Integrand size = 20, antiderivative size = 38 \[ \int \frac {(2+3 x)^2 (3+5 x)}{(1-2 x)^3} \, dx=\frac {539}{32 (1-2 x)^2}-\frac {707}{16 (1-2 x)}-\frac {45 x}{8}-\frac {309}{16} \log (1-2 x) \]
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Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {78} \[ \int \frac {(2+3 x)^2 (3+5 x)}{(1-2 x)^3} \, dx=-\frac {45 x}{8}-\frac {707}{16 (1-2 x)}+\frac {539}{32 (1-2 x)^2}-\frac {309}{16} \log (1-2 x) \]
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Rule 78
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {45}{8}-\frac {539}{8 (-1+2 x)^3}-\frac {707}{8 (-1+2 x)^2}-\frac {309}{8 (-1+2 x)}\right ) \, dx \\ & = \frac {539}{32 (1-2 x)^2}-\frac {707}{16 (1-2 x)}-\frac {45 x}{8}-\frac {309}{16} \log (1-2 x) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {(2+3 x)^2 (3+5 x)}{(1-2 x)^3} \, dx=\frac {1}{32} \left (-180 x+\frac {-785+2468 x+360 x^2}{(1-2 x)^2}-618 \log (1-2 x)\right ) \]
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Time = 2.72 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71
method | result | size |
risch | \(-\frac {45 x}{8}+\frac {\frac {707 x}{8}-\frac {875}{32}}{\left (-1+2 x \right )^{2}}-\frac {309 \ln \left (-1+2 x \right )}{16}\) | \(27\) |
default | \(-\frac {45 x}{8}-\frac {309 \ln \left (-1+2 x \right )}{16}+\frac {707}{16 \left (-1+2 x \right )}+\frac {539}{32 \left (-1+2 x \right )^{2}}\) | \(31\) |
norman | \(\frac {-\frac {213}{8} x +\frac {1055}{8} x^{2}-\frac {45}{2} x^{3}}{\left (-1+2 x \right )^{2}}-\frac {309 \ln \left (-1+2 x \right )}{16}\) | \(32\) |
parallelrisch | \(-\frac {1236 \ln \left (x -\frac {1}{2}\right ) x^{2}+360 x^{3}-1236 \ln \left (x -\frac {1}{2}\right ) x -2110 x^{2}+309 \ln \left (x -\frac {1}{2}\right )+426 x}{16 \left (-1+2 x \right )^{2}}\) | \(46\) |
meijerg | \(\frac {6 x \left (2-2 x \right )}{\left (1-2 x \right )^{2}}+\frac {28 x^{2}}{\left (1-2 x \right )^{2}}-\frac {29 x \left (-18 x +6\right )}{8 \left (1-2 x \right )^{2}}-\frac {309 \ln \left (1-2 x \right )}{16}-\frac {45 x \left (16 x^{2}-36 x +12\right )}{32 \left (1-2 x \right )^{2}}\) | \(72\) |
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Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.24 \[ \int \frac {(2+3 x)^2 (3+5 x)}{(1-2 x)^3} \, dx=-\frac {720 \, x^{3} - 720 \, x^{2} + 618 \, {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (2 \, x - 1\right ) - 2648 \, x + 875}{32 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \]
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Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {(2+3 x)^2 (3+5 x)}{(1-2 x)^3} \, dx=- \frac {45 x}{8} - \frac {875 - 2828 x}{128 x^{2} - 128 x + 32} - \frac {309 \log {\left (2 x - 1 \right )}}{16} \]
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Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {(2+3 x)^2 (3+5 x)}{(1-2 x)^3} \, dx=-\frac {45}{8} \, x + \frac {7 \, {\left (404 \, x - 125\right )}}{32 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} - \frac {309}{16} \, \log \left (2 \, x - 1\right ) \]
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Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71 \[ \int \frac {(2+3 x)^2 (3+5 x)}{(1-2 x)^3} \, dx=-\frac {45}{8} \, x + \frac {7 \, {\left (404 \, x - 125\right )}}{32 \, {\left (2 \, x - 1\right )}^{2}} - \frac {309}{16} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]
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Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.68 \[ \int \frac {(2+3 x)^2 (3+5 x)}{(1-2 x)^3} \, dx=\frac {\frac {707\,x}{32}-\frac {875}{128}}{x^2-x+\frac {1}{4}}-\frac {309\,\ln \left (x-\frac {1}{2}\right )}{16}-\frac {45\,x}{8} \]
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