Integrand size = 132, antiderivative size = 31 \[ \int \frac {e^{-e^{e+x}} \left (4 x \log (x)+e^{e^e} \left (-x+e^{e+x} x^2\right ) \log ^2(x)+\left (-4 x^2+4 e^{e+x} x^3\right ) \log ^2(x)+\left (-4 x-4 e^{e+x} x^2 \log (x)+e^{e^e} \left (-1-e^{e+x} x \log (x)\right )\right ) \log \left (\frac {1}{4} \left (e^{e^e}+4 x\right )\right )\right )}{e^{e^e} x \log ^2(x)+4 x^2 \log ^2(x)} \, dx=e^{-e^{e+x}} \left (-x+\frac {\log \left (\frac {e^{e^e}}{4}+x\right )}{\log (x)}\right ) \]
Time = 5.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-e^{e+x}} \left (4 x \log (x)+e^{e^e} \left (-x+e^{e+x} x^2\right ) \log ^2(x)+\left (-4 x^2+4 e^{e+x} x^3\right ) \log ^2(x)+\left (-4 x-4 e^{e+x} x^2 \log (x)+e^{e^e} \left (-1-e^{e+x} x \log (x)\right )\right ) \log \left (\frac {1}{4} \left (e^{e^e}+4 x\right )\right )\right )}{e^{e^e} x \log ^2(x)+4 x^2 \log ^2(x)} \, dx=e^{-e^{e+x}} \left (-x+\frac {\log \left (\frac {e^{e^e}}{4}+x\right )}{\log (x)}\right ) \]
Integrate[(4*x*Log[x] + E^E^E*(-x + E^(E + x)*x^2)*Log[x]^2 + (-4*x^2 + 4* E^(E + x)*x^3)*Log[x]^2 + (-4*x - 4*E^(E + x)*x^2*Log[x] + E^E^E*(-1 - E^( E + x)*x*Log[x]))*Log[(E^E^E + 4*x)/4])/(E^E^(E + x)*(E^E^E*x*Log[x]^2 + 4 *x^2*Log[x]^2)),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.
| \(\displaystyle \int \frac {e^{-e^{x+e}} \left (e^{e^e} \left (e^{x+e} x^2-x\right ) \log ^2(x)+\left (-4 e^{x+e} x^2 \log (x)-4 x+e^{e^e} \left (-e^{x+e} x \log (x)-1\right )\right ) \log \left (\frac {1}{4} \left (4 x+e^{e^e}\right )\right )+\left (4 e^{x+e} x^3-4 x^2\right ) \log ^2(x)+4 x \log (x)\right )}{4 x^2 \log ^2(x)+e^{e^e} x \log ^2(x)} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int e^{-e^{x+e}} \left (e^{x+e} x-\frac {\log \left (x+\frac {e^{e^e}}{4}\right )}{x \log ^2(x)}+\frac {\frac {4}{4 x+e^{e^e}}-e^{x+e} \log \left (x+\frac {e^{e^e}}{4}\right )}{\log (x)}-1\right )dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (e^{x-e^{x+e}+e} x-e^{-e^{x+e}}-\frac {e^{-e^{x+e}} \log \left (x+\frac {e^{e^e}}{4}\right )}{x \log ^2(x)}-\frac {e^{-e^{x+e}} \left (e^{x+e^e+e} \log \left (x+\frac {e^{e^e}}{4}\right )+4 e^{x+e} x \log \left (x+\frac {e^{e^e}}{4}\right )-4\right )}{\left (4 x+e^{e^e}\right ) \log (x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \int e^{x-e^{x+e}+e} xdx-\int \frac {e^{-e^{x+e}} \log \left (x+\frac {e^{e^e}}{4}\right )}{x \log ^2(x)}dx+4 \int \frac {e^{-e^{x+e}}}{\left (4 x+e^{e^e}\right ) \log (x)}dx-\int \frac {e^{x-e^{x+e}+e} \log \left (x+\frac {e^{e^e}}{4}\right )}{\log (x)}dx-\operatorname {ExpIntegralEi}\left (-e^{x+e}\right )\) |
Int[(4*x*Log[x] + E^E^E*(-x + E^(E + x)*x^2)*Log[x]^2 + (-4*x^2 + 4*E^(E + x)*x^3)*Log[x]^2 + (-4*x - 4*E^(E + x)*x^2*Log[x] + E^E^E*(-1 - E^(E + x) *x*Log[x]))*Log[(E^E^E + 4*x)/4])/(E^E^(E + x)*(E^E^E*x*Log[x]^2 + 4*x^2*L og[x]^2)),x]
3.22.44.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 87.33 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00
| method | result | size |
| risch | \(-\frac {\left (x \ln \left (x \right )-\ln \left (\frac {{\mathrm e}^{{\mathrm e}^{{\mathrm e}}}}{4}+x \right )\right ) {\mathrm e}^{-{\mathrm e}^{x +{\mathrm e}}}}{\ln \left (x \right )}\) | \(31\) |
| parallelrisch | \(\frac {\left (-16 \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}}} x \ln \left (x \right )+16 \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}}} \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{{\mathrm e}}}}{4}+x \right )\right ) {\mathrm e}^{-{\mathrm e}^{x +{\mathrm e}}} {\mathrm e}^{-{\mathrm e}^{{\mathrm e}}}}{16 \ln \left (x \right )}\) | \(46\) |
int((((-x*exp(x+exp(1))*ln(x)-1)*exp(exp(exp(1)))-4*x^2*exp(x+exp(1))*ln(x )-4*x)*ln(1/4*exp(exp(exp(1)))+x)+(x^2*exp(x+exp(1))-x)*ln(x)^2*exp(exp(ex p(1)))+(4*x^3*exp(x+exp(1))-4*x^2)*ln(x)^2+4*x*ln(x))/(x*ln(x)^2*exp(exp(e xp(1)))+4*x^2*ln(x)^2)/exp(exp(x+exp(1))),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {e^{-e^{e+x}} \left (4 x \log (x)+e^{e^e} \left (-x+e^{e+x} x^2\right ) \log ^2(x)+\left (-4 x^2+4 e^{e+x} x^3\right ) \log ^2(x)+\left (-4 x-4 e^{e+x} x^2 \log (x)+e^{e^e} \left (-1-e^{e+x} x \log (x)\right )\right ) \log \left (\frac {1}{4} \left (e^{e^e}+4 x\right )\right )\right )}{e^{e^e} x \log ^2(x)+4 x^2 \log ^2(x)} \, dx=-\frac {x e^{\left (-e^{\left (x + e\right )}\right )} \log \left (x\right ) - e^{\left (-e^{\left (x + e\right )}\right )} \log \left (x + \frac {1}{4} \, e^{\left (e^{e}\right )}\right )}{\log \left (x\right )} \]
integrate((((-x*exp(x+exp(1))*log(x)-1)*exp(exp(exp(1)))-4*x^2*exp(x+exp(1 ))*log(x)-4*x)*log(1/4*exp(exp(exp(1)))+x)+(x^2*exp(x+exp(1))-x)*log(x)^2* exp(exp(exp(1)))+(4*x^3*exp(x+exp(1))-4*x^2)*log(x)^2+4*x*log(x))/(x*log(x )^2*exp(exp(exp(1)))+4*x^2*log(x)^2)/exp(exp(x+exp(1))),x, algorithm=\
Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-e^{e+x}} \left (4 x \log (x)+e^{e^e} \left (-x+e^{e+x} x^2\right ) \log ^2(x)+\left (-4 x^2+4 e^{e+x} x^3\right ) \log ^2(x)+\left (-4 x-4 e^{e+x} x^2 \log (x)+e^{e^e} \left (-1-e^{e+x} x \log (x)\right )\right ) \log \left (\frac {1}{4} \left (e^{e^e}+4 x\right )\right )\right )}{e^{e^e} x \log ^2(x)+4 x^2 \log ^2(x)} \, dx=\frac {\left (- x \log {\left (x \right )} + \log {\left (x + \frac {e^{e^{e}}}{4} \right )}\right ) e^{- e^{x + e}}}{\log {\left (x \right )}} \]
integrate((((-x*exp(x+exp(1))*ln(x)-1)*exp(exp(exp(1)))-4*x**2*exp(x+exp(1 ))*ln(x)-4*x)*ln(1/4*exp(exp(exp(1)))+x)+(x**2*exp(x+exp(1))-x)*ln(x)**2*e xp(exp(exp(1)))+(4*x**3*exp(x+exp(1))-4*x**2)*ln(x)**2+4*x*ln(x))/(x*ln(x) **2*exp(exp(exp(1)))+4*x**2*ln(x)**2)/exp(exp(x+exp(1))),x)
Time = 0.34 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {e^{-e^{e+x}} \left (4 x \log (x)+e^{e^e} \left (-x+e^{e+x} x^2\right ) \log ^2(x)+\left (-4 x^2+4 e^{e+x} x^3\right ) \log ^2(x)+\left (-4 x-4 e^{e+x} x^2 \log (x)+e^{e^e} \left (-1-e^{e+x} x \log (x)\right )\right ) \log \left (\frac {1}{4} \left (e^{e^e}+4 x\right )\right )\right )}{e^{e^e} x \log ^2(x)+4 x^2 \log ^2(x)} \, dx=-\frac {{\left (x \log \left (x\right ) + 2 \, \log \left (2\right )\right )} e^{\left (-e^{\left (x + e\right )}\right )} - e^{\left (-e^{\left (x + e\right )}\right )} \log \left (4 \, x + e^{\left (e^{e}\right )}\right )}{\log \left (x\right )} \]
integrate((((-x*exp(x+exp(1))*log(x)-1)*exp(exp(exp(1)))-4*x^2*exp(x+exp(1 ))*log(x)-4*x)*log(1/4*exp(exp(exp(1)))+x)+(x^2*exp(x+exp(1))-x)*log(x)^2* exp(exp(exp(1)))+(4*x^3*exp(x+exp(1))-4*x^2)*log(x)^2+4*x*log(x))/(x*log(x )^2*exp(exp(exp(1)))+4*x^2*log(x)^2)/exp(exp(x+exp(1))),x, algorithm=\
\[ \int \frac {e^{-e^{e+x}} \left (4 x \log (x)+e^{e^e} \left (-x+e^{e+x} x^2\right ) \log ^2(x)+\left (-4 x^2+4 e^{e+x} x^3\right ) \log ^2(x)+\left (-4 x-4 e^{e+x} x^2 \log (x)+e^{e^e} \left (-1-e^{e+x} x \log (x)\right )\right ) \log \left (\frac {1}{4} \left (e^{e^e}+4 x\right )\right )\right )}{e^{e^e} x \log ^2(x)+4 x^2 \log ^2(x)} \, dx=\int { \frac {{\left ({\left (x^{2} e^{\left (x + e\right )} - x\right )} e^{\left (e^{e}\right )} \log \left (x\right )^{2} + 4 \, {\left (x^{3} e^{\left (x + e\right )} - x^{2}\right )} \log \left (x\right )^{2} - {\left (4 \, x^{2} e^{\left (x + e\right )} \log \left (x\right ) + {\left (x e^{\left (x + e\right )} \log \left (x\right ) + 1\right )} e^{\left (e^{e}\right )} + 4 \, x\right )} \log \left (x + \frac {1}{4} \, e^{\left (e^{e}\right )}\right ) + 4 \, x \log \left (x\right )\right )} e^{\left (-e^{\left (x + e\right )}\right )}}{4 \, x^{2} \log \left (x\right )^{2} + x e^{\left (e^{e}\right )} \log \left (x\right )^{2}} \,d x } \]
integrate((((-x*exp(x+exp(1))*log(x)-1)*exp(exp(exp(1)))-4*x^2*exp(x+exp(1 ))*log(x)-4*x)*log(1/4*exp(exp(exp(1)))+x)+(x^2*exp(x+exp(1))-x)*log(x)^2* exp(exp(exp(1)))+(4*x^3*exp(x+exp(1))-4*x^2)*log(x)^2+4*x*log(x))/(x*log(x )^2*exp(exp(exp(1)))+4*x^2*log(x)^2)/exp(exp(x+exp(1))),x, algorithm=\
integrate(((x^2*e^(x + e) - x)*e^(e^e)*log(x)^2 + 4*(x^3*e^(x + e) - x^2)* log(x)^2 - (4*x^2*e^(x + e)*log(x) + (x*e^(x + e)*log(x) + 1)*e^(e^e) + 4* x)*log(x + 1/4*e^(e^e)) + 4*x*log(x))*e^(-e^(x + e))/(4*x^2*log(x)^2 + x*e ^(e^e)*log(x)^2), x)
Time = 11.79 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-e^{e+x}} \left (4 x \log (x)+e^{e^e} \left (-x+e^{e+x} x^2\right ) \log ^2(x)+\left (-4 x^2+4 e^{e+x} x^3\right ) \log ^2(x)+\left (-4 x-4 e^{e+x} x^2 \log (x)+e^{e^e} \left (-1-e^{e+x} x \log (x)\right )\right ) \log \left (\frac {1}{4} \left (e^{e^e}+4 x\right )\right )\right )}{e^{e^e} x \log ^2(x)+4 x^2 \log ^2(x)} \, dx=\frac {{\mathrm {e}}^{-{\mathrm {e}}^{\mathrm {e}}\,{\mathrm {e}}^x}\,\left (\ln \left (x+\frac {{\mathrm {e}}^{{\mathrm {e}}^{\mathrm {e}}}}{4}\right )-x\,\ln \left (x\right )\right )}{\ln \left (x\right )} \]
int(-(exp(-exp(x + exp(1)))*(log(x)^2*(4*x^2 - 4*x^3*exp(x + exp(1))) + lo g(x + exp(exp(exp(1)))/4)*(4*x + exp(exp(exp(1)))*(x*exp(x + exp(1))*log(x ) + 1) + 4*x^2*exp(x + exp(1))*log(x)) - 4*x*log(x) + exp(exp(exp(1)))*log (x)^2*(x - x^2*exp(x + exp(1)))))/(4*x^2*log(x)^2 + x*exp(exp(exp(1)))*log (x)^2),x)