Integrand size = 26, antiderivative size = 139 \[ \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\frac {(6 b c-5 a d) e \sqrt {e x} \sqrt [4]{a+b x^2}}{6 b^2}+\frac {d (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b e}+\frac {\sqrt {a} (6 b c-5 a d) \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2} \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{6 b^{3/2} \left (a+b x^2\right )^{3/4}} \]
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Time = 0.07 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {470, 327, 335, 243, 342, 281, 237} \[ \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\frac {\sqrt {a} (e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} (6 b c-5 a d) \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{6 b^{3/2} \left (a+b x^2\right )^{3/4}}+\frac {e \sqrt {e x} \sqrt [4]{a+b x^2} (6 b c-5 a d)}{6 b^2}+\frac {d (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b e} \]
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Rule 237
Rule 243
Rule 281
Rule 327
Rule 335
Rule 342
Rule 470
Rubi steps \begin{align*} \text {integral}& = \frac {d (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b e}-\frac {\left (-3 b c+\frac {5 a d}{2}\right ) \int \frac {(e x)^{3/2}}{\left (a+b x^2\right )^{3/4}} \, dx}{3 b} \\ & = \frac {(6 b c-5 a d) e \sqrt {e x} \sqrt [4]{a+b x^2}}{6 b^2}+\frac {d (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b e}-\frac {\left (a (6 b c-5 a d) e^2\right ) \int \frac {1}{\sqrt {e x} \left (a+b x^2\right )^{3/4}} \, dx}{12 b^2} \\ & = \frac {(6 b c-5 a d) e \sqrt {e x} \sqrt [4]{a+b x^2}}{6 b^2}+\frac {d (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b e}-\frac {(a (6 b c-5 a d) e) \text {Subst}\left (\int \frac {1}{\left (a+\frac {b x^4}{e^2}\right )^{3/4}} \, dx,x,\sqrt {e x}\right )}{6 b^2} \\ & = \frac {(6 b c-5 a d) e \sqrt {e x} \sqrt [4]{a+b x^2}}{6 b^2}+\frac {d (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b e}-\frac {\left (a (6 b c-5 a d) e \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a e^2}{b x^4}\right )^{3/4} x^3} \, dx,x,\sqrt {e x}\right )}{6 b^2 \left (a+b x^2\right )^{3/4}} \\ & = \frac {(6 b c-5 a d) e \sqrt {e x} \sqrt [4]{a+b x^2}}{6 b^2}+\frac {d (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b e}+\frac {\left (a (6 b c-5 a d) e \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2}\right ) \text {Subst}\left (\int \frac {x}{\left (1+\frac {a e^2 x^4}{b}\right )^{3/4}} \, dx,x,\frac {1}{\sqrt {e x}}\right )}{6 b^2 \left (a+b x^2\right )^{3/4}} \\ & = \frac {(6 b c-5 a d) e \sqrt {e x} \sqrt [4]{a+b x^2}}{6 b^2}+\frac {d (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b e}+\frac {\left (a (6 b c-5 a d) e \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a e^2 x^2}{b}\right )^{3/4}} \, dx,x,\frac {1}{e x}\right )}{12 b^2 \left (a+b x^2\right )^{3/4}} \\ & = \frac {(6 b c-5 a d) e \sqrt {e x} \sqrt [4]{a+b x^2}}{6 b^2}+\frac {d (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b e}+\frac {\sqrt {a} (6 b c-5 a d) \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{6 b^{3/2} \left (a+b x^2\right )^{3/4}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.08 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.70 \[ \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\frac {e \sqrt {e x} \left (-\left (\left (a+b x^2\right ) \left (5 a d-2 b \left (3 c+d x^2\right )\right )\right )+a (-6 b c+5 a d) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\frac {b x^2}{a}\right )\right )}{6 b^2 \left (a+b x^2\right )^{3/4}} \]
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\[\int \frac {\left (e x \right )^{\frac {3}{2}} \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x\]
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\[ \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}} \,d x } \]
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Result contains complex when optimal does not.
Time = 5.20 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.68 \[ \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\frac {c e^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{4}} \Gamma \left (\frac {9}{4}\right )} + \frac {d e^{\frac {3}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{4}} \Gamma \left (\frac {13}{4}\right )} \]
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\[ \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}} \,d x } \]
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\[ \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}} \,d x } \]
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Timed out. \[ \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\int \frac {{\left (e\,x\right )}^{3/2}\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{3/4}} \,d x \]
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