Optimal. Leaf size=44 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {f} \left (e-2 (d-f) x^3\right )}{\sqrt {d} e}\right )}{6 \sqrt {d} e \sqrt {f}} \]
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Rubi [A]
time = 0.05, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6, 1366, 632,
212} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {f} \left (e-2 x^3 (d-f)\right )}{\sqrt {d} e}\right )}{6 \sqrt {d} e \sqrt {f}} \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 212
Rule 632
Rule 1366
Rubi steps
\begin {align*} \int \frac {x^2}{e^2+4 e f x^3-4 d f x^6+4 f^2 x^6} \, dx &=\int \frac {x^2}{e^2+4 e f x^3+\left (-4 d f+4 f^2\right ) x^6} \, dx\\ &=\frac {1}{3} \text {Subst}\left (\int \frac {1}{e^2+4 e f x+\left (-4 d f+4 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 {\tanh ^{-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*}
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Mathematica [A]
time = 0.02, size = 46, normalized size = 1.05 \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {f} \left (e-2 d x^3+2 f x^3\right )}{\sqrt {d} e}\right )}{6 \sqrt {d} e \sqrt {f}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.02, size = 42, normalized size = 0.95
method | result | size |
default | \(\frac {\arctanh \left (\frac {2 \left (4 d f -4 f^{2}\right ) x^{3}-4 e f}{4 \sqrt {d f}\, e}\right )}{6 \sqrt {d f}\, e}\) | \(42\) |
risch | \(\frac {\ln \left (\left (-2 \sqrt {d f}-2 f \right ) x^{3}-e \right )}{12 \sqrt {d f}\, e}-\frac {\ln \left (\left (-2 \sqrt {d f}+2 f \right ) x^{3}+e \right )}{12 \sqrt {d f}\, e}\) | \(60\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.52, size = 70, normalized size = 1.59 \begin {gather*} \frac {e^{\left (-1\right )} \log \left (\frac {2 \, {\left (d f - f^{2}\right )} x^{3} - f e + \sqrt {d f} e}{2 \, {\left (d f - f^{2}\right )} x^{3} - f e - \sqrt {d f} e}\right )}{12 \, \sqrt {d f}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.36, size = 160, normalized size = 3.64 \begin {gather*} \left [\frac {\sqrt {d f} e^{\left (-1\right )} \log \left (-\frac {4 \, {\left (d^{2} f - 2 \, d f^{2} + f^{3}\right )} x^{6} - 4 \, {\left (d f - f^{2}\right )} x^{3} e + {\left (d + f\right )} e^{2} + 2 \, {\left (2 \, {\left (d - f\right )} x^{3} e - e^{2}\right )} \sqrt {d f}}{4 \, {\left (d f - f^{2}\right )} x^{6} - 4 \, f x^{3} e - e^{2}}\right )}{12 \, d f}, \frac {\sqrt {-d f} \arctan \left (-\frac {{\left (2 \, {\left (d - f\right )} x^{3} - e\right )} \sqrt {-d f} e^{\left (-1\right )}}{d}\right ) e^{\left (-1\right )}}{6 \, d f}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.48, size = 75, normalized size = 1.70 \begin {gather*} - \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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 5.46, size = 41, normalized size = 0.93 \begin {gather*} -\frac {\arctan \left (\frac {{\left (2 \, d f x^{3} - 2 \, f^{2} x^{3} - f e\right )} e^{\left (-1\right )}}{\sqrt {-d f}}\right ) e^{\left (-1\right )}}{6 \, \sqrt {-d f}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 3.53, size = 923, normalized size = 20.98 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\frac {\left (x^3\,\left (32\,d^3\,f^3-96\,d^2\,f^4+96\,d\,f^5-32\,f^6\right )+\frac {x^3\,\left (-64\,e\,d^3\,f^4+192\,e\,d^2\,f^5-192\,e\,d\,f^6+64\,e\,f^7\right )+16\,e^2\,f^6-48\,d\,e^2\,f^5+48\,d^2\,e^2\,f^4-16\,d^3\,e^2\,f^3-\frac {\frac {x^3\,\left (-384\,d^3\,e^2\,f^5+1152\,d^2\,e^2\,f^6-1152\,d\,e^2\,f^7+384\,e^2\,f^8\right )}{12}+16\,e^3\,f^7-48\,d\,e^3\,f^6+48\,d^2\,e^3\,f^5-16\,d^3\,e^3\,f^4}{\sqrt {d}\,e\,\sqrt {f}}}{\sqrt {d}\,e\,\sqrt {f}}\right )\,1{}\mathrm {i}}{\sqrt {d}\,e\,\sqrt {f}}+\frac {\left (x^3\,\left (32\,d^3\,f^3-96\,d^2\,f^4+96\,d\,f^5-32\,f^6\right )-\frac {x^3\,\left (-64\,e\,d^3\,f^4+192\,e\,d^2\,f^5-192\,e\,d\,f^6+64\,e\,f^7\right )+16\,e^2\,f^6-48\,d\,e^2\,f^5+48\,d^2\,e^2\,f^4-16\,d^3\,e^2\,f^3+\frac {\frac {x^3\,\left (-384\,d^3\,e^2\,f^5+1152\,d^2\,e^2\,f^6-1152\,d\,e^2\,f^7+384\,e^2\,f^8\right )}{12}+16\,e^3\,f^7-48\,d\,e^3\,f^6+48\,d^2\,e^3\,f^5-16\,d^3\,e^3\,f^4}{\sqrt {d}\,e\,\sqrt {f}}}{\sqrt {d}\,e\,\sqrt {f}}\right )\,1{}\mathrm {i}}{\sqrt {d}\,e\,\sqrt {f}}}{\frac {x^3\,\left (32\,d^3\,f^3-96\,d^2\,f^4+96\,d\,f^5-32\,f^6\right )+\frac {x^3\,\left (-64\,e\,d^3\,f^4+192\,e\,d^2\,f^5-192\,e\,d\,f^6+64\,e\,f^7\right )+16\,e^2\,f^6-48\,d\,e^2\,f^5+48\,d^2\,e^2\,f^4-16\,d^3\,e^2\,f^3-\frac {\frac {x^3\,\left (-384\,d^3\,e^2\,f^5+1152\,d^2\,e^2\,f^6-1152\,d\,e^2\,f^7+384\,e^2\,f^8\right )}{12}+16\,e^3\,f^7-48\,d\,e^3\,f^6+48\,d^2\,e^3\,f^5-16\,d^3\,e^3\,f^4}{\sqrt {d}\,e\,\sqrt {f}}}{\sqrt {d}\,e\,\sqrt {f}}}{\sqrt {d}\,e\,\sqrt {f}}-\frac {x^3\,\left (32\,d^3\,f^3-96\,d^2\,f^4+96\,d\,f^5-32\,f^6\right )-\frac {x^3\,\left (-64\,e\,d^3\,f^4+192\,e\,d^2\,f^5-192\,e\,d\,f^6+64\,e\,f^7\right )+16\,e^2\,f^6-48\,d\,e^2\,f^5+48\,d^2\,e^2\,f^4-16\,d^3\,e^2\,f^3+\frac {\frac {x^3\,\left (-384\,d^3\,e^2\,f^5+1152\,d^2\,e^2\,f^6-1152\,d\,e^2\,f^7+384\,e^2\,f^8\right )}{12}+16\,e^3\,f^7-48\,d\,e^3\,f^6+48\,d^2\,e^3\,f^5-16\,d^3\,e^3\,f^4}{\sqrt {d}\,e\,\sqrt {f}}}{\sqrt {d}\,e\,\sqrt {f}}}{\sqrt {d}\,e\,\sqrt {f}}}\right )\,1{}\mathrm {i}}{6\,\sqrt {d}\,e\,\sqrt {f}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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