∫ e−e 4 __________ log2 (x4) (8e 4 __________ log2 (x4) x+x log3(x 4 )+ee 4 __________ log2 (x4) ((−5+3x+2x2+(1−x) log(3)) log3(x 4 )−x log3(x 4 ) log(x))) ______________________________________________________________________________________________________________________________________

Optimal. Leaf size=34 \[ e^{-e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}} x+(5+x-\log (3)) (-1+x-\log (x)) \]

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Rubi [F] time = 2.93, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

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]

x + x^2 + x*(3 - Log[3]) - x*Log[x] - (5 - Log[3])*Log[x] + 4*Defer[Subst][Defer[Int][E^(-E^(4/Log[x]^2)), x],

x, x/4] + 32*Defer[Subst][Defer[Int][E^(-E^(4/Log[x]^2) + 4/Log[x]^2)/Log[x]^3, x], x, x/4]

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (3+e^{-e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}}-\frac {5}{x}+2 x-\log (3)+\frac {\log (3)}{x}+\frac {8 e^{-e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}+\frac {4}{\log ^2\left (\frac {x}{4}\right )}}}{\log ^3\left (\frac {x}{4}\right )}-\log (x)\right ) \, dx\\ &=\int \left (3+e^{-e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}}+2 x+\frac {-5+\log (3)}{x}-\log (3)+\frac {8 e^{-e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}+\frac {4}{\log ^2\left (\frac {x}{4}\right )}}}{\log ^3\left (\frac {x}{4}\right )}-\log (x)\right ) \, dx\\ &=x^2+x (3-\log (3))-(5-\log (3)) \log (x)+8 \int \frac {e^{-e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}+\frac {4}{\log ^2\left (\frac {x}{4}\right )}}}{\log ^3\left (\frac {x}{4}\right )} \, dx+\int e^{-e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}} \, dx-\int \log (x) \, dx\\ &=x+x^2+x (3-\log (3))-x \log (x)-(5-\log (3)) \log (x)+4 \operatorname {Subst}\left (\int e^{-e^{\frac {4}{\log ^2(x)}}} \, dx,x,\frac {x}{4}\right )+32 \operatorname {Subst}\left (\int \frac {e^{-e^{\frac {4}{\log ^2(x)}}+\frac {4}{\log ^2(x)}}}{\log ^3(x)} \, dx,x,\frac {x}{4}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.35, size = 36, normalized size = 1.06 \begin {gather*} x \left (4+e^{-e^{\frac {4}{\log ^2\left (\frac {x}{4}\right )}}}+x-\log (3)\right )+(-5-x+\log (3)) \log (x) \end {gather*}

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]

x*(4 + E^(-E^(4/Log[x/4]^2)) + x - Log[3]) + (-5 - x + Log[3])*Log[x]

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fricas [A] time = 0.85, size = 56, normalized size = 1.65 \begin {gather*} {\left ({\left (x^{2} - x \log \relax (3) - 2 \, x \log \relax (2) - {\left (x - \log \relax (3) + 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 )} \end {gather*}

integrate(((-x*log(1/4*x)^3*log(x)+((-x+1)*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")

((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/

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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

integrate(((-x*log(1/4*x)^3*log(x)+((-x+1)*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")

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method	result	size

risch	\(-x \ln \relax (x )+\ln \relax (3) \ln \relax (x )-x \ln \relax (3)+x^{2}-5 \ln \relax (x )+4 x +x \,{\mathrm e}^{-{\mathrm e}^{\frac {4}{\left (2 \ln \relax (2)-\ln \relax (x )\right )^{2}}}}\)	\(46\)

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(exp(4/ln(1/4*x)^2)),x,method=_RETURNVERBOSE)

-x*ln(x)+ln(3)*ln(x)-x*ln(3)+x^2-5*ln(x)+4*x+x*exp(-exp(4/(2*ln(2)-ln(x))^2))

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maxima [A] time = 0.61, size = 53, normalized size = 1.56 \begin {gather*} x^{2} + x e^{\left (-e^{\left (\frac {4}{4 \, \log \relax (2)^{2} - 4 \, \log \relax (2) \log \relax (x) + \log \relax (x)^{2}}\right )}\right )} - x \log \relax (3) - x \log \relax (x) + \log \relax (3) \log \relax (x) + 4 \, x - 5 \, \log \relax (x) \end {gather*}

integrate(((-x*log(1/4*x)^3*log(x)+((-x+1)*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")

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*l

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mupad [B] time = 3.42, size = 50, normalized size = 1.47 \begin {gather*} \ln \relax (x)\,\left (\ln \relax (3)-5\right )+x\,{\mathrm {e}}^{-{\mathrm {e}}^{\frac {4}{{\ln \relax (x)}^2-4\,\ln \relax (2)\,\ln \relax (x)+4\,{\ln \relax (2)}^2}}}-x\,\left (\ln \relax (3)-4\right )-x\,\ln \relax (x)+x^2 \end {gather*}

int((exp(-exp(4/log(x/4)^2))*(8*x*exp(4/log(x/4)^2) + x*log(x/4)^3 + exp(exp(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)

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

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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

integrate(((-x*ln(1/4*x)**3*ln(x)+((-x+1)*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)

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3.43.59 \(\int \frac {e^{-e^{\frac {4}{\log ^2(\frac {x}{4})}}} (8 e^{\frac {4}{\log ^2(\frac {x}{4})}} x+x \log ^3(\frac {x}{4})+e^{e^{\frac {4}{\log ^2(\frac {x}{4})}}} ((-5+3 x+2 x^2+(1-x) \log (3)) \log ^3(\frac {x}{4})-x \log ^3(\frac {x}{4}) \log (x)))}{x \log ^3(\frac {x}{4})} \, dx\)