Optimal. Leaf size=27 \[ 16 \log \left (2 x-\log \left (\log \left (\frac {5}{4 \left (4+2 x+x^2+\log (2)\right )}\right )\right )\right ) \]
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Rubi [A]
time = 0.16, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 3, integrand size = 132, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {6820, 12,
6816} \begin {gather*} 16 \log \left (2 x-\log \left (\log \left (\frac {5}{4 \left (x^2+2 x+4+\log (2)\right )}\right )\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 6816
Rule 6820
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {32 \left (1+x+\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{4 \left (4+2 x+x^2+\log (2)\right )}\right )\right )}{\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{4 \left (4+2 x+x^2+\log (2)\right )}\right ) \left (2 x-\log \left (\log \left (\frac {5}{4 \left (4+2 x+x^2+\log (2)\right )}\right )\right )\right )} \, dx\\ &=32 \int \frac {1+x+\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{4 \left (4+2 x+x^2+\log (2)\right )}\right )}{\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{4 \left (4+2 x+x^2+\log (2)\right )}\right ) \left (2 x-\log \left (\log \left (\frac {5}{4 \left (4+2 x+x^2+\log (2)\right )}\right )\right )\right )} \, dx\\ &=16 \log \left (2 x-\log \left (\log \left (\frac {5}{4 \left (4+2 x+x^2+\log (2)\right )}\right )\right )\right )\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.05, size = 27, normalized size = 1.00 \begin {gather*} 16 \log \left (2 x-\log \left (\log \left (\frac {5}{4 \left (4+2 x+x^2+\log (2)\right )}\right )\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 3.61, size = 0, normalized size = 0.00 \[\int \frac {\left (-32 \ln \left (2\right )-32 x^{2}-64 x -128\right ) \ln \left (\frac {5}{4 \ln \left (2\right )+4 x^{2}+8 x +16}\right )-32 x -32}{\left (\ln \left (2\right )+x^{2}+2 x +4\right ) \ln \left (\frac {5}{4 \ln \left (2\right )+4 x^{2}+8 x +16}\right ) \ln \left (\ln \left (\frac {5}{4 \ln \left (2\right )+4 x^{2}+8 x +16}\right )\right )+\left (-2 x \ln \left (2\right )-2 x^{3}-4 x^{2}-8 x \right ) \ln \left (\frac {5}{4 \ln \left (2\right )+4 x^{2}+8 x +16}\right )}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.54, size = 28, normalized size = 1.04 \begin {gather*} 16 \, \log \left (-2 \, x + \log \left (\log \left (5\right ) - 2 \, \log \left (2\right ) - \log \left (x^{2} + 2 \, x + \log \left (2\right ) + 4\right )\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.40, size = 23, normalized size = 0.85 \begin {gather*} 16 \, \log \left (-2 \, x + \log \left (\log \left (\frac {5}{4 \, {\left (x^{2} + 2 \, x + \log \left (2\right ) + 4\right )}}\right )\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.43, size = 26, normalized size = 0.96 \begin {gather*} 16 \log {\left (- 2 x + \log {\left (\log {\left (\frac {5}{4 x^{2} + 8 x + 4 \log {\left (2 \right )} + 16} \right )} \right )} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.58, size = 30, normalized size = 1.11 \begin {gather*} 16 \, \log \left (2 \, x - \log \left (\log \left (5\right ) - 2 \, \log \left (2\right ) - \log \left (x^{2} + 2 \, x + \log \left (2\right ) + 4\right )\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \text {Hanged} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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