Integrand size = 20, antiderivative size = 54 \[ \int \frac {2+3 x}{(1-2 x)^3 (3+5 x)^2} \, dx=\frac {7}{242 (1-2 x)^2}+\frac {37}{1331 (1-2 x)}-\frac {5}{1331 (3+5 x)}-\frac {195 \log (1-2 x)}{14641}+\frac {195 \log (3+5 x)}{14641} \]
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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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {78} \[ \int \frac {2+3 x}{(1-2 x)^3 (3+5 x)^2} \, dx=\frac {37}{1331 (1-2 x)}-\frac {5}{1331 (5 x+3)}+\frac {7}{242 (1-2 x)^2}-\frac {195 \log (1-2 x)}{14641}+\frac {195 \log (5 x+3)}{14641} \]
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Rule 78
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {14}{121 (-1+2 x)^3}+\frac {74}{1331 (-1+2 x)^2}-\frac {390}{14641 (-1+2 x)}+\frac {25}{1331 (3+5 x)^2}+\frac {975}{14641 (3+5 x)}\right ) \, dx \\ & = \frac {7}{242 (1-2 x)^2}+\frac {37}{1331 (1-2 x)}-\frac {5}{1331 (3+5 x)}-\frac {195 \log (1-2 x)}{14641}+\frac {195 \log (3+5 x)}{14641} \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.15 \[ \int \frac {2+3 x}{(1-2 x)^3 (3+5 x)^2} \, dx=\frac {10}{1331 (-11+5 (1-2 x))}+\frac {7}{242 (1-2 x)^2}+\frac {37}{1331 (1-2 x)}+\frac {195 \log (11-5 (1-2 x))}{14641}-\frac {195 \log (1-2 x)}{14641} \]
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Time = 0.88 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81
method | result | size |
risch | \(\frac {-\frac {390}{1331} x^{2}+\frac {351}{2662} x +\frac {443}{2662}}{\left (-1+2 x \right )^{2} \left (3+5 x \right )}-\frac {195 \ln \left (-1+2 x \right )}{14641}+\frac {195 \ln \left (3+5 x \right )}{14641}\) | \(44\) |
default | \(-\frac {5}{1331 \left (3+5 x \right )}+\frac {195 \ln \left (3+5 x \right )}{14641}+\frac {7}{242 \left (-1+2 x \right )^{2}}-\frac {37}{1331 \left (-1+2 x \right )}-\frac {195 \ln \left (-1+2 x \right )}{14641}\) | \(45\) |
norman | \(\frac {\frac {602}{3993} x^{2}+\frac {2077}{3993} x -\frac {4430}{3993} x^{3}}{\left (-1+2 x \right )^{2} \left (3+5 x \right )}-\frac {195 \ln \left (-1+2 x \right )}{14641}+\frac {195 \ln \left (3+5 x \right )}{14641}\) | \(47\) |
parallelrisch | \(\frac {11700 \ln \left (x +\frac {3}{5}\right ) x^{3}-11700 \ln \left (x -\frac {1}{2}\right ) x^{3}-4680 \ln \left (x +\frac {3}{5}\right ) x^{2}+4680 \ln \left (x -\frac {1}{2}\right ) x^{2}-48730 x^{3}-4095 \ln \left (x +\frac {3}{5}\right ) x +4095 \ln \left (x -\frac {1}{2}\right ) x +6622 x^{2}+1755 \ln \left (x +\frac {3}{5}\right )-1755 \ln \left (x -\frac {1}{2}\right )+22847 x}{43923 \left (-1+2 x \right )^{2} \left (3+5 x \right )}\) | \(93\) |
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none
Time = 0.22 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.39 \[ \int \frac {2+3 x}{(1-2 x)^3 (3+5 x)^2} \, dx=-\frac {8580 \, x^{2} - 390 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )} \log \left (5 \, x + 3\right ) + 390 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )} \log \left (2 \, x - 1\right ) - 3861 \, x - 4873}{29282 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81 \[ \int \frac {2+3 x}{(1-2 x)^3 (3+5 x)^2} \, dx=- \frac {780 x^{2} - 351 x - 443}{53240 x^{3} - 21296 x^{2} - 18634 x + 7986} - \frac {195 \log {\left (x - \frac {1}{2} \right )}}{14641} + \frac {195 \log {\left (x + \frac {3}{5} \right )}}{14641} \]
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Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {2+3 x}{(1-2 x)^3 (3+5 x)^2} \, dx=-\frac {780 \, x^{2} - 351 \, x - 443}{2662 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )}} + \frac {195}{14641} \, \log \left (5 \, x + 3\right ) - \frac {195}{14641} \, \log \left (2 \, x - 1\right ) \]
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Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.94 \[ \int \frac {2+3 x}{(1-2 x)^3 (3+5 x)^2} \, dx=-\frac {5}{1331 \, {\left (5 \, x + 3\right )}} + \frac {10 \, {\left (\frac {792}{5 \, x + 3} - 109\right )}}{14641 \, {\left (\frac {11}{5 \, x + 3} - 2\right )}^{2}} - \frac {195}{14641} \, \log \left ({\left | -\frac {11}{5 \, x + 3} + 2 \right |}\right ) \]
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Time = 0.05 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.70 \[ \int \frac {2+3 x}{(1-2 x)^3 (3+5 x)^2} \, dx=\frac {390\,\mathrm {atanh}\left (\frac {20\,x}{11}+\frac {1}{11}\right )}{14641}-\frac {-\frac {39\,x^2}{2662}+\frac {351\,x}{53240}+\frac {443}{53240}}{-x^3+\frac {2\,x^2}{5}+\frac {7\,x}{20}-\frac {3}{20}} \]
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