Integrand size = 15, antiderivative size = 125 \[ \int x^5 \left (a+b x^4\right )^{3/4} \, dx=-\frac {2 a^2 x^2}{15 b \sqrt [4]{a+b x^4}}+\frac {a x^2 \left (a+b x^4\right )^{3/4}}{15 b}+\frac {1}{9} x^6 \left (a+b x^4\right )^{3/4}+\frac {2 a^{5/2} \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{15 b^{3/2} \sqrt [4]{a+b x^4}} \]
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Time = 0.06 (sec) , antiderivative size = 125, 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 = {281, 285, 327, 235, 233, 202} \[ \int x^5 \left (a+b x^4\right )^{3/4} \, dx=\frac {2 a^{5/2} \sqrt [4]{\frac {b x^4}{a}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{15 b^{3/2} \sqrt [4]{a+b x^4}}-\frac {2 a^2 x^2}{15 b \sqrt [4]{a+b x^4}}+\frac {1}{9} x^6 \left (a+b x^4\right )^{3/4}+\frac {a x^2 \left (a+b x^4\right )^{3/4}}{15 b} \]
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Rule 202
Rule 233
Rule 235
Rule 281
Rule 285
Rule 327
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int x^2 \left (a+b x^2\right )^{3/4} \, dx,x,x^2\right ) \\ & = \frac {1}{9} x^6 \left (a+b x^4\right )^{3/4}+\frac {1}{6} a \text {Subst}\left (\int \frac {x^2}{\sqrt [4]{a+b x^2}} \, dx,x,x^2\right ) \\ & = \frac {a x^2 \left (a+b x^4\right )^{3/4}}{15 b}+\frac {1}{9} x^6 \left (a+b x^4\right )^{3/4}-\frac {a^2 \text {Subst}\left (\int \frac {1}{\sqrt [4]{a+b x^2}} \, dx,x,x^2\right )}{15 b} \\ & = \frac {a x^2 \left (a+b x^4\right )^{3/4}}{15 b}+\frac {1}{9} x^6 \left (a+b x^4\right )^{3/4}-\frac {\left (a^2 \sqrt [4]{1+\frac {b x^4}{a}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1+\frac {b x^2}{a}}} \, dx,x,x^2\right )}{15 b \sqrt [4]{a+b x^4}} \\ & = -\frac {2 a^2 x^2}{15 b \sqrt [4]{a+b x^4}}+\frac {a x^2 \left (a+b x^4\right )^{3/4}}{15 b}+\frac {1}{9} x^6 \left (a+b x^4\right )^{3/4}+\frac {\left (a^2 \sqrt [4]{1+\frac {b x^4}{a}}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/4}} \, dx,x,x^2\right )}{15 b \sqrt [4]{a+b x^4}} \\ & = -\frac {2 a^2 x^2}{15 b \sqrt [4]{a+b x^4}}+\frac {a x^2 \left (a+b x^4\right )^{3/4}}{15 b}+\frac {1}{9} x^6 \left (a+b x^4\right )^{3/4}+\frac {2 a^{5/2} \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{15 b^{3/2} \sqrt [4]{a+b x^4}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.50 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.51 \[ \int x^5 \left (a+b x^4\right )^{3/4} \, dx=\frac {x^2 \left (a+b x^4\right )^{3/4} \left (a+b x^4-\frac {a \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^4}{a}\right )}{\left (1+\frac {b x^4}{a}\right )^{3/4}}\right )}{9 b} \]
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\[\int x^{5} \left (b \,x^{4}+a \right )^{\frac {3}{4}}d x\]
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\[ \int x^5 \left (a+b x^4\right )^{3/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{5} \,d x } \]
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Result contains complex when optimal does not.
Time = 0.66 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.23 \[ \int x^5 \left (a+b x^4\right )^{3/4} \, dx=\frac {a^{\frac {3}{4}} x^{6} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{6} \]
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\[ \int x^5 \left (a+b x^4\right )^{3/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{5} \,d x } \]
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\[ \int x^5 \left (a+b x^4\right )^{3/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{5} \,d x } \]
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Timed out. \[ \int x^5 \left (a+b x^4\right )^{3/4} \, dx=\int x^5\,{\left (b\,x^4+a\right )}^{3/4} \,d x \]
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