Integrand size = 20, antiderivative size = 44 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^4} \, dx=-\frac {7}{243 (2+3 x)^3}+\frac {4}{9 (2+3 x)^2}-\frac {65}{27 (2+3 x)}-\frac {50}{81} \log (2+3 x) \]
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Time = 0.01 (sec) , antiderivative size = 44, 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} \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^4} \, dx=-\frac {65}{27 (3 x+2)}+\frac {4}{9 (3 x+2)^2}-\frac {7}{243 (3 x+2)^3}-\frac {50}{81} \log (3 x+2) \]
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Rule 78
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {7}{27 (2+3 x)^4}-\frac {8}{3 (2+3 x)^3}+\frac {65}{9 (2+3 x)^2}-\frac {50}{27 (2+3 x)}\right ) \, dx \\ & = -\frac {7}{243 (2+3 x)^3}+\frac {4}{9 (2+3 x)^2}-\frac {65}{27 (2+3 x)}-\frac {50}{81} \log (2+3 x) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^4} \, dx=-\frac {2131+6696 x+5265 x^2+150 (2+3 x)^3 \log (2+3 x)}{243 (2+3 x)^3} \]
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Time = 2.17 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.66
method | result | size |
risch | \(\frac {-\frac {65}{3} x^{2}-\frac {248}{9} x -\frac {2131}{243}}{\left (2+3 x \right )^{3}}-\frac {50 \ln \left (2+3 x \right )}{81}\) | \(29\) |
norman | \(\frac {\frac {643}{54} x +\frac {1351}{36} x^{2}+\frac {2131}{72} x^{3}}{\left (2+3 x \right )^{3}}-\frac {50 \ln \left (2+3 x \right )}{81}\) | \(32\) |
default | \(-\frac {7}{243 \left (2+3 x \right )^{3}}+\frac {4}{9 \left (2+3 x \right )^{2}}-\frac {65}{27 \left (2+3 x \right )}-\frac {50 \ln \left (2+3 x \right )}{81}\) | \(37\) |
parallelrisch | \(-\frac {10800 \ln \left (\frac {2}{3}+x \right ) x^{3}+21600 \ln \left (\frac {2}{3}+x \right ) x^{2}-19179 x^{3}+14400 \ln \left (\frac {2}{3}+x \right ) x -24318 x^{2}+3200 \ln \left (\frac {2}{3}+x \right )-7716 x}{648 \left (2+3 x \right )^{3}}\) | \(55\) |
meijerg | \(\frac {3 x \left (\frac {9}{4} x^{2}+\frac {9}{2} x +3\right )}{16 \left (1+\frac {3 x}{2}\right )^{3}}+\frac {x^{2} \left (3+\frac {3 x}{2}\right )}{8 \left (1+\frac {3 x}{2}\right )^{3}}-\frac {35 x^{3}}{48 \left (1+\frac {3 x}{2}\right )^{3}}+\frac {25 x \left (\frac {99}{2} x^{2}+45 x +12\right )}{324 \left (1+\frac {3 x}{2}\right )^{3}}-\frac {50 \ln \left (1+\frac {3 x}{2}\right )}{81}\) | \(79\) |
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Time = 0.22 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.18 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^4} \, dx=-\frac {5265 \, x^{2} + 150 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \log \left (3 \, x + 2\right ) + 6696 \, x + 2131}{243 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} \]
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Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^4} \, dx=- \frac {5265 x^{2} + 6696 x + 2131}{6561 x^{3} + 13122 x^{2} + 8748 x + 1944} - \frac {50 \log {\left (3 x + 2 \right )}}{81} \]
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Time = 0.20 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^4} \, dx=-\frac {5265 \, x^{2} + 6696 \, x + 2131}{243 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} - \frac {50}{81} \, \log \left (3 \, x + 2\right ) \]
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Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.66 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^4} \, dx=-\frac {5265 \, x^{2} + 6696 \, x + 2131}{243 \, {\left (3 \, x + 2\right )}^{3}} - \frac {50}{81} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]
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Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.77 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^4} \, dx=-\frac {50\,\ln \left (x+\frac {2}{3}\right )}{81}-\frac {\frac {65\,x^2}{81}+\frac {248\,x}{243}+\frac {2131}{6561}}{x^3+2\,x^2+\frac {4\,x}{3}+\frac {8}{27}} \]
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