Integrand size = 15, antiderivative size = 105 \[ \int \frac {x^8}{\left (a+b x^4\right )^{3/4}} \, dx=-\frac {5 a x \sqrt [4]{a+b x^4}}{12 b^2}+\frac {x^5 \sqrt [4]{a+b x^4}}{6 b}-\frac {5 a^{3/2} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{12 b^{3/2} \left (a+b x^4\right )^{3/4}} \]
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Time = 0.03 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {327, 243, 342, 281, 237} \[ \int \frac {x^8}{\left (a+b x^4\right )^{3/4}} \, dx=-\frac {5 a^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{12 b^{3/2} \left (a+b x^4\right )^{3/4}}-\frac {5 a x \sqrt [4]{a+b x^4}}{12 b^2}+\frac {x^5 \sqrt [4]{a+b x^4}}{6 b} \]
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Rule 237
Rule 243
Rule 281
Rule 327
Rule 342
Rubi steps \begin{align*} \text {integral}& = \frac {x^5 \sqrt [4]{a+b x^4}}{6 b}-\frac {(5 a) \int \frac {x^4}{\left (a+b x^4\right )^{3/4}} \, dx}{6 b} \\ & = -\frac {5 a x \sqrt [4]{a+b x^4}}{12 b^2}+\frac {x^5 \sqrt [4]{a+b x^4}}{6 b}+\frac {\left (5 a^2\right ) \int \frac {1}{\left (a+b x^4\right )^{3/4}} \, dx}{12 b^2} \\ & = -\frac {5 a x \sqrt [4]{a+b x^4}}{12 b^2}+\frac {x^5 \sqrt [4]{a+b x^4}}{6 b}+\frac {\left (5 a^2 \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \int \frac {1}{\left (1+\frac {a}{b x^4}\right )^{3/4} x^3} \, dx}{12 b^2 \left (a+b x^4\right )^{3/4}} \\ & = -\frac {5 a x \sqrt [4]{a+b x^4}}{12 b^2}+\frac {x^5 \sqrt [4]{a+b x^4}}{6 b}-\frac {\left (5 a^2 \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \text {Subst}\left (\int \frac {x}{\left (1+\frac {a x^4}{b}\right )^{3/4}} \, dx,x,\frac {1}{x}\right )}{12 b^2 \left (a+b x^4\right )^{3/4}} \\ & = -\frac {5 a x \sqrt [4]{a+b x^4}}{12 b^2}+\frac {x^5 \sqrt [4]{a+b x^4}}{6 b}-\frac {\left (5 a^2 \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{3/4}} \, dx,x,\frac {1}{x^2}\right )}{24 b^2 \left (a+b x^4\right )^{3/4}} \\ & = -\frac {5 a x \sqrt [4]{a+b x^4}}{12 b^2}+\frac {x^5 \sqrt [4]{a+b x^4}}{6 b}-\frac {5 a^{3/2} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{12 b^{3/2} \left (a+b x^4\right )^{3/4}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.15 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.75 \[ \int \frac {x^8}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {-5 a^2 x-3 a b x^5+2 b^2 x^9+5 a^2 x \left (1+\frac {b x^4}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\frac {b x^4}{a}\right )}{12 b^2 \left (a+b x^4\right )^{3/4}} \]
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\[\int \frac {x^{8}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}}}d x\]
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\[ \int \frac {x^8}{\left (a+b x^4\right )^{3/4}} \, dx=\int { \frac {x^{8}}{{\left (b x^{4} + a\right )}^{\frac {3}{4}}} \,d x } \]
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Result contains complex when optimal does not.
Time = 0.54 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.35 \[ \int \frac {x^8}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{4}} \Gamma \left (\frac {13}{4}\right )} \]
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\[ \int \frac {x^8}{\left (a+b x^4\right )^{3/4}} \, dx=\int { \frac {x^{8}}{{\left (b x^{4} + a\right )}^{\frac {3}{4}}} \,d x } \]
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\[ \int \frac {x^8}{\left (a+b x^4\right )^{3/4}} \, dx=\int { \frac {x^{8}}{{\left (b x^{4} + a\right )}^{\frac {3}{4}}} \,d x } \]
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Timed out. \[ \int \frac {x^8}{\left (a+b x^4\right )^{3/4}} \, dx=\int \frac {x^8}{{\left (b\,x^4+a\right )}^{3/4}} \,d x \]
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