Integrand size = 35, antiderivative size = 84 \[ \int \frac {5+x+3 x^2+2 x^3}{x^2 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=-\frac {5}{2 x}+\frac {5}{12} \sqrt {\frac {5}{3}} \arctan \left (\frac {1-4 x}{\sqrt {15}}\right )-\frac {10 \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {3 \log (x)}{4}+\frac {1}{3} \log \left (1+x+x^2\right )+\frac {1}{24} \log \left (2-x+2 x^2\right ) \]
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Time = 0.09 (sec) , antiderivative size = 84, 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} \[ \int \frac {5+x+3 x^2+2 x^3}{x^2 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=\frac {5}{12} \sqrt {\frac {5}{3}} \arctan \left (\frac {1-4 x}{\sqrt {15}}\right )-\frac {10 \arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {1}{3} \log \left (x^2+x+1\right )+\frac {1}{24} \log \left (2 x^2-x+2\right )-\frac {5}{2 x}-\frac {3 \log (x)}{4} \]
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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^2 \left (4-2 x+4 x^2\right )} \, dx\right )+\frac {1}{3} \int \frac {24+16 x}{x^2 \left (4+4 x+4 x^2\right )} \, dx \\ & = \frac {1}{3} \int \left (\frac {6}{x^2}-\frac {2}{x}+\frac {2 (-2+x)}{1+x+x^2}\right ) \, dx-\frac {1}{3} \int \left (-\frac {3}{2 x^2}+\frac {1}{4 x}+\frac {13-2 x}{4 \left (2-x+2 x^2\right )}\right ) \, dx \\ & = -\frac {5}{2 x}-\frac {3 \log (x)}{4}-\frac {1}{12} \int \frac {13-2 x}{2-x+2 x^2} \, dx+\frac {2}{3} \int \frac {-2+x}{1+x+x^2} \, dx \\ & = -\frac {5}{2 x}-\frac {3 \log (x)}{4}+\frac {1}{24} \int \frac {-1+4 x}{2-x+2 x^2} \, dx+\frac {1}{3} \int \frac {1+2 x}{1+x+x^2} \, dx-\frac {25}{24} \int \frac {1}{2-x+2 x^2} \, dx-\frac {5}{3} \int \frac {1}{1+x+x^2} \, dx \\ & = -\frac {5}{2 x}-\frac {3 \log (x)}{4}+\frac {1}{3} \log \left (1+x+x^2\right )+\frac {1}{24} \log \left (2-x+2 x^2\right )+\frac {25}{12} \text {Subst}\left (\int \frac {1}{-15-x^2} \, dx,x,-1+4 x\right )+\frac {10}{3} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right ) \\ & = -\frac {5}{2 x}+\frac {5}{12} \sqrt {\frac {5}{3}} \tan ^{-1}\left (\frac {1-4 x}{\sqrt {15}}\right )-\frac {10 \tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {3 \log (x)}{4}+\frac {1}{3} \log \left (1+x+x^2\right )+\frac {1}{24} \log \left (2-x+2 x^2\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.93 \[ \int \frac {5+x+3 x^2+2 x^3}{x^2 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=-\frac {180+80 \sqrt {3} x \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )+10 \sqrt {15} x \arctan \left (\frac {-1+4 x}{\sqrt {15}}\right )+54 x \log (x)-24 x \log \left (1+x+x^2\right )-3 x \log \left (2-x+2 x^2\right )}{72 x} \]
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Time = 0.09 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.77
method | result | size |
default | \(-\frac {5}{2 x}-\frac {3 \ln \left (x \right )}{4}+\frac {\ln \left (x^{2}+x +1\right )}{3}-\frac {10 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{9}+\frac {\ln \left (2 x^{2}-x +2\right )}{24}-\frac {5 \sqrt {15}\, \arctan \left (\frac {\left (-1+4 x \right ) \sqrt {15}}{15}\right )}{36}\) | \(65\) |
risch | \(-\frac {5}{2 x}-\frac {5 \sqrt {15}\, \arctan \left (\frac {\left (-1+4 x \right ) \sqrt {15}}{15}\right )}{36}+\frac {\ln \left (16 x^{2}-8 x +16\right )}{24}-\frac {3 \ln \left (x \right )}{4}+\frac {\ln \left (25 x^{2}+25 x +25\right )}{3}-\frac {10 \sqrt {3}\, \arctan \left (\frac {2 \left (5 x +\frac {5}{2}\right ) \sqrt {3}}{15}\right )}{9}\) | \(69\) |
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Time = 0.30 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.90 \[ \int \frac {5+x+3 x^2+2 x^3}{x^2 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=-\frac {10 \, \sqrt {5} \sqrt {3} x \arctan \left (\frac {1}{15} \, \sqrt {5} \sqrt {3} {\left (4 \, x - 1\right )}\right ) + 80 \, \sqrt {3} x \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - 3 \, x \log \left (2 \, x^{2} - x + 2\right ) - 24 \, x \log \left (x^{2} + x + 1\right ) + 54 \, x \log \left (x\right ) + 180}{72 \, x} \]
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Time = 0.16 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.04 \[ \int \frac {5+x+3 x^2+2 x^3}{x^2 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=- \frac {3 \log {\left (x \right )}}{4} + \frac {\log {\left (x^{2} - \frac {x}{2} + 1 \right )}}{24} + \frac {\log {\left (x^{2} + x + 1 \right )}}{3} - \frac {5 \sqrt {15} \operatorname {atan}{\left (\frac {4 \sqrt {15} x}{15} - \frac {\sqrt {15}}{15} \right )}}{36} - \frac {10 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{9} - \frac {5}{2 x} \]
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Time = 0.28 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.76 \[ \int \frac {5+x+3 x^2+2 x^3}{x^2 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=-\frac {5}{36} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, x - 1\right )}\right ) - \frac {10}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {5}{2 \, x} + \frac {1}{24} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac {1}{3} \, \log \left (x^{2} + x + 1\right ) - \frac {3}{4} \, \log \left (x\right ) \]
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Time = 0.29 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.77 \[ \int \frac {5+x+3 x^2+2 x^3}{x^2 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=-\frac {5}{36} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, x - 1\right )}\right ) - \frac {10}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {5}{2 \, x} + \frac {1}{24} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac {1}{3} \, \log \left (x^{2} + x + 1\right ) - \frac {3}{4} \, \log \left ({\left | x \right |}\right ) \]
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Time = 9.09 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.05 \[ \int \frac {5+x+3 x^2+2 x^3}{x^2 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=-\frac {3\,\ln \left (x\right )}{4}+\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{3}+\frac {\sqrt {3}\,5{}\mathrm {i}}{9}\right )-\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{3}+\frac {\sqrt {3}\,5{}\mathrm {i}}{9}\right )+\ln \left (x-\frac {1}{4}-\frac {\sqrt {15}\,1{}\mathrm {i}}{4}\right )\,\left (\frac {1}{24}+\frac {\sqrt {15}\,5{}\mathrm {i}}{72}\right )-\ln \left (x-\frac {1}{4}+\frac {\sqrt {15}\,1{}\mathrm {i}}{4}\right )\,\left (-\frac {1}{24}+\frac {\sqrt {15}\,5{}\mathrm {i}}{72}\right )-\frac {5}{2\,x} \]
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