3.2.68 \(\int \frac {(-2+(4+2 x) \log (4 x)+2 x \log (4 x) \log (\frac {e^{-x} \log (4 x)}{x^2})) \log (-2+x-\log (\log (\frac {e^{-x} \log (4 x)}{x^2})))}{(2 x-x^2) \log (4 x) \log (\frac {e^{-x} \log (4 x)}{x^2})+x \log (4 x) \log (\frac {e^{-x} \log (4 x)}{x^2}) \log (\log (\frac {e^{-x} \log (4 x)}{x^2}))} \, dx\)

Optimal. Leaf size=27 \[ 5-\log ^2\left (-2+x-\log \left (\log \left (\frac {e^{-x} \log (4 x)}{x^2}\right )\right )\right ) \]

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Rubi [F]  time = 1.97, 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 {\left (-2+(4+2 x) \log (4 x)+2 x \log (4 x) \log \left (\frac {e^{-x} \log (4 x)}{x^2}\right )\right ) \log \left (-2+x-\log \left (\log \left (\frac {e^{-x} \log (4 x)}{x^2}\right )\right )\right )}{\left (2 x-x^2\right ) \log (4 x) \log \left (\frac {e^{-x} \log (4 x)}{x^2}\right )+x \log (4 x) \log \left (\frac {e^{-x} \log (4 x)}{x^2}\right ) \log \left (\log \left (\frac {e^{-x} \log (4 x)}{x^2}\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 25, normalized size = 0.93 \begin {gather*} -\log ^2\left (-2+x-\log \left (\log \left (\frac {e^{-x} \log (4 x)}{x^2}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.89, size = 24, normalized size = 0.89 \begin {gather*} -\log \left (x - \log \left (\log \left (\frac {e^{\left (-x\right )} \log \left (4 \, x\right )}{x^{2}}\right )\right ) - 2\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(4*x)*log(log(4*x)/exp(x)/x^2)+(2*x+4)*log(4*x)-2)*log(-log(log(log(4*x)/exp(x)/x^2))+x-2)/(
x*log(4*x)*log(log(4*x)/exp(x)/x^2)*log(log(log(4*x)/exp(x)/x^2))+(-x^2+2*x)*log(4*x)*log(log(4*x)/exp(x)/x^2)
),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left (x \log \left (4 \, x\right ) \log \left (\frac {e^{\left (-x\right )} \log \left (4 \, x\right )}{x^{2}}\right ) + {\left (x + 2\right )} \log \left (4 \, x\right ) - 1\right )} \log \left (x - \log \left (\log \left (\frac {e^{\left (-x\right )} \log \left (4 \, x\right )}{x^{2}}\right )\right ) - 2\right )}{x \log \left (4 \, x\right ) \log \left (\frac {e^{\left (-x\right )} \log \left (4 \, x\right )}{x^{2}}\right ) \log \left (\log \left (\frac {e^{\left (-x\right )} \log \left (4 \, x\right )}{x^{2}}\right )\right ) - {\left (x^{2} - 2 \, x\right )} \log \left (4 \, x\right ) \log \left (\frac {e^{\left (-x\right )} \log \left (4 \, x\right )}{x^{2}}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(4*x)*log(log(4*x)/exp(x)/x^2)+(2*x+4)*log(4*x)-2)*log(-log(log(log(4*x)/exp(x)/x^2))+x-2)/(
x*log(4*x)*log(log(4*x)/exp(x)/x^2)*log(log(log(4*x)/exp(x)/x^2))+(-x^2+2*x)*log(4*x)*log(log(4*x)/exp(x)/x^2)
),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (2 x \ln \left (4 x \right ) \ln \left (\frac {\ln \left (4 x \right ) {\mathrm e}^{-x}}{x^{2}}\right )+\left (2 x +4\right ) \ln \left (4 x \right )-2\right ) \ln \left (-\ln \left (\ln \left (\frac {\ln \left (4 x \right ) {\mathrm e}^{-x}}{x^{2}}\right )\right )+x -2\right )}{x \ln \left (4 x \right ) \ln \left (\frac {\ln \left (4 x \right ) {\mathrm e}^{-x}}{x^{2}}\right ) \ln \left (\ln \left (\frac {\ln \left (4 x \right ) {\mathrm e}^{-x}}{x^{2}}\right )\right )+\left (-x^{2}+2 x \right ) \ln \left (4 x \right ) \ln \left (\frac {\ln \left (4 x \right ) {\mathrm e}^{-x}}{x^{2}}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*ln(4*x)*ln(ln(4*x)/exp(x)/x^2)+(2*x+4)*ln(4*x)-2)*ln(-ln(ln(ln(4*x)/exp(x)/x^2))+x-2)/(x*ln(4*x)*ln(l
n(4*x)/exp(x)/x^2)*ln(ln(ln(4*x)/exp(x)/x^2))+(-x^2+2*x)*ln(4*x)*ln(ln(4*x)/exp(x)/x^2)),x)

[Out]

int((2*x*ln(4*x)*ln(ln(4*x)/exp(x)/x^2)+(2*x+4)*ln(4*x)-2)*ln(-ln(ln(ln(4*x)/exp(x)/x^2))+x-2)/(x*ln(4*x)*ln(l
n(4*x)/exp(x)/x^2)*ln(ln(ln(4*x)/exp(x)/x^2))+(-x^2+2*x)*ln(4*x)*ln(ln(4*x)/exp(x)/x^2)),x)

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maxima [B]  time = 1.08, size = 71, normalized size = 2.63 \begin {gather*} -2 \, \log \left (x - \log \left (\log \left (\frac {e^{\left (-x\right )} \log \left (4 \, x\right )}{x^{2}}\right )\right ) - 2\right ) \log \left (-x + \log \left (-x - 2 \, \log \relax (x) + \log \left (2 \, \log \relax (2) + \log \relax (x)\right )\right ) + 2\right ) + \log \left (-x + \log \left (-x - 2 \, \log \relax (x) + \log \left (2 \, \log \relax (2) + \log \relax (x)\right )\right ) + 2\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(4*x)*log(log(4*x)/exp(x)/x^2)+(2*x+4)*log(4*x)-2)*log(-log(log(log(4*x)/exp(x)/x^2))+x-2)/(
x*log(4*x)*log(log(4*x)/exp(x)/x^2)*log(log(log(4*x)/exp(x)/x^2))+(-x^2+2*x)*log(4*x)*log(log(4*x)/exp(x)/x^2)
),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.49, size = 24, normalized size = 0.89 \begin {gather*} -{\ln \left (x-\ln \left (\ln \left (\frac {\ln \left (4\,x\right )\,{\mathrm {e}}^{-x}}{x^2}\right )\right )-2\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [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((2*x*ln(4*x)*ln(ln(4*x)/exp(x)/x**2)+(2*x+4)*ln(4*x)-2)*ln(-ln(ln(ln(4*x)/exp(x)/x**2))+x-2)/(x*ln(4
*x)*ln(ln(4*x)/exp(x)/x**2)*ln(ln(ln(4*x)/exp(x)/x**2))+(-x**2+2*x)*ln(4*x)*ln(ln(4*x)/exp(x)/x**2)),x)

[Out]

Timed out

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