Integrand size = 110, antiderivative size = 34 \[ \int \frac {e^{-e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}} \left (8 e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}} x+x \log ^3\left (\frac {x}{4}\right )+e^{e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}} \left (\left (-5+3 x+2 x^2+(1-x) \log (3)\right ) \log ^3\left (\frac {x}{4}\right )-x \log ^3\left (\frac {x}{4}\right ) \log (x)\right )\right )}{x \log ^3\left (\frac {x}{4}\right )} \, dx=e^{-e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}} x+(5+x-\log (3)) (-1+x-\log (x)) \] Output:
(x+5-ln(3))*(x-ln(x)-1)+x/exp(exp(4/ln(1/4*x)^2))
Time = 0.23 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}} \left (8 e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}} x+x \log ^3\left (\frac {x}{4}\right )+e^{e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}} \left (\left (-5+3 x+2 x^2+(1-x) \log (3)\right ) \log ^3\left (\frac {x}{4}\right )-x \log ^3\left (\frac {x}{4}\right ) \log (x)\right )\right )}{x \log ^3\left (\frac {x}{4}\right )} \, dx=x \left (4+e^{-e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}}+x-\log (3)\right )+(-5-x+\log (3)) \log (x) \] Input:
Integrate[(8*E^(4/Log[x/4]^2)*x + x*Log[x/4]^3 + E^E^(4/Log[x/4]^2)*((-5 + 3*x + 2*x^2 + (1 - x)*Log[3])*Log[x/4]^3 - x*Log[x/4]^3*Log[x]))/(E^E^(4/ Log[x/4]^2)*x*Log[x/4]^3),x]
Output:
x*(4 + E^(-E^(4/Log[x/4]^2)) + x - Log[3]) + (-5 - x + Log[3])*Log[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.
| \(\displaystyle \int \frac {e^{-e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}} \left (e^{e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}} \left (\left (2 x^2+3 x+(1-x) \log (3)-5\right ) \log ^3\left (\frac {x}{4}\right )-x \log ^3\left (\frac {x}{4}\right ) \log (x)\right )+x \log ^3\left (\frac {x}{4}\right )+8 x e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}\right )}{x \log ^3\left (\frac {x}{4}\right )} \, dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {e^{-e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}} \left (2 x^2 e^{e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}}+x-x e^{e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}} \log (x)+3 x \left (1-\frac {\log (3)}{3}\right ) e^{e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}}-5 \left (1-\frac {\log (3)}{5}\right ) e^{e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}}\right )}{x}+\frac {8 e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}-e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}}}{\log ^3\left (\frac {x}{4}\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 4 \text {Subst}\left (\int e^{-e^{\frac {4}{\log ^2(x)}}}dx,x,\frac {x}{4}\right )+32 \text {Subst}\left (\int \frac {e^{\frac {4}{\log ^2(x)}-e^{\frac {4}{\log ^2(x)}}}}{\log ^3(x)}dx,x,\frac {x}{4}\right )+x^2+x-x \log (x)+x (3-\log (3))-(5-\log (3)) \log (x)\) |
Input:
Int[(8*E^(4/Log[x/4]^2)*x + x*Log[x/4]^3 + E^E^(4/Log[x/4]^2)*((-5 + 3*x + 2*x^2 + (1 - x)*Log[3])*Log[x/4]^3 - x*Log[x/4]^3*Log[x]))/(E^E^(4/Log[x/ 4]^2)*x*Log[x/4]^3),x]
Output:
$Aborted
Time = 241.66 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.35
| method | result | size |
| risch | \(-x \ln \left (x \right )+\ln \left (x \right ) \ln \left (3\right )-x \ln \left (3\right )+x^{2}-5 \ln \left (x \right )+4 x +{\mathrm e}^{-{\mathrm e}^{\frac {4}{\left (2 \ln \left (2\right )-\ln \left (x \right )\right )^{2}}}} x\) | \(46\) |
| parallelrisch | \(\left (-x \,{\mathrm e}^{{\mathrm e}^{\frac {4}{\ln \left (\frac {x}{4}\right )^{2}}}} \ln \left (3\right )+\ln \left (x \right ) {\mathrm e}^{{\mathrm e}^{\frac {4}{\ln \left (\frac {x}{4}\right )^{2}}}} \ln \left (3\right )+x^{2} {\mathrm e}^{{\mathrm e}^{\frac {4}{\ln \left (\frac {x}{4}\right )^{2}}}}-{\mathrm e}^{{\mathrm e}^{\frac {4}{\ln \left (\frac {x}{4}\right )^{2}}}} \ln \left (x \right ) x +4 x \,{\mathrm e}^{{\mathrm e}^{\frac {4}{\ln \left (\frac {x}{4}\right )^{2}}}}-5 \ln \left (x \right ) {\mathrm e}^{{\mathrm e}^{\frac {4}{\ln \left (\frac {x}{4}\right )^{2}}}}+x \right ) {\mathrm e}^{-{\mathrm e}^{\frac {4}{\ln \left (\frac {x}{4}\right )^{2}}}}\) | \(102\) |
Input:
int(((-x*ln(1/4*x)^3*ln(x)+((1-x)*ln(3)+2*x^2+3*x-5)*ln(1/4*x)^3)*exp(exp( 4/ln(1/4*x)^2))+8*x*exp(4/ln(1/4*x)^2)+x*ln(1/4*x)^3)/x/ln(1/4*x)^3/exp(ex p(4/ln(1/4*x)^2)),x,method=_RETURNVERBOSE)
Output:
-x*ln(x)+ln(x)*ln(3)-x*ln(3)+x^2-5*ln(x)+4*x+exp(-exp(4/(2*ln(2)-ln(x))^2) )*x
Time = 0.09 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.65 \[ \int \frac {e^{-e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}} \left (8 e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}} x+x \log ^3\left (\frac {x}{4}\right )+e^{e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}} \left (\left (-5+3 x+2 x^2+(1-x) \log (3)\right ) \log ^3\left (\frac {x}{4}\right )-x \log ^3\left (\frac {x}{4}\right ) \log (x)\right )\right )}{x \log ^3\left (\frac {x}{4}\right )} \, dx={\left ({\left (x^{2} - x \log \left (3\right ) - 2 \, x \log \left (2\right ) - {\left (x - \log \left (3\right ) + 5\right )} \log \left (\frac {1}{4} \, x\right ) + 4 \, x\right )} e^{\left (e^{\left (\frac {4}{\log \left (\frac {1}{4} \, x\right )^{2}}\right )}\right )} + x\right )} e^{\left (-e^{\left (\frac {4}{\log \left (\frac {1}{4} \, x\right )^{2}}\right )}\right )} \] Input:
integrate(((-x*log(1/4*x)^3*log(x)+((1-x)*log(3)+2*x^2+3*x-5)*log(1/4*x)^3 )*exp(exp(4/log(1/4*x)^2))+8*x*exp(4/log(1/4*x)^2)+x*log(1/4*x)^3)/x/log(1 /4*x)^3/exp(exp(4/log(1/4*x)^2)),x, algorithm="fricas")
Output:
((x^2 - x*log(3) - 2*x*log(2) - (x - log(3) + 5)*log(1/4*x) + 4*x)*e^(e^(4 /log(1/4*x)^2)) + x)*e^(-e^(4/log(1/4*x)^2))
Timed out. \[ \int \frac {e^{-e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}} \left (8 e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}} x+x \log ^3\left (\frac {x}{4}\right )+e^{e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}} \left (\left (-5+3 x+2 x^2+(1-x) \log (3)\right ) \log ^3\left (\frac {x}{4}\right )-x \log ^3\left (\frac {x}{4}\right ) \log (x)\right )\right )}{x \log ^3\left (\frac {x}{4}\right )} \, dx=\text {Timed out} \] Input:
integrate(((-x*ln(1/4*x)**3*ln(x)+((1-x)*ln(3)+2*x**2+3*x-5)*ln(1/4*x)**3) *exp(exp(4/ln(1/4*x)**2))+8*x*exp(4/ln(1/4*x)**2)+x*ln(1/4*x)**3)/x/ln(1/4 *x)**3/exp(exp(4/ln(1/4*x)**2)),x)
Output:
Timed out
Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.56 \[ \int \frac {e^{-e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}} \left (8 e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}} x+x \log ^3\left (\frac {x}{4}\right )+e^{e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}} \left (\left (-5+3 x+2 x^2+(1-x) \log (3)\right ) \log ^3\left (\frac {x}{4}\right )-x \log ^3\left (\frac {x}{4}\right ) \log (x)\right )\right )}{x \log ^3\left (\frac {x}{4}\right )} \, dx=x^{2} + x e^{\left (-e^{\left (\frac {4}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) \log \left (x\right ) + \log \left (x\right )^{2}}\right )}\right )} - x \log \left (3\right ) - x \log \left (x\right ) + \log \left (3\right ) \log \left (x\right ) + 4 \, x - 5 \, \log \left (x\right ) \] Input:
integrate(((-x*log(1/4*x)^3*log(x)+((1-x)*log(3)+2*x^2+3*x-5)*log(1/4*x)^3 )*exp(exp(4/log(1/4*x)^2))+8*x*exp(4/log(1/4*x)^2)+x*log(1/4*x)^3)/x/log(1 /4*x)^3/exp(exp(4/log(1/4*x)^2)),x, algorithm="maxima")
Output:
x^2 + x*e^(-e^(4/(4*log(2)^2 - 4*log(2)*log(x) + log(x)^2))) - x*log(3) - x*log(x) + log(3)*log(x) + 4*x - 5*log(x)
\[ \int \frac {e^{-e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}} \left (8 e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}} x+x \log ^3\left (\frac {x}{4}\right )+e^{e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}} \left (\left (-5+3 x+2 x^2+(1-x) \log (3)\right ) \log ^3\left (\frac {x}{4}\right )-x \log ^3\left (\frac {x}{4}\right ) \log (x)\right )\right )}{x \log ^3\left (\frac {x}{4}\right )} \, dx=\int { \frac {{\left (x \log \left (\frac {1}{4} \, x\right )^{3} + 8 \, x e^{\left (\frac {4}{\log \left (\frac {1}{4} \, x\right )^{2}}\right )} - {\left (x \log \left (\frac {1}{4} \, x\right )^{3} \log \left (x\right ) - {\left (2 \, x^{2} - {\left (x - 1\right )} \log \left (3\right ) + 3 \, x - 5\right )} \log \left (\frac {1}{4} \, x\right )^{3}\right )} e^{\left (e^{\left (\frac {4}{\log \left (\frac {1}{4} \, x\right )^{2}}\right )}\right )}\right )} e^{\left (-e^{\left (\frac {4}{\log \left (\frac {1}{4} \, x\right )^{2}}\right )}\right )}}{x \log \left (\frac {1}{4} \, x\right )^{3}} \,d x } \] Input:
integrate(((-x*log(1/4*x)^3*log(x)+((1-x)*log(3)+2*x^2+3*x-5)*log(1/4*x)^3 )*exp(exp(4/log(1/4*x)^2))+8*x*exp(4/log(1/4*x)^2)+x*log(1/4*x)^3)/x/log(1 /4*x)^3/exp(exp(4/log(1/4*x)^2)),x, algorithm="giac")
Output:
undef
Time = 3.79 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.47 \[ \int \frac {e^{-e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}} \left (8 e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}} x+x \log ^3\left (\frac {x}{4}\right )+e^{e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}} \left (\left (-5+3 x+2 x^2+(1-x) \log (3)\right ) \log ^3\left (\frac {x}{4}\right )-x \log ^3\left (\frac {x}{4}\right ) \log (x)\right )\right )}{x \log ^3\left (\frac {x}{4}\right )} \, dx=\ln \left (x\right )\,\left (\ln \left (3\right )-5\right )+x\,{\mathrm {e}}^{-{\mathrm {e}}^{\frac {4}{{\ln \left (x\right )}^2-4\,\ln \left (2\right )\,\ln \left (x\right )+4\,{\ln \left (2\right )}^2}}}-x\,\left (\ln \left (3\right )-4\right )-x\,\ln \left (x\right )+x^2 \] Input:
int((exp(-exp(4/log(x/4)^2))*(8*x*exp(4/log(x/4)^2) + x*log(x/4)^3 + exp(e xp(4/log(x/4)^2))*(log(x/4)^3*(3*x - log(3)*(x - 1) + 2*x^2 - 5) - x*log(x /4)^3*log(x))))/(x*log(x/4)^3),x)
Output:
log(x)*(log(3) - 5) + x*exp(-exp(4/(log(x)^2 - 4*log(2)*log(x) + 4*log(2)^ 2))) - x*(log(3) - 4) - x*log(x) + x^2
Time = 0.20 (sec) , antiderivative size = 115, normalized size of antiderivative = 3.38 \[ \int \frac {e^{-e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}} \left (8 e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}} x+x \log ^3\left (\frac {x}{4}\right )+e^{e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}} \left (\left (-5+3 x+2 x^2+(1-x) \log (3)\right ) \log ^3\left (\frac {x}{4}\right )-x \log ^3\left (\frac {x}{4}\right ) \log (x)\right )\right )}{x \log ^3\left (\frac {x}{4}\right )} \, dx=\frac {e^{e^{\frac {4}{\mathrm {log}\left (\frac {x}{4}\right )^{2}}}} \mathrm {log}\left (x \right ) \mathrm {log}\left (3\right )-e^{e^{\frac {4}{\mathrm {log}\left (\frac {x}{4}\right )^{2}}}} \mathrm {log}\left (x \right ) x -5 e^{e^{\frac {4}{\mathrm {log}\left (\frac {x}{4}\right )^{2}}}} \mathrm {log}\left (x \right )-e^{e^{\frac {4}{\mathrm {log}\left (\frac {x}{4}\right )^{2}}}} \mathrm {log}\left (3\right ) x +e^{e^{\frac {4}{\mathrm {log}\left (\frac {x}{4}\right )^{2}}}} x^{2}+4 e^{e^{\frac {4}{\mathrm {log}\left (\frac {x}{4}\right )^{2}}}} x +x}{e^{e^{\frac {4}{\mathrm {log}\left (\frac {x}{4}\right )^{2}}}}} \] Input:
int(((-x*log(1/4*x)^3*log(x)+((1-x)*log(3)+2*x^2+3*x-5)*log(1/4*x)^3)*exp( exp(4/log(1/4*x)^2))+8*x*exp(4/log(1/4*x)^2)+x*log(1/4*x)^3)/x/log(1/4*x)^ 3/exp(exp(4/log(1/4*x)^2)),x)
Output:
(e**(e**(4/log(x/4)**2))*log(x)*log(3) - e**(e**(4/log(x/4)**2))*log(x)*x - 5*e**(e**(4/log(x/4)**2))*log(x) - e**(e**(4/log(x/4)**2))*log(3)*x + e* *(e**(4/log(x/4)**2))*x**2 + 4*e**(e**(4/log(x/4)**2))*x + x)/e**(e**(4/lo g(x/4)**2))