Optimal. Leaf size=29 \[ x-\log (x)+\frac {1}{4} x \left (5+x+x^2-\frac {5 \left (3+\log \left (x^2\right )\right )}{\log (x)}\right ) \]
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Rubi [A] time = 0.39, antiderivative size = 46, normalized size of antiderivative = 1.59, number of steps used = 16, number of rules used = 8, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {12, 6742, 6688, 2297, 2298, 2360, 2361, 6496} \begin {gather*} \frac {x^3}{4}+\frac {x^2}{4}-\frac {5 x \log \left (x^2\right )}{4 \log (x)}+\frac {9 x}{4}-\frac {15 x}{4 \log (x)}-\log (x) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2297
Rule 2298
Rule 2360
Rule 2361
Rule 6496
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {15 x-25 x \log (x)+\left (-4+9 x+2 x^2+3 x^3\right ) \log ^2(x)+(5 x-5 x \log (x)) \log \left (x^2\right )}{x \log ^2(x)} \, dx\\ &=\frac {1}{4} \int \left (\frac {15 x-25 x \log (x)-4 \log ^2(x)+9 x \log ^2(x)+2 x^2 \log ^2(x)+3 x^3 \log ^2(x)}{x \log ^2(x)}-\frac {5 (-1+\log (x)) \log \left (x^2\right )}{\log ^2(x)}\right ) \, dx\\ &=\frac {1}{4} \int \frac {15 x-25 x \log (x)-4 \log ^2(x)+9 x \log ^2(x)+2 x^2 \log ^2(x)+3 x^3 \log ^2(x)}{x \log ^2(x)} \, dx-\frac {5}{4} \int \frac {(-1+\log (x)) \log \left (x^2\right )}{\log ^2(x)} \, dx\\ &=\frac {1}{4} \int \left (9-\frac {4}{x}+2 x+3 x^2+\frac {15}{\log ^2(x)}-\frac {25}{\log (x)}\right ) \, dx-\frac {5}{4} \int \left (-\frac {\log \left (x^2\right )}{\log ^2(x)}+\frac {\log \left (x^2\right )}{\log (x)}\right ) \, dx\\ &=\frac {9 x}{4}+\frac {x^2}{4}+\frac {x^3}{4}-\log (x)+\frac {5}{4} \int \frac {\log \left (x^2\right )}{\log ^2(x)} \, dx-\frac {5}{4} \int \frac {\log \left (x^2\right )}{\log (x)} \, dx+\frac {15}{4} \int \frac {1}{\log ^2(x)} \, dx-\frac {25}{4} \int \frac {1}{\log (x)} \, dx\\ &=\frac {9 x}{4}+\frac {x^2}{4}+\frac {x^3}{4}-\frac {15 x}{4 \log (x)}-\log (x)-\frac {5 x \log \left (x^2\right )}{4 \log (x)}-\frac {25 \text {li}(x)}{4}+\frac {5}{2} \int \frac {\text {li}(x)}{x} \, dx-\frac {5}{2} \int \left (-\frac {1}{\log (x)}+\frac {\text {li}(x)}{x}\right ) \, dx+\frac {15}{4} \int \frac {1}{\log (x)} \, dx\\ &=-\frac {x}{4}+\frac {x^2}{4}+\frac {x^3}{4}-\frac {15 x}{4 \log (x)}-\log (x)-\frac {5 x \log \left (x^2\right )}{4 \log (x)}-\frac {5 \text {li}(x)}{2}+\frac {5}{2} \log (x) \text {li}(x)+\frac {5}{2} \int \frac {1}{\log (x)} \, dx-\frac {5}{2} \int \frac {\text {li}(x)}{x} \, dx\\ &=\frac {9 x}{4}+\frac {x^2}{4}+\frac {x^3}{4}-\frac {15 x}{4 \log (x)}-\log (x)-\frac {5 x \log \left (x^2\right )}{4 \log (x)}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.03, size = 30, normalized size = 1.03 \begin {gather*} \frac {1}{4} \left (x \left (9+x+x^2\right )-4 \log (x)-\frac {5 x \left (3+\log \left (x^2\right )\right )}{\log (x)}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.68, size = 29, normalized size = 1.00 \begin {gather*} \frac {{\left (x^{3} + x^{2} - x\right )} \log \relax (x) - 4 \, \log \relax (x)^{2} - 15 \, x}{4 \, \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.20, size = 25, normalized size = 0.86 \begin {gather*} \frac {1}{4} \, x^{3} + \frac {1}{4} \, x^{2} - \frac {1}{4} \, x - \frac {15 \, x}{4 \, \log \relax (x)} - \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.05, size = 74, normalized size = 2.55
method | result | size |
risch | \(\frac {x^{3}}{4}+\frac {x^{2}}{4}-\frac {x}{4}-\ln \relax (x )+\frac {5 i x \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+6 i\right )}{8 \ln \relax (x )}\) | \(74\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.38, size = 69, normalized size = 2.38 \begin {gather*} \frac {1}{4} \, x^{3} + \frac {1}{4} \, x^{2} - \frac {5}{4} \, {\rm Ei}\left (\log \relax (x)\right ) \log \left (x^{2}\right ) + \frac {5}{4} \, \Gamma \left (-1, -\log \relax (x)\right ) \log \left (x^{2}\right ) + \frac {5}{2} \, {\rm Ei}\left (\log \relax (x)\right ) \log \relax (x) - \frac {5}{2} \, \Gamma \left (-1, -\log \relax (x)\right ) \log \relax (x) - \frac {1}{4} \, x - \frac {15}{4} \, {\rm Ei}\left (\log \relax (x)\right ) + \frac {15}{4} \, \Gamma \left (-1, -\log \relax (x)\right ) - \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.31, size = 35, normalized size = 1.21 \begin {gather*} \frac {9\,x}{4}-\ln \relax (x)+\frac {x^2}{4}+\frac {x^3}{4}-\frac {\frac {15\,x}{4}+\frac {5\,x\,\ln \left (x^2\right )}{4}}{\ln \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.24, size = 24, normalized size = 0.83 \begin {gather*} \frac {x^{3}}{4} + \frac {x^{2}}{4} - \frac {x}{4} - \frac {15 x}{4 \log {\relax (x )}} - \log {\relax (x )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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