Optimal. Leaf size=20 \[ \log \left (2 \left (5+\log \left (\frac {25 (-2+x)^2 \log (5)}{16 e^5}\right )\right )\right ) \]
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Rubi [A]
time = 0.05, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {12, 6873, 2437,
2339, 29} \begin {gather*} \log \left (\log \left (\frac {25 (2-x)^2 \log (5)}{16 e^5}\right )+5\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 29
Rule 2339
Rule 2437
Rule 6873
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=2 \int \frac {1}{-10+5 x+(-2+x) \log \left (\frac {\left (100-100 x+25 x^2\right ) \log (5)}{16 e^5}\right )} \, dx\\ &=2 \int \frac {1}{(-2+x) \left (5+\log \left (\frac {25 (-2+x)^2 \log (5)}{16 e^5}\right )\right )} \, dx\\ &=2 \text {Subst}\left (\int \frac {1}{x \left (5+\log \left (\frac {25 x^2 \log (5)}{16 e^5}\right )\right )} \, dx,x,-2+x\right )\\ &=\text {Subst}\left (\int \frac {1}{x} \, dx,x,5+\log \left (\frac {25 (-2+x)^2 \log (5)}{16 e^5}\right )\right )\\ &=\log \left (5+\log \left (\frac {25 (-2+x)^2 \log (5)}{16 e^5}\right )\right )\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.03, size = 18, normalized size = 0.90 \begin {gather*} \log \left (5+\log \left (\frac {25 (-2+x)^2 \log (5)}{16 e^5}\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.58, size = 23, normalized size = 1.15
method | result | size |
risch | \(\ln \left (\ln \left (\frac {\left (25 x^{2}-100 x +100\right ) \ln \left (5\right ) {\mathrm e}^{-5}}{16}\right )+5\right )\) | \(21\) |
default | \(\ln \left (2 \ln \left (5\right )-4 \ln \left (2\right )+\ln \left (\ln \left (5\right )\right )+\ln \left (x^{2}-4 x +4\right )\right )\) | \(23\) |
norman | \(\ln \left (\ln \left (\frac {\left (25 x^{2}-100 x +100\right ) \ln \left (5\right ) {\mathrm e}^{-5}}{16}\right )+5\right )\) | \(23\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.46, size = 17, normalized size = 0.85 \begin {gather*} \log \left (\log \left (5\right ) - 2 \, \log \left (2\right ) + \log \left (x - 2\right ) + \frac {1}{2} \, \log \left (\log \left (5\right )\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.34, size = 18, normalized size = 0.90 \begin {gather*} \log \left (\log \left (\frac {25}{16} \, {\left (x^{2} - 4 \, x + 4\right )} e^{\left (-5\right )} \log \left (5\right )\right ) + 5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.06, size = 26, normalized size = 1.30 \begin {gather*} \log {\left (\log {\left (\frac {\left (\frac {25 x^{2}}{16} - \frac {25 x}{4} + \frac {25}{4}\right ) \log {\left (5 \right )}}{e^{5}} \right )} + 5 \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.43, size = 27, normalized size = 1.35 \begin {gather*} \log \left (2 \, \log \left (5\right ) - 4 \, \log \left (2\right ) + \log \left (x^{2} \log \left (5\right ) - 4 \, x \log \left (5\right ) + 4 \, \log \left (5\right )\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.02, size = 20, normalized size = 1.00 \begin {gather*} \ln \left (\ln \left (\frac {{\mathrm {e}}^{-5}\,\ln \left (5\right )\,\left (25\,x^2-100\,x+100\right )}{16}\right )+5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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