Optimal. Leaf size=20 \[ \frac {-4+x+\log (x)-4 x \log (x)}{\frac {1}{5}-5 x} \]
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Rubi [A] time = 0.32, antiderivative size = 28, normalized size of antiderivative = 1.40, number of steps used = 10, number of rules used = 8, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {1594, 27, 6741, 12, 6742, 893, 2314, 31} \begin {gather*} -\frac {99}{5 (1-25 x)}+\frac {105 x \log (x)}{1-25 x}+5 \log (x) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 27
Rule 31
Rule 893
Rule 1594
Rule 2314
Rule 6741
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5-640 x+500 x^2+105 x \log (x)}{x \left (1-50 x+625 x^2\right )} \, dx\\ &=\int \frac {5-640 x+500 x^2+105 x \log (x)}{x (-1+25 x)^2} \, dx\\ &=\int \frac {5 \left (1-128 x+100 x^2+21 x \log (x)\right )}{(1-25 x)^2 x} \, dx\\ &=5 \int \frac {1-128 x+100 x^2+21 x \log (x)}{(1-25 x)^2 x} \, dx\\ &=5 \int \left (\frac {1-128 x+100 x^2}{x (-1+25 x)^2}+\frac {21 \log (x)}{(-1+25 x)^2}\right ) \, dx\\ &=5 \int \frac {1-128 x+100 x^2}{x (-1+25 x)^2} \, dx+105 \int \frac {\log (x)}{(-1+25 x)^2} \, dx\\ &=\frac {105 x \log (x)}{1-25 x}+5 \int \left (\frac {1}{x}-\frac {99}{(-1+25 x)^2}-\frac {21}{-1+25 x}\right ) \, dx+105 \int \frac {1}{-1+25 x} \, dx\\ &=-\frac {99}{5 (1-25 x)}+5 \log (x)+\frac {105 x \log (x)}{1-25 x}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.04, size = 20, normalized size = 1.00 \begin {gather*} \frac {5 (99+25 (-1+4 x) \log (x))}{-25+625 x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.73, size = 20, normalized size = 1.00 \begin {gather*} \frac {25 \, {\left (4 \, x - 1\right )} \log \relax (x) + 99}{5 \, {\left (25 \, x - 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.17, size = 25, normalized size = 1.25 \begin {gather*} -\frac {21 \, \log \relax (x)}{5 \, {\left (25 \, x - 1\right )}} + \frac {99}{5 \, {\left (25 \, x - 1\right )}} + \frac {4}{5} \, \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 22, normalized size = 1.10
method | result | size |
norman | \(\frac {-5 \ln \relax (x )+495 x +20 x \ln \relax (x )}{-1+25 x}\) | \(22\) |
default | \(\frac {99}{5 \left (-1+25 x \right )}+5 \ln \relax (x )-\frac {105 \ln \relax (x ) x}{-1+25 x}\) | \(27\) |
risch | \(-\frac {21 \ln \relax (x )}{5 \left (-1+25 x \right )}+\frac {100 x \ln \relax (x )-4 \ln \relax (x )+99}{-5+125 x}\) | \(33\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.48, size = 25, normalized size = 1.25 \begin {gather*} -\frac {21 \, \log \relax (x)}{5 \, {\left (25 \, x - 1\right )}} + \frac {99}{5 \, {\left (25 \, x - 1\right )}} + \frac {4}{5} \, \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.14, size = 20, normalized size = 1.00 \begin {gather*} \frac {4\,\ln \relax (x)}{5}-\frac {\frac {21\,\ln \relax (x)}{5}-\frac {99}{5}}{25\,x-1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.15, size = 22, normalized size = 1.10 \begin {gather*} \frac {4 \log {\relax (x )}}{5} - \frac {21 \log {\relax (x )}}{125 x - 5} + \frac {99}{125 x - 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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