Integrand size = 32, antiderivative size = 42 \[ \int \frac {x^2}{e^2+4 e f x^3+4 d f x^6+4 f^2 x^6} \, dx=\frac {\arctan \left (\frac {\sqrt {f} \left (e+2 (d+f) x^3\right )}{\sqrt {d} e}\right )}{6 \sqrt {d} e \sqrt {f}} \]
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Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6, 1366, 632, 210} \[ \int \frac {x^2}{e^2+4 e f x^3+4 d f x^6+4 f^2 x^6} \, dx=\frac {\arctan \left (\frac {\sqrt {f} \left (2 x^3 (d+f)+e\right )}{\sqrt {d} e}\right )}{6 \sqrt {d} e \sqrt {f}} \]
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Rule 6
Rule 210
Rule 632
Rule 1366
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2}{e^2+4 e f x^3+4 \left (d f+f^2\right ) x^6} \, dx \\ & = \frac {1}{3} \text {Subst}\left (\int \frac {1}{e^2+4 e f x+4 \left (d f+f^2\right ) x^2} \, dx,x,x^3\right ) \\ & = -\left (\frac {2}{3} \text {Subst}\left (\int \frac {1}{-16 d e^2 f-x^2} \, dx,x,4 f \left (e+2 (d+f) x^3\right )\right )\right ) \\ & = \frac {\tan ^{-1}\left (\frac {\sqrt {f} \left (e+2 (d+f) x^3\right )}{\sqrt {d} e}\right )}{6 \sqrt {d} e \sqrt {f}} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{e^2+4 e f x^3+4 d f x^6+4 f^2 x^6} \, dx=\frac {\arctan \left (\frac {\sqrt {f} \left (e+2 (d+f) x^3\right )}{\sqrt {d} e}\right )}{6 \sqrt {d} e \sqrt {f}} \]
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Time = 0.06 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00
method | result | size |
default | \(\frac {\arctan \left (\frac {2 \left (4 d f +4 f^{2}\right ) x^{3}+4 e f}{4 \sqrt {d f}\, e}\right )}{6 \sqrt {d f}\, e}\) | \(42\) |
risch | \(-\frac {\ln \left (\left (2 \sqrt {-d f}-2 f \right ) x^{3}-e \right )}{12 \sqrt {-d f}\, e}+\frac {\ln \left (\left (2 \sqrt {-d f}+2 f \right ) x^{3}+e \right )}{12 \sqrt {-d f}\, e}\) | \(64\) |
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Time = 0.31 (sec) , antiderivative size = 155, normalized size of antiderivative = 3.69 \[ \int \frac {x^2}{e^2+4 e f x^3+4 d f x^6+4 f^2 x^6} \, dx=\left [-\frac {\sqrt {-d f} \log \left (\frac {4 \, {\left (d^{2} f + 2 \, d f^{2} + f^{3}\right )} x^{6} + 4 \, {\left (d e f + e f^{2}\right )} x^{3} - d e^{2} + e^{2} f - 2 \, {\left (2 \, {\left (d e + e f\right )} x^{3} + e^{2}\right )} \sqrt {-d f}}{4 \, {\left (d f + f^{2}\right )} x^{6} + 4 \, e f x^{3} + e^{2}}\right )}{12 \, d e f}, \frac {\sqrt {d f} \arctan \left (\frac {{\left (2 \, {\left (d + f\right )} x^{3} + e\right )} \sqrt {d f}}{d e}\right )}{6 \, d e f}\right ] \]
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Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (37) = 74\).
Time = 0.45 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.86 \[ \int \frac {x^2}{e^2+4 e f x^3+4 d f x^6+4 f^2 x^6} \, dx=\frac {- \frac {\sqrt {- \frac {1}{d f}} \log {\left (x^{3} + \frac {- d e \sqrt {- \frac {1}{d f}} + e}{2 d + 2 f} \right )}}{12} + \frac {\sqrt {- \frac {1}{d f}} \log {\left (x^{3} + \frac {d e \sqrt {- \frac {1}{d f}} + e}{2 d + 2 f} \right )}}{12}}{e} \]
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Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86 \[ \int \frac {x^2}{e^2+4 e f x^3+4 d f x^6+4 f^2 x^6} \, dx=\frac {\arctan \left (\frac {2 \, {\left (d f + f^{2}\right )} x^{3} + e f}{\sqrt {d f} e}\right )}{6 \, \sqrt {d f} e} \]
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Time = 0.36 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.93 \[ \int \frac {x^2}{e^2+4 e f x^3+4 d f x^6+4 f^2 x^6} \, dx=\frac {\arctan \left (\frac {2 \, d f x^{3} + 2 \, f^{2} x^{3} + e f}{\sqrt {d f} e}\right )}{6 \, \sqrt {d f} e} \]
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Time = 17.47 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{e^2+4 e f x^3+4 d f x^6+4 f^2 x^6} \, dx=\frac {\mathrm {atan}\left (\frac {e\,\sqrt {f}+2\,f^{3/2}\,x^3+2\,d\,\sqrt {f}\,x^3}{\sqrt {d}\,e}\right )}{6\,\sqrt {d}\,e\,\sqrt {f}} \]
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