Optimal. Leaf size=22 \[ -3+\log (4-x)+\frac {1}{e^{10} (1-\log (-1+x))} \]
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Rubi [A]
time = 0.24, antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps
used = 7, number of rules used = 6, integrand size = 88, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {6820, 12, 6860,
2437, 2339, 30} \begin {gather*} \log (4-x)+\frac {1}{e^{10} (1-\log (x-1))} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 30
Rule 2339
Rule 2437
Rule 6820
Rule 6860
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4+e^{10} (-1+x)+x-2 e^{10} (-1+x) \log (-1+x)+e^{10} (-1+x) \log ^2(-1+x)}{e^{10} \left (4-5 x+x^2\right ) (1-\log (-1+x))^2} \, dx\\ &=\frac {\int \frac {-4+e^{10} (-1+x)+x-2 e^{10} (-1+x) \log (-1+x)+e^{10} (-1+x) \log ^2(-1+x)}{\left (4-5 x+x^2\right ) (1-\log (-1+x))^2} \, dx}{e^{10}}\\ &=\frac {\int \left (\frac {e^{10}}{-4+x}+\frac {1}{(-1+x) (-1+\log (-1+x))^2}\right ) \, dx}{e^{10}}\\ &=\log (4-x)+\frac {\int \frac {1}{(-1+x) (-1+\log (-1+x))^2} \, dx}{e^{10}}\\ &=\log (4-x)+\frac {\text {Subst}\left (\int \frac {1}{x (-1+\log (x))^2} \, dx,x,-1+x\right )}{e^{10}}\\ &=\log (4-x)+\frac {\text {Subst}\left (\int \frac {1}{x^2} \, dx,x,-1+\log (-1+x)\right )}{e^{10}}\\ &=\log (4-x)+\frac {1}{e^{10} (1-\log (-1+x))}\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.04, size = 25, normalized size = 1.14 \begin {gather*} \frac {e^{10} \log (4-x)-\frac {1}{-1+\log (-1+x)}}{e^{10}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 1.84, size = 26, normalized size = 1.18
method | result | size |
risch | \(-\frac {{\mathrm e}^{-10}}{\ln \left (x -1\right )-1}+\ln \left (x -4\right )\) | \(18\) |
norman | \(-\frac {{\mathrm e}^{-10}}{\ln \left (x -1\right )-1}+\ln \left (x -4\right )\) | \(20\) |
derivativedivides | \({\mathrm e}^{-10} \left (-\frac {1}{\ln \left (x -1\right )-1}+{\mathrm e}^{10} \ln \left (x -4\right )\right )\) | \(26\) |
default | \({\mathrm e}^{-10} \left (-\frac {1}{\ln \left (x -1\right )-1}+{\mathrm e}^{10} \ln \left (x -4\right )\right )\) | \(26\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.41, size = 21, normalized size = 0.95 \begin {gather*} -\frac {1}{e^{10} \log \left (x - 1\right ) - e^{10}} + \log \left (x - 4\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.40, size = 36, normalized size = 1.64 \begin {gather*} \frac {e^{10} \log \left (x - 1\right ) \log \left (x - 4\right ) - e^{10} \log \left (x - 4\right ) - 1}{e^{10} \log \left (x - 1\right ) - e^{10}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.05, size = 17, normalized size = 0.77 \begin {gather*} \log {\left (x - 4 \right )} - \frac {1}{e^{10} \log {\left (x - 1 \right )} - e^{10}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.40, size = 36, normalized size = 1.64 \begin {gather*} \frac {e^{10} \log \left (x - 1\right ) \log \left (x - 4\right ) - e^{10} \log \left (x - 4\right ) - 1}{e^{10} \log \left (x - 1\right ) - e^{10}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.26, size = 17, normalized size = 0.77 \begin {gather*} \ln \left (x-4\right )-\frac {{\mathrm {e}}^{-10}}{\ln \left (x-1\right )-1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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