Optimal. Leaf size=34 \[ 3+\frac {5 \left (\frac {5}{x}+\frac {1}{5} (1-x) x\right ) \log ^2(5)}{4 (-4+x) x} \]
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Rubi [A] time = 0.06, antiderivative size = 38, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {12, 1594, 27, 1620} \begin {gather*} -\frac {25 \log ^2(5)}{16 x^2}-\frac {25 \log ^2(5)}{64 x}+\frac {23 \log ^2(5)}{64 (4-x)} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 27
Rule 1594
Rule 1620
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\log ^2(5) \int \frac {200-75 x+3 x^3}{64 x^3-32 x^4+4 x^5} \, dx\\ &=\log ^2(5) \int \frac {200-75 x+3 x^3}{x^3 \left (64-32 x+4 x^2\right )} \, dx\\ &=\log ^2(5) \int \frac {200-75 x+3 x^3}{4 (-4+x)^2 x^3} \, dx\\ &=\frac {1}{4} \log ^2(5) \int \frac {200-75 x+3 x^3}{(-4+x)^2 x^3} \, dx\\ &=\frac {1}{4} \log ^2(5) \int \left (\frac {23}{16 (-4+x)^2}+\frac {25}{2 x^3}+\frac {25}{16 x^2}\right ) \, dx\\ &=\frac {23 \log ^2(5)}{64 (4-x)}-\frac {25 \log ^2(5)}{16 x^2}-\frac {25 \log ^2(5)}{64 x}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.01, size = 23, normalized size = 0.68 \begin {gather*} \frac {\left (25-3 x^2\right ) \log ^2(5)}{4 (-4+x) x^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.48, size = 24, normalized size = 0.71 \begin {gather*} -\frac {{\left (3 \, x^{2} - 25\right )} \log \relax (5)^{2}}{4 \, {\left (x^{3} - 4 \, x^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.26, size = 22, normalized size = 0.65 \begin {gather*} -\frac {1}{64} \, {\left (\frac {23}{x - 4} + \frac {25 \, {\left (x + 4\right )}}{x^{2}}\right )} \log \relax (5)^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 21, normalized size = 0.62
method | result | size |
risch | \(\frac {\ln \relax (5)^{2} \left (-\frac {3 x^{2}}{4}+\frac {25}{4}\right )}{\left (x -4\right ) x^{2}}\) | \(21\) |
gosper | \(-\frac {\ln \relax (5)^{2} \left (3 x^{2}-25\right )}{4 x^{2} \left (x -4\right )}\) | \(22\) |
default | \(\frac {\ln \relax (5)^{2} \left (-\frac {25}{4 x^{2}}-\frac {25}{16 x}-\frac {23}{16 \left (x -4\right )}\right )}{4}\) | \(25\) |
norman | \(\frac {-\frac {3 x^{2} \ln \relax (5)^{2}}{4}+\frac {25 \ln \relax (5)^{2}}{4}}{\left (x -4\right ) x^{2}}\) | \(26\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.42, size = 24, normalized size = 0.71 \begin {gather*} -\frac {{\left (3 \, x^{2} - 25\right )} \log \relax (5)^{2}}{4 \, {\left (x^{3} - 4 \, x^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.54, size = 30, normalized size = 0.88 \begin {gather*} \frac {3\,x^2\,{\ln \relax (5)}^2-25\,{\ln \relax (5)}^2}{16\,x^2-4\,x^3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.19, size = 26, normalized size = 0.76 \begin {gather*} \frac {- 3 x^{2} \log {\relax (5 )}^{2} + 25 \log {\relax (5 )}^{2}}{4 x^{3} - 16 x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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