Optimal. Leaf size=24 \[ e^{-4 e^{3 e^9}} (x+(5+2 x) \log (x))^2 \]
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Rubi [B] time = 0.11, antiderivative size = 142, normalized size of antiderivative = 5.92, number of steps used = 14, number of rules used = 9, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {12, 14, 2357, 2295, 2301, 2304, 2330, 2296, 2305} \begin {gather*} -2 e^{-4 e^{3 e^9}} x^2+4 e^{-4 e^{3 e^9}} x^2 \log ^2(x)+4 e^{-4 e^{3 e^9}} x^2 \log (x)-10 e^{-4 e^{3 e^9}} x+\frac {1}{3} e^{-4 e^{3 e^9}} (3 x+5)^2+20 e^{-4 e^{3 e^9}} x \log ^2(x)+25 e^{-4 e^{3 e^9}} \log ^2(x)+10 e^{-4 e^{3 e^9}} x \log (x) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 14
Rule 2295
Rule 2296
Rule 2301
Rule 2304
Rule 2305
Rule 2330
Rule 2357
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=e^{-4 e^{3 e^9}} \int \frac {10 x+6 x^2+\left (50+50 x+16 x^2\right ) \log (x)+\left (20 x+8 x^2\right ) \log ^2(x)}{x} \, dx\\ &=e^{-4 e^{3 e^9}} \int \left (2 (5+3 x)+\frac {2 \left (25+25 x+8 x^2\right ) \log (x)}{x}+4 (5+2 x) \log ^2(x)\right ) \, dx\\ &=\frac {1}{3} e^{-4 e^{3 e^9}} (5+3 x)^2+\left (2 e^{-4 e^{3 e^9}}\right ) \int \frac {\left (25+25 x+8 x^2\right ) \log (x)}{x} \, dx+\left (4 e^{-4 e^{3 e^9}}\right ) \int (5+2 x) \log ^2(x) \, dx\\ &=\frac {1}{3} e^{-4 e^{3 e^9}} (5+3 x)^2+\left (2 e^{-4 e^{3 e^9}}\right ) \int \left (25 \log (x)+\frac {25 \log (x)}{x}+8 x \log (x)\right ) \, dx+\left (4 e^{-4 e^{3 e^9}}\right ) \int \left (5 \log ^2(x)+2 x \log ^2(x)\right ) \, dx\\ &=\frac {1}{3} e^{-4 e^{3 e^9}} (5+3 x)^2+\left (8 e^{-4 e^{3 e^9}}\right ) \int x \log ^2(x) \, dx+\left (16 e^{-4 e^{3 e^9}}\right ) \int x \log (x) \, dx+\left (20 e^{-4 e^{3 e^9}}\right ) \int \log ^2(x) \, dx+\left (50 e^{-4 e^{3 e^9}}\right ) \int \log (x) \, dx+\left (50 e^{-4 e^{3 e^9}}\right ) \int \frac {\log (x)}{x} \, dx\\ &=-50 e^{-4 e^{3 e^9}} x-4 e^{-4 e^{3 e^9}} x^2+\frac {1}{3} e^{-4 e^{3 e^9}} (5+3 x)^2+50 e^{-4 e^{3 e^9}} x \log (x)+8 e^{-4 e^{3 e^9}} x^2 \log (x)+25 e^{-4 e^{3 e^9}} \log ^2(x)+20 e^{-4 e^{3 e^9}} x \log ^2(x)+4 e^{-4 e^{3 e^9}} x^2 \log ^2(x)-\left (8 e^{-4 e^{3 e^9}}\right ) \int x \log (x) \, dx-\left (40 e^{-4 e^{3 e^9}}\right ) \int \log (x) \, dx\\ &=-10 e^{-4 e^{3 e^9}} x-2 e^{-4 e^{3 e^9}} x^2+\frac {1}{3} e^{-4 e^{3 e^9}} (5+3 x)^2+10 e^{-4 e^{3 e^9}} x \log (x)+4 e^{-4 e^{3 e^9}} x^2 \log (x)+25 e^{-4 e^{3 e^9}} \log ^2(x)+20 e^{-4 e^{3 e^9}} x \log ^2(x)+4 e^{-4 e^{3 e^9}} x^2 \log ^2(x)\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 0.03, size = 57, normalized size = 2.38 \begin {gather*} 2 e^{-4 e^{3 e^9}} \left (\frac {x^2}{2}+5 x \log (x)+2 x^2 \log (x)+\frac {25 \log ^2(x)}{2}+10 x \log ^2(x)+2 x^2 \log ^2(x)\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.51, size = 41, normalized size = 1.71 \begin {gather*} {\left ({\left (4 \, x^{2} + 20 \, x + 25\right )} \log \relax (x)^{2} + x^{2} + 2 \, {\left (2 \, x^{2} + 5 \, x\right )} \log \relax (x)\right )} e^{\left (-4 \, e^{\left (3 \, e^{9}\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.16, size = 41, normalized size = 1.71 \begin {gather*} {\left ({\left (4 \, x^{2} + 20 \, x + 25\right )} \log \relax (x)^{2} + x^{2} + 2 \, {\left (2 \, x^{2} + 5 \, x\right )} \log \relax (x)\right )} e^{\left (-4 \, e^{\left (3 \, e^{9}\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.05, size = 50, normalized size = 2.08
method | result | size |
default | \({\mathrm e}^{-4 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} \left (4 x^{2} \ln \relax (x )^{2}+4 x^{2} \ln \relax (x )+x^{2}+20 x \ln \relax (x )^{2}+10 x \ln \relax (x )+25 \ln \relax (x )^{2}\right )\) | \(50\) |
risch | \({\mathrm e}^{-4 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} \left (4 x^{2}+20 x +25\right ) \ln \relax (x )^{2}+{\mathrm e}^{-4 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} \left (4 x^{2}+10 x \right ) \ln \relax (x )+{\mathrm e}^{-4 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} x^{2}\) | \(57\) |
norman | \(\left ({\mathrm e}^{-2 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} x^{2}+25 \,{\mathrm e}^{-2 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} \ln \relax (x )^{2}+10 \,{\mathrm e}^{-2 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} x \ln \relax (x )+20 \,{\mathrm e}^{-2 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} x \ln \relax (x )^{2}+4 \,{\mathrm e}^{-2 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} x^{2} \ln \relax (x )+4 \,{\mathrm e}^{-2 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} x^{2} \ln \relax (x )^{2}\right ) {\mathrm e}^{-2 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}}\) | \(99\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.37, size = 66, normalized size = 2.75 \begin {gather*} {\left (2 \, {\left (2 \, \log \relax (x)^{2} - 2 \, \log \relax (x) + 1\right )} x^{2} + 8 \, x^{2} \log \relax (x) + 20 \, {\left (\log \relax (x)^{2} - 2 \, \log \relax (x) + 2\right )} x - x^{2} + 50 \, x \log \relax (x) + 25 \, \log \relax (x)^{2} - 40 \, x\right )} e^{\left (-4 \, e^{\left (3 \, e^{9}\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.37, size = 22, normalized size = 0.92 \begin {gather*} {\mathrm {e}}^{-4\,{\mathrm {e}}^{3\,{\mathrm {e}}^9}}\,{\left (x+5\,\ln \relax (x)+2\,x\,\ln \relax (x)\right )}^2 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.24, size = 60, normalized size = 2.50 \begin {gather*} \frac {x^{2}}{e^{4 e^{3 e^{9}}}} + \frac {\left (4 x^{2} + 10 x\right ) \log {\relax (x )}}{e^{4 e^{3 e^{9}}}} + \frac {\left (4 x^{2} + 20 x + 25\right ) \log {\relax (x )}^{2}}{e^{4 e^{3 e^{9}}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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