Integrand size = 26, antiderivative size = 142 \[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {(6 b c-7 a d) e (e x)^{3/2}}{6 b^2 \sqrt [4]{a+b x^2}}+\frac {d (e x)^{7/2}}{3 b e \sqrt [4]{a+b x^2}}+\frac {\sqrt {a} (6 b c-7 a d) e^2 \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 b^{5/2} \sqrt [4]{a+b x^2}} \]
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Time = 0.04 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {470, 291, 290, 342, 202} \[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {\sqrt {a} e^2 \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} (6 b c-7 a d) E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 b^{5/2} \sqrt [4]{a+b x^2}}+\frac {e (e x)^{3/2} (6 b c-7 a d)}{6 b^2 \sqrt [4]{a+b x^2}}+\frac {d (e x)^{7/2}}{3 b e \sqrt [4]{a+b x^2}} \]
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Rule 202
Rule 290
Rule 291
Rule 342
Rule 470
Rubi steps \begin{align*} \text {integral}& = \frac {d (e x)^{7/2}}{3 b e \sqrt [4]{a+b x^2}}-\frac {\left (-3 b c+\frac {7 a d}{2}\right ) \int \frac {(e x)^{5/2}}{\left (a+b x^2\right )^{5/4}} \, dx}{3 b} \\ & = \frac {(6 b c-7 a d) e (e x)^{3/2}}{6 b^2 \sqrt [4]{a+b x^2}}+\frac {d (e x)^{7/2}}{3 b e \sqrt [4]{a+b x^2}}-\frac {\left (a (6 b c-7 a d) e^2\right ) \int \frac {\sqrt {e x}}{\left (a+b x^2\right )^{5/4}} \, dx}{4 b^2} \\ & = \frac {(6 b c-7 a d) e (e x)^{3/2}}{6 b^2 \sqrt [4]{a+b x^2}}+\frac {d (e x)^{7/2}}{3 b e \sqrt [4]{a+b x^2}}-\frac {\left (a (6 b c-7 a d) e^2 \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x}\right ) \int \frac {1}{\left (1+\frac {a}{b x^2}\right )^{5/4} x^2} \, dx}{4 b^3 \sqrt [4]{a+b x^2}} \\ & = \frac {(6 b c-7 a d) e (e x)^{3/2}}{6 b^2 \sqrt [4]{a+b x^2}}+\frac {d (e x)^{7/2}}{3 b e \sqrt [4]{a+b x^2}}+\frac {\left (a (6 b c-7 a d) e^2 \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{5/4}} \, dx,x,\frac {1}{x}\right )}{4 b^3 \sqrt [4]{a+b x^2}} \\ & = \frac {(6 b c-7 a d) e (e x)^{3/2}}{6 b^2 \sqrt [4]{a+b x^2}}+\frac {d (e x)^{7/2}}{3 b e \sqrt [4]{a+b x^2}}+\frac {\sqrt {a} (6 b c-7 a d) e^2 \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 b^{5/2} \sqrt [4]{a+b x^2}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.08 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.60 \[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {e (e x)^{3/2} \left (6 b c-7 a d+2 b d x^2+(-6 b c+7 a d) \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},-\frac {b x^2}{a}\right )\right )}{6 b^2 \sqrt [4]{a+b x^2}} \]
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\[\int \frac {\left (e x \right )^{\frac {5}{2}} \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{\frac {5}{4}}}d x\]
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\[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \]
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Result contains complex when optimal does not.
Time = 42.94 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.66 \[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {c e^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} \Gamma \left (\frac {11}{4}\right )} + \frac {d e^{\frac {5}{2}} x^{\frac {11}{2}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} \Gamma \left (\frac {15}{4}\right )} \]
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\[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \]
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\[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \]
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Timed out. \[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int \frac {{\left (e\,x\right )}^{5/2}\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{5/4}} \,d x \]
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