Integrand size = 16, antiderivative size = 34 \[ \int \frac {1}{x^3 \left (13+\frac {2}{x}+15 x\right )} \, dx=-\frac {1}{2 x}-\frac {13 \log (x)}{4}-\frac {9}{28} \log (2+3 x)+\frac {25}{7} \log (1+5 x) \]
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Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {1400, 723, 814} \[ \int \frac {1}{x^3 \left (13+\frac {2}{x}+15 x\right )} \, dx=-\frac {1}{2 x}-\frac {13 \log (x)}{4}-\frac {9}{28} \log (3 x+2)+\frac {25}{7} \log (5 x+1) \]
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Rule 723
Rule 814
Rule 1400
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{x^2 \left (2+13 x+15 x^2\right )} \, dx \\ & = -\frac {1}{2 x}+\frac {1}{2} \int \frac {-13-15 x}{x \left (2+13 x+15 x^2\right )} \, dx \\ & = -\frac {1}{2 x}+\frac {1}{2} \int \left (-\frac {13}{2 x}-\frac {27}{14 (2+3 x)}+\frac {250}{7 (1+5 x)}\right ) \, dx \\ & = -\frac {1}{2 x}-\frac {13 \log (x)}{4}-\frac {9}{28} \log (2+3 x)+\frac {25}{7} \log (1+5 x) \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^3 \left (13+\frac {2}{x}+15 x\right )} \, dx=-\frac {1}{2 x}-\frac {13 \log (x)}{4}-\frac {9}{28} \log (2+3 x)+\frac {25}{7} \log (1+5 x) \]
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Time = 0.05 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79
method | result | size |
default | \(-\frac {1}{2 x}-\frac {13 \ln \left (x \right )}{4}-\frac {9 \ln \left (3 x +2\right )}{28}+\frac {25 \ln \left (1+5 x \right )}{7}\) | \(27\) |
norman | \(-\frac {1}{2 x}-\frac {13 \ln \left (x \right )}{4}-\frac {9 \ln \left (3 x +2\right )}{28}+\frac {25 \ln \left (1+5 x \right )}{7}\) | \(27\) |
risch | \(-\frac {1}{2 x}-\frac {13 \ln \left (x \right )}{4}-\frac {9 \ln \left (3 x +2\right )}{28}+\frac {25 \ln \left (1+5 x \right )}{7}\) | \(27\) |
parallelrisch | \(-\frac {91 \ln \left (x \right ) x -100 \ln \left (x +\frac {1}{5}\right ) x +9 \ln \left (x +\frac {2}{3}\right ) x +14}{28 x}\) | \(27\) |
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Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^3 \left (13+\frac {2}{x}+15 x\right )} \, dx=\frac {100 \, x \log \left (5 \, x + 1\right ) - 9 \, x \log \left (3 \, x + 2\right ) - 91 \, x \log \left (x\right ) - 14}{28 \, x} \]
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Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^3 \left (13+\frac {2}{x}+15 x\right )} \, dx=- \frac {13 \log {\left (x \right )}}{4} + \frac {25 \log {\left (x + \frac {1}{5} \right )}}{7} - \frac {9 \log {\left (x + \frac {2}{3} \right )}}{28} - \frac {1}{2 x} \]
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Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x^3 \left (13+\frac {2}{x}+15 x\right )} \, dx=-\frac {1}{2 \, x} + \frac {25}{7} \, \log \left (5 \, x + 1\right ) - \frac {9}{28} \, \log \left (3 \, x + 2\right ) - \frac {13}{4} \, \log \left (x\right ) \]
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Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^3 \left (13+\frac {2}{x}+15 x\right )} \, dx=-\frac {1}{2 \, x} + \frac {25}{7} \, \log \left ({\left | 5 \, x + 1 \right |}\right ) - \frac {9}{28} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - \frac {13}{4} \, \log \left ({\left | x \right |}\right ) \]
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Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.65 \[ \int \frac {1}{x^3 \left (13+\frac {2}{x}+15 x\right )} \, dx=\frac {25\,\ln \left (x+\frac {1}{5}\right )}{7}-\frac {9\,\ln \left (x+\frac {2}{3}\right )}{28}-\frac {13\,\ln \left (x\right )}{4}-\frac {1}{2\,x} \]
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