Integrand size = 15, antiderivative size = 43 \[ \int \frac {x^6}{\sqrt {16-x^4}} \, dx=-\frac {1}{5} x^3 \sqrt {16-x^4}+\frac {96}{5} E\left (\left .\arcsin \left (\frac {x}{2}\right )\right |-1\right )-\frac {96}{5} \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{2}\right ),-1\right ) \]
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Time = 0.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {327, 313, 227, 1195, 21, 435} \[ \int \frac {x^6}{\sqrt {16-x^4}} \, dx=-\frac {96}{5} \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{2}\right ),-1\right )+\frac {96}{5} E\left (\left .\arcsin \left (\frac {x}{2}\right )\right |-1\right )-\frac {1}{5} \sqrt {16-x^4} x^3 \]
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Rule 21
Rule 227
Rule 313
Rule 327
Rule 435
Rule 1195
Rubi steps \begin{align*} \text {integral}& = -\frac {1}{5} x^3 \sqrt {16-x^4}+\frac {48}{5} \int \frac {x^2}{\sqrt {16-x^4}} \, dx \\ & = -\frac {1}{5} x^3 \sqrt {16-x^4}-\frac {192}{5} \int \frac {1}{\sqrt {16-x^4}} \, dx+\frac {192}{5} \int \frac {1+\frac {x^2}{4}}{\sqrt {16-x^4}} \, dx \\ & = -\frac {1}{5} x^3 \sqrt {16-x^4}-\frac {96}{5} F\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )+\frac {192}{5} \int \frac {1+\frac {x^2}{4}}{\sqrt {4-x^2} \sqrt {4+x^2}} \, dx \\ & = -\frac {1}{5} x^3 \sqrt {16-x^4}-\frac {96}{5} F\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )+\frac {48}{5} \int \frac {\sqrt {4+x^2}}{\sqrt {4-x^2}} \, dx \\ & = -\frac {1}{5} x^3 \sqrt {16-x^4}+\frac {96}{5} E\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )-\frac {96}{5} F\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right ) \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.88 \[ \int \frac {x^6}{\sqrt {16-x^4}} \, dx=-\frac {1}{5} x^3 \left (\sqrt {16-x^4}-4 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},\frac {x^4}{16}\right )\right ) \]
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Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 4.55 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.40
method | result | size |
meijerg | \(\frac {x^{7} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},\frac {7}{4};\frac {11}{4};\frac {x^{4}}{16}\right )}{28}\) | \(17\) |
default | \(-\frac {x^{3} \sqrt {-x^{4}+16}}{5}-\frac {96 \sqrt {-x^{2}+4}\, \sqrt {x^{2}+4}\, \left (F\left (\frac {x}{2}, i\right )-E\left (\frac {x}{2}, i\right )\right )}{5 \sqrt {-x^{4}+16}}\) | \(58\) |
elliptic | \(-\frac {x^{3} \sqrt {-x^{4}+16}}{5}-\frac {96 \sqrt {-x^{2}+4}\, \sqrt {x^{2}+4}\, \left (F\left (\frac {x}{2}, i\right )-E\left (\frac {x}{2}, i\right )\right )}{5 \sqrt {-x^{4}+16}}\) | \(58\) |
risch | \(\frac {x^{3} \left (x^{4}-16\right )}{5 \sqrt {-x^{4}+16}}-\frac {96 \sqrt {-x^{2}+4}\, \sqrt {x^{2}+4}\, \left (F\left (\frac {x}{2}, i\right )-E\left (\frac {x}{2}, i\right )\right )}{5 \sqrt {-x^{4}+16}}\) | \(63\) |
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none
Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.02 \[ \int \frac {x^6}{\sqrt {16-x^4}} \, dx=\frac {-384 i \, x E(\arcsin \left (\frac {2}{x}\right )\,|\,-1) + 384 i \, x F(\arcsin \left (\frac {2}{x}\right )\,|\,-1) - {\left (x^{4} + 48\right )} \sqrt {-x^{4} + 16}}{5 \, x} \]
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Time = 0.43 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.74 \[ \int \frac {x^6}{\sqrt {16-x^4}} \, dx=\frac {x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {x^{4} e^{2 i \pi }}{16}} \right )}}{16 \Gamma \left (\frac {11}{4}\right )} \]
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\[ \int \frac {x^6}{\sqrt {16-x^4}} \, dx=\int { \frac {x^{6}}{\sqrt {-x^{4} + 16}} \,d x } \]
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\[ \int \frac {x^6}{\sqrt {16-x^4}} \, dx=\int { \frac {x^{6}}{\sqrt {-x^{4} + 16}} \,d x } \]
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Timed out. \[ \int \frac {x^6}{\sqrt {16-x^4}} \, dx=\int \frac {x^6}{\sqrt {16-x^4}} \,d x \]
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