Integrand size = 93, antiderivative size = 27 \[ \int \frac {e^{\frac {2 \left (-x^2+(x+e x) \log (x)\right )}{1+e}+e^{\frac {2 \left (-x^2+(x+e x) \log (x)\right )}{1+e}} x \log ^2(\log (2))} \left (-1+e (-1-2 x)-2 x+4 x^2+(-2 x-2 e x) \log (x)\right ) \log ^2(\log (2))}{1+e} \, dx=-e^{e^{2 x \left (-\frac {x}{1+e}+\log (x)\right )} x \log ^2(\log (2))} \]
Time = 5.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^{\frac {2 \left (-x^2+(x+e x) \log (x)\right )}{1+e}+e^{\frac {2 \left (-x^2+(x+e x) \log (x)\right )}{1+e}} x \log ^2(\log (2))} \left (-1+e (-1-2 x)-2 x+4 x^2+(-2 x-2 e x) \log (x)\right ) \log ^2(\log (2))}{1+e} \, dx=-e^{e^{-\frac {2 x^2}{1+e}} x^{1+2 x} \log ^2(\log (2))} \]
Integrate[(E^((2*(-x^2 + (x + E*x)*Log[x]))/(1 + E) + E^((2*(-x^2 + (x + E *x)*Log[x]))/(1 + E))*x*Log[Log[2]]^2)*(-1 + E*(-1 - 2*x) - 2*x + 4*x^2 + (-2*x - 2*E*x)*Log[x])*Log[Log[2]]^2)/(1 + E),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.
\(\displaystyle \int \frac {\log ^2(\log (2)) \left (4 x^2-2 x+e (-2 x-1)+(-2 e x-2 x) \log (x)-1\right ) \exp \left (x \log ^2(\log (2)) e^{\frac {2 \left ((e x+x) \log (x)-x^2\right )}{1+e}}+\frac {2 \left ((e x+x) \log (x)-x^2\right )}{1+e}\right )}{1+e} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\log ^2(\log (2)) \int -\exp \left (e^{-\frac {2 x^2}{1+e}} x^{\frac {2 (e x+x)}{1+e}+1} \log ^2(\log (2))-\frac {2 \left (x^2-(1+e) x \log (x)\right )}{1+e}\right ) \left (-4 x^2+2 (1+e) \log (x) x+2 x+e (2 x+1)+1\right )dx}{1+e}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\log ^2(\log (2)) \int \exp \left (e^{-\frac {2 x^2}{1+e}} x^{2 x+1} \log ^2(\log (2))-\frac {2 \left (x^2-(1+e) x \log (x)\right )}{1+e}\right ) \left (-4 x^2+2 (1+e) \log (x) x+2 x+e (2 x+1)+1\right )dx}{1+e}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {\log ^2(\log (2)) \int \left (-4 \exp \left (e^{-\frac {2 x^2}{1+e}} x^{2 x+1} \log ^2(\log (2))-\frac {2 \left (x^2-(1+e) x \log (x)\right )}{1+e}\right ) x^2+2 \exp \left (e^{-\frac {2 x^2}{1+e}} x^{2 x+1} \log ^2(\log (2))-\frac {2 \left (x^2-(1+e) x \log (x)\right )}{1+e}\right ) (1+e) \log (x) x+2 \exp \left (e^{-\frac {2 x^2}{1+e}} x^{2 x+1} \log ^2(\log (2))-\frac {2 \left (x^2-(1+e) x \log (x)\right )}{1+e}\right ) (1+e) x+\exp \left (e^{-\frac {2 x^2}{1+e}} x^{2 x+1} \log ^2(\log (2))-\frac {2 \left (x^2-(1+e) x \log (x)\right )}{1+e}\right ) (1+e)\right )dx}{1+e}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\log ^2(\log (2)) \left ((1+e) \int \exp \left (e^{-\frac {2 x^2}{1+e}} x^{2 x+1} \log ^2(\log (2))-\frac {2 \left (x^2-(1+e) x \log (x)\right )}{1+e}\right )dx+2 (1+e) \int \exp \left (e^{-\frac {2 x^2}{1+e}} x^{2 x+1} \log ^2(\log (2))-\frac {2 \left (x^2-(1+e) x \log (x)\right )}{1+e}\right ) xdx-4 \int \exp \left (e^{-\frac {2 x^2}{1+e}} x^{2 x+1} \log ^2(\log (2))-\frac {2 \left (x^2-(1+e) x \log (x)\right )}{1+e}\right ) x^2dx+2 (1+e) \int \exp \left (e^{-\frac {2 x^2}{1+e}} x^{2 x+1} \log ^2(\log (2))-\frac {2 \left (x^2-(1+e) x \log (x)\right )}{1+e}\right ) x \log (x)dx\right )}{1+e}\) |
Int[(E^((2*(-x^2 + (x + E*x)*Log[x]))/(1 + E) + E^((2*(-x^2 + (x + E*x)*Lo g[x]))/(1 + E))*x*Log[Log[2]]^2)*(-1 + E*(-1 - 2*x) - 2*x + 4*x^2 + (-2*x - 2*E*x)*Log[x])*Log[Log[2]]^2)/(1 + E),x]
3.9.13.3.1 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 = 2.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19
method | result | size |
risch | \(-{\mathrm e}^{x \ln \left (\ln \left (2\right )\right )^{2} {\mathrm e}^{\frac {2 x \left ({\mathrm e} \ln \left (x \right )+\ln \left (x \right )-x \right )}{1+{\mathrm e}}}}\) | \(32\) |
parallelrisch | \(\frac {-{\mathrm e}^{x \ln \left (\ln \left (2\right )\right )^{2} {\mathrm e}^{\frac {2 x \left ({\mathrm e} \ln \left (x \right )+\ln \left (x \right )-x \right )}{1+{\mathrm e}}}} {\mathrm e}-{\mathrm e}^{x \ln \left (\ln \left (2\right )\right )^{2} {\mathrm e}^{\frac {2 x \left ({\mathrm e} \ln \left (x \right )+\ln \left (x \right )-x \right )}{1+{\mathrm e}}}}}{1+{\mathrm e}}\) | \(75\) |
int(((-2*x*exp(1)-2*x)*ln(x)+(-1-2*x)*exp(1)+4*x^2-2*x-1)*ln(ln(2))^2*exp( ((x*exp(1)+x)*ln(x)-x^2)/(1+exp(1)))^2*exp(x*ln(ln(2))^2*exp(((x*exp(1)+x) *ln(x)-x^2)/(1+exp(1)))^2)/(1+exp(1)),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 84 vs. \(2 (27) = 54\).
Time = 0.26 (sec) , antiderivative size = 84, normalized size of antiderivative = 3.11 \[ \int \frac {e^{\frac {2 \left (-x^2+(x+e x) \log (x)\right )}{1+e}+e^{\frac {2 \left (-x^2+(x+e x) \log (x)\right )}{1+e}} x \log ^2(\log (2))} \left (-1+e (-1-2 x)-2 x+4 x^2+(-2 x-2 e x) \log (x)\right ) \log ^2(\log (2))}{1+e} \, dx=-e^{\left (\frac {{\left (x e + x\right )} e^{\left (-\frac {2 \, {\left (x^{2} - {\left (x e + x\right )} \log \left (x\right )\right )}}{e + 1}\right )} \log \left (\log \left (2\right )\right )^{2} - 2 \, x^{2} + 2 \, {\left (x e + x\right )} \log \left (x\right )}{e + 1} + \frac {2 \, {\left (x^{2} - {\left (x e + x\right )} \log \left (x\right )\right )}}{e + 1}\right )} \]
integrate(((-2*x*exp(1)-2*x)*log(x)+(-1-2*x)*exp(1)+4*x^2-2*x-1)*log(log(2 ))^2*exp(((x*exp(1)+x)*log(x)-x^2)/(1+exp(1)))^2*exp(x*log(log(2))^2*exp(( (x*exp(1)+x)*log(x)-x^2)/(1+exp(1)))^2)/(1+exp(1)),x, algorithm=\
-e^(((x*e + x)*e^(-2*(x^2 - (x*e + x)*log(x))/(e + 1))*log(log(2))^2 - 2*x ^2 + 2*(x*e + x)*log(x))/(e + 1) + 2*(x^2 - (x*e + x)*log(x))/(e + 1))
Time = 1.68 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {e^{\frac {2 \left (-x^2+(x+e x) \log (x)\right )}{1+e}+e^{\frac {2 \left (-x^2+(x+e x) \log (x)\right )}{1+e}} x \log ^2(\log (2))} \left (-1+e (-1-2 x)-2 x+4 x^2+(-2 x-2 e x) \log (x)\right ) \log ^2(\log (2))}{1+e} \, dx=- e^{x e^{\frac {2 \left (- x^{2} + \left (x + e x\right ) \log {\left (x \right )}\right )}{1 + e}} \log {\left (\log {\left (2 \right )} \right )}^{2}} \]
integrate(((-2*x*exp(1)-2*x)*ln(x)+(-1-2*x)*exp(1)+4*x**2-2*x-1)*ln(ln(2)) **2*exp(((x*exp(1)+x)*ln(x)-x**2)/(1+exp(1)))**2*exp(x*ln(ln(2))**2*exp((( x*exp(1)+x)*ln(x)-x**2)/(1+exp(1)))**2)/(1+exp(1)),x)
Time = 0.49 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74 \[ \int \frac {e^{\frac {2 \left (-x^2+(x+e x) \log (x)\right )}{1+e}+e^{\frac {2 \left (-x^2+(x+e x) \log (x)\right )}{1+e}} x \log ^2(\log (2))} \left (-1+e (-1-2 x)-2 x+4 x^2+(-2 x-2 e x) \log (x)\right ) \log ^2(\log (2))}{1+e} \, dx=-e^{\left (x e^{\left (\frac {2 \, x e \log \left (x\right )}{e + 1} - \frac {2 \, x^{2}}{e + 1} + \frac {2 \, x \log \left (x\right )}{e + 1}\right )} \log \left (\log \left (2\right )\right )^{2}\right )} \]
integrate(((-2*x*exp(1)-2*x)*log(x)+(-1-2*x)*exp(1)+4*x^2-2*x-1)*log(log(2 ))^2*exp(((x*exp(1)+x)*log(x)-x^2)/(1+exp(1)))^2*exp(x*log(log(2))^2*exp(( (x*exp(1)+x)*log(x)-x^2)/(1+exp(1)))^2)/(1+exp(1)),x, algorithm=\
\[ \int \frac {e^{\frac {2 \left (-x^2+(x+e x) \log (x)\right )}{1+e}+e^{\frac {2 \left (-x^2+(x+e x) \log (x)\right )}{1+e}} x \log ^2(\log (2))} \left (-1+e (-1-2 x)-2 x+4 x^2+(-2 x-2 e x) \log (x)\right ) \log ^2(\log (2))}{1+e} \, dx=\int { \frac {{\left (4 \, x^{2} - {\left (2 \, x + 1\right )} e - 2 \, {\left (x e + x\right )} \log \left (x\right ) - 2 \, x - 1\right )} e^{\left (x e^{\left (-\frac {2 \, {\left (x^{2} - {\left (x e + x\right )} \log \left (x\right )\right )}}{e + 1}\right )} \log \left (\log \left (2\right )\right )^{2} - \frac {2 \, {\left (x^{2} - {\left (x e + x\right )} \log \left (x\right )\right )}}{e + 1}\right )} \log \left (\log \left (2\right )\right )^{2}}{e + 1} \,d x } \]
integrate(((-2*x*exp(1)-2*x)*log(x)+(-1-2*x)*exp(1)+4*x^2-2*x-1)*log(log(2 ))^2*exp(((x*exp(1)+x)*log(x)-x^2)/(1+exp(1)))^2*exp(x*log(log(2))^2*exp(( (x*exp(1)+x)*log(x)-x^2)/(1+exp(1)))^2)/(1+exp(1)),x, algorithm=\
integrate((4*x^2 - (2*x + 1)*e - 2*(x*e + x)*log(x) - 2*x - 1)*e^(x*e^(-2* (x^2 - (x*e + x)*log(x))/(e + 1))*log(log(2))^2 - 2*(x^2 - (x*e + x)*log(x ))/(e + 1))*log(log(2))^2/(e + 1), x)
Time = 8.81 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.70 \[ \int \frac {e^{\frac {2 \left (-x^2+(x+e x) \log (x)\right )}{1+e}+e^{\frac {2 \left (-x^2+(x+e x) \log (x)\right )}{1+e}} x \log ^2(\log (2))} \left (-1+e (-1-2 x)-2 x+4 x^2+(-2 x-2 e x) \log (x)\right ) \log ^2(\log (2))}{1+e} \, dx=-{\mathrm {e}}^{x\,x^{\frac {2\,x\,\mathrm {e}}{\mathrm {e}+1}}\,x^{\frac {2\,x}{\mathrm {e}+1}}\,{\mathrm {e}}^{-\frac {2\,x^2}{\mathrm {e}+1}}\,{\ln \left (\ln \left (2\right )\right )}^2} \]
int(-(exp(x*exp((2*(log(x)*(x + x*exp(1)) - x^2))/(exp(1) + 1))*log(log(2) )^2)*exp((2*(log(x)*(x + x*exp(1)) - x^2))/(exp(1) + 1))*log(log(2))^2*(2* x + log(x)*(2*x + 2*x*exp(1)) - 4*x^2 + exp(1)*(2*x + 1) + 1))/(exp(1) + 1 ),x)