Integrand size = 16, antiderivative size = 27 \[ \int \frac {1}{x \left (2+13 x+15 x^2\right )} \, dx=\frac {\log (x)}{2}+\frac {3}{14} \log (2+3 x)-\frac {5}{7} \log (1+5 x) \]
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Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {719, 29, 646, 31} \[ \int \frac {1}{x \left (2+13 x+15 x^2\right )} \, dx=\frac {\log (x)}{2}+\frac {3}{14} \log (3 x+2)-\frac {5}{7} \log (5 x+1) \]
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Rule 29
Rule 31
Rule 646
Rule 719
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {1}{x} \, dx+\frac {1}{2} \int \frac {-13-15 x}{2+13 x+15 x^2} \, dx \\ & = \frac {\log (x)}{2}+\frac {45}{14} \int \frac {1}{10+15 x} \, dx-\frac {75}{7} \int \frac {1}{3+15 x} \, dx \\ & = \frac {\log (x)}{2}+\frac {3}{14} \log (2+3 x)-\frac {5}{7} \log (1+5 x) \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (2+13 x+15 x^2\right )} \, dx=\frac {\log (x)}{2}+\frac {3}{14} \log (2+3 x)-\frac {5}{7} \log (1+5 x) \]
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Time = 25.12 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67
method | result | size |
parallelrisch | \(\frac {\ln \left (x \right )}{2}+\frac {3 \ln \left (\frac {2}{3}+x \right )}{14}-\frac {5 \ln \left (x +\frac {1}{5}\right )}{7}\) | \(18\) |
default | \(\frac {\ln \left (x \right )}{2}+\frac {3 \ln \left (2+3 x \right )}{14}-\frac {5 \ln \left (1+5 x \right )}{7}\) | \(22\) |
norman | \(\frac {\ln \left (x \right )}{2}+\frac {3 \ln \left (2+3 x \right )}{14}-\frac {5 \ln \left (1+5 x \right )}{7}\) | \(22\) |
risch | \(\frac {\ln \left (x \right )}{2}+\frac {3 \ln \left (2+3 x \right )}{14}-\frac {5 \ln \left (1+5 x \right )}{7}\) | \(22\) |
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Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x \left (2+13 x+15 x^2\right )} \, dx=-\frac {5}{7} \, \log \left (5 \, x + 1\right ) + \frac {3}{14} \, \log \left (3 \, x + 2\right ) + \frac {1}{2} \, \log \left (x\right ) \]
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Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x \left (2+13 x+15 x^2\right )} \, dx=\frac {\log {\left (x \right )}}{2} - \frac {5 \log {\left (x + \frac {1}{5} \right )}}{7} + \frac {3 \log {\left (x + \frac {2}{3} \right )}}{14} \]
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Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x \left (2+13 x+15 x^2\right )} \, dx=-\frac {5}{7} \, \log \left (5 \, x + 1\right ) + \frac {3}{14} \, \log \left (3 \, x + 2\right ) + \frac {1}{2} \, \log \left (x\right ) \]
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Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x \left (2+13 x+15 x^2\right )} \, dx=-\frac {5}{7} \, \log \left ({\left | 5 \, x + 1 \right |}\right ) + \frac {3}{14} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + \frac {1}{2} \, \log \left ({\left | x \right |}\right ) \]
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Time = 9.97 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \frac {1}{x \left (2+13 x+15 x^2\right )} \, dx=\frac {3\,\ln \left (x+\frac {2}{3}\right )}{14}-\frac {5\,\ln \left (x+\frac {1}{5}\right )}{7}+\frac {\ln \left (x\right )}{2} \]
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