Optimal. Leaf size=36 \[ \log \left (-3+e^3-x+\frac {x-\frac {5+\log \left (-x+\frac {5}{\log (5)}\right )}{4 x}}{\log (x)}\right ) \]
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Rubi [A] time = 6.35, antiderivative size = 48, normalized size of antiderivative = 1.33, number of steps used = 7, number of rules used = 5, integrand size = 200, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {6688, 6742, 2365, 43, 6684} \begin {gather*} \log \left (-4 x^2+4 x^2 \log (x)+4 \left (3-e^3\right ) x \log (x)+\log \left (\frac {5}{\log (5)}-x\right )+5\right )-\log (x)-\log (\log (x)) \end {gather*}
Antiderivative was successfully verified.
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Rule 43
Rule 2365
Rule 6684
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4 x^2 (-5+x \log (5)) \log ^2(x)-(-5+x \log (5)) \left (-5+4 x^2-\log \left (-x+\frac {5}{\log (5)}\right )\right )-\log (x) \left (25+20 x^2-4 x \log (5)-4 x^3 \log (5)+(5-x \log (5)) \log \left (-x+\frac {5}{\log (5)}\right )\right )}{x (5-x \log (5)) \log (x) \left (5-4 x^2+4 x \left (3-e^3+x\right ) \log (x)+\log \left (-x+\frac {5}{\log (5)}\right )\right )} \, dx\\ &=\int \left (\frac {-1-\log (x)}{x \log (x)}+\frac {60 \left (1-\frac {e^3}{3}-\frac {\log (5)}{60}\right )+4 x^2 \log (5)-20 x \left (1-\frac {1}{5} \left (-3+e^3\right ) \log (5)\right )+60 \left (1-\frac {e^3}{3}\right ) \log (x)-8 x^2 \log (5) \log (x)+40 x \left (1+\frac {1}{10} \left (-3+e^3\right ) \log (5)\right ) \log (x)}{(5-x \log (5)) \left (5-4 x^2+12 \left (1-\frac {e^3}{3}\right ) x \log (x)+4 x^2 \log (x)+\log \left (-x+\frac {5}{\log (5)}\right )\right )}\right ) \, dx\\ &=\int \frac {-1-\log (x)}{x \log (x)} \, dx+\int \frac {60 \left (1-\frac {e^3}{3}-\frac {\log (5)}{60}\right )+4 x^2 \log (5)-20 x \left (1-\frac {1}{5} \left (-3+e^3\right ) \log (5)\right )+60 \left (1-\frac {e^3}{3}\right ) \log (x)-8 x^2 \log (5) \log (x)+40 x \left (1+\frac {1}{10} \left (-3+e^3\right ) \log (5)\right ) \log (x)}{(5-x \log (5)) \left (5-4 x^2+12 \left (1-\frac {e^3}{3}\right ) x \log (x)+4 x^2 \log (x)+\log \left (-x+\frac {5}{\log (5)}\right )\right )} \, dx\\ &=\log \left (5-4 x^2+4 \left (3-e^3\right ) x \log (x)+4 x^2 \log (x)+\log \left (-x+\frac {5}{\log (5)}\right )\right )+\operatorname {Subst}\left (\int \frac {-1-x}{x} \, dx,x,\log (x)\right )\\ &=\log \left (5-4 x^2+4 \left (3-e^3\right ) x \log (x)+4 x^2 \log (x)+\log \left (-x+\frac {5}{\log (5)}\right )\right )+\operatorname {Subst}\left (\int \left (-1-\frac {1}{x}\right ) \, dx,x,\log (x)\right )\\ &=-\log (x)-\log (\log (x))+\log \left (5-4 x^2+4 \left (3-e^3\right ) x \log (x)+4 x^2 \log (x)+\log \left (-x+\frac {5}{\log (5)}\right )\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.11, size = 49, normalized size = 1.36 \begin {gather*} -\log (x)-\log (\log (x))+\log \left (5-4 x^2+12 x \log (x)-4 e^3 x \log (x)+4 x^2 \log (x)+\log \left (-x+\frac {5}{\log (5)}\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.66, size = 68, normalized size = 1.89 \begin {gather*} \log \left (x - e^{3} + 3\right ) + \log \left (\frac {4 \, x^{2} - 4 \, {\left (x^{2} - x e^{3} + 3 \, x\right )} \log \relax (x) - \log \left (-\frac {x \log \relax (5) - 5}{\log \relax (5)}\right ) - 5}{x^{2} - x e^{3} + 3 \, x}\right ) - \log \left (\log \relax (x)\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.46, size = 50, normalized size = 1.39 \begin {gather*} \log \left (4 \, x^{2} \log \relax (x) - 4 \, x e^{3} \log \relax (x) - 4 \, x^{2} + 12 \, x \log \relax (x) + \log \left (-x \log \relax (5) + 5\right ) - \log \left (\log \relax (5)\right ) + 5\right ) - \log \relax (x) - \log \left (\log \relax (x)\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 51, normalized size = 1.42
method | result | size |
risch | \(-\ln \relax (x )-\ln \left (\ln \relax (x )\right )+\ln \left (-4 x \,{\mathrm e}^{3} \ln \relax (x )+4 x^{2} \ln \relax (x )-4 x^{2}+12 x \ln \relax (x )+\ln \left (\frac {-x \ln \relax (5)+5}{\ln \relax (5)}\right )+5\right )\) | \(51\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.53, size = 46, normalized size = 1.28 \begin {gather*} \log \left (-4 \, x^{2} + 4 \, {\left (x^{2} - x {\left (e^{3} - 3\right )}\right )} \log \relax (x) + \log \left (-x \log \relax (5) + 5\right ) - \log \left (\log \relax (5)\right ) + 5\right ) - \log \relax (x) - \log \left (\log \relax (x)\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.84, size = 50, normalized size = 1.39 \begin {gather*} \ln \left (\ln \left (5-x\,\ln \relax (5)\right )-\ln \left (\ln \relax (5)\right )+4\,x^2\,\ln \relax (x)+12\,x\,\ln \relax (x)-4\,x^2-4\,x\,{\mathrm {e}}^3\,\ln \relax (x)+5\right )-\ln \left (\ln \relax (x)\right )-\ln \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.00, size = 53, normalized size = 1.47 \begin {gather*} - \log {\relax (x )} + \log {\left (4 x^{2} \log {\relax (x )} - 4 x^{2} - 4 x e^{3} \log {\relax (x )} + 12 x \log {\relax (x )} + \log {\left (\frac {- x \log {\relax (5 )} + 5}{\log {\relax (5 )}} \right )} + 5 \right )} - \log {\left (\log {\relax (x )} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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