Integrand size = 104, antiderivative size = 28 \[ \int \frac {-\frac {240}{e^3}+e^{e^{-3+e^{e^x}}} \left (-\frac {60}{e^3}+4 e^{-6+e^{e^x}+e^x+x} x^2\right )+\left (-\frac {32 x}{e^3}-8 e^{-3+e^{-3+e^{e^x}}} x\right ) \log \left (4+e^{e^{-3+e^{e^x}}}\right )}{4 x^4+e^{e^{-3+e^{e^x}}} x^4} \, dx=\frac {4 \left (\frac {5}{x}+\log \left (4+e^{e^{-3+e^{e^x}}}\right )\right )}{e^3 x^2} \]
Time = 0.12 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {-\frac {240}{e^3}+e^{e^{-3+e^{e^x}}} \left (-\frac {60}{e^3}+4 e^{-6+e^{e^x}+e^x+x} x^2\right )+\left (-\frac {32 x}{e^3}-8 e^{-3+e^{-3+e^{e^x}}} x\right ) \log \left (4+e^{e^{-3+e^{e^x}}}\right )}{4 x^4+e^{e^{-3+e^{e^x}}} x^4} \, dx=-\frac {4 \left (-\frac {5 e^3}{x^3}-\frac {e^3 \log \left (4+e^{e^{-3+e^{e^x}}}\right )}{x^2}\right )}{e^6} \]
Integrate[(-240/E^3 + E^E^(-3 + E^E^x)*(-60/E^3 + 4*E^(-6 + E^E^x + E^x + x)*x^2) + ((-32*x)/E^3 - 8*E^(-3 + E^(-3 + E^E^x))*x)*Log[4 + E^E^(-3 + E^ E^x)])/(4*x^4 + E^E^(-3 + E^E^x)*x^4),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.
\(\displaystyle \int \frac {e^{e^{e^{e^x}-3}} \left (4 e^{x+e^{e^x}+e^x-6} x^2-\frac {60}{e^3}\right )+\left (-8 e^{e^{e^{e^x}-3}-3} x-\frac {32 x}{e^3}\right ) \log \left (e^{e^{e^{e^x}-3}}+4\right )-\frac {240}{e^3}}{e^{e^{e^{e^x}-3}} x^4+4 x^4} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {4 e^{x+e^{e^x}+e^{e^{e^x}-3}+e^x-6}}{\left (e^{e^{e^{e^x}-3}}+4\right ) x^2}-\frac {4 \left (2 x \log \left (e^{e^{e^{e^x}-3}}+4\right )+15\right )}{e^3 x^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \int \frac {e^{x+e^{e^x}+e^{-3+e^{e^x}}+e^x-6}}{\left (4+e^{e^{-3+e^{e^x}}}\right ) x^2}dx-\frac {4 \int \frac {e^{x+e^{e^x}+e^{-3+e^{e^x}}+e^x-3}}{\left (4+e^{e^{-3+e^{e^x}}}\right ) x^2}dx}{e^3}+\frac {20}{e^3 x^3}+\frac {4 \log \left (e^{e^{e^{e^x}-3}}+4\right )}{e^3 x^2}\) |
Int[(-240/E^3 + E^E^(-3 + E^E^x)*(-60/E^3 + 4*E^(-6 + E^E^x + E^x + x)*x^2 ) + ((-32*x)/E^3 - 8*E^(-3 + E^(-3 + E^E^x))*x)*Log[4 + E^E^(-3 + E^E^x)]) /(4*x^4 + E^E^(-3 + E^E^x)*x^4),x]
3.14.43.3.1 Defintions of rubi rules used
Time = 124.32 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93
method | result | size |
risch | \(\frac {4 \,{\mathrm e}^{-3} \ln \left ({\mathrm e}^{{\mathrm e}^{-3+{\mathrm e}^{{\mathrm e}^{x}}}}+4\right )}{x^{2}}+\frac {20 \,{\mathrm e}^{-3}}{x^{3}}\) | \(26\) |
parallelrisch | \(\frac {{\mathrm e}^{2 \ln \left (2\right )-3} \ln \left ({\mathrm e}^{{\mathrm e}^{-3+{\mathrm e}^{{\mathrm e}^{x}}}}+4\right ) x +5 \,{\mathrm e}^{2 \ln \left (2\right )-3}}{x^{3}}\) | \(34\) |
int(((-2*x*exp(2*ln(2)-3)*exp(exp(-3+exp(exp(x))))-8*x*exp(2*ln(2)-3))*ln( exp(exp(-3+exp(exp(x))))+4)+(x^2*exp(x)*exp(2*ln(2)-3)*exp(exp(x))*exp(-3+ exp(exp(x)))-15*exp(2*ln(2)-3))*exp(exp(-3+exp(exp(x))))-60*exp(2*ln(2)-3) )/(x^4*exp(exp(-3+exp(exp(x))))+4*x^4),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 89 vs. \(2 (27) = 54\).
Time = 0.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 3.18 \[ \int \frac {-\frac {240}{e^3}+e^{e^{-3+e^{e^x}}} \left (-\frac {60}{e^3}+4 e^{-6+e^{e^x}+e^x+x} x^2\right )+\left (-\frac {32 x}{e^3}-8 e^{-3+e^{-3+e^{e^x}}} x\right ) \log \left (4+e^{e^{-3+e^{e^x}}}\right )}{4 x^4+e^{e^{-3+e^{e^x}}} x^4} \, dx=\frac {x e^{\left (2 \, \log \left (2\right ) - 3\right )} \log \left ({\left (e^{\left ({\left ({\left (2 \, \log \left (2\right ) - 3\right )} e^{\left (x + e^{x} + 2 \, \log \left (2\right ) - 3\right )} + e^{\left (x + e^{x} + e^{\left (e^{x}\right )} + 2 \, \log \left (2\right ) - 6\right )}\right )} e^{\left (-x - e^{x} - 2 \, \log \left (2\right ) + 3\right )}\right )} + 4 \, e^{\left (2 \, \log \left (2\right ) - 3\right )}\right )} e^{\left (-2 \, \log \left (2\right ) + 3\right )}\right ) + 5 \, e^{\left (2 \, \log \left (2\right ) - 3\right )}}{x^{3}} \]
integrate(((-2*x*exp(2*log(2)-3)*exp(exp(-3+exp(exp(x))))-8*x*exp(2*log(2) -3))*log(exp(exp(-3+exp(exp(x))))+4)+(x^2*exp(x)*exp(2*log(2)-3)*exp(exp(x ))*exp(-3+exp(exp(x)))-15*exp(2*log(2)-3))*exp(exp(-3+exp(exp(x))))-60*exp (2*log(2)-3))/(x^4*exp(exp(-3+exp(exp(x))))+4*x^4),x, algorithm=\
(x*e^(2*log(2) - 3)*log((e^(((2*log(2) - 3)*e^(x + e^x + 2*log(2) - 3) + e ^(x + e^x + e^(e^x) + 2*log(2) - 6))*e^(-x - e^x - 2*log(2) + 3)) + 4*e^(2 *log(2) - 3))*e^(-2*log(2) + 3)) + 5*e^(2*log(2) - 3))/x^3
Time = 2.75 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {-\frac {240}{e^3}+e^{e^{-3+e^{e^x}}} \left (-\frac {60}{e^3}+4 e^{-6+e^{e^x}+e^x+x} x^2\right )+\left (-\frac {32 x}{e^3}-8 e^{-3+e^{-3+e^{e^x}}} x\right ) \log \left (4+e^{e^{-3+e^{e^x}}}\right )}{4 x^4+e^{e^{-3+e^{e^x}}} x^4} \, dx=\frac {4 \log {\left (e^{e^{e^{e^{x}} - 3}} + 4 \right )}}{x^{2} e^{3}} + \frac {20}{x^{3} e^{3}} \]
integrate(((-2*x*exp(2*ln(2)-3)*exp(exp(-3+exp(exp(x))))-8*x*exp(2*ln(2)-3 ))*ln(exp(exp(-3+exp(exp(x))))+4)+(x**2*exp(x)*exp(2*ln(2)-3)*exp(exp(x))* exp(-3+exp(exp(x)))-15*exp(2*ln(2)-3))*exp(exp(-3+exp(exp(x))))-60*exp(2*l n(2)-3))/(x**4*exp(exp(-3+exp(exp(x))))+4*x**4),x)
Time = 0.32 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {-\frac {240}{e^3}+e^{e^{-3+e^{e^x}}} \left (-\frac {60}{e^3}+4 e^{-6+e^{e^x}+e^x+x} x^2\right )+\left (-\frac {32 x}{e^3}-8 e^{-3+e^{-3+e^{e^x}}} x\right ) \log \left (4+e^{e^{-3+e^{e^x}}}\right )}{4 x^4+e^{e^{-3+e^{e^x}}} x^4} \, dx=\frac {4 \, {\left (x \log \left (e^{\left (e^{\left (e^{\left (e^{x}\right )} - 3\right )}\right )} + 4\right ) + 5\right )} e^{\left (-3\right )}}{x^{3}} \]
integrate(((-2*x*exp(2*log(2)-3)*exp(exp(-3+exp(exp(x))))-8*x*exp(2*log(2) -3))*log(exp(exp(-3+exp(exp(x))))+4)+(x^2*exp(x)*exp(2*log(2)-3)*exp(exp(x ))*exp(-3+exp(exp(x)))-15*exp(2*log(2)-3))*exp(exp(-3+exp(exp(x))))-60*exp (2*log(2)-3))/(x^4*exp(exp(-3+exp(exp(x))))+4*x^4),x, algorithm=\
Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.86 \[ \int \frac {-\frac {240}{e^3}+e^{e^{-3+e^{e^x}}} \left (-\frac {60}{e^3}+4 e^{-6+e^{e^x}+e^x+x} x^2\right )+\left (-\frac {32 x}{e^3}-8 e^{-3+e^{-3+e^{e^x}}} x\right ) \log \left (4+e^{e^{-3+e^{e^x}}}\right )}{4 x^4+e^{e^{-3+e^{e^x}}} x^4} \, dx=\frac {4 \, {\left (x \log \left ({\left (e^{\left (x + e^{x} + e^{\left (e^{\left (e^{x}\right )} - 3\right )} + e^{\left (e^{x}\right )}\right )} + 4 \, e^{\left (x + e^{x} + e^{\left (e^{x}\right )}\right )}\right )} e^{\left (-x - e^{x} - e^{\left (e^{x}\right )}\right )}\right ) + 5\right )} e^{\left (-3\right )}}{x^{3}} \]
integrate(((-2*x*exp(2*log(2)-3)*exp(exp(-3+exp(exp(x))))-8*x*exp(2*log(2) -3))*log(exp(exp(-3+exp(exp(x))))+4)+(x^2*exp(x)*exp(2*log(2)-3)*exp(exp(x ))*exp(-3+exp(exp(x)))-15*exp(2*log(2)-3))*exp(exp(-3+exp(exp(x))))-60*exp (2*log(2)-3))/(x^4*exp(exp(-3+exp(exp(x))))+4*x^4),x, algorithm=\
4*(x*log((e^(x + e^x + e^(e^(e^x) - 3) + e^(e^x)) + 4*e^(x + e^x + e^(e^x) ))*e^(-x - e^x - e^(e^x))) + 5)*e^(-3)/x^3
Time = 0.33 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {-\frac {240}{e^3}+e^{e^{-3+e^{e^x}}} \left (-\frac {60}{e^3}+4 e^{-6+e^{e^x}+e^x+x} x^2\right )+\left (-\frac {32 x}{e^3}-8 e^{-3+e^{-3+e^{e^x}}} x\right ) \log \left (4+e^{e^{-3+e^{e^x}}}\right )}{4 x^4+e^{e^{-3+e^{e^x}}} x^4} \, dx=\frac {4\,{\mathrm {e}}^{-3}\,\left (x\,\ln \left ({\mathrm {e}}^{{\mathrm {e}}^{-3}\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^x}}}+4\right )+5\right )}{x^3} \]
int(-(60*exp(2*log(2) - 3) + exp(exp(exp(exp(x)) - 3))*(15*exp(2*log(2) - 3) - x^2*exp(exp(exp(x)) - 3)*exp(exp(x))*exp(2*log(2) - 3)*exp(x)) + log( exp(exp(exp(exp(x)) - 3)) + 4)*(8*x*exp(2*log(2) - 3) + 2*x*exp(exp(exp(ex p(x)) - 3))*exp(2*log(2) - 3)))/(x^4*exp(exp(exp(exp(x)) - 3)) + 4*x^4),x)