Optimal. Leaf size=27 \[ x \left (\frac {x}{12}-\log (2)-\log \left (x+x \log \left (\frac {4+x}{x}\right )\right )\right ) \]
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Rubi [F]
time = 0.55, 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 {-2 x+x^2+(-24-6 x) \log (2)+\left (-24-2 x+x^2+(-24-6 x) \log (2)\right ) \log \left (\frac {4+x}{x}\right )+\left (-24-6 x+(-24-6 x) \log \left (\frac {4+x}{x}\right )\right ) \log \left (x+x \log \left (\frac {4+x}{x}\right )\right )}{24+6 x+(24+6 x) \log \left (\frac {4+x}{x}\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2 x+x^2+(-24-6 x) \log (2)+\left (-24-2 x+x^2+(-24-6 x) \log (2)\right ) \log \left (\frac {4+x}{x}\right )+\left (-24-6 x+(-24-6 x) \log \left (\frac {4+x}{x}\right )\right ) \log \left (x+x \log \left (\frac {4+x}{x}\right )\right )}{6 (4+x) \left (1+\log \left (\frac {4+x}{x}\right )\right )} \, dx\\ &=\frac {1}{6} \int \frac {-2 x+x^2+(-24-6 x) \log (2)+\left (-24-2 x+x^2+(-24-6 x) \log (2)\right ) \log \left (\frac {4+x}{x}\right )+\left (-24-6 x+(-24-6 x) \log \left (\frac {4+x}{x}\right )\right ) \log \left (x+x \log \left (\frac {4+x}{x}\right )\right )}{(4+x) \left (1+\log \left (\frac {4+x}{x}\right )\right )} \, dx\\ &=\frac {1}{6} \int \left (\frac {x^2-24 \log (2)-2 x (1+\log (8))+x^2 \log \left (\frac {4+x}{x}\right )-24 (1+\log (2)) \log \left (\frac {4+x}{x}\right )-2 x (1+\log (8)) \log \left (\frac {4+x}{x}\right )}{(4+x) \left (1+\log \left (\frac {4+x}{x}\right )\right )}-6 \log \left (x+x \log \left (\frac {4+x}{x}\right )\right )\right ) \, dx\\ &=\frac {1}{6} \int \frac {x^2-24 \log (2)-2 x (1+\log (8))+x^2 \log \left (\frac {4+x}{x}\right )-24 (1+\log (2)) \log \left (\frac {4+x}{x}\right )-2 x (1+\log (8)) \log \left (\frac {4+x}{x}\right )}{(4+x) \left (1+\log \left (\frac {4+x}{x}\right )\right )} \, dx-\int \log \left (x+x \log \left (\frac {4+x}{x}\right )\right ) \, dx\\ &=\frac {1}{6} \int \frac {x^2-24 \log (2)-2 x (1+\log (8))+\left (x^2-24 (1+\log (2))-2 x (1+\log (8))\right ) \log \left (\frac {4+x}{x}\right )}{(4+x) \left (1+\log \left (\frac {4+x}{x}\right )\right )} \, dx-\int \log \left (x+x \log \left (\frac {4+x}{x}\right )\right ) \, dx\\ &=\frac {1}{6} \int \left (\frac {x^2-24 (1+\log (2))-2 x (1+\log (8))}{4+x}+\frac {24}{(4+x) \left (1+\log \left (\frac {4+x}{x}\right )\right )}\right ) \, dx-\int \log \left (x+x \log \left (\frac {4+x}{x}\right )\right ) \, dx\\ &=\frac {1}{6} \int \frac {x^2-24 (1+\log (2))-2 x (1+\log (8))}{4+x} \, dx+4 \int \frac {1}{(4+x) \left (1+\log \left (\frac {4+x}{x}\right )\right )} \, dx-\int \log \left (x+x \log \left (\frac {4+x}{x}\right )\right ) \, dx\\ &=\frac {1}{6} \int (x-2 (3+\log (8))) \, dx+4 \int \frac {1}{(4+x) \left (1+\log \left (1+\frac {4}{x}\right )\right )} \, dx-\int \log \left (x+x \log \left (\frac {4+x}{x}\right )\right ) \, dx\\ &=\frac {x^2}{12}-\frac {1}{3} x (3+\log (8))+4 \int \frac {1}{(4+x) \left (1+\log \left (1+\frac {4}{x}\right )\right )} \, dx-\int \log \left (x+x \log \left (\frac {4+x}{x}\right )\right ) \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.04, size = 33, normalized size = 1.22 \begin {gather*} \frac {1}{6} \left (\frac {x^2}{2}-6 x \log (2)-6 x \log \left (x \left (1+\log \left (\frac {4+x}{x}\right )\right )\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 7.50, size = 28, normalized size = 1.04
method | result | size |
norman | \(\frac {x^{2}}{12}-x \ln \left (2\right )-\ln \left (\ln \left (\frac {4+x}{x}\right ) x +x \right ) x\) | \(28\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.49, size = 30, normalized size = 1.11 \begin {gather*} \frac {1}{12} \, x^{2} - x \log \left (2\right ) - x \log \left (x\right ) - x \log \left (\log \left (x + 4\right ) - \log \left (x\right ) + 1\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.37, size = 27, normalized size = 1.00 \begin {gather*} \frac {1}{12} \, x^{2} - x \log \left (2\right ) - x \log \left (x \log \left (\frac {x + 4}{x}\right ) + x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 48 vs.
\(2 (19) = 38\).
time = 0.36, size = 48, normalized size = 1.78 \begin {gather*} \frac {x^{2}}{12} - x \log {\left (2 \right )} + \left (- x - \frac {2}{3}\right ) \log {\left (x \log {\left (\frac {x + 4}{x} \right )} + x \right )} + \frac {2 \log {\left (x \right )}}{3} + \frac {2 \log {\left (\log {\left (\frac {x + 4}{x} \right )} + 1 \right )}}{3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.47, size = 30, normalized size = 1.11 \begin {gather*} \frac {1}{12} \, x^{2} - x \log \left (2\right ) - x \log \left (x\right ) - x \log \left (\log \left (\frac {x + 4}{x}\right ) + 1\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 2.41, size = 27, normalized size = 1.00 \begin {gather*} \frac {x^2}{12}-x\,\ln \left (x+x\,\ln \left (\frac {x+4}{x}\right )\right )-x\,\ln \left (2\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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