Optimal. Leaf size=18 \[ \frac {x}{-2-x+x \log \left (x+\log \left (\frac {5}{4}\right )\right )} \]
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Rubi [F] time = 0.42, 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 x-x^2-2 \log \left (\frac {5}{4}\right )}{4 x+4 x^2+x^3+\left (4+4 x+x^2\right ) \log \left (\frac {5}{4}\right )+\left (-4 x^2-2 x^3+\left (-4 x-2 x^2\right ) \log \left (\frac {5}{4}\right )\right ) \log \left (x+\log \left (\frac {5}{4}\right )\right )+\left (x^3+x^2 \log \left (\frac {5}{4}\right )\right ) \log ^2\left (x+\log \left (\frac {5}{4}\right )\right )} \, 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 x-x^2-\log \left (\frac {25}{16}\right )}{\left (x+\log \left (\frac {5}{4}\right )\right ) \left (2+x-x \log \left (x+\log \left (\frac {5}{4}\right )\right )\right )^2} \, dx\\ &=\int \left (-\frac {x}{\left (-2-x+x \log \left (x+\log \left (\frac {5}{4}\right )\right )\right )^2}-\frac {\log ^2\left (\frac {5}{4}\right )}{\left (x+\log \left (\frac {5}{4}\right )\right ) \left (-2-x+x \log \left (x+\log \left (\frac {5}{4}\right )\right )\right )^2}-\frac {2 \left (1+\log \left (\frac {2}{\sqrt {5}}\right )\right )}{\left (-2-x+x \log \left (x+\log \left (\frac {5}{4}\right )\right )\right )^2}\right ) \, dx\\ &=-\left (\log ^2\left (\frac {5}{4}\right ) \int \frac {1}{\left (x+\log \left (\frac {5}{4}\right )\right ) \left (-2-x+x \log \left (x+\log \left (\frac {5}{4}\right )\right )\right )^2} \, dx\right )-\left (2 \left (1+\log \left (\frac {2}{\sqrt {5}}\right )\right )\right ) \int \frac {1}{\left (-2-x+x \log \left (x+\log \left (\frac {5}{4}\right )\right )\right )^2} \, dx-\int \frac {x}{\left (-2-x+x \log \left (x+\log \left (\frac {5}{4}\right )\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 1.66, size = 18, normalized size = 1.00 \begin {gather*} -\frac {x}{2+x-x \log \left (x+\log \left (\frac {5}{4}\right )\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.86, size = 16, normalized size = 0.89 \begin {gather*} \frac {x}{x \log \left (x + \log \left (\frac {5}{4}\right )\right ) - x - 2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.40, size = 247, normalized size = 13.72 \begin {gather*} \frac {x^{4} + x^{3} \log \left (\frac {5}{4}\right ) + 2 \, x^{3} + 2 \, x^{2} \log \relax (5) - 4 \, x^{2} \log \relax (2) + 2 \, x^{2} \log \left (\frac {5}{4}\right ) + 2 \, x \log \relax (5) \log \left (\frac {5}{4}\right ) - 4 \, x \log \relax (2) \log \left (\frac {5}{4}\right )}{x^{4} \log \left (x + \log \left (\frac {5}{4}\right )\right ) + x^{3} \log \relax (5) \log \left (x + \log \left (\frac {5}{4}\right )\right ) - 2 \, x^{3} \log \relax (2) \log \left (x + \log \left (\frac {5}{4}\right )\right ) - x^{4} - x^{3} \log \relax (5) + 2 \, x^{3} \log \relax (2) + 2 \, x^{3} \log \left (x + \log \left (\frac {5}{4}\right )\right ) + 2 \, x^{2} \log \relax (5) \log \left (x + \log \left (\frac {5}{4}\right )\right ) - 4 \, x^{2} \log \relax (2) \log \left (x + \log \left (\frac {5}{4}\right )\right ) + 2 \, x^{2} \log \left (\frac {5}{4}\right ) \log \left (x + \log \left (\frac {5}{4}\right )\right ) + 2 \, x \log \relax (5) \log \left (\frac {5}{4}\right ) \log \left (x + \log \left (\frac {5}{4}\right )\right ) - 4 \, x \log \relax (2) \log \left (\frac {5}{4}\right ) \log \left (x + \log \left (\frac {5}{4}\right )\right ) - 4 \, x^{3} - 4 \, x^{2} \log \relax (5) + 8 \, x^{2} \log \relax (2) - 2 \, x^{2} \log \left (\frac {5}{4}\right ) - 2 \, x \log \relax (5) \log \left (\frac {5}{4}\right ) + 4 \, x \log \relax (2) \log \left (\frac {5}{4}\right ) - 4 \, x^{2} - 4 \, x \log \relax (5) + 8 \, x \log \relax (2) - 4 \, x \log \left (\frac {5}{4}\right ) - 4 \, \log \relax (5) \log \left (\frac {5}{4}\right ) + 8 \, \log \relax (2) \log \left (\frac {5}{4}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.20, size = 17, normalized size = 0.94
method | result | size |
norman | \(\frac {x}{\ln \left (\ln \left (\frac {5}{4}\right )+x \right ) x -2-x}\) | \(17\) |
risch | \(\frac {x}{x \ln \left (\ln \relax (5)-2 \ln \relax (2)+x \right )-x -2}\) | \(21\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.49, size = 20, normalized size = 1.11 \begin {gather*} \frac {x}{x \log \left (x + \log \relax (5) - 2 \, \log \relax (2)\right ) - x - 2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.06 \begin {gather*} \int -\frac {x^2+2\,x+2\,\ln \left (\frac {5}{4}\right )}{4\,x-\ln \left (x+\ln \left (\frac {5}{4}\right )\right )\,\left (\ln \left (\frac {5}{4}\right )\,\left (2\,x^2+4\,x\right )+4\,x^2+2\,x^3\right )+{\ln \left (x+\ln \left (\frac {5}{4}\right )\right )}^2\,\left (x^3+\ln \left (\frac {5}{4}\right )\,x^2\right )+4\,x^2+x^3+\ln \left (\frac {5}{4}\right )\,\left (x^2+4\,x+4\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.21, size = 14, normalized size = 0.78 \begin {gather*} \frac {x}{x \log {\left (x + \log {\left (\frac {5}{4} \right )} \right )} - x - 2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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