Optimal. Leaf size=27 \[ \frac {4+x+\log \left (\frac {e^{-x} x}{2 \left (e^{98}-x\right )}\right )}{x} \]
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Rubi [A] time = 0.30, antiderivative size = 30, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 4, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {1593, 6742, 893, 2551} \begin {gather*} \frac {4}{x}+\frac {\log \left (\frac {e^{-x} x}{2 \left (e^{98}-x\right )}\right )}{x} \end {gather*}
Antiderivative was successfully verified.
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Rule 893
Rule 1593
Rule 2551
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{98} (-3-x)+4 x+x^2+\left (-e^{98}+x\right ) \log \left (\frac {e^{-x} x}{2 e^{98}-2 x}\right )}{\left (e^{98}-x\right ) x^2} \, dx\\ &=\int \left (\frac {-3 e^{98}+\left (4-e^{98}\right ) x+x^2}{\left (e^{98}-x\right ) x^2}-\frac {\log \left (\frac {e^{-x} x}{2 e^{98}-2 x}\right )}{x^2}\right ) \, dx\\ &=\int \frac {-3 e^{98}+\left (4-e^{98}\right ) x+x^2}{\left (e^{98}-x\right ) x^2} \, dx-\int \frac {\log \left (\frac {e^{-x} x}{2 e^{98}-2 x}\right )}{x^2} \, dx\\ &=\frac {\log \left (\frac {e^{-x} x}{2 \left (e^{98}-x\right )}\right )}{x}+\int \left (\frac {1}{e^{98} \left (e^{98}-x\right )}-\frac {3}{x^2}+\frac {1-e^{98}}{e^{98} x}\right ) \, dx-\int \frac {e^{98}-e^{98} x+x^2}{\left (e^{98}-x\right ) x^2} \, dx\\ &=\frac {3}{x}-\frac {\log \left (e^{98}-x\right )}{e^{98}}-\left (1-\frac {1}{e^{98}}\right ) \log (x)+\frac {\log \left (\frac {e^{-x} x}{2 \left (e^{98}-x\right )}\right )}{x}-\int \left (\frac {1}{e^{98} \left (e^{98}-x\right )}+\frac {1}{x^2}+\frac {1-e^{98}}{e^{98} x}\right ) \, dx\\ &=\frac {4}{x}+\frac {\log \left (\frac {e^{-x} x}{2 \left (e^{98}-x\right )}\right )}{x}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.09, size = 29, normalized size = 1.07 \begin {gather*} \frac {4}{x}+\frac {\log \left (\frac {e^{-x} x}{2 e^{98}-2 x}\right )}{x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.70, size = 22, normalized size = 0.81 \begin {gather*} \frac {\log \left (-\frac {x e^{\left (-x\right )}}{2 \, {\left (x - e^{98}\right )}}\right ) + 4}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.28, size = 18, normalized size = 0.67 \begin {gather*} \frac {\log \left (-\frac {x}{2 \, {\left (x - e^{98}\right )}}\right ) + 4}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.66, size = 26, normalized size = 0.96
method | result | size |
norman | \(\frac {4+\ln \left (\frac {x \,{\mathrm e}^{-x}}{2 \,{\mathrm e}^{98}-2 x}\right )}{x}\) | \(26\) |
default | \(\frac {\ln \left (\frac {x \,{\mathrm e}^{-x}}{2 \,{\mathrm e}^{98}-2 x}\right )}{x}+\ln \left (2 \,{\mathrm e}^{98}-2 x \right )-2 \,{\mathrm e}^{98} {\mathrm e}^{-196} \ln \relax (x )+\frac {\left ({\mathrm e}^{98}\right )^{2} {\mathrm e}^{-196}}{x}+2 \,{\mathrm e}^{98} {\mathrm e}^{-196} \ln \left (-{\mathrm e}^{98}+x \right )+\frac {1}{{\mathrm e}^{98}-x}+2 \,{\mathrm e}^{-98} \ln \relax (x )-2 \,{\mathrm e}^{-98} \ln \left (-{\mathrm e}^{98}+x \right )+\frac {\left ({\mathrm e}^{98}\right )^{2} {\mathrm e}^{-196}}{-{\mathrm e}^{98}+x}-{\mathrm e}^{98} {\mathrm e}^{-98} \ln \left (-{\mathrm e}^{98}+x \right )+\frac {3 \,{\mathrm e}^{-98} {\mathrm e}^{98}}{x}\) | \(197\) |
risch | \(-\frac {\ln \left ({\mathrm e}^{x}\right )}{x}-\frac {i \pi \mathrm {csgn}\left (\frac {i x \,{\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right )^{3}-i \pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right ) \mathrm {csgn}\left (\frac {i x \,{\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right )^{2}-i \pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{98}-x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right )^{2}-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x \,{\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right )^{2}-i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right )^{2}+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right ) \mathrm {csgn}\left (\frac {i x \,{\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right )+i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right )^{3}+i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{98}-x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right )-8+2 \ln \relax (2)-2 \ln \relax (x )+2 \ln \left ({\mathrm e}^{98}-x \right )}{2 x}\) | \(292\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.51, size = 264, normalized size = 9.78 \begin {gather*} {\left (e^{\left (-196\right )} \log \left (x - e^{98}\right ) - e^{\left (-196\right )} \log \relax (x) + \frac {e^{\left (-98\right )}}{x}\right )} e^{98} \log \left (-\frac {x}{2 \, {\left (x e^{x} - e^{\left (x + 98\right )}\right )}}\right ) + {\left (e^{\left (-98\right )} \log \left (x - e^{98}\right ) - e^{\left (-98\right )} \log \relax (x)\right )} e^{98} + 3 \, {\left (e^{\left (-196\right )} \log \left (x - e^{98}\right ) - e^{\left (-196\right )} \log \relax (x) + \frac {e^{\left (-98\right )}}{x}\right )} e^{98} - \frac {1}{2} \, {\left (2 \, {\left (x - e^{98} - \log \relax (x)\right )} \log \left (x - e^{98}\right ) + \log \left (x - e^{98}\right )^{2} - 2 \, x \log \relax (x) + \log \relax (x)^{2}\right )} e^{\left (-98\right )} - 4 \, e^{\left (-98\right )} \log \left (x - e^{98}\right ) + 4 \, e^{\left (-98\right )} \log \relax (x) - {\left (e^{\left (-98\right )} \log \left (x - e^{98}\right ) - e^{\left (-98\right )} \log \relax (x)\right )} \log \left (-\frac {x}{2 \, {\left (x e^{x} - e^{\left (x + 98\right )}\right )}}\right ) + \frac {{\left (x \log \left (x - e^{98}\right )^{2} + x \log \relax (x)^{2} + 2 \, {\left (x^{2} - x {\left (e^{98} - 1\right )} - x \log \relax (x)\right )} \log \left (x - e^{98}\right ) - 2 \, {\left (x^{2} - x {\left (e^{98} - 1\right )}\right )} \log \relax (x) + 2 \, e^{98}\right )} e^{\left (-98\right )}}{2 \, x} - \log \left (x - e^{98}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.46, size = 24, normalized size = 0.89 \begin {gather*} \frac {\ln \left (-\frac {x\,{\mathrm {e}}^{-x}}{2\,\left (x-{\mathrm {e}}^{98}\right )}\right )+4}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.25, size = 19, normalized size = 0.70 \begin {gather*} \frac {\log {\left (\frac {x e^{- x}}{- 2 x + 2 e^{98}} \right )}}{x} + \frac {4}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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