Integrand size = 20, antiderivative size = 32 \[ \int \frac {a+a x+c x^2}{1-x^3} \, dx=-\frac {1}{3} (2 a+c) \log (1-x)+\frac {1}{3} (a-c) \log \left (1+x+x^2\right ) \]
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Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1889, 31, 642} \[ \int \frac {a+a x+c x^2}{1-x^3} \, dx=\frac {1}{3} (a-c) \log \left (x^2+x+1\right )-\frac {1}{3} (2 a+c) \log (1-x) \]
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Rule 31
Rule 642
Rule 1889
Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \int \frac {a-c+(2 a-2 c) x}{1+x+x^2} \, dx+\frac {1}{3} (2 a+c) \int \frac {1}{1-x} \, dx \\ & = -\frac {1}{3} (2 a+c) \log (1-x)+\frac {1}{3} (a-c) \log \left (1+x+x^2\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {a+a x+c x^2}{1-x^3} \, dx=\frac {1}{3} \left (-((2 a+c) \log (1-x))+(a-c) \log \left (1+x+x^2\right )\right ) \]
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Time = 1.72 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88
method | result | size |
default | \(\left (-\frac {2 a}{3}-\frac {c}{3}\right ) \ln \left (-1+x \right )+\frac {\left (a -c \right ) \ln \left (x^{2}+x +1\right )}{3}\) | \(28\) |
norman | \(\left (-\frac {2 a}{3}-\frac {c}{3}\right ) \ln \left (-1+x \right )+\left (\frac {a}{3}-\frac {c}{3}\right ) \ln \left (x^{2}+x +1\right )\) | \(29\) |
parallelrisch | \(-\frac {2 \ln \left (-1+x \right ) a}{3}-\frac {\ln \left (-1+x \right ) c}{3}+\frac {\ln \left (x^{2}+x +1\right ) a}{3}-\frac {\ln \left (x^{2}+x +1\right ) c}{3}\) | \(36\) |
risch | \(-\frac {2 \ln \left (-1+x \right ) a}{3}-\frac {\ln \left (-1+x \right ) c}{3}+\frac {\ln \left (-x^{2}-x -1\right ) a}{3}-\frac {\ln \left (-x^{2}-x -1\right ) c}{3}\) | \(44\) |
meijerg | \(-\frac {c \ln \left (-x^{3}+1\right )}{3}-\frac {a \,x^{2} \left (\ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}\right )-\frac {\ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{2}+\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2+\left (x^{3}\right )^{\frac {1}{3}}}\right )\right )}{3 \left (x^{3}\right )^{\frac {2}{3}}}-\frac {a x \left (\ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}\right )-\frac {\ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2+\left (x^{3}\right )^{\frac {1}{3}}}\right )\right )}{3 \left (x^{3}\right )^{\frac {1}{3}}}\) | \(138\) |
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Time = 0.49 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {a+a x+c x^2}{1-x^3} \, dx=\frac {1}{3} \, {\left (a - c\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{3} \, {\left (2 \, a + c\right )} \log \left (x - 1\right ) \]
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Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {a+a x+c x^2}{1-x^3} \, dx=\frac {\left (a - c\right ) \log {\left (x^{2} + x + 1 \right )}}{3} - \frac {\left (2 a + c\right ) \log {\left (x - 1 \right )}}{3} \]
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Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {a+a x+c x^2}{1-x^3} \, dx=\frac {1}{3} \, {\left (a - c\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{3} \, {\left (2 \, a + c\right )} \log \left (x - 1\right ) \]
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Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {a+a x+c x^2}{1-x^3} \, dx=\frac {1}{3} \, {\left (a - c\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{3} \, {\left (2 \, a + c\right )} \log \left ({\left | x - 1 \right |}\right ) \]
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Time = 9.46 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09 \[ \int \frac {a+a x+c x^2}{1-x^3} \, dx=\frac {a\,\ln \left (x^2+x+1\right )}{3}-\frac {c\,\ln \left (x-1\right )}{3}-\frac {2\,a\,\ln \left (x-1\right )}{3}-\frac {c\,\ln \left (x^2+x+1\right )}{3} \]
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