Optimal. Leaf size=33 \[ \frac {4+\log \left (-3+4 \left (\frac {3}{x}-\frac {x}{2}\right )+\frac {7 x}{4}-\frac {5}{\log (2)}\right )}{x} \]
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Rubi [A] time = 1.13, antiderivative size = 64, normalized size of antiderivative = 1.94, number of steps used = 17, number of rules used = 9, integrand size = 90, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6741, 6728, 1628, 634, 618, 206, 628, 2525, 12} \begin {gather*} \frac {\log \left (\frac {x^2 (-\log (2))-4 x (5+\log (8))+48 \log (2)}{x \log (16)}\right )}{x}+\frac {4}{x}-\frac {(5+\log (8)) \log (x)}{\log (4096)}+\frac {(5+\log (8)) \log (x)}{12 \log (2)} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 206
Rule 618
Rule 628
Rule 634
Rule 1628
Rule 2525
Rule 6728
Rule 6741
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {80 x-\left (240-48 x-3 x^2\right ) \log (2)-\left (-20 x+\left (48-12 x-x^2\right ) \log (2)\right ) \log \left (\frac {-20 x+\left (48-12 x-x^2\right ) \log (2)}{4 x \log (2)}\right )}{x^2 \left (48 \log (2)-x^2 \log (2)-4 x (5+\log (8))\right )} \, dx\\ &=\int \left (\frac {-240 \log (2)+x^2 \log (8)+16 x (5+\log (8))}{x^2 \left (48 \log (2)-x^2 \log (2)-4 x (5+\log (8))\right )}-\frac {\log \left (\frac {48 \log (2)-x^2 \log (2)-4 x (5+\log (8))}{x \log (16)}\right )}{x^2}\right ) \, dx\\ &=\int \frac {-240 \log (2)+x^2 \log (8)+16 x (5+\log (8))}{x^2 \left (48 \log (2)-x^2 \log (2)-4 x (5+\log (8))\right )} \, dx-\int \frac {\log \left (\frac {48 \log (2)-x^2 \log (2)-4 x (5+\log (8))}{x \log (16)}\right )}{x^2} \, dx\\ &=\frac {\log \left (\frac {48 \log (2)-x^2 \log (2)-4 x (5+\log (8))}{x \log (16)}\right )}{x}-\int \frac {\left (-48-x^2\right ) \log (2)}{x^2 \left (48 \log (2)-x^2 \log (2)-4 x (5+\log (8))\right )} \, dx+\int \left (-\frac {5}{x^2}+\frac {-x \log (2) (5+\log (8))-20 \left (5+3 \log ^2(2)+\log (64)\right )}{12 \log (2) \left (48 \log (2)-x^2 \log (2)-4 x (5+\log (8))\right )}+\frac {-5-\log (8)}{x \log (4096)}\right ) \, dx\\ &=\frac {5}{x}-\frac {(5+\log (8)) \log (x)}{\log (4096)}+\frac {\log \left (\frac {48 \log (2)-x^2 \log (2)-4 x (5+\log (8))}{x \log (16)}\right )}{x}+\frac {\int \frac {-x \log (2) (5+\log (8))-20 \left (5+3 \log ^2(2)+\log (64)\right )}{48 \log (2)-x^2 \log (2)-4 x (5+\log (8))} \, dx}{12 \log (2)}-\log (2) \int \frac {-48-x^2}{x^2 \left (48 \log (2)-x^2 \log (2)-4 x (5+\log (8))\right )} \, dx\\ &=\frac {5}{x}-\frac {(5+\log (8)) \log (x)}{\log (4096)}+\frac {\log \left (\frac {48 \log (2)-x^2 \log (2)-4 x (5+\log (8))}{x \log (16)}\right )}{x}-\log (2) \int \left (-\frac {1}{x^2 \log (2)}+\frac {-5-\log (8)}{12 x \log ^2(2)}+\frac {-x \log (2) (5+\log (8))-4 \left (6 \log ^2(2)+(5+\log (8))^2\right )}{12 \log ^2(2) \left (48 \log (2)-x^2 \log (2)-4 x (5+\log (8))\right )}\right ) \, dx+\frac {(5+\log (8)) \int \frac {-2 x \log (2)-4 (5+\log (8))}{48 \log (2)-x^2 \log (2)-4 x (5+\log (8))} \, dx}{24 \log (2)}+\frac {\left ((5+\log (8))^2-10 \left (5+3 \log ^2(2)+\log (64)\right )\right ) \int \frac {1}{48 \log (2)-x^2 \log (2)-4 x (5+\log (8))} \, dx}{6 \log (2)}\\ &=\frac {4}{x}+\frac {(5+\log (8)) \log (x)}{12 \log (2)}-\frac {(5+\log (8)) \log (x)}{\log (4096)}+\frac {(5+\log (8)) \log \left (48 \log (2)-x^2 \log (2)-4 x (5+\log (8))\right )}{24 \log (2)}+\frac {\log \left (\frac {48 \log (2)-x^2 \log (2)-4 x (5+\log (8))}{x \log (16)}\right )}{x}-\frac {\int \frac {-x \log (2) (5+\log (8))-4 \left (6 \log ^2(2)+(5+\log (8))^2\right )}{48 \log (2)-x^2 \log (2)-4 x (5+\log (8))} \, dx}{12 \log (2)}-\frac {\left ((5+\log (8))^2-10 \left (5+3 \log ^2(2)+\log (64)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-x^2+16 \left (12 \log ^2(2)+(5+\log (8))^2\right )} \, dx,x,-2 x \log (2)-4 (5+\log (8))\right )}{3 \log (2)}\\ &=\frac {4}{x}-\frac {\tanh ^{-1}\left (\frac {10+x \log (2)+\log (64)}{2 \sqrt {12 \log ^2(2)+(5+\log (8))^2}}\right ) \sqrt {12 \log ^2(2)+(5+\log (8))^2}}{12 \log (2)}+\frac {(5+\log (8)) \log (x)}{12 \log (2)}-\frac {(5+\log (8)) \log (x)}{\log (4096)}+\frac {(5+\log (8)) \log \left (48 \log (2)-x^2 \log (2)-4 x (5+\log (8))\right )}{24 \log (2)}+\frac {\log \left (\frac {48 \log (2)-x^2 \log (2)-4 x (5+\log (8))}{x \log (16)}\right )}{x}-\frac {(5+\log (8)) \int \frac {-2 x \log (2)-4 (5+\log (8))}{48 \log (2)-x^2 \log (2)-4 x (5+\log (8))} \, dx}{24 \log (2)}+\frac {\left (12 \log ^2(2)+(5+\log (8))^2\right ) \int \frac {1}{48 \log (2)-x^2 \log (2)-4 x (5+\log (8))} \, dx}{6 \log (2)}\\ &=\frac {4}{x}-\frac {\tanh ^{-1}\left (\frac {10+x \log (2)+\log (64)}{2 \sqrt {12 \log ^2(2)+(5+\log (8))^2}}\right ) \sqrt {12 \log ^2(2)+(5+\log (8))^2}}{12 \log (2)}+\frac {(5+\log (8)) \log (x)}{12 \log (2)}-\frac {(5+\log (8)) \log (x)}{\log (4096)}+\frac {\log \left (\frac {48 \log (2)-x^2 \log (2)-4 x (5+\log (8))}{x \log (16)}\right )}{x}-\frac {\left (12 \log ^2(2)+(5+\log (8))^2\right ) \operatorname {Subst}\left (\int \frac {1}{-x^2+16 \left (12 \log ^2(2)+(5+\log (8))^2\right )} \, dx,x,-2 x \log (2)-4 (5+\log (8))\right )}{3 \log (2)}\\ &=\frac {4}{x}+\frac {(5+\log (8)) \log (x)}{12 \log (2)}-\frac {(5+\log (8)) \log (x)}{\log (4096)}+\frac {\log \left (\frac {48 \log (2)-x^2 \log (2)-4 x (5+\log (8))}{x \log (16)}\right )}{x}\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 0.17, size = 96, normalized size = 2.91 \begin {gather*} \frac {4}{x}+\frac {\tanh ^{-1}\left (\frac {10+x \log (2)+\log (64)}{2 \sqrt {12 \log ^2(2)+(5+\log (8))^2}}\right ) (-3 \log (2)+\log (8))}{2 \sqrt {12 \log ^2(2)+(5+\log (8))^2}}+\frac {\log \left (\frac {48 \log (2)-x^2 \log (2)-4 x (5+\log (8))}{x \log (16)}\right )}{x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.68, size = 31, normalized size = 0.94 \begin {gather*} \frac {\log \left (-\frac {{\left (x^{2} + 12 \, x - 48\right )} \log \relax (2) + 20 \, x}{4 \, x \log \relax (2)}\right ) + 4}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.39, size = 45, normalized size = 1.36 \begin {gather*} -\frac {2 \, {\left (\log \relax (2) - 2\right )}}{x} + \frac {\log \left (-x^{2} \log \relax (2) - 12 \, x \log \relax (2) - 20 \, x + 48 \, \log \relax (2)\right )}{x} - \frac {\log \left (x \log \relax (2)\right )}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.15, size = 34, normalized size = 1.03
method | result | size |
norman | \(\frac {4+\ln \left (\frac {\left (-x^{2}-12 x +48\right ) \ln \relax (2)-20 x}{4 x \ln \relax (2)}\right )}{x}\) | \(34\) |
risch | \(\frac {\ln \left (\frac {\left (-x^{2}-12 x +48\right ) \ln \relax (2)-20 x}{4 x \ln \relax (2)}\right )}{x}+\frac {4}{x}\) | \(38\) |
default | \(-\frac {\ln \left (\ln \relax (2)\right )}{x}-\frac {2 \ln \relax (2)}{x}+\frac {4}{x}+\frac {\ln \left (\frac {-x^{2} \ln \relax (2)-12 x \ln \relax (2)+48 \ln \relax (2)-20 x}{x}\right )}{x}\) | \(51\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.93, size = 317, normalized size = 9.61 \begin {gather*} \frac {5}{24} \, {\left (\frac {{\left (3 \, \log \relax (2) + 5\right )} \log \left (x^{2} \log \relax (2) + 4 \, x {\left (3 \, \log \relax (2) + 5\right )} - 48 \, \log \relax (2)\right )}{\log \relax (2)^{2}} - \frac {2 \, {\left (3 \, \log \relax (2) + 5\right )} \log \relax (x)}{\log \relax (2)^{2}} + \frac {5 \, {\left (3 \, \log \relax (2)^{2} + 6 \, \log \relax (2) + 5\right )} \log \left (\frac {x \log \relax (2) - 2 \, \sqrt {21 \, \log \relax (2)^{2} + 30 \, \log \relax (2) + 25} + 6 \, \log \relax (2) + 10}{x \log \relax (2) + 2 \, \sqrt {21 \, \log \relax (2)^{2} + 30 \, \log \relax (2) + 25} + 6 \, \log \relax (2) + 10}\right )}{\sqrt {21 \, \log \relax (2)^{2} + 30 \, \log \relax (2) + 25} \log \relax (2)^{2}} + \frac {24}{x \log \relax (2)}\right )} \log \relax (2) - \frac {25 \, {\left (3 \, \log \relax (2)^{2} + 6 \, \log \relax (2) + 5\right )} \log \left (\frac {x \log \relax (2) - 2 \, \sqrt {21 \, \log \relax (2)^{2} + 30 \, \log \relax (2) + 25} + 6 \, \log \relax (2) + 10}{x \log \relax (2) + 2 \, \sqrt {21 \, \log \relax (2)^{2} + 30 \, \log \relax (2) + 25} + 6 \, \log \relax (2) + 10}\right )}{24 \, \sqrt {21 \, \log \relax (2)^{2} + 30 \, \log \relax (2) + 25} \log \relax (2)} - \frac {24 \, {\left (\log \left (\log \relax (2)\right ) + 1\right )} \log \relax (2) + 48 \, \log \relax (2)^{2} + {\left (5 \, x {\left (3 \, \log \relax (2) + 5\right )} - 24 \, \log \relax (2)\right )} \log \left (-x^{2} \log \relax (2) - 4 \, x {\left (3 \, \log \relax (2) + 5\right )} + 48 \, \log \relax (2)\right ) - 2 \, {\left (5 \, x {\left (3 \, \log \relax (2) + 5\right )} - 12 \, \log \relax (2)\right )} \log \relax (x)}{24 \, x \log \relax (2)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.06, size = 32, normalized size = 0.97 \begin {gather*} \frac {\ln \left (-\frac {5\,x+\frac {\ln \relax (2)\,\left (x^2+12\,x-48\right )}{4}}{x\,\ln \relax (2)}\right )+4}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.25, size = 27, normalized size = 0.82 \begin {gather*} \frac {\log {\left (\frac {- 5 x + \frac {\left (- x^{2} - 12 x + 48\right ) \log {\relax (2 )}}{4}}{x \log {\relax (2 )}} \right )}}{x} + \frac {4}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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