Integrand size = 112, antiderivative size = 30 \[ \int \frac {1}{5} \left (10 x+e^{e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )}} \left (-5+e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )} \left (-10 x-4 x^2+e^x \left (2 x+2 x^2\right )+e^{e^x} \left (2 x+2 e^x x^2\right )\right )\right )\right ) \, dx=x \left (-e^{e^{\frac {2}{5} x \left (5-e^{e^x}-e^x+x\right )}}+x\right ) \]
Time = 0.45 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {1}{5} \left (10 x+e^{e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )}} \left (-5+e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )} \left (-10 x-4 x^2+e^x \left (2 x+2 x^2\right )+e^{e^x} \left (2 x+2 e^x x^2\right )\right )\right )\right ) \, dx=x \left (-e^{e^{\frac {2}{5} x \left (5-e^{e^x}-e^x+x\right )}}+x\right ) \]
Integrate[(10*x + E^E^((10*x - 2*E^E^x*x - 2*E^x*x + 2*x^2)/5)*(-5 + E^((1 0*x - 2*E^E^x*x - 2*E^x*x + 2*x^2)/5)*(-10*x - 4*x^2 + E^x*(2*x + 2*x^2) + E^E^x*(2*x + 2*E^x*x^2))))/5,x]
Leaf count is larger than twice the leaf count of optimal. \(111\) vs. \(2(30)=60\).
Time = 0.40 (sec) , antiderivative size = 111, normalized size of antiderivative = 3.70, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {27, 2009}
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.
\(\displaystyle \int \frac {1}{5} \left (e^{e^{\frac {1}{5} \left (2 x^2-2 e^{e^x} x-2 e^x x+10 x\right )}} \left (e^{\frac {1}{5} \left (2 x^2-2 e^{e^x} x-2 e^x x+10 x\right )} \left (-4 x^2+e^x \left (2 x^2+2 x\right )+e^{e^x} \left (2 e^x x^2+2 x\right )-10 x\right )-5\right )+10 x\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \left (10 x-e^{e^{\frac {2}{5} \left (x^2-e^{e^x} x-e^x x+5 x\right )}} \left (2 e^{\frac {2}{5} \left (x^2-e^{e^x} x-e^x x+5 x\right )} \left (2 x^2+5 x-e^x \left (x^2+x\right )-e^{e^x} \left (e^x x^2+x\right )\right )+5\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (5 x^2-\frac {5 e^{e^{\frac {2}{5} \left (x^2-e^{e^x} x-e^x x+5 x\right )}} \left (2 x^2-e^x \left (x^2+x\right )-e^{e^x} \left (e^x x^2+x\right )+5 x\right )}{-e^x x-e^{x+e^x} x+2 x-e^{e^x}-e^x+5}\right )\) |
Int[(10*x + E^E^((10*x - 2*E^E^x*x - 2*E^x*x + 2*x^2)/5)*(-5 + E^((10*x - 2*E^E^x*x - 2*E^x*x + 2*x^2)/5)*(-10*x - 4*x^2 + E^x*(2*x + 2*x^2) + E^E^x *(2*x + 2*E^x*x^2))))/5,x]
(5*x^2 - (5*E^E^((2*(5*x - E^E^x*x - E^x*x + x^2))/5)*(5*x + 2*x^2 - E^x*( x + x^2) - E^E^x*(x + E^x*x^2)))/(5 - E^E^x - E^x + 2*x - E^x*x - E^(E^x + x)*x))/5
3.25.2.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.70 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77
method | result | size |
risch | \(-x \,{\mathrm e}^{{\mathrm e}^{-\frac {2 x \left ({\mathrm e}^{{\mathrm e}^{x}}+{\mathrm e}^{x}-x -5\right )}{5}}}+x^{2}\) | \(23\) |
parallelrisch | \(-x \,{\mathrm e}^{{\mathrm e}^{-\frac {2 x \left ({\mathrm e}^{{\mathrm e}^{x}}+{\mathrm e}^{x}-x -5\right )}{5}}}+x^{2}\) | \(23\) |
int(1/5*(((2*exp(x)*x^2+2*x)*exp(exp(x))+(2*x^2+2*x)*exp(x)-4*x^2-10*x)*ex p(-2/5*x*exp(exp(x))-2/5*exp(x)*x+2/5*x^2+2*x)-5)*exp(exp(-2/5*x*exp(exp(x ))-2/5*exp(x)*x+2/5*x^2+2*x))+2*x,x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {1}{5} \left (10 x+e^{e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )}} \left (-5+e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )} \left (-10 x-4 x^2+e^x \left (2 x+2 x^2\right )+e^{e^x} \left (2 x+2 e^x x^2\right )\right )\right )\right ) \, dx=x^{2} - x e^{\left (e^{\left (\frac {2}{5} \, x^{2} - \frac {2}{5} \, x e^{x} - \frac {2}{5} \, x e^{\left (e^{x}\right )} + 2 \, x\right )}\right )} \]
integrate(1/5*(((2*exp(x)*x^2+2*x)*exp(exp(x))+(2*x^2+2*x)*exp(x)-4*x^2-10 *x)*exp(-2/5*x*exp(exp(x))-2/5*exp(x)*x+2/5*x^2+2*x)-5)*exp(exp(-2/5*x*exp (exp(x))-2/5*exp(x)*x+2/5*x^2+2*x))+2*x,x, algorithm=\
Time = 4.38 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {1}{5} \left (10 x+e^{e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )}} \left (-5+e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )} \left (-10 x-4 x^2+e^x \left (2 x+2 x^2\right )+e^{e^x} \left (2 x+2 e^x x^2\right )\right )\right )\right ) \, dx=x^{2} - x e^{e^{\frac {2 x^{2}}{5} - \frac {2 x e^{x}}{5} - \frac {2 x e^{e^{x}}}{5} + 2 x}} \]
integrate(1/5*(((2*exp(x)*x**2+2*x)*exp(exp(x))+(2*x**2+2*x)*exp(x)-4*x**2 -10*x)*exp(-2/5*x*exp(exp(x))-2/5*exp(x)*x+2/5*x**2+2*x)-5)*exp(exp(-2/5*x *exp(exp(x))-2/5*exp(x)*x+2/5*x**2+2*x))+2*x,x)
\[ \int \frac {1}{5} \left (10 x+e^{e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )}} \left (-5+e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )} \left (-10 x-4 x^2+e^x \left (2 x+2 x^2\right )+e^{e^x} \left (2 x+2 e^x x^2\right )\right )\right )\right ) \, dx=\int { -\frac {1}{5} \, {\left (2 \, {\left (2 \, x^{2} - {\left (x^{2} + x\right )} e^{x} - {\left (x^{2} e^{x} + x\right )} e^{\left (e^{x}\right )} + 5 \, x\right )} e^{\left (\frac {2}{5} \, x^{2} - \frac {2}{5} \, x e^{x} - \frac {2}{5} \, x e^{\left (e^{x}\right )} + 2 \, x\right )} + 5\right )} e^{\left (e^{\left (\frac {2}{5} \, x^{2} - \frac {2}{5} \, x e^{x} - \frac {2}{5} \, x e^{\left (e^{x}\right )} + 2 \, x\right )}\right )} + 2 \, x \,d x } \]
integrate(1/5*(((2*exp(x)*x^2+2*x)*exp(exp(x))+(2*x^2+2*x)*exp(x)-4*x^2-10 *x)*exp(-2/5*x*exp(exp(x))-2/5*exp(x)*x+2/5*x^2+2*x)-5)*exp(exp(-2/5*x*exp (exp(x))-2/5*exp(x)*x+2/5*x^2+2*x))+2*x,x, algorithm=\
x^2 - 1/5*integrate(-(2*((x^2 + x)*e^(3*x) - (2*x^2 + 5*x)*e^(2*x) + (x^2* e^(3*x) + x*e^(2*x))*e^(e^x))*e^(2/5*x^2) - 5*e^(2/5*x*e^x + 2/5*x*e^(e^x) ))*e^(-2/5*x*e^x - 2/5*x*e^(e^x) + e^(2/5*x^2 - 2/5*x*e^x - 2/5*x*e^(e^x) + 2*x)), x)
\[ \int \frac {1}{5} \left (10 x+e^{e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )}} \left (-5+e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )} \left (-10 x-4 x^2+e^x \left (2 x+2 x^2\right )+e^{e^x} \left (2 x+2 e^x x^2\right )\right )\right )\right ) \, dx=\int { -\frac {1}{5} \, {\left (2 \, {\left (2 \, x^{2} - {\left (x^{2} + x\right )} e^{x} - {\left (x^{2} e^{x} + x\right )} e^{\left (e^{x}\right )} + 5 \, x\right )} e^{\left (\frac {2}{5} \, x^{2} - \frac {2}{5} \, x e^{x} - \frac {2}{5} \, x e^{\left (e^{x}\right )} + 2 \, x\right )} + 5\right )} e^{\left (e^{\left (\frac {2}{5} \, x^{2} - \frac {2}{5} \, x e^{x} - \frac {2}{5} \, x e^{\left (e^{x}\right )} + 2 \, x\right )}\right )} + 2 \, x \,d x } \]
integrate(1/5*(((2*exp(x)*x^2+2*x)*exp(exp(x))+(2*x^2+2*x)*exp(x)-4*x^2-10 *x)*exp(-2/5*x*exp(exp(x))-2/5*exp(x)*x+2/5*x^2+2*x)-5)*exp(exp(-2/5*x*exp (exp(x))-2/5*exp(x)*x+2/5*x^2+2*x))+2*x,x, algorithm=\
integrate(-1/5*(2*(2*x^2 - (x^2 + x)*e^x - (x^2*e^x + x)*e^(e^x) + 5*x)*e^ (2/5*x^2 - 2/5*x*e^x - 2/5*x*e^(e^x) + 2*x) + 5)*e^(e^(2/5*x^2 - 2/5*x*e^x - 2/5*x*e^(e^x) + 2*x)) + 2*x, x)
Time = 13.70 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {1}{5} \left (10 x+e^{e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )}} \left (-5+e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )} \left (-10 x-4 x^2+e^x \left (2 x+2 x^2\right )+e^{e^x} \left (2 x+2 e^x x^2\right )\right )\right )\right ) \, dx=x\,\left (x-{\mathrm {e}}^{{\mathrm {e}}^{-\frac {2\,x\,{\mathrm {e}}^x}{5}}\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-\frac {2\,x\,{\mathrm {e}}^{{\mathrm {e}}^x}}{5}}\,{\mathrm {e}}^{\frac {2\,x^2}{5}}}\right ) \]