Integrand size = 35, antiderivative size = 75 \[ \int \frac {5+x+3 x^2+2 x^3}{x \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=\frac {1}{6} \sqrt {\frac {5}{3}} \arctan \left (\frac {1-4 x}{\sqrt {15}}\right )+\frac {2 \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {5 \log (x)}{2}-\log \left (1+x+x^2\right )-\frac {1}{4} \log \left (2-x+2 x^2\right ) \]
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Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {2112, 814, 648, 632, 210, 642} \[ \int \frac {5+x+3 x^2+2 x^3}{x \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=\frac {1}{6} \sqrt {\frac {5}{3}} \arctan \left (\frac {1-4 x}{\sqrt {15}}\right )+\frac {2 \arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{3 \sqrt {3}}-\log \left (x^2+x+1\right )-\frac {1}{4} \log \left (2 x^2-x+2\right )+\frac {5 \log (x)}{2} \]
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Rule 210
Rule 632
Rule 642
Rule 648
Rule 814
Rule 2112
Rubi steps \begin{align*} \text {integral}& = -\left (\frac {1}{3} \int \frac {-6+4 x}{x \left (4-2 x+4 x^2\right )} \, dx\right )+\frac {1}{3} \int \frac {24+16 x}{x \left (4+4 x+4 x^2\right )} \, dx \\ & = \frac {1}{3} \int \left (\frac {6}{x}-\frac {2 (1+3 x)}{1+x+x^2}\right ) \, dx-\frac {1}{3} \int \left (-\frac {3}{2 x}+\frac {1+6 x}{2 \left (2-x+2 x^2\right )}\right ) \, dx \\ & = \frac {5 \log (x)}{2}-\frac {1}{6} \int \frac {1+6 x}{2-x+2 x^2} \, dx-\frac {2}{3} \int \frac {1+3 x}{1+x+x^2} \, dx \\ & = \frac {5 \log (x)}{2}-\frac {1}{4} \int \frac {-1+4 x}{2-x+2 x^2} \, dx+\frac {1}{3} \int \frac {1}{1+x+x^2} \, dx-\frac {5}{12} \int \frac {1}{2-x+2 x^2} \, dx-\int \frac {1+2 x}{1+x+x^2} \, dx \\ & = \frac {5 \log (x)}{2}-\log \left (1+x+x^2\right )-\frac {1}{4} \log \left (2-x+2 x^2\right )-\frac {2}{3} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right )+\frac {5}{6} \text {Subst}\left (\int \frac {1}{-15-x^2} \, dx,x,-1+4 x\right ) \\ & = \frac {1}{6} \sqrt {\frac {5}{3}} \tan ^{-1}\left (\frac {1-4 x}{\sqrt {15}}\right )+\frac {2 \tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {5 \log (x)}{2}-\log \left (1+x+x^2\right )-\frac {1}{4} \log \left (2-x+2 x^2\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.92 \[ \int \frac {5+x+3 x^2+2 x^3}{x \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=\frac {1}{36} \left (8 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )-2 \sqrt {15} \arctan \left (\frac {-1+4 x}{\sqrt {15}}\right )+90 \log (x)-36 \log \left (1+x+x^2\right )-9 \log \left (2-x+2 x^2\right )\right ) \]
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Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.80
method | result | size |
default | \(\frac {5 \ln \left (x \right )}{2}-\ln \left (x^{2}+x +1\right )+\frac {2 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{9}-\frac {\ln \left (2 x^{2}-x +2\right )}{4}-\frac {\sqrt {15}\, \arctan \left (\frac {\left (-1+4 x \right ) \sqrt {15}}{15}\right )}{18}\) | \(60\) |
risch | \(-\ln \left (4 x^{2}+4 x +4\right )+\frac {2 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{9}-\frac {\sqrt {15}\, \arctan \left (\frac {\left (-1+4 x \right ) \sqrt {15}}{15}\right )}{18}-\frac {\ln \left (16 x^{2}-8 x +16\right )}{4}+\frac {5 \ln \left (x \right )}{2}\) | \(64\) |
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Time = 0.29 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.87 \[ \int \frac {5+x+3 x^2+2 x^3}{x \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=-\frac {1}{18} \, \sqrt {5} \sqrt {3} \arctan \left (\frac {1}{15} \, \sqrt {5} \sqrt {3} {\left (4 \, x - 1\right )}\right ) + \frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{4} \, \log \left (2 \, x^{2} - x + 2\right ) - \log \left (x^{2} + x + 1\right ) + \frac {5}{2} \, \log \left (x\right ) \]
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Time = 0.14 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.04 \[ \int \frac {5+x+3 x^2+2 x^3}{x \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=\frac {5 \log {\left (x \right )}}{2} - \frac {\log {\left (x^{2} - \frac {x}{2} + 1 \right )}}{4} - \log {\left (x^{2} + x + 1 \right )} - \frac {\sqrt {15} \operatorname {atan}{\left (\frac {4 \sqrt {15} x}{15} - \frac {\sqrt {15}}{15} \right )}}{18} + \frac {2 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{9} \]
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Time = 0.29 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.79 \[ \int \frac {5+x+3 x^2+2 x^3}{x \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=-\frac {1}{18} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, x - 1\right )}\right ) + \frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{4} \, \log \left (2 \, x^{2} - x + 2\right ) - \log \left (x^{2} + x + 1\right ) + \frac {5}{2} \, \log \left (x\right ) \]
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Time = 0.34 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.80 \[ \int \frac {5+x+3 x^2+2 x^3}{x \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=-\frac {1}{18} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, x - 1\right )}\right ) + \frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{4} \, \log \left (2 \, x^{2} - x + 2\right ) - \log \left (x^{2} + x + 1\right ) + \frac {5}{2} \, \log \left ({\left | x \right |}\right ) \]
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Time = 9.30 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.11 \[ \int \frac {5+x+3 x^2+2 x^3}{x \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=\frac {5\,\ln \left (x\right )}{2}-\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (1+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right )+\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-1+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right )+\ln \left (x-\frac {1}{4}-\frac {\sqrt {15}\,1{}\mathrm {i}}{4}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {15}\,1{}\mathrm {i}}{36}\right )-\ln \left (x-\frac {1}{4}+\frac {\sqrt {15}\,1{}\mathrm {i}}{4}\right )\,\left (\frac {1}{4}+\frac {\sqrt {15}\,1{}\mathrm {i}}{36}\right ) \]
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