Optimal. Leaf size=25 \[ \frac {5 e^{\frac {2}{x}+x^2} \left (\log \left (\frac {1}{x}\right )-\log (x)\right )}{x} \]
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Rubi [B] time = 1.06, antiderivative size = 71, normalized size of antiderivative = 2.84, number of steps used = 9, number of rules used = 5, integrand size = 78, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.064, Rules used = {6688, 12, 6742, 2288, 2554} \begin {gather*} \frac {5 e^{x^2+\frac {2}{x}} \left (1-x^3\right ) \log \left (\frac {1}{x}\right )}{\left (\frac {1}{x^2}-x\right ) x^3}-\frac {5 e^{x^2+\frac {2}{x}} \left (1-x^3\right ) \log (x)}{\left (\frac {1}{x^2}-x\right ) x^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2288
Rule 2554
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 e^{\frac {2}{x}+x^2} \left (-2 x+\left (-2-x+2 x^3\right ) \log \left (\frac {1}{x}\right )+\left (2+x-2 x^3\right ) \log (x)\right )}{x^3} \, dx\\ &=5 \int \frac {e^{\frac {2}{x}+x^2} \left (-2 x+\left (-2-x+2 x^3\right ) \log \left (\frac {1}{x}\right )+\left (2+x-2 x^3\right ) \log (x)\right )}{x^3} \, dx\\ &=5 \int \left (\frac {e^{\frac {2}{x}+x^2} \left (-2 x-2 \log \left (\frac {1}{x}\right )-x \log \left (\frac {1}{x}\right )+2 x^3 \log \left (\frac {1}{x}\right )\right )}{x^3}-\frac {e^{\frac {2}{x}+x^2} \left (-2-x+2 x^3\right ) \log (x)}{x^3}\right ) \, dx\\ &=5 \int \frac {e^{\frac {2}{x}+x^2} \left (-2 x-2 \log \left (\frac {1}{x}\right )-x \log \left (\frac {1}{x}\right )+2 x^3 \log \left (\frac {1}{x}\right )\right )}{x^3} \, dx-5 \int \frac {e^{\frac {2}{x}+x^2} \left (-2-x+2 x^3\right ) \log (x)}{x^3} \, dx\\ &=-\frac {5 e^{\frac {2}{x}+x^2} \left (1-x^3\right ) \log (x)}{\left (\frac {1}{x^2}-x\right ) x^3}+5 \int \frac {e^{\frac {2}{x}+x^2}}{x^2} \, dx+5 \int \frac {e^{\frac {2}{x}+x^2} \left (-2 x+\left (-2-x+2 x^3\right ) \log \left (\frac {1}{x}\right )\right )}{x^3} \, dx\\ &=-\frac {5 e^{\frac {2}{x}+x^2} \left (1-x^3\right ) \log (x)}{\left (\frac {1}{x^2}-x\right ) x^3}+5 \int \frac {e^{\frac {2}{x}+x^2}}{x^2} \, dx+5 \int \left (-\frac {2 e^{\frac {2}{x}+x^2}}{x^2}+\frac {e^{\frac {2}{x}+x^2} \left (-2-x+2 x^3\right ) \log \left (\frac {1}{x}\right )}{x^3}\right ) \, dx\\ &=-\frac {5 e^{\frac {2}{x}+x^2} \left (1-x^3\right ) \log (x)}{\left (\frac {1}{x^2}-x\right ) x^3}+5 \int \frac {e^{\frac {2}{x}+x^2}}{x^2} \, dx+5 \int \frac {e^{\frac {2}{x}+x^2} \left (-2-x+2 x^3\right ) \log \left (\frac {1}{x}\right )}{x^3} \, dx-10 \int \frac {e^{\frac {2}{x}+x^2}}{x^2} \, dx\\ &=\frac {5 e^{\frac {2}{x}+x^2} \left (1-x^3\right ) \log \left (\frac {1}{x}\right )}{\left (\frac {1}{x^2}-x\right ) x^3}-\frac {5 e^{\frac {2}{x}+x^2} \left (1-x^3\right ) \log (x)}{\left (\frac {1}{x^2}-x\right ) x^3}+2 \left (5 \int \frac {e^{\frac {2}{x}+x^2}}{x^2} \, dx\right )-10 \int \frac {e^{\frac {2}{x}+x^2}}{x^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.23, size = 25, normalized size = 1.00 \begin {gather*} \frac {5 e^{\frac {2}{x}+x^2} \left (\log \left (\frac {1}{x}\right )-\log (x)\right )}{x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.65, size = 22, normalized size = 0.88 \begin {gather*} e^{\left (\frac {x^{3} + x \log \left (\frac {10 \, \log \left (\frac {1}{x}\right )}{x}\right ) + 2}{x}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.15, size = 18, normalized size = 0.72 \begin {gather*} e^{\left (x^{2} + \frac {2}{x} + \log \left (-\frac {10 \, \log \relax (x)}{x}\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.40, size = 141, normalized size = 5.64
method | result | size |
risch | \(-{\mathrm e}^{\frac {i x \pi \mathrm {csgn}\left (\frac {i \ln \relax (x )}{x}\right )^{3}+i x \pi \mathrm {csgn}\left (\frac {i \ln \relax (x )}{x}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right )+i x \pi \mathrm {csgn}\left (\frac {i \ln \relax (x )}{x}\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )\right )-i x \pi \,\mathrm {csgn}\left (\frac {i \ln \relax (x )}{x}\right ) \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \ln \relax (x )\right )-2 i x \pi \mathrm {csgn}\left (\frac {i \ln \relax (x )}{x}\right )^{2}+2 x^{3}-2 x \ln \relax (x )+2 x \ln \left (\ln \relax (x )\right )+2 x \ln \relax (5)+2 x \ln \relax (2)+4}{2 x}}\) | \(141\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.58, size = 17, normalized size = 0.68 \begin {gather*} -\frac {10 \, e^{\left (x^{2} + \frac {2}{x}\right )} \log \relax (x)}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{\frac {x\,\ln \left (\frac {5\,\ln \left (\frac {1}{x}\right )-5\,\ln \relax (x)}{x}\right )+x^3+2}{x}}\,\left (2\,x-\ln \relax (x)\,\left (-2\,x^3+x+2\right )+\ln \left (\frac {1}{x}\right )\,\left (-2\,x^3+x+2\right )\right )}{x^2\,\ln \relax (x)-x^2\,\ln \left (\frac {1}{x}\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.55, size = 19, normalized size = 0.76 \begin {gather*} e^{\frac {x^{3} + x \log {\left (- \frac {10 \log {\relax (x )}}{x} \right )} + 2}{x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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