Integrand size = 26, antiderivative size = 173 \[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\frac {(8 b c-7 a d) e (e x)^{3/2} \sqrt [4]{a+b x^2}}{16 b^2}+\frac {d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e}+\frac {3 a (8 b c-7 a d) e^{5/2} \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{32 b^{11/4}}-\frac {3 a (8 b c-7 a d) e^{5/2} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{32 b^{11/4}} \]
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Time = 0.08 (sec) , antiderivative size = 173, 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, 338, 304, 211, 214} \[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\frac {3 a e^{5/2} (8 b c-7 a d) \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{32 b^{11/4}}-\frac {3 a e^{5/2} (8 b c-7 a d) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{32 b^{11/4}}+\frac {e (e x)^{3/2} \sqrt [4]{a+b x^2} (8 b c-7 a d)}{16 b^2}+\frac {d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e} \]
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Rule 211
Rule 214
Rule 304
Rule 327
Rule 335
Rule 338
Rule 470
Rubi steps \begin{align*} \text {integral}& = \frac {d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e}-\frac {\left (-4 b c+\frac {7 a d}{2}\right ) \int \frac {(e x)^{5/2}}{\left (a+b x^2\right )^{3/4}} \, dx}{4 b} \\ & = \frac {(8 b c-7 a d) e (e x)^{3/2} \sqrt [4]{a+b x^2}}{16 b^2}+\frac {d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e}-\frac {\left (3 a (8 b c-7 a d) e^2\right ) \int \frac {\sqrt {e x}}{\left (a+b x^2\right )^{3/4}} \, dx}{32 b^2} \\ & = \frac {(8 b c-7 a d) e (e x)^{3/2} \sqrt [4]{a+b x^2}}{16 b^2}+\frac {d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e}-\frac {(3 a (8 b c-7 a d) e) \text {Subst}\left (\int \frac {x^2}{\left (a+\frac {b x^4}{e^2}\right )^{3/4}} \, dx,x,\sqrt {e x}\right )}{16 b^2} \\ & = \frac {(8 b c-7 a d) e (e x)^{3/2} \sqrt [4]{a+b x^2}}{16 b^2}+\frac {d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e}-\frac {(3 a (8 b c-7 a d) e) \text {Subst}\left (\int \frac {x^2}{1-\frac {b x^4}{e^2}} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{a+b x^2}}\right )}{16 b^2} \\ & = \frac {(8 b c-7 a d) e (e x)^{3/2} \sqrt [4]{a+b x^2}}{16 b^2}+\frac {d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e}-\frac {\left (3 a (8 b c-7 a d) e^3\right ) \text {Subst}\left (\int \frac {1}{e-\sqrt {b} x^2} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{a+b x^2}}\right )}{32 b^{5/2}}+\frac {\left (3 a (8 b c-7 a d) e^3\right ) \text {Subst}\left (\int \frac {1}{e+\sqrt {b} x^2} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{a+b x^2}}\right )}{32 b^{5/2}} \\ & = \frac {(8 b c-7 a d) e (e x)^{3/2} \sqrt [4]{a+b x^2}}{16 b^2}+\frac {d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e}+\frac {3 a (8 b c-7 a d) e^{5/2} \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{32 b^{11/4}}-\frac {3 a (8 b c-7 a d) e^{5/2} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{32 b^{11/4}} \\ \end{align*}
Time = 0.74 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.76 \[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\frac {(e x)^{5/2} \left (2 b^{3/4} x^{3/2} \sqrt [4]{a+b x^2} \left (8 b c-7 a d+4 b d x^2\right )-3 a (-8 b c+7 a d) \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )+3 a (-8 b c+7 a d) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )\right )}{32 b^{11/4} x^{5/2}} \]
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\[\int \frac {\left (e x \right )^{\frac {5}{2}} \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x\]
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Timed out. \[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\text {Timed out} \]
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Result contains complex when optimal does not.
Time = 15.84 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.54 \[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\frac {c e^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{4}} \Gamma \left (\frac {11}{4}\right )} + \frac {d e^{\frac {5}{2}} x^{\frac {11}{2}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{4}} \Gamma \left (\frac {15}{4}\right )} \]
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\[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}} \,d x } \]
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\[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}} \,d x } \]
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Timed out. \[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\int \frac {{\left (e\,x\right )}^{5/2}\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{3/4}} \,d x \]
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