Integrand size = 29, antiderivative size = 30 \[ \int \frac {1+2 x+3 x^2+x^3}{(-3+x) (-2+x) (-1+x)} \, dx=x+\frac {7}{2} \log (1-x)-25 \log (2-x)+\frac {61}{2} \log (3-x) \]
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Time = 0.04 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {1626} \[ \int \frac {1+2 x+3 x^2+x^3}{(-3+x) (-2+x) (-1+x)} \, dx=x+\frac {7}{2} \log (1-x)-25 \log (2-x)+\frac {61}{2} \log (3-x) \]
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Rule 1626
Rubi steps \begin{align*} \text {integral}& = \int \left (1+\frac {61}{2 (-3+x)}-\frac {25}{-2+x}+\frac {7}{2 (-1+x)}\right ) \, dx \\ & = x+\frac {7}{2} \log (1-x)-25 \log (2-x)+\frac {61}{2} \log (3-x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {1+2 x+3 x^2+x^3}{(-3+x) (-2+x) (-1+x)} \, dx=x+\frac {61}{2} \log (-3+x)-25 \log (-2+x)+\frac {7}{2} \log (-1+x) \]
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Time = 0.85 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70
method | result | size |
default | \(x +\frac {61 \ln \left (-3+x \right )}{2}+\frac {7 \ln \left (x -1\right )}{2}-25 \ln \left (x -2\right )\) | \(21\) |
norman | \(x +\frac {61 \ln \left (-3+x \right )}{2}+\frac {7 \ln \left (x -1\right )}{2}-25 \ln \left (x -2\right )\) | \(21\) |
risch | \(x +\frac {61 \ln \left (-3+x \right )}{2}+\frac {7 \ln \left (x -1\right )}{2}-25 \ln \left (x -2\right )\) | \(21\) |
parallelrisch | \(x +\frac {61 \ln \left (-3+x \right )}{2}+\frac {7 \ln \left (x -1\right )}{2}-25 \ln \left (x -2\right )\) | \(21\) |
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none
Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {1+2 x+3 x^2+x^3}{(-3+x) (-2+x) (-1+x)} \, dx=x + \frac {7}{2} \, \log \left (x - 1\right ) - 25 \, \log \left (x - 2\right ) + \frac {61}{2} \, \log \left (x - 3\right ) \]
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Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {1+2 x+3 x^2+x^3}{(-3+x) (-2+x) (-1+x)} \, dx=x + \frac {61 \log {\left (x - 3 \right )}}{2} - 25 \log {\left (x - 2 \right )} + \frac {7 \log {\left (x - 1 \right )}}{2} \]
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none
Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {1+2 x+3 x^2+x^3}{(-3+x) (-2+x) (-1+x)} \, dx=x + \frac {7}{2} \, \log \left (x - 1\right ) - 25 \, \log \left (x - 2\right ) + \frac {61}{2} \, \log \left (x - 3\right ) \]
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none
Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {1+2 x+3 x^2+x^3}{(-3+x) (-2+x) (-1+x)} \, dx=x + \frac {7}{2} \, \log \left ({\left | x - 1 \right |}\right ) - 25 \, \log \left ({\left | x - 2 \right |}\right ) + \frac {61}{2} \, \log \left ({\left | x - 3 \right |}\right ) \]
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Time = 9.51 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {1+2 x+3 x^2+x^3}{(-3+x) (-2+x) (-1+x)} \, dx=x+\frac {7\,\ln \left (x-1\right )}{2}-25\,\ln \left (x-2\right )+\frac {61\,\ln \left (x-3\right )}{2} \]
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