Optimal. Leaf size=22 \[ 3+\frac {\log (x \log (2))}{6 e^{20} x^2 (2+x)} \]
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Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(58\) vs. \(2(22)=44\).
time = 0.26, antiderivative size = 58, normalized size of antiderivative = 2.64, number of steps
used = 14, number of rules used = 9, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {12, 1608, 27,
6874, 46, 2404, 2341, 2351, 31} \begin {gather*} \frac {\log (x \log (2))}{12 e^{20} x^2}+\frac {\log (x)}{48 e^{20}}-\frac {\log (x \log (2))}{24 e^{20} x}-\frac {x \log (x \log (2))}{48 e^{20} (x+2)} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 27
Rule 31
Rule 46
Rule 1608
Rule 2341
Rule 2351
Rule 2404
Rule 6874
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {2+x+(-4-3 x) \log (x \log (2))}{24 x^3+24 x^4+6 x^5} \, dx}{e^{20}}\\ &=\frac {\int \frac {2+x+(-4-3 x) \log (x \log (2))}{x^3 \left (24+24 x+6 x^2\right )} \, dx}{e^{20}}\\ &=\frac {\int \frac {2+x+(-4-3 x) \log (x \log (2))}{6 x^3 (2+x)^2} \, dx}{e^{20}}\\ &=\frac {\int \frac {2+x+(-4-3 x) \log (x \log (2))}{x^3 (2+x)^2} \, dx}{6 e^{20}}\\ &=\frac {\int \left (\frac {1}{x^3 (2+x)}-\frac {(4+3 x) \log (x \log (2))}{x^3 (2+x)^2}\right ) \, dx}{6 e^{20}}\\ &=\frac {\int \frac {1}{x^3 (2+x)} \, dx}{6 e^{20}}-\frac {\int \frac {(4+3 x) \log (x \log (2))}{x^3 (2+x)^2} \, dx}{6 e^{20}}\\ &=\frac {\int \left (\frac {1}{2 x^3}-\frac {1}{4 x^2}+\frac {1}{8 x}-\frac {1}{8 (2+x)}\right ) \, dx}{6 e^{20}}-\frac {\int \left (\frac {\log (x \log (2))}{x^3}-\frac {\log (x \log (2))}{4 x^2}+\frac {\log (x \log (2))}{4 (2+x)^2}\right ) \, dx}{6 e^{20}}\\ &=-\frac {1}{24 e^{20} x^2}+\frac {1}{24 e^{20} x}+\frac {\log (x)}{48 e^{20}}-\frac {\log (2+x)}{48 e^{20}}+\frac {\int \frac {\log (x \log (2))}{x^2} \, dx}{24 e^{20}}-\frac {\int \frac {\log (x \log (2))}{(2+x)^2} \, dx}{24 e^{20}}-\frac {\int \frac {\log (x \log (2))}{x^3} \, dx}{6 e^{20}}\\ &=\frac {\log (x)}{48 e^{20}}-\frac {\log (2+x)}{48 e^{20}}+\frac {\log (x \log (2))}{12 e^{20} x^2}-\frac {\log (x \log (2))}{24 e^{20} x}-\frac {x \log (x \log (2))}{48 e^{20} (2+x)}+\frac {\int \frac {1}{2+x} \, dx}{48 e^{20}}\\ &=\frac {\log (x)}{48 e^{20}}+\frac {\log (x \log (2))}{12 e^{20} x^2}-\frac {\log (x \log (2))}{24 e^{20} x}-\frac {x \log (x \log (2))}{48 e^{20} (2+x)}\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.06, size = 20, normalized size = 0.91 \begin {gather*} \frac {\log (x \log (2))}{6 e^{20} x^2 (2+x)} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(67\) vs.
\(2(21)=42\).
time = 1.37, size = 68, normalized size = 3.09
method | result | size |
risch | \(\frac {\ln \left (x \ln \left (2\right )\right ) {\mathrm e}^{-20}}{6 x^{2} \left (2+x \right )}\) | \(18\) |
norman | \(\frac {\ln \left (x \ln \left (2\right )\right ) {\mathrm e}^{-20}}{6 x^{2} \left (2+x \right )}\) | \(20\) |
derivativedivides | \(\frac {{\mathrm e}^{-20} \left (\frac {\ln \left (x \ln \left (2\right )\right ) \ln \left (2\right )}{8}-\frac {\ln \left (2\right ) \ln \left (x \ln \left (2\right )\right )}{4 x}-\frac {\ln \left (2\right )^{2} \ln \left (x \ln \left (2\right )\right ) x}{8 \left (x \ln \left (2\right )+2 \ln \left (2\right )\right )}+\frac {\ln \left (2\right ) \ln \left (x \ln \left (2\right )\right )}{2 x^{2}}\right )}{6 \ln \left (2\right )}\) | \(68\) |
default | \(\frac {{\mathrm e}^{-20} \left (\frac {\ln \left (x \ln \left (2\right )\right ) \ln \left (2\right )}{8}-\frac {\ln \left (2\right ) \ln \left (x \ln \left (2\right )\right )}{4 x}-\frac {\ln \left (2\right )^{2} \ln \left (x \ln \left (2\right )\right ) x}{8 \left (x \ln \left (2\right )+2 \ln \left (2\right )\right )}+\frac {\ln \left (2\right ) \ln \left (x \ln \left (2\right )\right )}{2 x^{2}}\right )}{6 \ln \left (2\right )}\) | \(68\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 85 vs.
\(2 (19) = 38\).
time = 0.56, size = 85, normalized size = 3.86 \begin {gather*} \frac {1}{48} \, {\left (\frac {2 \, {\left (3 \, x^{2} + 3 \, x - 2\right )}}{x^{3} + 2 \, x^{2}} - \frac {2 \, x^{2} + {\left (x^{3} + 2 \, x^{2} - 8\right )} \log \left (x\right ) + 2 \, x - 8 \, \log \left (\log \left (2\right )\right ) - 4}{x^{3} + 2 \, x^{2}} - \frac {4 \, {\left (x + 1\right )}}{x^{2} + 2 \, x} + \log \left (x\right )\right )} e^{\left (-20\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.53, size = 20, normalized size = 0.91 \begin {gather*} \frac {e^{\left (-20\right )} \log \left (x \log \left (2\right )\right )}{6 \, {\left (x^{3} + 2 \, x^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.09, size = 22, normalized size = 1.00 \begin {gather*} \frac {\log {\left (x \log {\left (2 \right )} \right )}}{6 x^{3} e^{20} + 12 x^{2} e^{20}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.41, size = 23, normalized size = 1.05 \begin {gather*} \frac {1}{24} \, {\left (\frac {1}{x + 2} - \frac {x - 2}{x^{2}}\right )} e^{\left (-20\right )} \log \left (x \log \left (2\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 4.94, size = 24, normalized size = 1.09 \begin {gather*} \frac {\ln \left (\ln \left (2\right )\right )+\ln \left (x\right )}{6\,{\mathrm {e}}^{20}\,x^3+12\,{\mathrm {e}}^{20}\,x^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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