Integrand size = 35, antiderivative size = 90 \[ \int \frac {x^3 \left (5+x+3 x^2+2 x^3\right )}{2+x+3 x^2+x^3+2 x^4} \, dx=-\frac {3 x}{2}+\frac {x^2}{2}+\frac {x^3}{3}+\frac {5}{12} \sqrt {\frac {5}{3}} \arctan \left (\frac {1-4 x}{\sqrt {15}}\right )+\frac {8 \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {2}{3} \log \left (1+x+x^2\right )-\frac {1}{24} \log \left (2-x+2 x^2\right ) \]
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Time = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2100, 648, 632, 210, 642} \[ \int \frac {x^3 \left (5+x+3 x^2+2 x^3\right )}{2+x+3 x^2+x^3+2 x^4} \, dx=\frac {5}{12} \sqrt {\frac {5}{3}} \arctan \left (\frac {1-4 x}{\sqrt {15}}\right )+\frac {8 \arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {x^3}{3}+\frac {x^2}{2}+\frac {2}{3} \log \left (x^2+x+1\right )-\frac {1}{24} \log \left (2 x^2-x+2\right )-\frac {3 x}{2} \]
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Rule 210
Rule 632
Rule 642
Rule 648
Rule 2100
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {3}{2}+x+x^2+\frac {2 (3+2 x)}{3 \left (1+x+x^2\right )}+\frac {-6-x}{6 \left (2-x+2 x^2\right )}\right ) \, dx \\ & = -\frac {3 x}{2}+\frac {x^2}{2}+\frac {x^3}{3}+\frac {1}{6} \int \frac {-6-x}{2-x+2 x^2} \, dx+\frac {2}{3} \int \frac {3+2 x}{1+x+x^2} \, dx \\ & = -\frac {3 x}{2}+\frac {x^2}{2}+\frac {x^3}{3}-\frac {1}{24} \int \frac {-1+4 x}{2-x+2 x^2} \, dx+\frac {2}{3} \int \frac {1+2 x}{1+x+x^2} \, dx-\frac {25}{24} \int \frac {1}{2-x+2 x^2} \, dx+\frac {4}{3} \int \frac {1}{1+x+x^2} \, dx \\ & = -\frac {3 x}{2}+\frac {x^2}{2}+\frac {x^3}{3}+\frac {2}{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 {8}{3} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right ) \\ & = -\frac {3 x}{2}+\frac {x^2}{2}+\frac {x^3}{3}+\frac {5}{12} \sqrt {\frac {5}{3}} \tan ^{-1}\left (\frac {1-4 x}{\sqrt {15}}\right )+\frac {8 \tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {2}{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 = 0.87 \[ \int \frac {x^3 \left (5+x+3 x^2+2 x^3\right )}{2+x+3 x^2+x^3+2 x^4} \, dx=\frac {1}{72} \left (-108 x+36 x^2+24 x^3+64 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )-10 \sqrt {15} \arctan \left (\frac {-1+4 x}{\sqrt {15}}\right )+48 \log \left (1+x+x^2\right )-3 \log \left (2-x+2 x^2\right )\right ) \]
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Time = 0.07 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.77
method | result | size |
default | \(\frac {x^{3}}{3}+\frac {x^{2}}{2}-\frac {3 x}{2}+\frac {2 \ln \left (x^{2}+x +1\right )}{3}+\frac {8 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{9}-\frac {\ln \left (2 x^{2}-x +2\right )}{24}-\frac {5 \sqrt {15}\, \arctan \left (\frac {\left (-1+4 x \right ) \sqrt {15}}{15}\right )}{36}\) | \(69\) |
risch | \(\frac {x^{3}}{3}+\frac {x^{2}}{2}-\frac {3 x}{2}-\frac {\ln \left (16 x^{2}-8 x +16\right )}{24}-\frac {5 \sqrt {15}\, \arctan \left (\frac {\left (-1+4 x \right ) \sqrt {15}}{15}\right )}{36}+\frac {2 \ln \left (4 x^{2}+4 x +4\right )}{3}+\frac {8 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{9}\) | \(73\) |
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Time = 0.34 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.82 \[ \int \frac {x^3 \left (5+x+3 x^2+2 x^3\right )}{2+x+3 x^2+x^3+2 x^4} \, dx=\frac {1}{3} \, x^{3} + \frac {1}{2} \, x^{2} - \frac {5}{36} \, \sqrt {5} \sqrt {3} \arctan \left (\frac {1}{15} \, \sqrt {5} \sqrt {3} {\left (4 \, x - 1\right )}\right ) + \frac {8}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {3}{2} \, x - \frac {1}{24} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac {2}{3} \, \log \left (x^{2} + x + 1\right ) \]
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Time = 0.12 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \frac {x^3 \left (5+x+3 x^2+2 x^3\right )}{2+x+3 x^2+x^3+2 x^4} \, dx=\frac {x^{3}}{3} + \frac {x^{2}}{2} - \frac {3 x}{2} - \frac {\log {\left (x^{2} - \frac {x}{2} + 1 \right )}}{24} + \frac {2 \log {\left (x^{2} + x + 1 \right )}}{3} - \frac {5 \sqrt {15} \operatorname {atan}{\left (\frac {4 \sqrt {15} x}{15} - \frac {\sqrt {15}}{15} \right )}}{36} + \frac {8 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{9} \]
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Time = 0.27 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.76 \[ \int \frac {x^3 \left (5+x+3 x^2+2 x^3\right )}{2+x+3 x^2+x^3+2 x^4} \, dx=\frac {1}{3} \, x^{3} + \frac {1}{2} \, x^{2} - \frac {5}{36} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, x - 1\right )}\right ) + \frac {8}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {3}{2} \, x - \frac {1}{24} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac {2}{3} \, \log \left (x^{2} + x + 1\right ) \]
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Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.76 \[ \int \frac {x^3 \left (5+x+3 x^2+2 x^3\right )}{2+x+3 x^2+x^3+2 x^4} \, dx=\frac {1}{3} \, x^{3} + \frac {1}{2} \, x^{2} - \frac {5}{36} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, x - 1\right )}\right ) + \frac {8}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {3}{2} \, x - \frac {1}{24} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac {2}{3} \, \log \left (x^{2} + x + 1\right ) \]
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Time = 0.10 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \frac {x^3 \left (5+x+3 x^2+2 x^3\right )}{2+x+3 x^2+x^3+2 x^4} \, dx=\frac {x^2}{2}-\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {2}{3}+\frac {\sqrt {3}\,4{}\mathrm {i}}{9}\right )+\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {2}{3}+\frac {\sqrt {3}\,4{}\mathrm {i}}{9}\right )+\ln \left (x-\frac {1}{4}-\frac {\sqrt {15}\,1{}\mathrm {i}}{4}\right )\,\left (-\frac {1}{24}+\frac {\sqrt {15}\,5{}\mathrm {i}}{72}\right )-\ln \left (x-\frac {1}{4}+\frac {\sqrt {15}\,1{}\mathrm {i}}{4}\right )\,\left (\frac {1}{24}+\frac {\sqrt {15}\,5{}\mathrm {i}}{72}\right )-\frac {3\,x}{2}+\frac {x^3}{3} \]
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