Integrand size = 20, antiderivative size = 32 \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)^2} \, dx=-\frac {1}{55 (3+5 x)}-\frac {7}{121} \log (1-2 x)+\frac {7}{121} \log (3+5 x) \]
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Time = 0.01 (sec) , antiderivative size = 32, 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} \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)^2} \, dx=-\frac {1}{55 (5 x+3)}-\frac {7}{121} \log (1-2 x)+\frac {7}{121} \log (5 x+3) \]
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Rule 78
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {14}{121 (-1+2 x)}+\frac {1}{11 (3+5 x)^2}+\frac {35}{121 (3+5 x)}\right ) \, dx \\ & = -\frac {1}{55 (3+5 x)}-\frac {7}{121} \log (1-2 x)+\frac {7}{121} \log (3+5 x) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)^2} \, dx=\frac {1}{605} \left (-\frac {11}{3+5 x}-35 \log (5-10 x)+35 \log (3+5 x)\right ) \]
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Time = 2.56 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78
method | result | size |
risch | \(-\frac {1}{275 \left (x +\frac {3}{5}\right )}-\frac {7 \ln \left (-1+2 x \right )}{121}+\frac {7 \ln \left (3+5 x \right )}{121}\) | \(25\) |
default | \(-\frac {1}{55 \left (3+5 x \right )}+\frac {7 \ln \left (3+5 x \right )}{121}-\frac {7 \ln \left (-1+2 x \right )}{121}\) | \(27\) |
norman | \(\frac {x}{99+165 x}-\frac {7 \ln \left (-1+2 x \right )}{121}+\frac {7 \ln \left (3+5 x \right )}{121}\) | \(28\) |
parallelrisch | \(\frac {105 \ln \left (x +\frac {3}{5}\right ) x -105 \ln \left (x -\frac {1}{2}\right ) x +63 \ln \left (x +\frac {3}{5}\right )-63 \ln \left (x -\frac {1}{2}\right )+11 x}{1089+1815 x}\) | \(40\) |
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Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)^2} \, dx=\frac {35 \, {\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) - 35 \, {\left (5 \, x + 3\right )} \log \left (2 \, x - 1\right ) - 11}{605 \, {\left (5 \, x + 3\right )}} \]
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Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)^2} \, dx=- \frac {7 \log {\left (x - \frac {1}{2} \right )}}{121} + \frac {7 \log {\left (x + \frac {3}{5} \right )}}{121} - \frac {1}{275 x + 165} \]
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Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)^2} \, dx=-\frac {1}{55 \, {\left (5 \, x + 3\right )}} + \frac {7}{121} \, \log \left (5 \, x + 3\right ) - \frac {7}{121} \, \log \left (2 \, x - 1\right ) \]
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Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)^2} \, dx=-\frac {1}{55 \, {\left (5 \, x + 3\right )}} - \frac {7}{121} \, \log \left ({\left | -\frac {11}{5 \, x + 3} + 2 \right |}\right ) \]
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Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.56 \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)^2} \, dx=\frac {14\,\mathrm {atanh}\left (\frac {20\,x}{11}+\frac {1}{11}\right )}{121}-\frac {1}{275\,\left (x+\frac {3}{5}\right )} \]
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