Integrand size = 130, antiderivative size = 25 \[ \int \frac {-2-2 e^{2 e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}+6 x^2+e^{e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}} \left (-4+6 x^2+e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )} \left (-x^2-2 x^3\right )\right )}{2+4 e^{e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}+2 e^{2 e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}} \, dx=-x+\frac {x^3}{1+e^{e^{1+x} \sqrt {x}}} \] Output:
x^3/(exp(exp(1/2*ln(exp(2)*x)+x))+1)-x
Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {-2-2 e^{2 e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}+6 x^2+e^{e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}} \left (-4+6 x^2+e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )} \left (-x^2-2 x^3\right )\right )}{2+4 e^{e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}+2 e^{2 e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}} \, dx=\frac {1}{2} \left (-2 x+\frac {2 x^3}{1+e^{e^{1+x} \sqrt {x}}}\right ) \] Input:
Integrate[(-2 - 2*E^(2*E^((2*x + Log[E^2*x])/2)) + 6*x^2 + E^E^((2*x + Log [E^2*x])/2)*(-4 + 6*x^2 + E^((2*x + Log[E^2*x])/2)*(-x^2 - 2*x^3)))/(2 + 4 *E^E^((2*x + Log[E^2*x])/2) + 2*E^(2*E^((2*x + Log[E^2*x])/2))),x]
Output:
(-2*x + (2*x^3)/(1 + E^(E^(1 + x)*Sqrt[x])))/2
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.
| \(\displaystyle \int \frac {6 x^2+e^{e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}} \left (6 x^2+\left (-2 x^3-x^2\right ) e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}-4\right )-2 e^{2 e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}-2}{4 e^{e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}+2 e^{2 e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}+2} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {6 x^2+e^{e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}} \left (6 x^2+\left (-2 x^3-x^2\right ) e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}-4\right )-2 e^{2 e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}-2}{2 \left (e^{e^{x+1} \sqrt {x}}+1\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int -\frac {-6 x^2+2 e^{2 e^{x+1} \sqrt {x}}+e^{e^{x+1} \sqrt {x}} \left (-6 x^2+e^{x+1} \left (2 x^3+x^2\right ) \sqrt {x}+4\right )+2}{\left (1+e^{e^{x+1} \sqrt {x}}\right )^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{2} \int \frac {-6 x^2+2 e^{2 e^{x+1} \sqrt {x}}+e^{e^{x+1} \sqrt {x}} \left (-6 x^2+e^{x+1} \left (2 x^3+x^2\right ) \sqrt {x}+4\right )+2}{\left (1+e^{e^{x+1} \sqrt {x}}\right )^2}dx\) |
| \(\Big \downarrow \) 7267 |
| \(\displaystyle -\int \frac {\sqrt {x} \left (-6 x^2+2 e^{2 e^{x+1} \sqrt {x}}+e^{e^{x+1} \sqrt {x}} \left (e^{x+1} (2 x+1) x^{5/2}-6 x^2+4\right )+2\right )}{\left (1+e^{e^{x+1} \sqrt {x}}\right )^2}d\sqrt {x}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\int \left (\frac {e^{x+e^{x+1} \sqrt {x}+1} (2 x+1) x^3}{\left (1+e^{e^{x+1} \sqrt {x}}\right )^2}-\frac {6 e^{e^{x+1} \sqrt {x}} x^{5/2}}{\left (1+e^{e^{x+1} \sqrt {x}}\right )^2}-\frac {6 x^{5/2}}{\left (1+e^{e^{x+1} \sqrt {x}}\right )^2}+\frac {4 e^{e^{x+1} \sqrt {x}} \sqrt {x}}{\left (1+e^{e^{x+1} \sqrt {x}}\right )^2}+\frac {2 e^{2 e^{x+1} \sqrt {x}} \sqrt {x}}{\left (1+e^{e^{x+1} \sqrt {x}}\right )^2}+\frac {2 \sqrt {x}}{\left (1+e^{e^{x+1} \sqrt {x}}\right )^2}\right )d\sqrt {x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 6 \int \frac {x^{5/2}}{\left (1+e^{e^{x+1} \sqrt {x}}\right )^2}d\sqrt {x}+6 \int \frac {e^{e^{x+1} \sqrt {x}} x^{5/2}}{\left (1+e^{e^{x+1} \sqrt {x}}\right )^2}d\sqrt {x}-2 \int \frac {e^{x+e^{x+1} \sqrt {x}+1} x^4}{\left (1+e^{e^{x+1} \sqrt {x}}\right )^2}d\sqrt {x}-\int \frac {e^{x+e^{x+1} \sqrt {x}+1} x^3}{\left (1+e^{e^{x+1} \sqrt {x}}\right )^2}d\sqrt {x}-2 \int \frac {\sqrt {x}}{\left (1+e^{e^{x+1} \sqrt {x}}\right )^2}d\sqrt {x}-4 \int \frac {e^{e^{x+1} \sqrt {x}} \sqrt {x}}{\left (1+e^{e^{x+1} \sqrt {x}}\right )^2}d\sqrt {x}-2 \int \frac {e^{2 e^{x+1} \sqrt {x}} \sqrt {x}}{\left (1+e^{e^{x+1} \sqrt {x}}\right )^2}d\sqrt {x}\) |
Input:
Int[(-2 - 2*E^(2*E^((2*x + Log[E^2*x])/2)) + 6*x^2 + E^E^((2*x + Log[E^2*x ])/2)*(-4 + 6*x^2 + E^((2*x + Log[E^2*x])/2)*(-x^2 - 2*x^3)))/(2 + 4*E^E^( (2*x + Log[E^2*x])/2) + 2*E^(2*E^((2*x + Log[E^2*x])/2))),x]
Output:
$Aborted
Time = 0.57 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92
| method | result | size |
| risch | \(\frac {x^{3}}{{\mathrm e}^{\sqrt {{\mathrm e}^{2} x}\, {\mathrm e}^{x}}+1}-x\) | \(23\) |
| parallelrisch | \(-\frac {-2 x^{3}+2 \,{\mathrm e}^{{\mathrm e}^{\frac {\ln \left ({\mathrm e}^{2} x \right )}{2}+x}} x +2 x}{2 \left ({\mathrm e}^{{\mathrm e}^{\frac {\ln \left ({\mathrm e}^{2} x \right )}{2}+x}}+1\right )}\) | \(41\) |
Input:
int((-2*exp(exp(1/2*ln(exp(2)*x)+x))^2+((-2*x^3-x^2)*exp(1/2*ln(exp(2)*x)+ x)+6*x^2-4)*exp(exp(1/2*ln(exp(2)*x)+x))+6*x^2-2)/(2*exp(exp(1/2*ln(exp(2) *x)+x))^2+4*exp(exp(1/2*ln(exp(2)*x)+x))+2),x,method=_RETURNVERBOSE)
Output:
x^3/(exp((exp(2)*x)^(1/2)*exp(x))+1)-x
Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {-2-2 e^{2 e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}+6 x^2+e^{e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}} \left (-4+6 x^2+e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )} \left (-x^2-2 x^3\right )\right )}{2+4 e^{e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}+2 e^{2 e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}} \, dx=\frac {x^{3} - x e^{\left (e^{\left (x + \frac {1}{2} \, \log \left (x e^{2}\right )\right )}\right )} - x}{e^{\left (e^{\left (x + \frac {1}{2} \, \log \left (x e^{2}\right )\right )}\right )} + 1} \] Input:
integrate((-2*exp(exp(1/2*log(exp(2)*x)+x))^2+((-2*x^3-x^2)*exp(1/2*log(ex p(2)*x)+x)+6*x^2-4)*exp(exp(1/2*log(exp(2)*x)+x))+6*x^2-2)/(2*exp(exp(1/2* log(exp(2)*x)+x))^2+4*exp(exp(1/2*log(exp(2)*x)+x))+2),x, algorithm="frica s")
Output:
(x^3 - x*e^(e^(x + 1/2*log(x*e^2))) - x)/(e^(e^(x + 1/2*log(x*e^2))) + 1)
Timed out. \[ \int \frac {-2-2 e^{2 e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}+6 x^2+e^{e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}} \left (-4+6 x^2+e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )} \left (-x^2-2 x^3\right )\right )}{2+4 e^{e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}+2 e^{2 e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}} \, dx=\text {Timed out} \] Input:
integrate((-2*exp(exp(1/2*ln(exp(2)*x)+x))**2+((-2*x**3-x**2)*exp(1/2*ln(e xp(2)*x)+x)+6*x**2-4)*exp(exp(1/2*ln(exp(2)*x)+x))+6*x**2-2)/(2*exp(exp(1/ 2*ln(exp(2)*x)+x))**2+4*exp(exp(1/2*ln(exp(2)*x)+x))+2),x)
Output:
Timed out
Time = 0.08 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.80 \[ \int \frac {-2-2 e^{2 e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}+6 x^2+e^{e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}} \left (-4+6 x^2+e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )} \left (-x^2-2 x^3\right )\right )}{2+4 e^{e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}+2 e^{2 e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}} \, dx=\frac {{\left (x^{3} e^{6} - x e^{6} - x e^{\left (\sqrt {x} e^{\left (x + 1\right )} + 6\right )}\right )} e^{\left (-2\right )}}{e^{4} + e^{\left (\sqrt {x} e^{\left (x + 1\right )} + 4\right )}} \] Input:
integrate((-2*exp(exp(1/2*log(exp(2)*x)+x))^2+((-2*x^3-x^2)*exp(1/2*log(ex p(2)*x)+x)+6*x^2-4)*exp(exp(1/2*log(exp(2)*x)+x))+6*x^2-2)/(2*exp(exp(1/2* log(exp(2)*x)+x))^2+4*exp(exp(1/2*log(exp(2)*x)+x))+2),x, algorithm="maxim a")
Output:
(x^3*e^6 - x*e^6 - x*e^(sqrt(x)*e^(x + 1) + 6))*e^(-2)/(e^4 + e^(sqrt(x)*e ^(x + 1) + 4))
\[ \int \frac {-2-2 e^{2 e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}+6 x^2+e^{e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}} \left (-4+6 x^2+e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )} \left (-x^2-2 x^3\right )\right )}{2+4 e^{e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}+2 e^{2 e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}} \, dx=\int { \frac {6 \, x^{2} + {\left (6 \, x^{2} - {\left (2 \, x^{3} + x^{2}\right )} e^{\left (x + \frac {1}{2} \, \log \left (x e^{2}\right )\right )} - 4\right )} e^{\left (e^{\left (x + \frac {1}{2} \, \log \left (x e^{2}\right )\right )}\right )} - 2 \, e^{\left (2 \, e^{\left (x + \frac {1}{2} \, \log \left (x e^{2}\right )\right )}\right )} - 2}{2 \, {\left (e^{\left (2 \, e^{\left (x + \frac {1}{2} \, \log \left (x e^{2}\right )\right )}\right )} + 2 \, e^{\left (e^{\left (x + \frac {1}{2} \, \log \left (x e^{2}\right )\right )}\right )} + 1\right )}} \,d x } \] Input:
integrate((-2*exp(exp(1/2*log(exp(2)*x)+x))^2+((-2*x^3-x^2)*exp(1/2*log(ex p(2)*x)+x)+6*x^2-4)*exp(exp(1/2*log(exp(2)*x)+x))+6*x^2-2)/(2*exp(exp(1/2* log(exp(2)*x)+x))^2+4*exp(exp(1/2*log(exp(2)*x)+x))+2),x, algorithm="giac" )
Output:
integrate(1/2*(6*x^2 + (6*x^2 - (2*x^3 + x^2)*e^(x + 1/2*log(x*e^2)) - 4)* e^(e^(x + 1/2*log(x*e^2))) - 2*e^(2*e^(x + 1/2*log(x*e^2))) - 2)/(e^(2*e^( x + 1/2*log(x*e^2))) + 2*e^(e^(x + 1/2*log(x*e^2))) + 1), x)
Time = 7.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {-2-2 e^{2 e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}+6 x^2+e^{e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}} \left (-4+6 x^2+e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )} \left (-x^2-2 x^3\right )\right )}{2+4 e^{e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}+2 e^{2 e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}} \, dx=\frac {x^3}{{\mathrm {e}}^{\sqrt {x}\,\mathrm {e}\,{\mathrm {e}}^x}+1}-x \] Input:
int(-(2*exp(2*exp(x + log(x*exp(2))/2)) + exp(exp(x + log(x*exp(2))/2))*(e xp(x + log(x*exp(2))/2)*(x^2 + 2*x^3) - 6*x^2 + 4) - 6*x^2 + 2)/(2*exp(2*e xp(x + log(x*exp(2))/2)) + 4*exp(exp(x + log(x*exp(2))/2)) + 2),x)
Output:
x^3/(exp(x^(1/2)*exp(1)*exp(x)) + 1) - x
Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {-2-2 e^{2 e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}+6 x^2+e^{e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}} \left (-4+6 x^2+e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )} \left (-x^2-2 x^3\right )\right )}{2+4 e^{e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}+2 e^{2 e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}} \, dx=\frac {x \left (-e^{\sqrt {x}\, e^{x} e}+x^{2}-1\right )}{e^{\sqrt {x}\, e^{x} e}+1} \] Input:
int((-2*exp(exp(1/2*log(exp(2)*x)+x))^2+((-2*x^3-x^2)*exp(1/2*log(exp(2)*x )+x)+6*x^2-4)*exp(exp(1/2*log(exp(2)*x)+x))+6*x^2-2)/(2*exp(exp(1/2*log(ex p(2)*x)+x))^2+4*exp(exp(1/2*log(exp(2)*x)+x))+2),x)
Output:
(x*( - e**(sqrt(x)*e**x*e) + x**2 - 1))/(e**(sqrt(x)*e**x*e) + 1)