Optimal. Leaf size=25 \[ \frac {1}{5} \left (x+\frac {16 (x+\log (x-5 (2+2 x)))}{3 x}\right ) \]
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Rubi [A] time = 0.24, antiderivative size = 19, normalized size of antiderivative = 0.76, number of steps used = 9, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {1593, 6742, 893, 2395, 36, 31, 29} \begin {gather*} \frac {x}{5}+\frac {16 \log (-9 x-10)}{15 x} \end {gather*}
Antiderivative was successfully verified.
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Rule 29
Rule 31
Rule 36
Rule 893
Rule 1593
Rule 2395
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {144 x+30 x^2+27 x^3+(-160-144 x) \log (-10-9 x)}{x^2 (150+135 x)} \, dx\\ &=\int \left (\frac {48+10 x+9 x^2}{5 x (10+9 x)}-\frac {16 \log (-10-9 x)}{15 x^2}\right ) \, dx\\ &=\frac {1}{5} \int \frac {48+10 x+9 x^2}{x (10+9 x)} \, dx-\frac {16}{15} \int \frac {\log (-10-9 x)}{x^2} \, dx\\ &=\frac {16 \log (-10-9 x)}{15 x}+\frac {1}{5} \int \left (1+\frac {24}{5 x}-\frac {216}{5 (10+9 x)}\right ) \, dx+\frac {48}{5} \int \frac {1}{(-10-9 x) x} \, dx\\ &=\frac {x}{5}+\frac {16 \log (-10-9 x)}{15 x}+\frac {24 \log (x)}{25}-\frac {24}{25} \log (10+9 x)-\frac {24}{25} \int \frac {1}{x} \, dx-\frac {216}{25} \int \frac {1}{-10-9 x} \, dx\\ &=\frac {x}{5}+\frac {16 \log (-10-9 x)}{15 x}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.02, size = 19, normalized size = 0.76 \begin {gather*} \frac {1}{15} \left (3 x+\frac {16 \log (-10-9 x)}{x}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.51, size = 19, normalized size = 0.76 \begin {gather*} \frac {3 \, x^{2} + 16 \, \log \left (-9 \, x - 10\right )}{15 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.14, size = 15, normalized size = 0.60 \begin {gather*} \frac {1}{5} \, x + \frac {16 \, \log \left (-9 \, x - 10\right )}{15 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.16, size = 16, normalized size = 0.64
method | result | size |
risch | \(\frac {16 \ln \left (-9 x -10\right )}{15 x}+\frac {x}{5}\) | \(16\) |
norman | \(\frac {\frac {x^{2}}{5}+\frac {16 \ln \left (-9 x -10\right )}{15}}{x}\) | \(19\) |
derivativedivides | \(-\frac {8 \ln \left (-9 x -10\right ) \left (-9 x -10\right )}{75 x}+\frac {x}{5}+\frac {2}{9}-\frac {24 \ln \left (-9 x -10\right )}{25}\) | \(30\) |
default | \(-\frac {8 \ln \left (-9 x -10\right ) \left (-9 x -10\right )}{75 x}+\frac {x}{5}+\frac {2}{9}-\frac {24 \ln \left (-9 x -10\right )}{25}\) | \(30\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.41, size = 19, normalized size = 0.76 \begin {gather*} \frac {3 \, x^{2} + 16 \, \log \left (-9 \, x - 10\right )}{15 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.22, size = 15, normalized size = 0.60 \begin {gather*} \frac {x}{5}+\frac {16\,\ln \left (-9\,x-10\right )}{15\,x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.13, size = 15, normalized size = 0.60 \begin {gather*} \frac {x}{5} + \frac {16 \log {\left (- 9 x - 10 \right )}}{15 x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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