Integrand size = 20, antiderivative size = 46 \[ \int \frac {1-2 x}{(2+3 x)^3 (3+5 x)^2} \, dx=-\frac {7}{2 (2+3 x)^2}-\frac {68}{2+3 x}-\frac {55}{3+5 x}+505 \log (2+3 x)-505 \log (3+5 x) \]
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Time = 0.02 (sec) , antiderivative size = 46, 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} \[ \int \frac {1-2 x}{(2+3 x)^3 (3+5 x)^2} \, dx=-\frac {68}{3 x+2}-\frac {55}{5 x+3}-\frac {7}{2 (3 x+2)^2}+505 \log (3 x+2)-505 \log (5 x+3) \]
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Rule 78
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {21}{(2+3 x)^3}+\frac {204}{(2+3 x)^2}+\frac {1515}{2+3 x}+\frac {275}{(3+5 x)^2}-\frac {2525}{3+5 x}\right ) \, dx \\ & = -\frac {7}{2 (2+3 x)^2}-\frac {68}{2+3 x}-\frac {55}{3+5 x}+505 \log (2+3 x)-505 \log (3+5 x) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.04 \[ \int \frac {1-2 x}{(2+3 x)^3 (3+5 x)^2} \, dx=-\frac {7}{2 (2+3 x)^2}-\frac {68}{2+3 x}-\frac {55}{3+5 x}+505 \log (2+3 x)-505 \log (-3 (3+5 x)) \]
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Time = 1.83 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.96
method | result | size |
risch | \(\frac {-1515 x^{2}-\frac {3939}{2} x -\frac {1277}{2}}{\left (2+3 x \right )^{2} \left (3+5 x \right )}+505 \ln \left (2+3 x \right )-505 \ln \left (3+5 x \right )\) | \(44\) |
default | \(-\frac {7}{2 \left (2+3 x \right )^{2}}-\frac {68}{2+3 x}-\frac {55}{3+5 x}+505 \ln \left (2+3 x \right )-505 \ln \left (3+5 x \right )\) | \(45\) |
norman | \(\frac {\frac {6061}{6} x +\frac {19155}{8} x^{3}+\frac {24913}{8} x^{2}}{\left (2+3 x \right )^{2} \left (3+5 x \right )}+505 \ln \left (2+3 x \right )-505 \ln \left (3+5 x \right )\) | \(47\) |
parallelrisch | \(\frac {545400 \ln \left (\frac {2}{3}+x \right ) x^{3}-545400 \ln \left (x +\frac {3}{5}\right ) x^{3}+1054440 \ln \left (\frac {2}{3}+x \right ) x^{2}-1054440 \ln \left (x +\frac {3}{5}\right ) x^{2}+57465 x^{3}+678720 \ln \left (\frac {2}{3}+x \right ) x -678720 \ln \left (x +\frac {3}{5}\right ) x +74739 x^{2}+145440 \ln \left (\frac {2}{3}+x \right )-145440 \ln \left (x +\frac {3}{5}\right )+24244 x}{24 \left (2+3 x \right )^{2} \left (3+5 x \right )}\) | \(93\) |
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none
Time = 0.22 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.63 \[ \int \frac {1-2 x}{(2+3 x)^3 (3+5 x)^2} \, dx=-\frac {3030 \, x^{2} + 1010 \, {\left (45 \, x^{3} + 87 \, x^{2} + 56 \, x + 12\right )} \log \left (5 \, x + 3\right ) - 1010 \, {\left (45 \, x^{3} + 87 \, x^{2} + 56 \, x + 12\right )} \log \left (3 \, x + 2\right ) + 3939 \, x + 1277}{2 \, {\left (45 \, x^{3} + 87 \, x^{2} + 56 \, x + 12\right )}} \]
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Time = 0.08 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.89 \[ \int \frac {1-2 x}{(2+3 x)^3 (3+5 x)^2} \, dx=- \frac {3030 x^{2} + 3939 x + 1277}{90 x^{3} + 174 x^{2} + 112 x + 24} - 505 \log {\left (x + \frac {3}{5} \right )} + 505 \log {\left (x + \frac {2}{3} \right )} \]
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none
Time = 0.21 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int \frac {1-2 x}{(2+3 x)^3 (3+5 x)^2} \, dx=-\frac {3030 \, x^{2} + 3939 \, x + 1277}{2 \, {\left (45 \, x^{3} + 87 \, x^{2} + 56 \, x + 12\right )}} - 505 \, \log \left (5 \, x + 3\right ) + 505 \, \log \left (3 \, x + 2\right ) \]
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Time = 0.28 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.07 \[ \int \frac {1-2 x}{(2+3 x)^3 (3+5 x)^2} \, dx=-\frac {55}{5 \, x + 3} + \frac {15 \, {\left (\frac {206}{5 \, x + 3} + 513\right )}}{2 \, {\left (\frac {1}{5 \, x + 3} + 3\right )}^{2}} + 505 \, \log \left ({\left | -\frac {1}{5 \, x + 3} - 3 \right |}\right ) \]
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Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \frac {1-2 x}{(2+3 x)^3 (3+5 x)^2} \, dx=1010\,\mathrm {atanh}\left (30\,x+19\right )-\frac {\frac {101\,x^2}{3}+\frac {1313\,x}{30}+\frac {1277}{90}}{x^3+\frac {29\,x^2}{15}+\frac {56\,x}{45}+\frac {4}{15}} \]
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