Integrand size = 11, antiderivative size = 97 \[ \int \left (a+b x^4\right )^{5/4} \, dx=\frac {5}{12} a x \sqrt [4]{a+b x^4}+\frac {1}{6} x \left (a+b x^4\right )^{5/4}-\frac {5 a^{3/2} \sqrt {b} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{12 \left (a+b x^4\right )^{3/4}} \]
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Time = 0.03 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {201, 243, 342, 281, 237} \[ \int \left (a+b x^4\right )^{5/4} \, dx=-\frac {5 a^{3/2} \sqrt {b} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{12 \left (a+b x^4\right )^{3/4}}+\frac {1}{6} x \left (a+b x^4\right )^{5/4}+\frac {5}{12} a x \sqrt [4]{a+b x^4} \]
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Rule 201
Rule 237
Rule 243
Rule 281
Rule 342
Rubi steps \begin{align*} \text {integral}& = \frac {1}{6} x \left (a+b x^4\right )^{5/4}+\frac {1}{6} (5 a) \int \sqrt [4]{a+b x^4} \, dx \\ & = \frac {5}{12} a x \sqrt [4]{a+b x^4}+\frac {1}{6} x \left (a+b x^4\right )^{5/4}+\frac {1}{12} \left (5 a^2\right ) \int \frac {1}{\left (a+b x^4\right )^{3/4}} \, dx \\ & = \frac {5}{12} a x \sqrt [4]{a+b x^4}+\frac {1}{6} x \left (a+b x^4\right )^{5/4}+\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 \left (a+b x^4\right )^{3/4}} \\ & = \frac {5}{12} a x \sqrt [4]{a+b x^4}+\frac {1}{6} x \left (a+b x^4\right )^{5/4}-\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 \left (a+b x^4\right )^{3/4}} \\ & = \frac {5}{12} a x \sqrt [4]{a+b x^4}+\frac {1}{6} x \left (a+b x^4\right )^{5/4}-\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 \left (a+b x^4\right )^{3/4}} \\ & = \frac {5}{12} a x \sqrt [4]{a+b x^4}+\frac {1}{6} x \left (a+b x^4\right )^{5/4}-\frac {5 a^{3/2} \sqrt {b} \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 \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 = 8.48 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.48 \[ \int \left (a+b x^4\right )^{5/4} \, dx=\frac {a x \sqrt [4]{a+b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{4},\frac {5}{4},-\frac {b x^4}{a}\right )}{\sqrt [4]{1+\frac {b x^4}{a}}} \]
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\[\int \left (b \,x^{4}+a \right )^{\frac {5}{4}}d x\]
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\[ \int \left (a+b x^4\right )^{5/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {5}{4}} \,d x } \]
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Result contains complex when optimal does not.
Time = 0.60 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.38 \[ \int \left (a+b x^4\right )^{5/4} \, dx=\frac {a^{\frac {5}{4}} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \]
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\[ \int \left (a+b x^4\right )^{5/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {5}{4}} \,d x } \]
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\[ \int \left (a+b x^4\right )^{5/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {5}{4}} \,d x } \]
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Time = 5.58 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.38 \[ \int \left (a+b x^4\right )^{5/4} \, dx=\frac {x\,{\left (b\,x^4+a\right )}^{5/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{4};\ \frac {5}{4};\ -\frac {b\,x^4}{a}\right )}{{\left (\frac {b\,x^4}{a}+1\right )}^{5/4}} \]
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