Optimal. Leaf size=26 \[ 4+\frac {5}{x \log \left (x+\frac {x}{-9-\frac {x}{4+x}}\right )} \]
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Rubi [F] time = 1.25, 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 {-2880-1620 x-225 x^2+\left (-2880-1610 x-225 x^2\right ) \log \left (\frac {32 x+9 x^2}{36+10 x}\right )}{\left (576 x^2+322 x^3+45 x^4\right ) \log ^2\left (\frac {32 x+9 x^2}{36+10 x}\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 {-2880-1620 x-225 x^2+\left (-2880-1610 x-225 x^2\right ) \log \left (\frac {32 x+9 x^2}{36+10 x}\right )}{x^2 \left (576+322 x+45 x^2\right ) \log ^2\left (\frac {32 x+9 x^2}{36+10 x}\right )} \, dx\\ &=\int \left (-\frac {45 \left (64+36 x+5 x^2\right )}{x^2 (18+5 x) (32+9 x) \log ^2\left (\frac {x (32+9 x)}{36+10 x}\right )}-\frac {5}{x^2 \log \left (\frac {x (32+9 x)}{36+10 x}\right )}\right ) \, dx\\ &=-\left (5 \int \frac {1}{x^2 \log \left (\frac {x (32+9 x)}{36+10 x}\right )} \, dx\right )-45 \int \frac {64+36 x+5 x^2}{x^2 (18+5 x) (32+9 x) \log ^2\left (\frac {x (32+9 x)}{36+10 x}\right )} \, dx\\ &=-\left (5 \int \frac {1}{x^2 \log \left (\frac {x (32+9 x)}{36+10 x}\right )} \, dx\right )-45 \int \left (\frac {1}{9 x^2 \log ^2\left (\frac {x (32+9 x)}{36+10 x}\right )}+\frac {1}{2592 x \log ^2\left (\frac {x (32+9 x)}{36+10 x}\right )}+\frac {25}{162 (18+5 x) \log ^2\left (\frac {x (32+9 x)}{36+10 x}\right )}-\frac {9}{32 (32+9 x) \log ^2\left (\frac {x (32+9 x)}{36+10 x}\right )}\right ) \, dx\\ &=-\left (\frac {5}{288} \int \frac {1}{x \log ^2\left (\frac {x (32+9 x)}{36+10 x}\right )} \, dx\right )-5 \int \frac {1}{x^2 \log ^2\left (\frac {x (32+9 x)}{36+10 x}\right )} \, dx-5 \int \frac {1}{x^2 \log \left (\frac {x (32+9 x)}{36+10 x}\right )} \, dx-\frac {125}{18} \int \frac {1}{(18+5 x) \log ^2\left (\frac {x (32+9 x)}{36+10 x}\right )} \, dx+\frac {405}{32} \int \frac {1}{(32+9 x) \log ^2\left (\frac {x (32+9 x)}{36+10 x}\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.18, size = 22, normalized size = 0.85 \begin {gather*} \frac {5}{x \log \left (\frac {x (32+9 x)}{36+10 x}\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.71, size = 26, normalized size = 1.00 \begin {gather*} \frac {5}{x \log \left (\frac {9 \, x^{2} + 32 \, x}{2 \, {\left (5 \, x + 18\right )}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.39, size = 26, normalized size = 1.00 \begin {gather*} \frac {5}{x \log \left (\frac {9 \, x^{2} + 32 \, x}{2 \, {\left (5 \, x + 18\right )}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 26, normalized size = 1.00
method | result | size |
norman | \(\frac {5}{x \ln \left (\frac {9 x^{2}+32 x}{10 x +36}\right )}\) | \(26\) |
risch | \(\frac {5}{x \ln \left (\frac {9 x^{2}+32 x}{10 x +36}\right )}\) | \(26\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.72, size = 31, normalized size = 1.19 \begin {gather*} -\frac {5}{x \log \relax (2) - x \log \left (9 \, x + 32\right ) + x \log \left (5 \, x + 18\right ) - x \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.87, size = 25, normalized size = 0.96 \begin {gather*} \frac {5}{x\,\ln \left (\frac {9\,x^2+32\,x}{10\,x+36}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.19, size = 17, normalized size = 0.65 \begin {gather*} \frac {5}{x \log {\left (\frac {9 x^{2} + 32 x}{10 x + 36} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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