Optimal. Leaf size=21 \[ 5+5 \log \left (\frac {16 \log \left (\frac {1}{2} (5-e) x\right )}{x}\right ) \]
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Rubi [A] time = 0.14, antiderivative size = 19, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2444, 2365, 43} \begin {gather*} 5 \log \left (\log \left (\frac {1}{2} (5-e) x\right )\right )-5 \log (x) \end {gather*}
Antiderivative was successfully verified.
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Rule 43
Rule 2365
Rule 2444
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5-5 \log \left (\frac {1}{2} (5-e) x\right )}{x \log \left (\frac {1}{2} (5 x-e x)\right )} \, dx\\ &=\int \frac {5-5 \log \left (\frac {1}{2} (5-e) x\right )}{x \log \left (\frac {1}{2} (5-e) x\right )} \, dx\\ &=\operatorname {Subst}\left (\int \frac {5-5 x}{x} \, dx,x,\log \left (\frac {1}{2} (5-e) x\right )\right )\\ &=\operatorname {Subst}\left (\int \left (-5+\frac {5}{x}\right ) \, dx,x,\log \left (\frac {1}{2} (5-e) x\right )\right )\\ &=-5 \log (x)+5 \log \left (\log \left (\frac {1}{2} (5-e) x\right )\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.02, size = 17, normalized size = 0.81 \begin {gather*} -5 \log (x)+5 \log \left (\log \left (-\frac {1}{2} (-5+e) x\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.99, size = 26, normalized size = 1.24 \begin {gather*} -5 \, \log \left (-\frac {1}{2} \, x e + \frac {5}{2} \, x\right ) + 5 \, \log \left (\log \left (-\frac {1}{2} \, x e + \frac {5}{2} \, x\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.17, size = 18, normalized size = 0.86 \begin {gather*} -5 \, \log \relax (x) + 5 \, \log \left (\log \left (-\frac {1}{2} \, x e + \frac {5}{2} \, x\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 19, normalized size = 0.90
method | result | size |
risch | \(-5 \ln \relax (x )+5 \ln \left (\ln \left (-\frac {x \,{\mathrm e}}{2}+\frac {5 x}{2}\right )\right )\) | \(19\) |
derivativedivides | \(-5 \ln \left (\left (-\frac {{\mathrm e}}{2}+\frac {5}{2}\right ) x \right )+5 \ln \left (\ln \left (\left (-\frac {{\mathrm e}}{2}+\frac {5}{2}\right ) x \right )\right )\) | \(25\) |
default | \(-5 \ln \left (\left (-\frac {{\mathrm e}}{2}+\frac {5}{2}\right ) x \right )+5 \ln \left (\ln \left (\left (-\frac {{\mathrm e}}{2}+\frac {5}{2}\right ) x \right )\right )\) | \(25\) |
norman | \(-5 \ln \left (-\frac {x \,{\mathrm e}}{2}+\frac {5 x}{2}\right )+5 \ln \left (\ln \left (-\frac {x \,{\mathrm e}}{2}+\frac {5 x}{2}\right )\right )\) | \(27\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.34, size = 26, normalized size = 1.24 \begin {gather*} -5 \, \log \left (-\frac {1}{2} \, x e + \frac {5}{2} \, x\right ) + 5 \, \log \left (\log \left (-\frac {1}{2} \, x e + \frac {5}{2} \, x\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.26, size = 18, normalized size = 0.86 \begin {gather*} 5\,\ln \left (\ln \left (\frac {5\,x}{2}-\frac {x\,\mathrm {e}}{2}\right )\right )-5\,\ln \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.10, size = 20, normalized size = 0.95 \begin {gather*} - 5 \log {\relax (x )} + 5 \log {\left (\log {\left (- \frac {e x}{2} + \frac {5 x}{2} \right )} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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