3.3.71 \(\int \frac {-e^2+(-5 e^2-20 x) \log (\frac {-e^2-4 x}{x}) \log (\frac {e^{2-e^4}}{\log (\frac {-e^2-4 x}{x})})+(e^2+4 x) \log (\frac {-e^2-4 x}{x}) \log (\frac {e^{2-e^4}}{\log (\frac {-e^2-4 x}{x})}) \log (\log (\frac {e^{2-e^4}}{\log (\frac {-e^2-4 x}{x})}))}{(25 e^2+100 x) \log (\frac {-e^2-4 x}{x}) \log (\frac {e^{2-e^4}}{\log (\frac {-e^2-4 x}{x})})+(-10 e^2-40 x) \log (\frac {-e^2-4 x}{x}) \log (\frac {e^{2-e^4}}{\log (\frac {-e^2-4 x}{x})}) \log (\log (\frac {e^{2-e^4}}{\log (\frac {-e^2-4 x}{x})}))+(e^2+4 x) \log (\frac {-e^2-4 x}{x}) \log (\frac {e^{2-e^4}}{\log (\frac {-e^2-4 x}{x})}) \log ^2(\log (\frac {e^{2-e^4}}{\log (\frac {-e^2-4 x}{x})}))} \, dx\)

Optimal. Leaf size=31 \[ \frac {x}{-5+\log \left (\log \left (\frac {e^{2-e^4}}{\log \left (-4-\frac {e^2}{x}\right )}\right )\right )} \]

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Rubi [F]  time = 1.94, 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^2+\left (-5 e^2-20 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )}{\left (25 e^2+100 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (-10 e^2-40 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log ^2\left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-E^2 + (-5*E^2 - 20*x)*Log[(-E^2 - 4*x)/x]*Log[E^(2 - E^4)/Log[(-E^2 - 4*x)/x]] + (E^2 + 4*x)*Log[(-E^2 -
4*x)/x]*Log[E^(2 - E^4)/Log[(-E^2 - 4*x)/x]]*Log[Log[E^(2 - E^4)/Log[(-E^2 - 4*x)/x]]])/((25*E^2 + 100*x)*Log
[(-E^2 - 4*x)/x]*Log[E^(2 - E^4)/Log[(-E^2 - 4*x)/x]] + (-10*E^2 - 40*x)*Log[(-E^2 - 4*x)/x]*Log[E^(2 - E^4)/L
og[(-E^2 - 4*x)/x]]*Log[Log[E^(2 - E^4)/Log[(-E^2 - 4*x)/x]]] + (E^2 + 4*x)*Log[(-E^2 - 4*x)/x]*Log[E^(2 - E^4
)/Log[(-E^2 - 4*x)/x]]*Log[Log[E^(2 - E^4)/Log[(-E^2 - 4*x)/x]]]^2),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 28, normalized size = 0.90 \begin {gather*} \frac {x}{-5+\log \left (2-e^4+\log \left (\frac {1}{\log \left (-4-\frac {e^2}{x}\right )}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-E^2 + (-5*E^2 - 20*x)*Log[(-E^2 - 4*x)/x]*Log[E^(2 - E^4)/Log[(-E^2 - 4*x)/x]] + (E^2 + 4*x)*Log[(
-E^2 - 4*x)/x]*Log[E^(2 - E^4)/Log[(-E^2 - 4*x)/x]]*Log[Log[E^(2 - E^4)/Log[(-E^2 - 4*x)/x]]])/((25*E^2 + 100*
x)*Log[(-E^2 - 4*x)/x]*Log[E^(2 - E^4)/Log[(-E^2 - 4*x)/x]] + (-10*E^2 - 40*x)*Log[(-E^2 - 4*x)/x]*Log[E^(2 -
E^4)/Log[(-E^2 - 4*x)/x]]*Log[Log[E^(2 - E^4)/Log[(-E^2 - 4*x)/x]]] + (E^2 + 4*x)*Log[(-E^2 - 4*x)/x]*Log[E^(2
- E^4)/Log[(-E^2 - 4*x)/x]]*Log[Log[E^(2 - E^4)/Log[(-E^2 - 4*x)/x]]]^2),x]

[Out]

x/(-5 + Log[2 - E^4 + Log[Log[-4 - E^2/x]^(-1)]])

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fricas [A]  time = 0.60, size = 30, normalized size = 0.97 \begin {gather*} \frac {x}{\log \left (\log \left (\frac {e^{\left (-e^{4} + 2\right )}}{\log \left (-\frac {4 \, x + e^{2}}{x}\right )}\right )\right ) - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(2)+4*x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x))*log(log(1/exp(exp(4)-2)
/log((-exp(2)-4*x)/x)))+(-5*exp(2)-20*x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x))-exp(2)
)/((exp(2)+4*x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x))*log(log(1/exp(exp(4)-2)/log((-e
xp(2)-4*x)/x)))^2+(-10*exp(2)-40*x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x))*log(log(1/e
xp(exp(4)-2)/log((-exp(2)-4*x)/x)))+(25*exp(2)+100*x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*
x)/x))),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(2)+4*x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x))*log(log(1/exp(exp(4)-2)
/log((-exp(2)-4*x)/x)))+(-5*exp(2)-20*x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x))-exp(2)
)/((exp(2)+4*x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x))*log(log(1/exp(exp(4)-2)/log((-e
xp(2)-4*x)/x)))^2+(-10*exp(2)-40*x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x))*log(log(1/e
xp(exp(4)-2)/log((-exp(2)-4*x)/x)))+(25*exp(2)+100*x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*
x)/x))),x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 0.15, size = 0, normalized size = 0.00 \[\int \frac {\left ({\mathrm e}^{2}+4 x \right ) \ln \left (\frac {-{\mathrm e}^{2}-4 x}{x}\right ) \ln \left (\frac {{\mathrm e}^{2-{\mathrm e}^{4}}}{\ln \left (\frac {-{\mathrm e}^{2}-4 x}{x}\right )}\right ) \ln \left (\ln \left (\frac {{\mathrm e}^{2-{\mathrm e}^{4}}}{\ln \left (\frac {-{\mathrm e}^{2}-4 x}{x}\right )}\right )\right )+\left (-5 \,{\mathrm e}^{2}-20 x \right ) \ln \left (\frac {-{\mathrm e}^{2}-4 x}{x}\right ) \ln \left (\frac {{\mathrm e}^{2-{\mathrm e}^{4}}}{\ln \left (\frac {-{\mathrm e}^{2}-4 x}{x}\right )}\right )-{\mathrm e}^{2}}{\left ({\mathrm e}^{2}+4 x \right ) \ln \left (\frac {-{\mathrm e}^{2}-4 x}{x}\right ) \ln \left (\frac {{\mathrm e}^{2-{\mathrm e}^{4}}}{\ln \left (\frac {-{\mathrm e}^{2}-4 x}{x}\right )}\right ) \ln \left (\ln \left (\frac {{\mathrm e}^{2-{\mathrm e}^{4}}}{\ln \left (\frac {-{\mathrm e}^{2}-4 x}{x}\right )}\right )\right )^{2}+\left (-10 \,{\mathrm e}^{2}-40 x \right ) \ln \left (\frac {-{\mathrm e}^{2}-4 x}{x}\right ) \ln \left (\frac {{\mathrm e}^{2-{\mathrm e}^{4}}}{\ln \left (\frac {-{\mathrm e}^{2}-4 x}{x}\right )}\right ) \ln \left (\ln \left (\frac {{\mathrm e}^{2-{\mathrm e}^{4}}}{\ln \left (\frac {-{\mathrm e}^{2}-4 x}{x}\right )}\right )\right )+\left (25 \,{\mathrm e}^{2}+100 x \right ) \ln \left (\frac {-{\mathrm e}^{2}-4 x}{x}\right ) \ln \left (\frac {{\mathrm e}^{2-{\mathrm e}^{4}}}{\ln \left (\frac {-{\mathrm e}^{2}-4 x}{x}\right )}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(2)+4*x)*ln((-exp(2)-4*x)/x)*ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x))*ln(ln(1/exp(exp(4)-2)/ln((-exp(2
)-4*x)/x)))+(-5*exp(2)-20*x)*ln((-exp(2)-4*x)/x)*ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x))-exp(2))/((exp(2)+4*x)
*ln((-exp(2)-4*x)/x)*ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x))*ln(ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x)))^2+(-1
0*exp(2)-40*x)*ln((-exp(2)-4*x)/x)*ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x))*ln(ln(1/exp(exp(4)-2)/ln((-exp(2)-4
*x)/x)))+(25*exp(2)+100*x)*ln((-exp(2)-4*x)/x)*ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x))),x)

[Out]

int(((exp(2)+4*x)*ln((-exp(2)-4*x)/x)*ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x))*ln(ln(1/exp(exp(4)-2)/ln((-exp(2
)-4*x)/x)))+(-5*exp(2)-20*x)*ln((-exp(2)-4*x)/x)*ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x))-exp(2))/((exp(2)+4*x)
*ln((-exp(2)-4*x)/x)*ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x))*ln(ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x)))^2+(-1
0*exp(2)-40*x)*ln((-exp(2)-4*x)/x)*ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x))*ln(ln(1/exp(exp(4)-2)/ln((-exp(2)-4
*x)/x)))+(25*exp(2)+100*x)*ln((-exp(2)-4*x)/x)*ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x))),x)

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maxima [A]  time = 0.73, size = 30, normalized size = 0.97 \begin {gather*} \frac {x}{\log \left (-e^{4} - \log \left (-\log \relax (x) + \log \left (-4 \, x - e^{2}\right )\right ) + 2\right ) - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(2)+4*x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x))*log(log(1/exp(exp(4)-2)
/log((-exp(2)-4*x)/x)))+(-5*exp(2)-20*x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x))-exp(2)
)/((exp(2)+4*x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x))*log(log(1/exp(exp(4)-2)/log((-e
xp(2)-4*x)/x)))^2+(-10*exp(2)-40*x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x))*log(log(1/e
xp(exp(4)-2)/log((-exp(2)-4*x)/x)))+(25*exp(2)+100*x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*
x)/x))),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.38, size = 30, normalized size = 0.97 \begin {gather*} \frac {x}{\ln \left (\ln \left (\frac {{\mathrm {e}}^{-{\mathrm {e}}^4}\,{\mathrm {e}}^2}{\ln \left (-\frac {4\,x+{\mathrm {e}}^2}{x}\right )}\right )\right )-5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2) + log(-(4*x + exp(2))/x)*log(exp(2 - exp(4))/log(-(4*x + exp(2))/x))*(20*x + 5*exp(2)) - log(-(4*
x + exp(2))/x)*log(log(exp(2 - exp(4))/log(-(4*x + exp(2))/x)))*log(exp(2 - exp(4))/log(-(4*x + exp(2))/x))*(4
*x + exp(2)))/(log(-(4*x + exp(2))/x)*log(exp(2 - exp(4))/log(-(4*x + exp(2))/x))*(100*x + 25*exp(2)) - log(-(
4*x + exp(2))/x)*log(log(exp(2 - exp(4))/log(-(4*x + exp(2))/x)))*log(exp(2 - exp(4))/log(-(4*x + exp(2))/x))*
(40*x + 10*exp(2)) + log(-(4*x + exp(2))/x)*log(log(exp(2 - exp(4))/log(-(4*x + exp(2))/x)))^2*log(exp(2 - exp
(4))/log(-(4*x + exp(2))/x))*(4*x + exp(2))),x)

[Out]

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

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sympy [A]  time = 0.56, size = 26, normalized size = 0.84 \begin {gather*} \frac {x}{\log {\left (\log {\left (\frac {1}{e^{-2 + e^{4}} \log {\left (\frac {- 4 x - e^{2}}{x} \right )}} \right )} \right )} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(2)+4*x)*ln((-exp(2)-4*x)/x)*ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x))*ln(ln(1/exp(exp(4)-2)/ln((
-exp(2)-4*x)/x)))+(-5*exp(2)-20*x)*ln((-exp(2)-4*x)/x)*ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x))-exp(2))/((exp(2
)+4*x)*ln((-exp(2)-4*x)/x)*ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x))*ln(ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x)))
**2+(-10*exp(2)-40*x)*ln((-exp(2)-4*x)/x)*ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x))*ln(ln(1/exp(exp(4)-2)/ln((-e
xp(2)-4*x)/x)))+(25*exp(2)+100*x)*ln((-exp(2)-4*x)/x)*ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x))),x)

[Out]

x/(log(log(exp(2 - exp(4))/log((-4*x - exp(2))/x))) - 5)

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