Optimal. Leaf size=24 \[ \frac {1}{5} x \left (x^2+\left (5-x+\left (-8+x^2\right ) \log (x)\right )^2\right ) \]
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Rubi [B] time = 0.13, antiderivative size = 77, normalized size of antiderivative = 3.21, number of steps used = 17, number of rules used = 6, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {12, 2356, 2295, 2304, 2296, 2305} \begin {gather*} \frac {1}{5} x^5 \log ^2(x)-\frac {2}{5} x^4 \log (x)+\frac {2 x^3}{5}-\frac {16}{5} x^3 \log ^2(x)+2 x^3 \log (x)-2 x^2+\frac {16}{5} x^2 \log (x)+5 x+\frac {64}{5} x \log ^2(x)-16 x \log (x) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2295
Rule 2296
Rule 2304
Rule 2305
Rule 2356
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \left (-55-4 x+16 x^2-2 x^3+\left (48+32 x-2 x^2-8 x^3+2 x^4\right ) \log (x)+\left (64-48 x^2+5 x^4\right ) \log ^2(x)\right ) \, dx\\ &=-11 x-\frac {2 x^2}{5}+\frac {16 x^3}{15}-\frac {x^4}{10}+\frac {1}{5} \int \left (48+32 x-2 x^2-8 x^3+2 x^4\right ) \log (x) \, dx+\frac {1}{5} \int \left (64-48 x^2+5 x^4\right ) \log ^2(x) \, dx\\ &=-11 x-\frac {2 x^2}{5}+\frac {16 x^3}{15}-\frac {x^4}{10}+\frac {1}{5} \int \left (48 \log (x)+32 x \log (x)-2 x^2 \log (x)-8 x^3 \log (x)+2 x^4 \log (x)\right ) \, dx+\frac {1}{5} \int \left (64 \log ^2(x)-48 x^2 \log ^2(x)+5 x^4 \log ^2(x)\right ) \, dx\\ &=-11 x-\frac {2 x^2}{5}+\frac {16 x^3}{15}-\frac {x^4}{10}-\frac {2}{5} \int x^2 \log (x) \, dx+\frac {2}{5} \int x^4 \log (x) \, dx-\frac {8}{5} \int x^3 \log (x) \, dx+\frac {32}{5} \int x \log (x) \, dx+\frac {48}{5} \int \log (x) \, dx-\frac {48}{5} \int x^2 \log ^2(x) \, dx+\frac {64}{5} \int \log ^2(x) \, dx+\int x^4 \log ^2(x) \, dx\\ &=-\frac {103 x}{5}-2 x^2+\frac {10 x^3}{9}-\frac {2 x^5}{125}+\frac {48}{5} x \log (x)+\frac {16}{5} x^2 \log (x)-\frac {2}{15} x^3 \log (x)-\frac {2}{5} x^4 \log (x)+\frac {2}{25} x^5 \log (x)+\frac {64}{5} x \log ^2(x)-\frac {16}{5} x^3 \log ^2(x)+\frac {1}{5} x^5 \log ^2(x)-\frac {2}{5} \int x^4 \log (x) \, dx+\frac {32}{5} \int x^2 \log (x) \, dx-\frac {128}{5} \int \log (x) \, dx\\ &=5 x-2 x^2+\frac {2 x^3}{5}-16 x \log (x)+\frac {16}{5} x^2 \log (x)+2 x^3 \log (x)-\frac {2}{5} x^4 \log (x)+\frac {64}{5} x \log ^2(x)-\frac {16}{5} x^3 \log ^2(x)+\frac {1}{5} x^5 \log ^2(x)\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 0.01, size = 77, normalized size = 3.21 \begin {gather*} 5 x-2 x^2+\frac {2 x^3}{5}-16 x \log (x)+\frac {16}{5} x^2 \log (x)+2 x^3 \log (x)-\frac {2}{5} x^4 \log (x)+\frac {64}{5} x \log ^2(x)-\frac {16}{5} x^3 \log ^2(x)+\frac {1}{5} x^5 \log ^2(x) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.56, size = 53, normalized size = 2.21 \begin {gather*} \frac {2}{5} \, x^{3} + \frac {1}{5} \, {\left (x^{5} - 16 \, x^{3} + 64 \, x\right )} \log \relax (x)^{2} - 2 \, x^{2} - \frac {2}{5} \, {\left (x^{4} - 5 \, x^{3} - 8 \, x^{2} + 40 \, x\right )} \log \relax (x) + 5 \, x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.23, size = 65, normalized size = 2.71 \begin {gather*} \frac {1}{5} \, x^{5} \log \relax (x)^{2} - \frac {2}{5} \, x^{4} \log \relax (x) - \frac {16}{5} \, x^{3} \log \relax (x)^{2} + 2 \, x^{3} \log \relax (x) + \frac {2}{5} \, x^{3} + \frac {16}{5} \, x^{2} \log \relax (x) + \frac {64}{5} \, x \log \relax (x)^{2} - 2 \, x^{2} - 16 \, x \log \relax (x) + 5 \, x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.04, size = 66, normalized size = 2.75
method | result | size |
default | \(5 x -2 x^{2}+\frac {2 x^{3}}{5}+\frac {x^{5} \ln \relax (x )^{2}}{5}-\frac {16 x^{3} \ln \relax (x )^{2}}{5}+2 x^{3} \ln \relax (x )+\frac {64 x \ln \relax (x )^{2}}{5}-16 x \ln \relax (x )-\frac {2 x^{4} \ln \relax (x )}{5}+\frac {16 x^{2} \ln \relax (x )}{5}\) | \(66\) |
norman | \(5 x -2 x^{2}+\frac {2 x^{3}}{5}+\frac {x^{5} \ln \relax (x )^{2}}{5}-\frac {16 x^{3} \ln \relax (x )^{2}}{5}+2 x^{3} \ln \relax (x )+\frac {64 x \ln \relax (x )^{2}}{5}-16 x \ln \relax (x )-\frac {2 x^{4} \ln \relax (x )}{5}+\frac {16 x^{2} \ln \relax (x )}{5}\) | \(66\) |
risch | \(5 x -2 x^{2}+\frac {2 x^{3}}{5}+\frac {x^{5} \ln \relax (x )^{2}}{5}-\frac {16 x^{3} \ln \relax (x )^{2}}{5}+2 x^{3} \ln \relax (x )+\frac {64 x \ln \relax (x )^{2}}{5}-16 x \ln \relax (x )-\frac {2 x^{4} \ln \relax (x )}{5}+\frac {16 x^{2} \ln \relax (x )}{5}\) | \(66\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.70, size = 94, normalized size = 3.92 \begin {gather*} \frac {1}{125} \, {\left (25 \, \log \relax (x)^{2} - 10 \, \log \relax (x) + 2\right )} x^{5} - \frac {2}{125} \, x^{5} - \frac {16}{45} \, {\left (9 \, \log \relax (x)^{2} - 6 \, \log \relax (x) + 2\right )} x^{3} + \frac {10}{9} \, x^{3} + \frac {64}{5} \, {\left (\log \relax (x)^{2} - 2 \, \log \relax (x) + 2\right )} x - 2 \, x^{2} + \frac {2}{75} \, {\left (3 \, x^{5} - 15 \, x^{4} - 5 \, x^{3} + 120 \, x^{2} + 360 \, x\right )} \log \relax (x) - \frac {103}{5} \, x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.84, size = 59, normalized size = 2.46 \begin {gather*} \frac {x\,\left (x^4\,{\ln \relax (x)}^2-2\,x^3\,\ln \relax (x)-16\,x^2\,{\ln \relax (x)}^2+10\,x^2\,\ln \relax (x)+2\,x^2+16\,x\,\ln \relax (x)-10\,x+64\,{\ln \relax (x)}^2-80\,\ln \relax (x)+25\right )}{5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.16, size = 61, normalized size = 2.54 \begin {gather*} \frac {2 x^{3}}{5} - 2 x^{2} + 5 x + \left (\frac {x^{5}}{5} - \frac {16 x^{3}}{5} + \frac {64 x}{5}\right ) \log {\relax (x )}^{2} + \left (- \frac {2 x^{4}}{5} + 2 x^{3} + \frac {16 x^{2}}{5} - 16 x\right ) \log {\relax (x )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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