Optimal. Leaf size=25 \[ \left (5-\frac {2 (4-x)}{x}+x+\log (3)-\log \left (2 x^2\right )\right )^2 \]
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Rubi [B] time = 0.12, antiderivative size = 69, normalized size of antiderivative = 2.76, number of steps used = 9, number of rules used = 6, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {14, 1628, 2357, 2295, 2304, 2301} \begin {gather*} x^2+\frac {64}{x^2}+\log ^2\left (2 x^2\right )-2 x \log \left (2 x^2\right )+\frac {16 \log \left (2 x^2\right )}{x}+4 x+\frac {32}{x}+2 x (5+\log (3))-4 (7+\log (3)) \log (x)-\frac {16 (9+\log (3))}{x} \end {gather*}
Antiderivative was successfully verified.
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Rule 14
Rule 1628
Rule 2295
Rule 2301
Rule 2304
Rule 2357
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 \left (8-2 x+x^2\right ) \left (-8+x^2+x (7+\log (3))\right )}{x^3}-\frac {2 \left (8-2 x+x^2\right ) \log \left (2 x^2\right )}{x^2}\right ) \, dx\\ &=2 \int \frac {\left (8-2 x+x^2\right ) \left (-8+x^2+x (7+\log (3))\right )}{x^3} \, dx-2 \int \frac {\left (8-2 x+x^2\right ) \log \left (2 x^2\right )}{x^2} \, dx\\ &=2 \int \left (-\frac {64}{x^3}+x+5 \left (1+\frac {\log (3)}{5}\right )-\frac {2 (7+\log (3))}{x}+\frac {8 (9+\log (3))}{x^2}\right ) \, dx-2 \int \left (\log \left (2 x^2\right )+\frac {8 \log \left (2 x^2\right )}{x^2}-\frac {2 \log \left (2 x^2\right )}{x}\right ) \, dx\\ &=\frac {64}{x^2}+x^2+2 x (5+\log (3))-\frac {16 (9+\log (3))}{x}-4 (7+\log (3)) \log (x)-2 \int \log \left (2 x^2\right ) \, dx+4 \int \frac {\log \left (2 x^2\right )}{x} \, dx-16 \int \frac {\log \left (2 x^2\right )}{x^2} \, dx\\ &=\frac {64}{x^2}+\frac {32}{x}+4 x+x^2+2 x (5+\log (3))-\frac {16 (9+\log (3))}{x}-4 (7+\log (3)) \log (x)+\frac {16 \log \left (2 x^2\right )}{x}-2 x \log \left (2 x^2\right )+\log ^2\left (2 x^2\right )\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 0.07, size = 60, normalized size = 2.40 \begin {gather*} \frac {64}{x^2}+\frac {32}{x}+4 x+x^2+2 x (5+\log (3))+\frac {16 \left (-9+\log \left (\frac {2 x^2}{3}\right )\right )}{x}+\left (-7+\log \left (\frac {2 x^2}{3}\right )\right )^2-2 x \log \left (2 x^2\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.98, size = 66, normalized size = 2.64 \begin {gather*} \frac {x^{4} + x^{2} \log \left (2 \, x^{2}\right )^{2} + 14 \, x^{3} + 2 \, {\left (x^{3} - 8 \, x\right )} \log \relax (3) - 2 \, {\left (x^{3} + x^{2} \log \relax (3) + 7 \, x^{2} - 8 \, x\right )} \log \left (2 \, x^{2}\right ) - 112 \, x + 64}{x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left (x^{4} + 5 \, x^{3} - 14 \, x^{2} + {\left (x^{3} - 2 \, x^{2} + 8 \, x\right )} \log \relax (3) - {\left (x^{3} - 2 \, x^{2} + 8 \, x\right )} \log \left (2 \, x^{2}\right ) + 72 \, x - 64\right )}}{x^{3}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.05, size = 76, normalized size = 3.04
method | result | size |
norman | \(\frac {64+x^{4}+x^{2} \ln \left (2 x^{2}\right )^{2}+\left (-112-16 \ln \relax (3)\right ) x +\left (14+2 \ln \relax (3)\right ) x^{3}+\left (-14-2 \ln \relax (3)\right ) x^{2} \ln \left (2 x^{2}\right )+16 x \ln \left (2 x^{2}\right )-2 x^{3} \ln \left (2 x^{2}\right )}{x^{2}}\) | \(76\) |
default | \(14 x +x^{2}-28 \ln \relax (x )-\frac {112}{x}+\frac {64}{x^{2}}-2 x \ln \relax (2)+4 \ln \relax (2) \ln \relax (x )+\frac {16 \ln \relax (2)}{x}+2 x \ln \relax (3)-4 \ln \relax (3) \ln \relax (x )-\frac {16 \ln \relax (3)}{x}-2 x \ln \left (x^{2}\right )+4 \ln \relax (x ) \ln \left (x^{2}\right )-4 \ln \relax (x )^{2}+\frac {16 \ln \left (x^{2}\right )}{x}\) | \(88\) |
risch | \(14 x +x^{2}-28 \ln \relax (x )-\frac {112}{x}+\frac {64}{x^{2}}-2 x \ln \relax (2)+4 \ln \relax (2) \ln \relax (x )+\frac {16 \ln \relax (2)}{x}+2 x \ln \relax (3)-4 \ln \relax (3) \ln \relax (x )-\frac {16 \ln \relax (3)}{x}-2 x \ln \left (x^{2}\right )+4 \ln \relax (x ) \ln \left (x^{2}\right )-4 \ln \relax (x )^{2}+\frac {16 \ln \left (x^{2}\right )}{x}\) | \(88\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.36, size = 67, normalized size = 2.68 \begin {gather*} x^{2} + 2 \, x \log \relax (3) - 2 \, x \log \left (2 \, x^{2}\right ) + \log \left (2 \, x^{2}\right )^{2} - 4 \, \log \relax (3) \log \relax (x) + 14 \, x - \frac {16 \, \log \relax (3)}{x} + \frac {16 \, \log \left (2 \, x^{2}\right )}{x} - \frac {112}{x} + \frac {64}{x^{2}} - 28 \, \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.13, size = 63, normalized size = 2.52 \begin {gather*} {\ln \left (2\,x^2\right )}^2-\ln \left (x^2\right )\,\left (\ln \relax (9)+14\right )+x\,\left (\ln \relax (9)-2\,\ln \left (2\,x^2\right )+14\right )+\frac {64\,x-x^2\,\left (16\,\ln \relax (3)-16\,\ln \left (2\,x^2\right )+112\right )}{x^3}+x^2 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.33, size = 60, normalized size = 2.40 \begin {gather*} x^{2} + x \left (2 \log {\relax (3 )} + 14\right ) - 4 \left (\log {\relax (3 )} + 7\right ) \log {\relax (x )} + \log {\left (2 x^{2} \right )}^{2} + \frac {\left (16 - 2 x^{2}\right ) \log {\left (2 x^{2} \right )}}{x} + \frac {x \left (-112 - 16 \log {\relax (3 )}\right ) + 64}{x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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