Optimal. Leaf size=27 \[ 1-e^2+\frac {1}{x}-\frac {x}{5}+\log \left (\left (e^2-x\right ) x^2\right ) \]
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Rubi [A]
time = 0.04, antiderivative size = 21, normalized size of antiderivative = 0.78, number of steps
used = 3, number of rules used = 2, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {1607, 1634}
\begin {gather*} -\frac {x}{5}+\frac {1}{x}+\log \left (e^2-x\right )+2 \log (x) \end {gather*}
Antiderivative was successfully verified.
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Rule 1607
Rule 1634
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 x-15 x^2+x^3+e^2 \left (-5+10 x-x^2\right )}{\left (5 e^2-5 x\right ) x^2} \, dx\\ &=\int \left (-\frac {1}{5}-\frac {1}{x^2}+\frac {2}{x}+\frac {1}{-e^2+x}\right ) \, dx\\ &=\frac {1}{x}-\frac {x}{5}+\log \left (e^2-x\right )+2 \log (x)\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.01, size = 27, normalized size = 1.00 \begin {gather*} \frac {1}{5} \left (\frac {5}{x}-x+5 \log \left (e^2-x\right )+10 \log (x)\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.10, size = 19, normalized size = 0.70
method | result | size |
default | \(-\frac {x}{5}+\ln \left (x -{\mathrm e}^{2}\right )+\frac {1}{x}+2 \ln \left (x \right )\) | \(19\) |
risch | \(-\frac {x}{5}+\ln \left (x -{\mathrm e}^{2}\right )+\frac {1}{x}+2 \ln \left (x \right )\) | \(19\) |
norman | \(\frac {1-\frac {x^{2}}{5}}{x}+2 \ln \left (x \right )+\ln \left ({\mathrm e}^{2}-x \right )\) | \(24\) |
meijerg | \({\mathrm e}^{-2} \left (\ln \left (1-x \,{\mathrm e}^{-2}\right )-\ln \left (x \right )+2-i \pi +\frac {{\mathrm e}^{2}}{x}\right )-\left (-\frac {{\mathrm e}^{2}}{5}-3\right ) \ln \left (1-x \,{\mathrm e}^{-2}\right )+\left (2 \,{\mathrm e}^{2}+1\right ) {\mathrm e}^{-2} \left (-\ln \left (1-x \,{\mathrm e}^{-2}\right )+\ln \left (x \right )-2+i \pi \right )+\frac {{\mathrm e}^{2} \left (-x \,{\mathrm e}^{-2}-\ln \left (1-x \,{\mathrm e}^{-2}\right )\right )}{5}\) | \(92\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.27, size = 18, normalized size = 0.67 \begin {gather*} -\frac {1}{5} \, x + \frac {1}{x} + \log \left (x - e^{2}\right ) + 2 \, \log \left (x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.30, size = 25, normalized size = 0.93 \begin {gather*} -\frac {x^{2} - 5 \, x \log \left (x - e^{2}\right ) - 10 \, x \log \left (x\right ) - 5}{5 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.19, size = 17, normalized size = 0.63 \begin {gather*} - \frac {x}{5} + 2 \log {\left (x \right )} + \log {\left (x - e^{2} \right )} + \frac {1}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.41, size = 20, normalized size = 0.74 \begin {gather*} -\frac {1}{5} \, x + \frac {1}{x} + \log \left ({\left | x - e^{2} \right |}\right ) + 2 \, \log \left ({\left | x \right |}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.29, size = 18, normalized size = 0.67 \begin {gather*} \ln \left (x-{\mathrm {e}}^2\right )-\frac {x}{5}+2\,\ln \left (x\right )+\frac {1}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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