Optimal. Leaf size=364 \[ \frac {4 c^{5/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (A c+9 b B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {b x^2+c x^4}}-\frac {8 c^{5/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (A c+9 b B) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {b x^2+c x^4}}+\frac {8 c^{3/2} x^{3/2} \left (b+c x^2\right ) (A c+9 b B)}{15 b \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {2 \left (b x^2+c x^4\right )^{3/2} (A c+9 b B)}{45 b x^{11/2}}-\frac {4 c \sqrt {b x^2+c x^4} (A c+9 b B)}{15 b x^{3/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{19/2}} \]
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Rubi [A] time = 0.45, antiderivative size = 364, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2038, 2020, 2032, 329, 305, 220, 1196} \[ \frac {4 c^{5/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (A c+9 b B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {b x^2+c x^4}}-\frac {8 c^{5/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (A c+9 b B) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {b x^2+c x^4}}+\frac {8 c^{3/2} x^{3/2} \left (b+c x^2\right ) (A c+9 b B)}{15 b \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {2 \left (b x^2+c x^4\right )^{3/2} (A c+9 b B)}{45 b x^{11/2}}-\frac {4 c \sqrt {b x^2+c x^4} (A c+9 b B)}{15 b x^{3/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{19/2}} \]
Antiderivative was successfully verified.
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Rule 220
Rule 305
Rule 329
Rule 1196
Rule 2020
Rule 2032
Rule 2038
Rubi steps
\begin {align*} \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{17/2}} \, dx &=-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{19/2}}-\frac {\left (2 \left (-\frac {9 b B}{2}-\frac {A c}{2}\right )\right ) \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{13/2}} \, dx}{9 b}\\ &=-\frac {2 (9 b B+A c) \left (b x^2+c x^4\right )^{3/2}}{45 b x^{11/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{19/2}}+\frac {(2 c (9 b B+A c)) \int \frac {\sqrt {b x^2+c x^4}}{x^{5/2}} \, dx}{15 b}\\ &=-\frac {4 c (9 b B+A c) \sqrt {b x^2+c x^4}}{15 b x^{3/2}}-\frac {2 (9 b B+A c) \left (b x^2+c x^4\right )^{3/2}}{45 b x^{11/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{19/2}}+\frac {\left (4 c^2 (9 b B+A c)\right ) \int \frac {x^{3/2}}{\sqrt {b x^2+c x^4}} \, dx}{15 b}\\ &=-\frac {4 c (9 b B+A c) \sqrt {b x^2+c x^4}}{15 b x^{3/2}}-\frac {2 (9 b B+A c) \left (b x^2+c x^4\right )^{3/2}}{45 b x^{11/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{19/2}}+\frac {\left (4 c^2 (9 b B+A c) x \sqrt {b+c x^2}\right ) \int \frac {\sqrt {x}}{\sqrt {b+c x^2}} \, dx}{15 b \sqrt {b x^2+c x^4}}\\ &=-\frac {4 c (9 b B+A c) \sqrt {b x^2+c x^4}}{15 b x^{3/2}}-\frac {2 (9 b B+A c) \left (b x^2+c x^4\right )^{3/2}}{45 b x^{11/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{19/2}}+\frac {\left (8 c^2 (9 b B+A c) x \sqrt {b+c x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{15 b \sqrt {b x^2+c x^4}}\\ &=-\frac {4 c (9 b B+A c) \sqrt {b x^2+c x^4}}{15 b x^{3/2}}-\frac {2 (9 b B+A c) \left (b x^2+c x^4\right )^{3/2}}{45 b x^{11/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{19/2}}+\frac {\left (8 c^{3/2} (9 b B+A c) x \sqrt {b+c x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{15 \sqrt {b} \sqrt {b x^2+c x^4}}-\frac {\left (8 c^{3/2} (9 b B+A c) x \sqrt {b+c x^2}\right ) \operatorname {Subst}\left (\int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {b}}}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{15 \sqrt {b} \sqrt {b x^2+c x^4}}\\ &=\frac {8 c^{3/2} (9 b B+A c) x^{3/2} \left (b+c x^2\right )}{15 b \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {4 c (9 b B+A c) \sqrt {b x^2+c x^4}}{15 b x^{3/2}}-\frac {2 (9 b B+A c) \left (b x^2+c x^4\right )^{3/2}}{45 b x^{11/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{19/2}}-\frac {8 c^{5/4} (9 b B+A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {b x^2+c x^4}}+\frac {4 c^{5/4} (9 b B+A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {b x^2+c x^4}}\\ \end {align*}
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Mathematica [C] time = 0.06, size = 100, normalized size = 0.27 \[ -\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (b x^2 (A c+9 b B) \, _2F_1\left (-\frac {3}{2},-\frac {5}{4};-\frac {1}{4};-\frac {c x^2}{b}\right )+5 A \left (b+c x^2\right )^2 \sqrt {\frac {c x^2}{b}+1}\right )}{45 b x^{11/2} \sqrt {\frac {c x^2}{b}+1}} \]
Antiderivative was successfully verified.
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fricas [F] time = 1.06, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B c x^{4} + {\left (B b + A c\right )} x^{2} + A b\right )} \sqrt {c x^{4} + b x^{2}}}{x^{\frac {13}{2}}}, x\right ) \]
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 \[ \int \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{\frac {17}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 452, normalized size = 1.24 \[ \frac {2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (-12 A \,c^{3} x^{6}-63 B b \,c^{2} x^{6}+12 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, A b \,c^{2} x^{4} \EllipticE \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-6 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, A b \,c^{2} x^{4} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+108 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, B \,b^{2} c \,x^{4} \EllipticE \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-54 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, B \,b^{2} c \,x^{4} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-23 A b \,c^{2} x^{4}-72 B \,b^{2} c \,x^{4}-16 A \,b^{2} c \,x^{2}-9 B \,b^{3} x^{2}-5 A \,b^{3}\right )}{45 \left (c \,x^{2}+b \right )^{2} b \,x^{\frac {15}{2}}} \]
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 \[ \int \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{\frac {17}{2}}}\,{d x} \]
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 \[ \int \frac {\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{17/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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