Optimal. Leaf size=26 \[ \frac {x}{15}-\frac {4}{63} \log (2+3 x)+\frac {1}{175} \log (1+5 x) \]
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Rubi [A]
time = 0.01, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1400, 717, 646,
31} \begin {gather*} \frac {x}{15}-\frac {4}{63} \log (3 x+2)+\frac {1}{175} \log (5 x+1) \end {gather*}
Antiderivative was successfully verified.
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Rule 31
Rule 646
Rule 717
Rule 1400
Rubi steps
\begin {align*} \int \frac {x}{13+\frac {2}{x}+15 x} \, dx &=\int \frac {x^2}{2+13 x+15 x^2} \, dx\\ &=\frac {x}{15}+\frac {1}{15} \int \frac {-2-13 x}{2+13 x+15 x^2} \, dx\\ &=\frac {x}{15}+\frac {3}{35} \int \frac {1}{3+15 x} \, dx-\frac {20}{21} \int \frac {1}{10+15 x} \, dx\\ &=\frac {x}{15}-\frac {4}{63} \log (2+3 x)+\frac {1}{175} \log (1+5 x)\\ \end {align*}
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Mathematica [A]
time = 0.00, size = 26, normalized size = 1.00 \begin {gather*} \frac {x}{15}-\frac {4}{63} \log (2+3 x)+\frac {1}{175} \log (1+5 x) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.02, size = 21, normalized size = 0.81
method | result | size |
default | \(\frac {x}{15}-\frac {4 \ln \left (2+3 x \right )}{63}+\frac {\ln \left (1+5 x \right )}{175}\) | \(21\) |
norman | \(\frac {x}{15}-\frac {4 \ln \left (2+3 x \right )}{63}+\frac {\ln \left (1+5 x \right )}{175}\) | \(21\) |
risch | \(\frac {x}{15}-\frac {4 \ln \left (2+3 x \right )}{63}+\frac {\ln \left (1+5 x \right )}{175}\) | \(21\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.27, size = 20, normalized size = 0.77 \begin {gather*} \frac {1}{15} \, x + \frac {1}{175} \, \log \left (5 \, x + 1\right ) - \frac {4}{63} \, \log \left (3 \, x + 2\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.40, size = 20, normalized size = 0.77 \begin {gather*} \frac {1}{15} \, x + \frac {1}{175} \, \log \left (5 \, x + 1\right ) - \frac {4}{63} \, \log \left (3 \, x + 2\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.04, size = 20, normalized size = 0.77 \begin {gather*} \frac {x}{15} + \frac {\log {\left (x + \frac {1}{5} \right )}}{175} - \frac {4 \log {\left (x + \frac {2}{3} \right )}}{63} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 3.60, size = 22, normalized size = 0.85 \begin {gather*} \frac {1}{15} \, x + \frac {1}{175} \, \log \left ({\left | 5 \, x + 1 \right |}\right ) - \frac {4}{63} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 2.12, size = 16, normalized size = 0.62 \begin {gather*} \frac {x}{15}-\frac {4\,\ln \left (x+\frac {2}{3}\right )}{63}+\frac {\ln \left (x+\frac {1}{5}\right )}{175} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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