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

Optimal. Leaf size=21 \[ \log \left (\log \left (x+\frac {1}{4} (2-x) \log \left (\frac {5}{e^2}\right ) \log (x)\right )\right ) \]

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Rubi [F]  time = 1.37, 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 {-4 x+(-2+x) \log \left (\frac {5}{e^2}\right )+x \log \left (\frac {5}{e^2}\right ) \log (x)}{\left (-4 x^2+\left (-2 x+x^2\right ) \log \left (\frac {5}{e^2}\right ) \log (x)\right ) \log \left (\frac {1}{4} \left (4 x+(2-x) \log \left (\frac {5}{e^2}\right ) \log (x)\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(6 - Log[5])*Defer[Int][1/((4*x - (-2 + x)*(-2 + Log[5])*Log[x])*Log[x - ((-2 + x)*(-2 + Log[5])*Log[x])/4]),
x] - 2*(2 - Log[5])*Defer[Int][1/(x*(4*x - (-2 + x)*(-2 + Log[5])*Log[x])*Log[x - ((-2 + x)*(-2 + Log[5])*Log[
x])/4]), x] + (2 - Log[5])*Defer[Int][Log[x]/((4*x - (-2 + x)*(-2 + Log[5])*Log[x])*Log[x - ((-2 + x)*(-2 + Lo
g[5])*Log[x])/4]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {4 \left (1+\frac {1}{4} (2-\log (5))\right )}{\left (4 x-4 \left (1-\frac {\log (5)}{2}\right ) \log (x)+2 x \left (1-\frac {\log (5)}{2}\right ) \log (x)\right ) \log \left (x-\frac {1}{4} (-2+x) (-2+\log (5)) \log (x)\right )}+\frac {2 (-2+\log (5))}{x \left (4 x-4 \left (1-\frac {\log (5)}{2}\right ) \log (x)+2 x \left (1-\frac {\log (5)}{2}\right ) \log (x)\right ) \log \left (x-\frac {1}{4} (-2+x) (-2+\log (5)) \log (x)\right )}+\frac {(2-\log (5)) \log (x)}{\left (4 x-4 \left (1-\frac {\log (5)}{2}\right ) \log (x)+2 x \left (1-\frac {\log (5)}{2}\right ) \log (x)\right ) \log \left (x-\frac {1}{4} (-2+x) (-2+\log (5)) \log (x)\right )}\right ) \, dx\\ &=(2-\log (5)) \int \frac {\log (x)}{\left (4 x-4 \left (1-\frac {\log (5)}{2}\right ) \log (x)+2 x \left (1-\frac {\log (5)}{2}\right ) \log (x)\right ) \log \left (x-\frac {1}{4} (-2+x) (-2+\log (5)) \log (x)\right )} \, dx+(6-\log (5)) \int \frac {1}{\left (4 x-4 \left (1-\frac {\log (5)}{2}\right ) \log (x)+2 x \left (1-\frac {\log (5)}{2}\right ) \log (x)\right ) \log \left (x-\frac {1}{4} (-2+x) (-2+\log (5)) \log (x)\right )} \, dx+(2 (-2+\log (5))) \int \frac {1}{x \left (4 x-4 \left (1-\frac {\log (5)}{2}\right ) \log (x)+2 x \left (1-\frac {\log (5)}{2}\right ) \log (x)\right ) \log \left (x-\frac {1}{4} (-2+x) (-2+\log (5)) \log (x)\right )} \, dx\\ &=(2-\log (5)) \int \frac {\log (x)}{(4 x-(-2+x) (-2+\log (5)) \log (x)) \log \left (x-\frac {1}{4} (-2+x) (-2+\log (5)) \log (x)\right )} \, dx+(6-\log (5)) \int \frac {1}{(4 x-(-2+x) (-2+\log (5)) \log (x)) \log \left (x-\frac {1}{4} (-2+x) (-2+\log (5)) \log (x)\right )} \, dx+(2 (-2+\log (5))) \int \frac {1}{x (4 x-(-2+x) (-2+\log (5)) \log (x)) \log \left (x-\frac {1}{4} (-2+x) (-2+\log (5)) \log (x)\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.47, size = 17, normalized size = 0.81 \begin {gather*} \log \left (\log \left (x-\frac {1}{4} (-2+x) (-2+\log (5)) \log (x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.67, size = 19, normalized size = 0.90 \begin {gather*} \log \left (\log \left (-\frac {1}{4} \, {\left ({\left (x - 2\right )} \log \relax (5) - 2 \, x + 4\right )} \log \relax (x) + x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(5/exp(2))*log(x)+(x-2)*log(5/exp(2))-4*x)/((x^2-2*x)*log(5/exp(2))*log(x)-4*x^2)/log(1/4*(2-x
)*log(5/exp(2))*log(x)+x),x, algorithm="fricas")

[Out]

log(log(-1/4*((x - 2)*log(5) - 2*x + 4)*log(x) + x))

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giac [B]  time = 0.22, size = 35, normalized size = 1.67 \begin {gather*} \log \left (2 \, \log \relax (2) - \log \left (-x \log \relax (5) \log \relax (x) + 2 \, x \log \relax (x) + 2 \, \log \relax (5) \log \relax (x) + 4 \, x - 4 \, \log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(5/exp(2))*log(x)+(x-2)*log(5/exp(2))-4*x)/((x^2-2*x)*log(5/exp(2))*log(x)-4*x^2)/log(1/4*(2-x
)*log(5/exp(2))*log(x)+x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.26, size = 18, normalized size = 0.86

method	result	size
risch	\(\ln \left (\ln \left (\frac {\left (2-x \right ) \left (\ln \relax (5)-2\right ) \ln \relax (x )}{4}+x \right )\right )\)	\(18\)
norman	\(\ln \left (\ln \left (\frac {\left (2-x \right ) \ln \left (5 \,{\mathrm e}^{-2}\right ) \ln \relax (x )}{4}+x \right )\right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*ln(5/exp(2))*ln(x)+(x-2)*ln(5/exp(2))-4*x)/((x^2-2*x)*ln(5/exp(2))*ln(x)-4*x^2)/ln(1/4*(2-x)*ln(5/exp(2
))*ln(x)+x),x,method=_RETURNVERBOSE)

[Out]

ln(ln(1/4*(2-x)*(ln(5)-2)*ln(x)+x))

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maxima [A]  time = 0.52, size = 27, normalized size = 1.29 \begin {gather*} \log \left (-2 \, \log \relax (2) + \log \left (-{\left (x {\left (\log \relax (5) - 2\right )} - 2 \, \log \relax (5) + 4\right )} \log \relax (x) + 4 \, x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(5/exp(2))*log(x)+(x-2)*log(5/exp(2))-4*x)/((x^2-2*x)*log(5/exp(2))*log(x)-4*x^2)/log(1/4*(2-x
)*log(5/exp(2))*log(x)+x),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} -\int \frac {\ln \left (5\,{\mathrm {e}}^{-2}\right )\,\left (x-2\right )-4\,x+x\,\ln \left (5\,{\mathrm {e}}^{-2}\right )\,\ln \relax (x)}{\ln \left (x-\frac {\ln \left (5\,{\mathrm {e}}^{-2}\right )\,\ln \relax (x)\,\left (x-2\right )}{4}\right )\,\left (4\,x^2+\ln \left (5\,{\mathrm {e}}^{-2}\right )\,\ln \relax (x)\,\left (2\,x-x^2\right )\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.99, size = 20, normalized size = 0.95 \begin {gather*} \log {\left (\log {\left (x + \left (\frac {1}{2} - \frac {x}{4}\right ) \log {\relax (x )} \log {\left (\frac {5}{e^{2}} \right )} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*ln(5/exp(2))*ln(x)+(x-2)*ln(5/exp(2))-4*x)/((x**2-2*x)*ln(5/exp(2))*ln(x)-4*x**2)/ln(1/4*(2-x)*ln
(5/exp(2))*ln(x)+x),x)

[Out]

log(log(x + (1/2 - x/4)*log(x)*log(5*exp(-2))))

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