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

Optimal. Leaf size=23 \[ \left (\frac {x}{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}\right )^{\frac {1}{-1+e^2}} \]

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Rubi [F]  time = 2.58, 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 (\frac {x}{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}\right )^{\frac {1}{-1+e^2}} \left (-1+\log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )\right )}{\left (-x+e^2 x\right ) \log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.35, size = 23, normalized size = 1.00 \begin {gather*} \left (\frac {x}{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}\right )^{\frac {1}{-1+e^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.49, size = 18, normalized size = 0.78 \begin {gather*} \left (\frac {x}{\log \left (\frac {3}{4} \, \log \left (\frac {1}{2} \, x\right )\right )}\right )^{\left (\frac {1}{e^{2} - 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(1/2*x)*log(3/4*log(1/2*x))-1)*exp(log(x/log(3/4*log(1/2*x)))/(exp(2)-1))/(exp(2)*x-x)/log(1/2*x
)/log(3/4*log(1/2*x)),x, algorithm="fricas")

[Out]

(x/log(3/4*log(1/2*x)))^(1/(e^2 - 1))

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giac [A]  time = 0.20, size = 18, normalized size = 0.78 \begin {gather*} \left (\frac {x}{\log \left (\frac {3}{4} \, \log \left (\frac {1}{2} \, x\right )\right )}\right )^{\left (\frac {1}{e^{2} - 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(1/2*x)*log(3/4*log(1/2*x))-1)*exp(log(x/log(3/4*log(1/2*x)))/(exp(2)-1))/(exp(2)*x-x)/log(1/2*x
)/log(3/4*log(1/2*x)),x, algorithm="giac")

[Out]

(x/log(3/4*log(1/2*x)))^(1/(e^2 - 1))

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[\int \frac {\left (\ln \left (\frac {x}{2}\right ) \ln \left (\frac {3 \ln \left (\frac {x}{2}\right )}{4}\right )-1\right ) {\mathrm e}^{\frac {\ln \left (\frac {x}{\ln \left (\frac {3 \ln \left (\frac {x}{2}\right )}{4}\right )}\right )}{{\mathrm e}^{2}-1}}}{\left ({\mathrm e}^{2} x -x \right ) \ln \left (\frac {x}{2}\right ) \ln \left (\frac {3 \ln \left (\frac {x}{2}\right )}{4}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(1/2*x)*ln(3/4*ln(1/2*x))-1)*exp(ln(x/ln(3/4*ln(1/2*x)))/(exp(2)-1))/(exp(2)*x-x)/ln(1/2*x)/ln(3/4*ln(1
/2*x)),x)

[Out]

int((ln(1/2*x)*ln(3/4*ln(1/2*x))-1)*exp(ln(x/ln(3/4*ln(1/2*x)))/(exp(2)-1))/(exp(2)*x-x)/ln(1/2*x)/ln(3/4*ln(1
/2*x)),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (\log \left (\frac {1}{2} \, x\right ) \log \left (\frac {3}{4} \, \log \left (\frac {1}{2} \, x\right )\right ) - 1\right )} \left (\frac {x}{\log \left (\frac {3}{4} \, \log \left (\frac {1}{2} \, x\right )\right )}\right )^{\left (\frac {1}{e^{2} - 1}\right )}}{{\left (x e^{2} - x\right )} \log \left (\frac {1}{2} \, x\right ) \log \left (\frac {3}{4} \, \log \left (\frac {1}{2} \, x\right )\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(1/2*x)*log(3/4*log(1/2*x))-1)*exp(log(x/log(3/4*log(1/2*x)))/(exp(2)-1))/(exp(2)*x-x)/log(1/2*x
)/log(3/4*log(1/2*x)),x, algorithm="maxima")

[Out]

integrate((log(1/2*x)*log(3/4*log(1/2*x)) - 1)*(x/log(3/4*log(1/2*x)))^(1/(e^2 - 1))/((x*e^2 - x)*log(1/2*x)*l
og(3/4*log(1/2*x))), x)

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mupad [B]  time = 3.95, size = 18, normalized size = 0.78 \begin {gather*} {\left (\frac {x}{\ln \left (\frac {3\,\ln \left (\frac {x}{2}\right )}{4}\right )}\right )}^{\frac {1}{{\mathrm {e}}^2-1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((log((3*log(x/2))/4)*log(x/2) - 1)*(x/log((3*log(x/2))/4))^(1/(exp(2) - 1)))/(log((3*log(x/2))/4)*log(x/
2)*(x - x*exp(2))),x)

[Out]

(x/log((3*log(x/2))/4))^(1/(exp(2) - 1))

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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((ln(1/2*x)*ln(3/4*ln(1/2*x))-1)*exp(ln(x/ln(3/4*ln(1/2*x)))/(exp(2)-1))/(exp(2)*x-x)/ln(1/2*x)/ln(3/
4*ln(1/2*x)),x)

[Out]

Timed out

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