Integrand size = 22, antiderivative size = 37 \[ \int \frac {(2+3 x)^2 (3+5 x)^2}{1-2 x} \, dx=-\frac {5353 x}{16}-\frac {3529 x^2}{16}-\frac {455 x^3}{4}-\frac {225 x^4}{8}-\frac {5929}{32} \log (1-2 x) \]
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Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {90} \[ \int \frac {(2+3 x)^2 (3+5 x)^2}{1-2 x} \, dx=-\frac {225 x^4}{8}-\frac {455 x^3}{4}-\frac {3529 x^2}{16}-\frac {5353 x}{16}-\frac {5929}{32} \log (1-2 x) \]
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Rule 90
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {5353}{16}-\frac {3529 x}{8}-\frac {1365 x^2}{4}-\frac {225 x^3}{2}-\frac {5929}{16 (-1+2 x)}\right ) \, dx \\ & = -\frac {5353 x}{16}-\frac {3529 x^2}{16}-\frac {455 x^3}{4}-\frac {225 x^4}{8}-\frac {5929}{32} \log (1-2 x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86 \[ \int \frac {(2+3 x)^2 (3+5 x)^2}{1-2 x} \, dx=\frac {1}{128} \left (30515-42824 x-28232 x^2-14560 x^3-3600 x^4-23716 \log (1-2 x)\right ) \]
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Time = 2.52 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70
method | result | size |
parallelrisch | \(-\frac {225 x^{4}}{8}-\frac {455 x^{3}}{4}-\frac {3529 x^{2}}{16}-\frac {5353 x}{16}-\frac {5929 \ln \left (x -\frac {1}{2}\right )}{32}\) | \(26\) |
default | \(-\frac {225 x^{4}}{8}-\frac {455 x^{3}}{4}-\frac {3529 x^{2}}{16}-\frac {5353 x}{16}-\frac {5929 \ln \left (-1+2 x \right )}{32}\) | \(28\) |
norman | \(-\frac {225 x^{4}}{8}-\frac {455 x^{3}}{4}-\frac {3529 x^{2}}{16}-\frac {5353 x}{16}-\frac {5929 \ln \left (-1+2 x \right )}{32}\) | \(28\) |
risch | \(-\frac {225 x^{4}}{8}-\frac {455 x^{3}}{4}-\frac {3529 x^{2}}{16}-\frac {5353 x}{16}-\frac {5929 \ln \left (-1+2 x \right )}{32}\) | \(28\) |
meijerg | \(-\frac {5929 \ln \left (1-2 x \right )}{32}-114 x -\frac {541 x \left (6 x +6\right )}{24}-\frac {95 x \left (16 x^{2}+12 x +12\right )}{16}-\frac {15 x \left (120 x^{3}+80 x^{2}+60 x +60\right )}{64}\) | \(52\) |
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Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73 \[ \int \frac {(2+3 x)^2 (3+5 x)^2}{1-2 x} \, dx=-\frac {225}{8} \, x^{4} - \frac {455}{4} \, x^{3} - \frac {3529}{16} \, x^{2} - \frac {5353}{16} \, x - \frac {5929}{32} \, \log \left (2 \, x - 1\right ) \]
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Time = 0.04 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.97 \[ \int \frac {(2+3 x)^2 (3+5 x)^2}{1-2 x} \, dx=- \frac {225 x^{4}}{8} - \frac {455 x^{3}}{4} - \frac {3529 x^{2}}{16} - \frac {5353 x}{16} - \frac {5929 \log {\left (2 x - 1 \right )}}{32} \]
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Time = 0.23 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73 \[ \int \frac {(2+3 x)^2 (3+5 x)^2}{1-2 x} \, dx=-\frac {225}{8} \, x^{4} - \frac {455}{4} \, x^{3} - \frac {3529}{16} \, x^{2} - \frac {5353}{16} \, x - \frac {5929}{32} \, \log \left (2 \, x - 1\right ) \]
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Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76 \[ \int \frac {(2+3 x)^2 (3+5 x)^2}{1-2 x} \, dx=-\frac {225}{8} \, x^{4} - \frac {455}{4} \, x^{3} - \frac {3529}{16} \, x^{2} - \frac {5353}{16} \, x - \frac {5929}{32} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.68 \[ \int \frac {(2+3 x)^2 (3+5 x)^2}{1-2 x} \, dx=-\frac {5353\,x}{16}-\frac {5929\,\ln \left (x-\frac {1}{2}\right )}{32}-\frac {3529\,x^2}{16}-\frac {455\,x^3}{4}-\frac {225\,x^4}{8} \]
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