Optimal. Leaf size=23 \[ -e^{\frac {10}{x (-x+\log (2))}}+x-\log (5) \]
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Rubi [A] time = 0.93, antiderivative size = 19, normalized size of antiderivative = 0.83, number of steps used = 5, number of rules used = 4, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {1594, 27, 6688, 6706} \begin {gather*} x-e^{-\frac {10}{x (x-\log (2))}} \end {gather*}
Antiderivative was successfully verified.
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Rule 27
Rule 1594
Rule 6688
Rule 6706
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x^4-2 x^3 \log (2)+x^2 \log ^2(2)+e^{\frac {10}{-x^2+x \log (2)}} (-20 x+10 \log (2))}{x^2 \left (x^2-2 x \log (2)+\log ^2(2)\right )} \, dx\\ &=\int \frac {x^4-2 x^3 \log (2)+x^2 \log ^2(2)+e^{\frac {10}{-x^2+x \log (2)}} (-20 x+10 \log (2))}{x^2 (x-\log (2))^2} \, dx\\ &=\int \left (1+\frac {10 e^{-\frac {10}{x (x-\log (2))}} (-2 x+\log (2))}{x^2 (x-\log (2))^2}\right ) \, dx\\ &=x+10 \int \frac {e^{-\frac {10}{x (x-\log (2))}} (-2 x+\log (2))}{x^2 (x-\log (2))^2} \, dx\\ &=-e^{-\frac {10}{x (x-\log (2))}}+x\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.18, size = 30, normalized size = 1.30 \begin {gather*} -e^{\frac {10}{x \log (2)}-\frac {10}{(x-\log (2)) \log (2)}}+x \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.54, size = 18, normalized size = 0.78 \begin {gather*} x - e^{\left (-\frac {10}{x^{2} - x \log \relax (2)}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.38, size = 18, normalized size = 0.78 \begin {gather*} x - e^{\left (-\frac {10}{x^{2} - x \log \relax (2)}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.27, size = 19, normalized size = 0.83
method | result | size |
risch | \(-{\mathrm e}^{\frac {10}{x \left (\ln \relax (2)-x \right )}}+x\) | \(19\) |
norman | \(\frac {x^{2} {\mathrm e}^{\frac {10}{x \ln \relax (2)-x^{2}}}+x \ln \relax (2)^{2}-x^{3}-x \ln \relax (2) {\mathrm e}^{\frac {10}{x \ln \relax (2)-x^{2}}}}{\left (\ln \relax (2)-x \right ) x}\) | \(68\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.67, size = 80, normalized size = 3.48 \begin {gather*} 2 \, {\left (\frac {\log \relax (2)}{x - \log \relax (2)} - \log \left (x - \log \relax (2)\right )\right )} \log \relax (2) + 2 \, \log \relax (2) \log \left (x - \log \relax (2)\right ) + x - \frac {2 \, \log \relax (2)^{2}}{x - \log \relax (2)} - e^{\left (-\frac {10}{x \log \relax (2) - \log \relax (2)^{2}} + \frac {10}{x \log \relax (2)}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.27, size = 19, normalized size = 0.83 \begin {gather*} x-{\mathrm {e}}^{\frac {10}{x\,\ln \relax (2)-x^2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.26, size = 12, normalized size = 0.52 \begin {gather*} x - e^{\frac {10}{- x^{2} + x \log {\relax (2 )}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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