Integrand size = 20, antiderivative size = 32 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^2} \, dx=\frac {1}{21 (2+3 x)}-\frac {11}{49} \log (1-2 x)+\frac {11}{49} \log (2+3 x) \]
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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} \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^2} \, dx=\frac {1}{21 (3 x+2)}-\frac {11}{49} \log (1-2 x)+\frac {11}{49} \log (3 x+2) \]
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Rule 78
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {22}{49 (-1+2 x)}-\frac {1}{7 (2+3 x)^2}+\frac {33}{49 (2+3 x)}\right ) \, dx \\ & = \frac {1}{21 (2+3 x)}-\frac {11}{49} \log (1-2 x)+\frac {11}{49} \log (2+3 x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^2} \, dx=\frac {1}{147} \left (\frac {7}{2+3 x}-33 \log (3-6 x)+33 \log (2+3 x)\right ) \]
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Time = 2.48 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78
method | result | size |
risch | \(\frac {1}{42+63 x}-\frac {11 \ln \left (-1+2 x \right )}{49}+\frac {11 \ln \left (2+3 x \right )}{49}\) | \(25\) |
default | \(\frac {1}{42+63 x}-\frac {11 \ln \left (-1+2 x \right )}{49}+\frac {11 \ln \left (2+3 x \right )}{49}\) | \(27\) |
norman | \(-\frac {x}{14 \left (2+3 x \right )}-\frac {11 \ln \left (-1+2 x \right )}{49}+\frac {11 \ln \left (2+3 x \right )}{49}\) | \(28\) |
parallelrisch | \(\frac {66 \ln \left (\frac {2}{3}+x \right ) x -66 \ln \left (x -\frac {1}{2}\right ) x +44 \ln \left (\frac {2}{3}+x \right )-44 \ln \left (x -\frac {1}{2}\right )-7 x}{196+294 x}\) | \(40\) |
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none
Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^2} \, dx=\frac {33 \, {\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) - 33 \, {\left (3 \, x + 2\right )} \log \left (2 \, x - 1\right ) + 7}{147 \, {\left (3 \, x + 2\right )}} \]
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Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^2} \, dx=- \frac {11 \log {\left (x - \frac {1}{2} \right )}}{49} + \frac {11 \log {\left (x + \frac {2}{3} \right )}}{49} + \frac {1}{63 x + 42} \]
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none
Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^2} \, dx=\frac {1}{21 \, {\left (3 \, x + 2\right )}} + \frac {11}{49} \, \log \left (3 \, x + 2\right ) - \frac {11}{49} \, \log \left (2 \, x - 1\right ) \]
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Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^2} \, dx=\frac {1}{21 \, {\left (3 \, x + 2\right )}} - \frac {11}{49} \, \log \left ({\left | -\frac {7}{3 \, x + 2} + 2 \right |}\right ) \]
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Time = 0.04 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.50 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^2} \, dx=\frac {22\,\mathrm {atanh}\left (\frac {12\,x}{7}+\frac {1}{7}\right )}{49}+\frac {1}{63\,\left (x+\frac {2}{3}\right )} \]
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