Optimal. Leaf size=252 \[ -\frac {4 a^{11/4} e^{3/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (3 A b-a B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{231 b^{9/4} \sqrt {a+b x^2}}+\frac {8 a^2 e \sqrt {e x} \sqrt {a+b x^2} (3 A b-a B)}{231 b^2}+\frac {4 a (e x)^{5/2} \sqrt {a+b x^2} (3 A b-a B)}{77 b e}+\frac {2 (e x)^{5/2} \left (a+b x^2\right )^{3/2} (3 A b-a B)}{33 b e}+\frac {2 B (e x)^{5/2} \left (a+b x^2\right )^{5/2}}{15 b e} \]
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Rubi [A] time = 0.17, antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {459, 279, 321, 329, 220} \[ -\frac {4 a^{11/4} e^{3/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (3 A b-a B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{231 b^{9/4} \sqrt {a+b x^2}}+\frac {8 a^2 e \sqrt {e x} \sqrt {a+b x^2} (3 A b-a B)}{231 b^2}+\frac {4 a (e x)^{5/2} \sqrt {a+b x^2} (3 A b-a B)}{77 b e}+\frac {2 (e x)^{5/2} \left (a+b x^2\right )^{3/2} (3 A b-a B)}{33 b e}+\frac {2 B (e x)^{5/2} \left (a+b x^2\right )^{5/2}}{15 b e} \]
Antiderivative was successfully verified.
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Rule 220
Rule 279
Rule 321
Rule 329
Rule 459
Rubi steps
\begin {align*} \int (e x)^{3/2} \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx &=\frac {2 B (e x)^{5/2} \left (a+b x^2\right )^{5/2}}{15 b e}-\frac {\left (2 \left (-\frac {15 A b}{2}+\frac {5 a B}{2}\right )\right ) \int (e x)^{3/2} \left (a+b x^2\right )^{3/2} \, dx}{15 b}\\ &=\frac {2 (3 A b-a B) (e x)^{5/2} \left (a+b x^2\right )^{3/2}}{33 b e}+\frac {2 B (e x)^{5/2} \left (a+b x^2\right )^{5/2}}{15 b e}+\frac {(2 a (3 A b-a B)) \int (e x)^{3/2} \sqrt {a+b x^2} \, dx}{11 b}\\ &=\frac {4 a (3 A b-a B) (e x)^{5/2} \sqrt {a+b x^2}}{77 b e}+\frac {2 (3 A b-a B) (e x)^{5/2} \left (a+b x^2\right )^{3/2}}{33 b e}+\frac {2 B (e x)^{5/2} \left (a+b x^2\right )^{5/2}}{15 b e}+\frac {\left (4 a^2 (3 A b-a B)\right ) \int \frac {(e x)^{3/2}}{\sqrt {a+b x^2}} \, dx}{77 b}\\ &=\frac {8 a^2 (3 A b-a B) e \sqrt {e x} \sqrt {a+b x^2}}{231 b^2}+\frac {4 a (3 A b-a B) (e x)^{5/2} \sqrt {a+b x^2}}{77 b e}+\frac {2 (3 A b-a B) (e x)^{5/2} \left (a+b x^2\right )^{3/2}}{33 b e}+\frac {2 B (e x)^{5/2} \left (a+b x^2\right )^{5/2}}{15 b e}-\frac {\left (4 a^3 (3 A b-a B) e^2\right ) \int \frac {1}{\sqrt {e x} \sqrt {a+b x^2}} \, dx}{231 b^2}\\ &=\frac {8 a^2 (3 A b-a B) e \sqrt {e x} \sqrt {a+b x^2}}{231 b^2}+\frac {4 a (3 A b-a B) (e x)^{5/2} \sqrt {a+b x^2}}{77 b e}+\frac {2 (3 A b-a B) (e x)^{5/2} \left (a+b x^2\right )^{3/2}}{33 b e}+\frac {2 B (e x)^{5/2} \left (a+b x^2\right )^{5/2}}{15 b e}-\frac {\left (8 a^3 (3 A b-a B) e\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{231 b^2}\\ &=\frac {8 a^2 (3 A b-a B) e \sqrt {e x} \sqrt {a+b x^2}}{231 b^2}+\frac {4 a (3 A b-a B) (e x)^{5/2} \sqrt {a+b x^2}}{77 b e}+\frac {2 (3 A b-a B) (e x)^{5/2} \left (a+b x^2\right )^{3/2}}{33 b e}+\frac {2 B (e x)^{5/2} \left (a+b x^2\right )^{5/2}}{15 b e}-\frac {4 a^{11/4} (3 A b-a B) e^{3/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{231 b^{9/4} \sqrt {a+b x^2}}\\ \end {align*}
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Mathematica [C] time = 0.17, size = 114, normalized size = 0.45 \[ \frac {2 e \sqrt {e x} \sqrt {a+b x^2} \left (5 a^2 (a B-3 A b) \, _2F_1\left (-\frac {3}{2},\frac {1}{4};\frac {5}{4};-\frac {b x^2}{a}\right )-\left (a+b x^2\right )^2 \sqrt {\frac {b x^2}{a}+1} \left (5 a B-15 A b-11 b B x^2\right )\right )}{165 b^2 \sqrt {\frac {b x^2}{a}+1}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B b e x^{5} + {\left (B a + A b\right )} e x^{3} + A a e x\right )} \sqrt {b x^{2} + a} \sqrt {e x}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{\frac {3}{2}} \left (e x\right )^{\frac {3}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 300, normalized size = 1.19 \[ -\frac {2 \sqrt {e x}\, \left (-77 B \,b^{5} x^{9}-105 A \,b^{5} x^{7}-196 B a \,b^{4} x^{7}-300 A a \,b^{4} x^{5}-131 B \,a^{2} b^{3} x^{5}-255 A \,a^{2} b^{3} x^{3}+8 B \,a^{3} b^{2} x^{3}-60 A \,a^{3} b^{2} x +20 B \,a^{4} b x +30 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \sqrt {-a b}\, A \,a^{3} b \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-10 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \sqrt {-a b}\, B \,a^{4} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )\right ) e}{1155 \sqrt {b \,x^{2}+a}\, b^{3} x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{\frac {3}{2}} \left (e x\right )^{\frac {3}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (B\,x^2+A\right )\,{\left (e\,x\right )}^{3/2}\,{\left (b\,x^2+a\right )}^{3/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [C] time = 27.14, size = 199, normalized size = 0.79 \[ \frac {A a^{\frac {3}{2}} e^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {9}{4}\right )} + \frac {A \sqrt {a} b e^{\frac {3}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {13}{4}\right )} + \frac {B a^{\frac {3}{2}} e^{\frac {3}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {13}{4}\right )} + \frac {B \sqrt {a} b e^{\frac {3}{2}} x^{\frac {13}{2}} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {17}{4}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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