Integrand size = 18, antiderivative size = 23 \[ \int \frac {(2+3 x) (3+5 x)}{1-2 x} \, dx=-\frac {53 x}{4}-\frac {15 x^2}{4}-\frac {77}{8} \log (1-2 x) \]
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Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {78} \[ \int \frac {(2+3 x) (3+5 x)}{1-2 x} \, dx=-\frac {15 x^2}{4}-\frac {53 x}{4}-\frac {77}{8} \log (1-2 x) \]
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Rule 78
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {53}{4}-\frac {15 x}{2}-\frac {77}{4 (-1+2 x)}\right ) \, dx \\ & = -\frac {53 x}{4}-\frac {15 x^2}{4}-\frac {77}{8} \log (1-2 x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {(2+3 x) (3+5 x)}{1-2 x} \, dx=\frac {1}{16} \left (121-212 x-60 x^2-154 \log (1-2 x)\right ) \]
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Time = 2.47 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70
method | result | size |
parallelrisch | \(-\frac {15 x^{2}}{4}-\frac {53 x}{4}-\frac {77 \ln \left (x -\frac {1}{2}\right )}{8}\) | \(16\) |
default | \(-\frac {15 x^{2}}{4}-\frac {53 x}{4}-\frac {77 \ln \left (-1+2 x \right )}{8}\) | \(18\) |
norman | \(-\frac {15 x^{2}}{4}-\frac {53 x}{4}-\frac {77 \ln \left (-1+2 x \right )}{8}\) | \(18\) |
risch | \(-\frac {15 x^{2}}{4}-\frac {53 x}{4}-\frac {77 \ln \left (-1+2 x \right )}{8}\) | \(18\) |
meijerg | \(-\frac {77 \ln \left (1-2 x \right )}{8}-\frac {19 x}{2}-\frac {5 x \left (6 x +6\right )}{8}\) | \(21\) |
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Time = 0.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {(2+3 x) (3+5 x)}{1-2 x} \, dx=-\frac {15}{4} \, x^{2} - \frac {53}{4} \, x - \frac {77}{8} \, \log \left (2 \, x - 1\right ) \]
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Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {(2+3 x) (3+5 x)}{1-2 x} \, dx=- \frac {15 x^{2}}{4} - \frac {53 x}{4} - \frac {77 \log {\left (2 x - 1 \right )}}{8} \]
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Time = 0.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {(2+3 x) (3+5 x)}{1-2 x} \, dx=-\frac {15}{4} \, x^{2} - \frac {53}{4} \, x - \frac {77}{8} \, \log \left (2 \, x - 1\right ) \]
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Time = 0.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {(2+3 x) (3+5 x)}{1-2 x} \, dx=-\frac {15}{4} \, x^{2} - \frac {53}{4} \, x - \frac {77}{8} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {(2+3 x) (3+5 x)}{1-2 x} \, dx=-\frac {53\,x}{4}-\frac {77\,\ln \left (x-\frac {1}{2}\right )}{8}-\frac {15\,x^2}{4} \]
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