∫ −x log(2)+ee−4+x4 log(2) _________________ x log(2) (−5x log(2)+e−4+x4 log(2) _________________ x log(2) (20+15x4 log(2))) ___________________________________________________________________________________________

Optimal. Leaf size=29 \[ \frac {1}{5 x}+\frac {e^{e^{x^3-\frac {4}{x \log (2)}}}}{x} \]

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Rubi [F] time = 1.21, 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 {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx \end {gather*}

Int[(-(x*Log[2]) + E^E^((-4 + x^4*Log[2])/(x*Log[2]))*(-5*x*Log[2] + E^((-4 + x^4*Log[2])/(x*Log[2]))*(20 + 15

1/(5*x) + (4*Defer[Int][E^(E^(x^3 - 4/(x*Log[2])) + x^3 - 4/(x*Log[2]))/x^3, x])/Log[2] - Defer[Int][E^E^((x^4

- 4/Log[2])/x)/x^2, x] + (Log[8]*Defer[Int][E^(E^(x^3 - 4/(x*Log[2])) + x^3 - 4/(x*Log[2]))*x, x])/Log[2]

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{x^3} \, dx}{5 \log (2)}\\ &=\frac {\int \left (-\frac {\left (1+5 e^{e^{x^3-\frac {4}{x \log (2)}}}\right ) \log (2)}{x^2}+\frac {5 e^{e^{x^3-\frac {4}{x \log (2)}}+x^3-\frac {4}{x \log (2)}} \left (4+x^4 \log (8)\right )}{x^3}\right ) \, dx}{5 \log (2)}\\ &=-\left (\frac {1}{5} \int \frac {1+5 e^{e^{x^3-\frac {4}{x \log (2)}}}}{x^2} \, dx\right )+\frac {\int \frac {e^{e^{x^3-\frac {4}{x \log (2)}}+x^3-\frac {4}{x \log (2)}} \left (4+x^4 \log (8)\right )}{x^3} \, dx}{\log (2)}\\ &=-\left (\frac {1}{5} \int \left (\frac {1}{x^2}+\frac {5 e^{e^{\frac {x^4-\frac {4}{\log (2)}}{x}}}}{x^2}\right ) \, dx\right )+\frac {\int \left (\frac {4 e^{e^{x^3-\frac {4}{x \log (2)}}+x^3-\frac {4}{x \log (2)}}}{x^3}+e^{e^{x^3-\frac {4}{x \log (2)}}+x^3-\frac {4}{x \log (2)}} x \log (8)\right ) \, dx}{\log (2)}\\ &=\frac {1}{5 x}+\frac {4 \int \frac {e^{e^{x^3-\frac {4}{x \log (2)}}+x^3-\frac {4}{x \log (2)}}}{x^3} \, dx}{\log (2)}+\frac {\log (8) \int e^{e^{x^3-\frac {4}{x \log (2)}}+x^3-\frac {4}{x \log (2)}} x \, dx}{\log (2)}-\int \frac {e^{e^{\frac {x^4-\frac {4}{\log (2)}}{x}}}}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F] time = 1.43, size = 75, normalized size = 2.59 \begin {gather*} \frac {\int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{x^3} \, dx}{5 \log (2)} \end {gather*}

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

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fricas [A] time = 0.61, size = 27, normalized size = 0.93 \begin {gather*} \frac {5 \, e^{\left (e^{\left (\frac {x^{4} \log \relax (2) - 4}{x \log \relax (2)}\right )}\right )} + 1}{5 \, x} \end {gather*}

integrate(1/5*(((15*x^4*log(2)+20)*exp((x^4*log(2)-4)/x/log(2))-5*x*log(2))*exp(exp((x^4*log(2)-4)/x/log(2)))-

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

integrate(1/5*(((15*x^4*log(2)+20)*exp((x^4*log(2)-4)/x/log(2))-5*x*log(2))*exp(exp((x^4*log(2)-4)/x/log(2)))-

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method	result	size

norman	\(\frac {x \,{\mathrm e}^{{\mathrm e}^{\frac {x^{4} \ln \relax (2)-4}{x \ln \relax (2)}}}+\frac {x}{5}}{x^{2}}\)	\(29\)
risch	\(\frac {1}{5 x}+\frac {{\mathrm e}^{{\mathrm e}^{\frac {x^{4} \ln \relax (2)-4}{x \ln \relax (2)}}}}{x}\)	\(29\)

int(1/5*(((15*x^4*ln(2)+20)*exp((x^4*ln(2)-4)/x/ln(2))-5*x*ln(2))*exp(exp((x^4*ln(2)-4)/x/ln(2)))-x*ln(2))/x^3

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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {-\mathit {undef} + \frac {\log \relax (2)}{x}}{5 \, \log \relax (2)} \end {gather*}

integrate(1/5*(((15*x^4*log(2)+20)*exp((x^4*log(2)-4)/x/log(2))-5*x*log(2))*exp(exp((x^4*log(2)-4)/x/log(2)))-

1/5*(log(2)/x - integrate(5*(x*e^(4/(x*log(2)))*log(2) - (3*x^4*log(2) + 4)*e^(x^3))*e^(-4/(x*log(2)) + e^(x^3

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mupad [B] time = 4.47, size = 22, normalized size = 0.76 \begin {gather*} \frac {{\mathrm {e}}^{{\mathrm {e}}^{x^3}\,{\mathrm {e}}^{-\frac {4}{x\,\ln \relax (2)}}}+\frac {1}{5}}{x} \end {gather*}

int(-((x*log(2))/5 + (exp(exp((x^4*log(2) - 4)/(x*log(2))))*(5*x*log(2) - exp((x^4*log(2) - 4)/(x*log(2)))*(15

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sympy [A] time = 0.32, size = 22, normalized size = 0.76 \begin {gather*} \frac {e^{e^{\frac {x^{4} \log {\relax (2 )} - 4}{x \log {\relax (2 )}}}}}{x} + \frac {1}{5 x} \end {gather*}

integrate(1/5*(((15*x**4*ln(2)+20)*exp((x**4*ln(2)-4)/x/ln(2))-5*x*ln(2))*exp(exp((x**4*ln(2)-4)/x/ln(2)))-x*l

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3.67.31 \(\int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} (20+15 x^4 \log (2)))}{5 x^3 \log (2)} \, dx\)