Integrand size = 15, antiderivative size = 34 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2} \, dx=\frac {1331}{16 (1-2 x)}+\frac {175 x}{2}+\frac {125 x^2}{8}+\frac {1815}{16} \log (1-2 x) \]
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Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {45} \[ \int \frac {(3+5 x)^3}{(1-2 x)^2} \, dx=\frac {125 x^2}{8}+\frac {175 x}{2}+\frac {1331}{16 (1-2 x)}+\frac {1815}{16} \log (1-2 x) \]
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Rule 45
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {175}{2}+\frac {125 x}{4}+\frac {1331}{8 (-1+2 x)^2}+\frac {1815}{8 (-1+2 x)}\right ) \, dx \\ & = \frac {1331}{16 (1-2 x)}+\frac {175 x}{2}+\frac {125 x^2}{8}+\frac {1815}{16} \log (1-2 x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2} \, dx=\frac {-1137-5850 x+5100 x^2+1000 x^3+3630 (-1+2 x) \log (1-2 x)}{-32+64 x} \]
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Time = 2.66 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74
method | result | size |
risch | \(\frac {125 x^{2}}{8}+\frac {175 x}{2}-\frac {1331}{32 \left (x -\frac {1}{2}\right )}+\frac {1815 \ln \left (-1+2 x \right )}{16}\) | \(25\) |
default | \(\frac {125 x^{2}}{8}+\frac {175 x}{2}+\frac {1815 \ln \left (-1+2 x \right )}{16}-\frac {1331}{16 \left (-1+2 x \right )}\) | \(27\) |
norman | \(\frac {-\frac {2031}{8} x +\frac {1275}{8} x^{2}+\frac {125}{4} x^{3}}{-1+2 x}+\frac {1815 \ln \left (-1+2 x \right )}{16}\) | \(32\) |
parallelrisch | \(\frac {500 x^{3}+3630 \ln \left (x -\frac {1}{2}\right ) x +2550 x^{2}-1815 \ln \left (x -\frac {1}{2}\right )-4062 x}{-16+32 x}\) | \(37\) |
meijerg | \(\frac {189 x}{2 \left (1-2 x \right )}+\frac {1815 \ln \left (1-2 x \right )}{16}+\frac {75 x \left (-6 x +6\right )}{4 \left (1-2 x \right )}+\frac {125 x \left (-8 x^{2}-12 x +12\right )}{32 \left (1-2 x \right )}\) | \(55\) |
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Time = 0.21 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2} \, dx=\frac {500 \, x^{3} + 2550 \, x^{2} + 1815 \, {\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 1400 \, x - 1331}{16 \, {\left (2 \, x - 1\right )}} \]
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Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2} \, dx=\frac {125 x^{2}}{8} + \frac {175 x}{2} + \frac {1815 \log {\left (2 x - 1 \right )}}{16} - \frac {1331}{32 x - 16} \]
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Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2} \, dx=\frac {125}{8} \, x^{2} + \frac {175}{2} \, x - \frac {1331}{16 \, {\left (2 \, x - 1\right )}} + \frac {1815}{16} \, \log \left (2 \, x - 1\right ) \]
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Time = 0.29 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.41 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2} \, dx=\frac {25}{32} \, {\left (2 \, x - 1\right )}^{2} {\left (\frac {66}{2 \, x - 1} + 5\right )} - \frac {1331}{16 \, {\left (2 \, x - 1\right )}} - \frac {1815}{16} \, \log \left (\frac {{\left | 2 \, x - 1 \right |}}{2 \, {\left (2 \, x - 1\right )}^{2}}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2} \, dx=\frac {175\,x}{2}+\frac {1815\,\ln \left (x-\frac {1}{2}\right )}{16}-\frac {1331}{32\,\left (x-\frac {1}{2}\right )}+\frac {125\,x^2}{8} \]
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