Optimal. Leaf size=288 \[ -\frac {2 c^{7/4} e^{3/2} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (11 a^2 d^2+b c (3 b c-10 a d)\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{231 d^{13/4} \sqrt {c+d x^2}}+\frac {2 (e x)^{5/2} \sqrt {c+d x^2} \left (11 a^2 d^2+b c (3 b c-10 a d)\right )}{77 d^2 e}+\frac {4 c e \sqrt {e x} \sqrt {c+d x^2} \left (11 a^2 d^2+b c (3 b c-10 a d)\right )}{231 d^3}-\frac {2 b (e x)^{5/2} \left (c+d x^2\right )^{3/2} (3 b c-10 a d)}{55 d^2 e}+\frac {2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{3/2}}{15 d e^3} \]
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Rubi [A] time = 0.29, antiderivative size = 288, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {464, 459, 279, 321, 329, 220} \[ -\frac {2 c^{7/4} e^{3/2} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (11 a^2 d^2+b c (3 b c-10 a d)\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{231 d^{13/4} \sqrt {c+d x^2}}+\frac {2 (e x)^{5/2} \sqrt {c+d x^2} \left (11 a^2 d^2+b c (3 b c-10 a d)\right )}{77 d^2 e}+\frac {4 c e \sqrt {e x} \sqrt {c+d x^2} \left (11 a^2 d^2+b c (3 b c-10 a d)\right )}{231 d^3}-\frac {2 b (e x)^{5/2} \left (c+d x^2\right )^{3/2} (3 b c-10 a d)}{55 d^2 e}+\frac {2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{3/2}}{15 d e^3} \]
Antiderivative was successfully verified.
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Rule 220
Rule 279
Rule 321
Rule 329
Rule 459
Rule 464
Rubi steps
\begin {align*} \int (e x)^{3/2} \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx &=\frac {2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{3/2}}{15 d e^3}+\frac {2 \int (e x)^{3/2} \sqrt {c+d x^2} \left (\frac {15 a^2 d}{2}-\frac {3}{2} b (3 b c-10 a d) x^2\right ) \, dx}{15 d}\\ &=-\frac {2 b (3 b c-10 a d) (e x)^{5/2} \left (c+d x^2\right )^{3/2}}{55 d^2 e}+\frac {2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{3/2}}{15 d e^3}+\frac {1}{11} \left (11 a^2+\frac {b c (3 b c-10 a d)}{d^2}\right ) \int (e x)^{3/2} \sqrt {c+d x^2} \, dx\\ &=\frac {2 \left (11 a^2+\frac {b c (3 b c-10 a d)}{d^2}\right ) (e x)^{5/2} \sqrt {c+d x^2}}{77 e}-\frac {2 b (3 b c-10 a d) (e x)^{5/2} \left (c+d x^2\right )^{3/2}}{55 d^2 e}+\frac {2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{3/2}}{15 d e^3}+\frac {1}{77} \left (2 c \left (11 a^2+\frac {b c (3 b c-10 a d)}{d^2}\right )\right ) \int \frac {(e x)^{3/2}}{\sqrt {c+d x^2}} \, dx\\ &=\frac {4 c \left (11 a^2+\frac {b c (3 b c-10 a d)}{d^2}\right ) e \sqrt {e x} \sqrt {c+d x^2}}{231 d}+\frac {2 \left (11 a^2+\frac {b c (3 b c-10 a d)}{d^2}\right ) (e x)^{5/2} \sqrt {c+d x^2}}{77 e}-\frac {2 b (3 b c-10 a d) (e x)^{5/2} \left (c+d x^2\right )^{3/2}}{55 d^2 e}+\frac {2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{3/2}}{15 d e^3}-\frac {\left (2 c^2 \left (11 a^2+\frac {b c (3 b c-10 a d)}{d^2}\right ) e^2\right ) \int \frac {1}{\sqrt {e x} \sqrt {c+d x^2}} \, dx}{231 d}\\ &=\frac {4 c \left (11 a^2+\frac {b c (3 b c-10 a d)}{d^2}\right ) e \sqrt {e x} \sqrt {c+d x^2}}{231 d}+\frac {2 \left (11 a^2+\frac {b c (3 b c-10 a d)}{d^2}\right ) (e x)^{5/2} \sqrt {c+d x^2}}{77 e}-\frac {2 b (3 b c-10 a d) (e x)^{5/2} \left (c+d x^2\right )^{3/2}}{55 d^2 e}+\frac {2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{3/2}}{15 d e^3}-\frac {\left (4 c^2 \left (11 a^2+\frac {b c (3 b c-10 a d)}{d^2}\right ) e\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{231 d}\\ &=\frac {4 c \left (11 a^2+\frac {b c (3 b c-10 a d)}{d^2}\right ) e \sqrt {e x} \sqrt {c+d x^2}}{231 d}+\frac {2 \left (11 a^2+\frac {b c (3 b c-10 a d)}{d^2}\right ) (e x)^{5/2} \sqrt {c+d x^2}}{77 e}-\frac {2 b (3 b c-10 a d) (e x)^{5/2} \left (c+d x^2\right )^{3/2}}{55 d^2 e}+\frac {2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{3/2}}{15 d e^3}-\frac {2 c^{7/4} \left (11 a^2+\frac {b c (3 b c-10 a d)}{d^2}\right ) e^{3/2} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{231 d^{5/4} \sqrt {c+d x^2}}\\ \end {align*}
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Mathematica [C] time = 0.35, size = 225, normalized size = 0.78 \[ \frac {(e x)^{3/2} \left (\frac {2 \sqrt {x} \left (c+d x^2\right ) \left (55 a^2 d^2 \left (2 c+3 d x^2\right )+10 a b d \left (-10 c^2+6 c d x^2+21 d^2 x^4\right )+b^2 \left (30 c^3-18 c^2 d x^2+14 c d^2 x^4+77 d^3 x^6\right )\right )}{5 d^3}-\frac {4 i c^2 x \sqrt {\frac {c}{d x^2}+1} \left (11 a^2 d^2-10 a b c d+3 b^2 c^2\right ) F\left (\left .i \sinh ^{-1}\left (\frac {\sqrt {\frac {i \sqrt {c}}{\sqrt {d}}}}{\sqrt {x}}\right )\right |-1\right )}{d^3 \sqrt {\frac {i \sqrt {c}}{\sqrt {d}}}}\right )}{231 x^{3/2} \sqrt {c+d x^2}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{2} e x^{5} + 2 \, a b e x^{3} + a^{2} e x\right )} \sqrt {d x^{2} + c} \sqrt {e x}, x\right ) \]
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 \[ \int {\left (b x^{2} + a\right )}^{2} \sqrt {d x^{2} + c} \left (e x\right )^{\frac {3}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 448, normalized size = 1.56 \[ -\frac {2 \sqrt {e x}\, \left (-77 b^{2} d^{5} x^{9}-210 a b \,d^{5} x^{7}-91 b^{2} c \,d^{4} x^{7}-165 a^{2} d^{5} x^{5}-270 a b c \,d^{4} x^{5}+4 b^{2} c^{2} d^{3} x^{5}-275 a^{2} c \,d^{4} x^{3}+40 a b \,c^{2} d^{3} x^{3}-12 b^{2} c^{3} d^{2} x^{3}-110 a^{2} c^{2} d^{3} x +100 a b \,c^{3} d^{2} x -30 b^{2} c^{4} d x +55 \sqrt {-c d}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a^{2} c^{2} d^{2} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )-50 \sqrt {-c d}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a b \,c^{3} d \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )+15 \sqrt {-c d}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, b^{2} c^{4} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )\right ) e}{1155 \sqrt {d \,x^{2}+c}\, d^{4} x} \]
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 \[ \int {\left (b x^{2} + a\right )}^{2} \sqrt {d x^{2} + c} \left (e x\right )^{\frac {3}{2}}\,{d x} \]
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 \[ \int {\left (e\,x\right )}^{3/2}\,{\left (b\,x^2+a\right )}^2\,\sqrt {d\,x^2+c} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [C] time = 21.70, size = 150, normalized size = 0.52 \[ \frac {a^{2} \sqrt {c} e^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 \Gamma \left (\frac {9}{4}\right )} + \frac {a b \sqrt {c} e^{\frac {3}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{\Gamma \left (\frac {13}{4}\right )} + \frac {b^{2} \sqrt {c} e^{\frac {3}{2}} x^{\frac {13}{2}} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 \Gamma \left (\frac {17}{4}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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