log(x) + 2*x^3) + e^(-1/2*x + 1))*e^(-e^(2*x^3*e^x + 2*x^3*log(x) + 2*x^3) + e^(-1/2*x + 1)), x)
\[ \int \frac {1}{2} e^{e^{\frac {2-x}{2}}-e^{2 x^3+2 e^x x^3+2 x^3 \log (x)}} \left (-e^{\frac {2-x}{2}}+e^{2 x^3+2 e^x x^3+2 x^3 \log (x)} \left (-16 x^2+e^x \left (-12 x^2-4 x^3\right )-12 x^2 \log (x)\right )\right ) \, dx=\int { -\frac {1}{2} \, {\left (4 \, {\left (3 \, x^{2} \log \left (x\right ) + 4 \, x^{2} + {\left (x^{3} + 3 \, x^{2}\right )} e^{x}\right )} e^{\left (2 \, x^{3} e^{x} + 2 \, x^{3} \log \left (x\right ) + 2 \, x^{3}\right )} + e^{\left (-\frac {1}{2} \, x + 1\right )}\right )} e^{\left (-e^{\left (2 \, x^{3} e^{x} + 2 \, x^{3} \log \left (x\right ) + 2 \, x^{3}\right )} + e^{\left (-\frac {1}{2} \, x + 1\right )}\right )} \,d x } \] Input:
integrate(1/2*((-12*x^2*log(x)+(-4*x^3-12*x^2)*exp(x)-16*x^2)*exp(2*x^3*lo g(x)+2*exp(x)*x^3+2*x^3)-exp(1-1/2*x))*exp(-exp(2*x^3*log(x)+2*exp(x)*x^3+ 2*x^3)+exp(1-1/2*x)),x, algorithm="giac")
Output:
integrate(-1/2*(4*(3*x^2*log(x) + 4*x^2 + (x^3 + 3*x^2)*e^x)*e^(2*x^3*e^x + 2*x^3*log(x) + 2*x^3) + e^(-1/2*x + 1))*e^(-e^(2*x^3*e^x + 2*x^3*log(x) + 2*x^3) + e^(-1/2*x + 1)), x)
Time = 0.50 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {1}{2} e^{e^{\frac {2-x}{2}}-e^{2 x^3+2 e^x x^3+2 x^3 \log (x)}} \left (-e^{\frac {2-x}{2}}+e^{2 x^3+2 e^x x^3+2 x^3 \log (x)} \left (-16 x^2+e^x \left (-12 x^2-4 x^3\right )-12 x^2 \log (x)\right )\right ) \, dx={\mathrm {e}}^{{\mathrm {e}}^{-\frac {x}{2}}\,\mathrm {e}}\,{\mathrm {e}}^{-x^{2\,x^3}\,{\mathrm {e}}^{2\,x^3\,{\mathrm {e}}^x}\,{\mathrm {e}}^{2\,x^3}} \] Input:
int(-(exp(exp(1 - x/2) - exp(2*x^3*exp(x) + 2*x^3*log(x) + 2*x^3))*(exp(1 - x/2) + exp(2*x^3*exp(x) + 2*x^3*log(x) + 2*x^3)*(exp(x)*(12*x^2 + 4*x^3) + 12*x^2*log(x) + 16*x^2)))/2,x)
Output:
exp(exp(-x/2)*exp(1))*exp(-x^(2*x^3)*exp(2*x^3*exp(x))*exp(2*x^3))
\[ \int \frac {1}{2} e^{e^{\frac {2-x}{2}}-e^{2 x^3+2 e^x x^3+2 x^3 \log (x)}} \left (-e^{\frac {2-x}{2}}+e^{2 x^3+2 e^x x^3+2 x^3 \log (x)} \left (-16 x^2+e^x \left (-12 x^2-4 x^3\right )-12 x^2 \log (x)\right )\right ) \, dx=\int \frac {\left (\left (-12 \,\mathrm {log}\left (x \right ) x^{2}+\left (-4 x^{3}-12 x^{2}\right ) {\mathrm e}^{x}-16 x^{2}\right ) {\mathrm e}^{2 \,\mathrm {log}\left (x \right ) x^{3}+2 \,{\mathrm e}^{x} x^{3}+2 x^{3}}-{\mathrm e}^{1-\frac {x}{2}}\right ) {\mathrm e}^{-{\mathrm e}^{2 \,\mathrm {log}\left (x \right ) x^{3}+2 \,{\mathrm e}^{x} x^{3}+2 x^{3}}+{\mathrm e}^{1-\frac {x}{2}}}}{2}d x \] Input:
int(1/2*((-12*x^2*log(x)+(-4*x^3-12*x^2)*exp(x)-16*x^2)*exp(2*x^3*log(x)+2 *exp(x)*x^3+2*x^3)-exp(1-1/2*x))*exp(-exp(2*x^3*log(x)+2*exp(x)*x^3+2*x^3) +exp(1-1/2*x)),x)
Output:
int(1/2*((-12*x^2*log(x)+(-4*x^3-12*x^2)*exp(x)-16*x^2)*exp(2*x^3*log(x)+2 *exp(x)*x^3+2*x^3)-exp(1-1/2*x))*exp(-exp(2*x^3*log(x)+2*exp(x)*x^3+2*x^3) +exp(1-1/2*x)),x)