Optimal. Leaf size=18 \[ -5+\frac {\log (25+x+x \log (x))}{1-x} \]
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Rubi [F] time = 0.99, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {2-2 x+(1-x) \log (x)+(25+x+x \log (x)) \log (25+x+x \log (x))}{25-49 x+23 x^2+x^3+\left (x-2 x^2+x^3\right ) \log (x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2-2 x+(1-x) \log (x)+(25+x+x \log (x)) \log (25+x+x \log (x))}{(1-x)^2 (25+x+x \log (x))} \, dx\\ &=\int \left (\frac {2}{(-1+x)^2 (25+x+x \log (x))}-\frac {2 x}{(-1+x)^2 (25+x+x \log (x))}-\frac {\log (x)}{(-1+x) (25+x+x \log (x))}+\frac {\log (25+x+x \log (x))}{(-1+x)^2}\right ) \, dx\\ &=2 \int \frac {1}{(-1+x)^2 (25+x+x \log (x))} \, dx-2 \int \frac {x}{(-1+x)^2 (25+x+x \log (x))} \, dx-\int \frac {\log (x)}{(-1+x) (25+x+x \log (x))} \, dx+\int \frac {\log (25+x+x \log (x))}{(-1+x)^2} \, dx\\ &=2 \int \frac {1}{(-1+x)^2 (25+x+x \log (x))} \, dx-2 \int \left (\frac {1}{(-1+x)^2 (25+x+x \log (x))}+\frac {1}{(-1+x) (25+x+x \log (x))}\right ) \, dx-\int \left (\frac {1}{(-1+x) x}+\frac {-25-x}{(-1+x) x (25+x+x \log (x))}\right ) \, dx+\int \frac {\log (25+x+x \log (x))}{(-1+x)^2} \, dx\\ &=-\left (2 \int \frac {1}{(-1+x) (25+x+x \log (x))} \, dx\right )-\int \frac {1}{(-1+x) x} \, dx-\int \frac {-25-x}{(-1+x) x (25+x+x \log (x))} \, dx+\int \frac {\log (25+x+x \log (x))}{(-1+x)^2} \, dx\\ &=-\left (2 \int \frac {1}{(-1+x) (25+x+x \log (x))} \, dx\right )-\int \frac {1}{-1+x} \, dx+\int \frac {1}{x} \, dx-\int \left (-\frac {26}{(-1+x) (25+x+x \log (x))}+\frac {25}{x (25+x+x \log (x))}\right ) \, dx+\int \frac {\log (25+x+x \log (x))}{(-1+x)^2} \, dx\\ &=-\log (1-x)+\log (x)-2 \int \frac {1}{(-1+x) (25+x+x \log (x))} \, dx-25 \int \frac {1}{x (25+x+x \log (x))} \, dx+26 \int \frac {1}{(-1+x) (25+x+x \log (x))} \, dx+\int \frac {\log (25+x+x \log (x))}{(-1+x)^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.18, size = 15, normalized size = 0.83 \begin {gather*} -\frac {\log (25+x+x \log (x))}{-1+x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.65, size = 15, normalized size = 0.83 \begin {gather*} -\frac {\log \left (x \log \relax (x) + x + 25\right )}{x - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.19, size = 15, normalized size = 0.83 \begin {gather*} -\frac {\log \left (x \log \relax (x) + x + 25\right )}{x - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 16, normalized size = 0.89
method | result | size |
risch | \(-\frac {\ln \left (x \ln \relax (x )+x +25\right )}{x -1}\) | \(16\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.41, size = 15, normalized size = 0.83 \begin {gather*} -\frac {\log \left (x \log \relax (x) + x + 25\right )}{x - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.39, size = 15, normalized size = 0.83 \begin {gather*} -\frac {\ln \left (x+x\,\ln \relax (x)+25\right )}{x-1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.37, size = 14, normalized size = 0.78 \begin {gather*} - \frac {\log {\left (x \log {\relax (x )} + x + 25 \right )}}{x - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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