Integrand size = 22, antiderivative size = 54 \[ \int \frac {(3+5 x)^2}{(1-2 x)^3 (2+3 x)^2} \, dx=\frac {121}{196 (1-2 x)^2}-\frac {22}{343 (1-2 x)}-\frac {1}{343 (2+3 x)}+\frac {64 \log (1-2 x)}{2401}-\frac {64 \log (2+3 x)}{2401} \]
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Time = 0.02 (sec) , antiderivative size = 54, 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} \[ \int \frac {(3+5 x)^2}{(1-2 x)^3 (2+3 x)^2} \, dx=-\frac {22}{343 (1-2 x)}-\frac {1}{343 (3 x+2)}+\frac {121}{196 (1-2 x)^2}+\frac {64 \log (1-2 x)}{2401}-\frac {64 \log (3 x+2)}{2401} \]
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Rule 90
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {121}{49 (-1+2 x)^3}-\frac {44}{343 (-1+2 x)^2}+\frac {128}{2401 (-1+2 x)}+\frac {3}{343 (2+3 x)^2}-\frac {192}{2401 (2+3 x)}\right ) \, dx \\ & = \frac {121}{196 (1-2 x)^2}-\frac {22}{343 (1-2 x)}-\frac {1}{343 (2+3 x)}+\frac {64 \log (1-2 x)}{2401}-\frac {64 \log (2+3 x)}{2401} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.87 \[ \int \frac {(3+5 x)^2}{(1-2 x)^3 (2+3 x)^2} \, dx=\frac {\frac {7 \left (1514+2645 x+512 x^2\right )}{(1-2 x)^2 (2+3 x)}+256 \log (1-2 x)-256 \log (4+6 x)}{9604} \]
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Time = 0.88 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81
method | result | size |
risch | \(\frac {\frac {128}{343} x^{2}+\frac {2645}{1372} x +\frac {757}{686}}{\left (-1+2 x \right )^{2} \left (2+3 x \right )}+\frac {64 \ln \left (-1+2 x \right )}{2401}-\frac {64 \ln \left (2+3 x \right )}{2401}\) | \(44\) |
default | \(\frac {121}{196 \left (-1+2 x \right )^{2}}+\frac {22}{343 \left (-1+2 x \right )}+\frac {64 \ln \left (-1+2 x \right )}{2401}-\frac {1}{343 \left (2+3 x \right )}-\frac {64 \ln \left (2+3 x \right )}{2401}\) | \(45\) |
norman | \(\frac {\frac {885}{343} x^{2}-\frac {2271}{343} x^{3}+\frac {3215}{686} x}{\left (-1+2 x \right )^{2} \left (2+3 x \right )}+\frac {64 \ln \left (-1+2 x \right )}{2401}-\frac {64 \ln \left (2+3 x \right )}{2401}\) | \(47\) |
parallelrisch | \(-\frac {1536 \ln \left (\frac {2}{3}+x \right ) x^{3}-1536 \ln \left (x -\frac {1}{2}\right ) x^{3}-512 \ln \left (\frac {2}{3}+x \right ) x^{2}+512 \ln \left (x -\frac {1}{2}\right ) x^{2}+31794 x^{3}-640 \ln \left (\frac {2}{3}+x \right ) x +640 \ln \left (x -\frac {1}{2}\right ) x -12390 x^{2}+256 \ln \left (\frac {2}{3}+x \right )-256 \ln \left (x -\frac {1}{2}\right )-22505 x}{4802 \left (-1+2 x \right )^{2} \left (2+3 x \right )}\) | \(93\) |
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Time = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.39 \[ \int \frac {(3+5 x)^2}{(1-2 x)^3 (2+3 x)^2} \, dx=\frac {3584 \, x^{2} - 256 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )} \log \left (3 \, x + 2\right ) + 256 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )} \log \left (2 \, x - 1\right ) + 18515 \, x + 10598}{9604 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {(3+5 x)^2}{(1-2 x)^3 (2+3 x)^2} \, dx=- \frac {- 512 x^{2} - 2645 x - 1514}{16464 x^{3} - 5488 x^{2} - 6860 x + 2744} + \frac {64 \log {\left (x - \frac {1}{2} \right )}}{2401} - \frac {64 \log {\left (x + \frac {2}{3} \right )}}{2401} \]
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Time = 0.20 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {(3+5 x)^2}{(1-2 x)^3 (2+3 x)^2} \, dx=\frac {512 \, x^{2} + 2645 \, x + 1514}{1372 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )}} - \frac {64}{2401} \, \log \left (3 \, x + 2\right ) + \frac {64}{2401} \, \log \left (2 \, x - 1\right ) \]
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Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.94 \[ \int \frac {(3+5 x)^2}{(1-2 x)^3 (2+3 x)^2} \, dx=-\frac {1}{343 \, {\left (3 \, x + 2\right )}} + \frac {33 \, {\left (\frac {203}{3 \, x + 2} - 25\right )}}{2401 \, {\left (\frac {7}{3 \, x + 2} - 2\right )}^{2}} + \frac {64}{2401} \, \log \left ({\left | -\frac {7}{3 \, x + 2} + 2 \right |}\right ) \]
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Time = 0.05 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.70 \[ \int \frac {(3+5 x)^2}{(1-2 x)^3 (2+3 x)^2} \, dx=-\frac {128\,\mathrm {atanh}\left (\frac {12\,x}{7}+\frac {1}{7}\right )}{2401}-\frac {\frac {32\,x^2}{1029}+\frac {2645\,x}{16464}+\frac {757}{8232}}{-x^3+\frac {x^2}{3}+\frac {5\,x}{12}-\frac {1}{6}} \]
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