Integrand size = 20, antiderivative size = 42 \[ \int \frac {1+x^5}{-10 x-3 x^2+x^3} \, dx=19 x+\frac {3 x^2}{2}+\frac {x^3}{3}+\frac {3126}{35} \log (5-x)-\frac {\log (x)}{10}-\frac {31}{14} \log (2+x) \]
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Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1608, 1642} \[ \int \frac {1+x^5}{-10 x-3 x^2+x^3} \, dx=\frac {x^3}{3}+\frac {3 x^2}{2}+19 x+\frac {3126}{35} \log (5-x)-\frac {\log (x)}{10}-\frac {31}{14} \log (x+2) \]
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Rule 1608
Rule 1642
Rubi steps \begin{align*} \text {integral}& = \int \frac {1+x^5}{x \left (-10-3 x+x^2\right )} \, dx \\ & = \int \left (19+\frac {3126}{35 (-5+x)}-\frac {1}{10 x}+3 x+x^2-\frac {31}{14 (2+x)}\right ) \, dx \\ & = 19 x+\frac {3 x^2}{2}+\frac {x^3}{3}+\frac {3126}{35} \log (5-x)-\frac {\log (x)}{10}-\frac {31}{14} \log (2+x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \frac {1+x^5}{-10 x-3 x^2+x^3} \, dx=19 x+\frac {3 x^2}{2}+\frac {x^3}{3}+\frac {3126}{35} \log (5-x)-\frac {\log (x)}{10}-\frac {31}{14} \log (2+x) \]
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Time = 0.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.74
method | result | size |
default | \(\frac {x^{3}}{3}+\frac {3 x^{2}}{2}+19 x -\frac {\ln \left (x \right )}{10}+\frac {3126 \ln \left (-5+x \right )}{35}-\frac {31 \ln \left (x +2\right )}{14}\) | \(31\) |
norman | \(\frac {x^{3}}{3}+\frac {3 x^{2}}{2}+19 x -\frac {\ln \left (x \right )}{10}+\frac {3126 \ln \left (-5+x \right )}{35}-\frac {31 \ln \left (x +2\right )}{14}\) | \(31\) |
risch | \(\frac {x^{3}}{3}+\frac {3 x^{2}}{2}+19 x -\frac {\ln \left (x \right )}{10}+\frac {3126 \ln \left (-5+x \right )}{35}-\frac {31 \ln \left (x +2\right )}{14}\) | \(31\) |
parallelrisch | \(\frac {x^{3}}{3}+\frac {3 x^{2}}{2}+19 x -\frac {\ln \left (x \right )}{10}+\frac {3126 \ln \left (-5+x \right )}{35}-\frac {31 \ln \left (x +2\right )}{14}\) | \(31\) |
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Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.71 \[ \int \frac {1+x^5}{-10 x-3 x^2+x^3} \, dx=\frac {1}{3} \, x^{3} + \frac {3}{2} \, x^{2} + 19 \, x - \frac {31}{14} \, \log \left (x + 2\right ) + \frac {3126}{35} \, \log \left (x - 5\right ) - \frac {1}{10} \, \log \left (x\right ) \]
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Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86 \[ \int \frac {1+x^5}{-10 x-3 x^2+x^3} \, dx=\frac {x^{3}}{3} + \frac {3 x^{2}}{2} + 19 x - \frac {\log {\left (x \right )}}{10} + \frac {3126 \log {\left (x - 5 \right )}}{35} - \frac {31 \log {\left (x + 2 \right )}}{14} \]
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Time = 0.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.71 \[ \int \frac {1+x^5}{-10 x-3 x^2+x^3} \, dx=\frac {1}{3} \, x^{3} + \frac {3}{2} \, x^{2} + 19 \, x - \frac {31}{14} \, \log \left (x + 2\right ) + \frac {3126}{35} \, \log \left (x - 5\right ) - \frac {1}{10} \, \log \left (x\right ) \]
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Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.79 \[ \int \frac {1+x^5}{-10 x-3 x^2+x^3} \, dx=\frac {1}{3} \, x^{3} + \frac {3}{2} \, x^{2} + 19 \, x - \frac {31}{14} \, \log \left ({\left | x + 2 \right |}\right ) + \frac {3126}{35} \, \log \left ({\left | x - 5 \right |}\right ) - \frac {1}{10} \, \log \left ({\left | x \right |}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.71 \[ \int \frac {1+x^5}{-10 x-3 x^2+x^3} \, dx=19\,x-\frac {31\,\ln \left (x+2\right )}{14}+\frac {3126\,\ln \left (x-5\right )}{35}-\frac {\ln \left (x\right )}{10}+\frac {3\,x^2}{2}+\frac {x^3}{3} \]
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