Optimal. Leaf size=528 \[ \frac {2 \sqrt {d+e x} \left (-A \left (2 a c e+b^2 (-e)+b c d\right )+c x (-2 a B e+A b e-2 A c d+b B d)+a B (2 c d-b e)\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}-\frac {\sqrt {2} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (-2 a B e+A b e-2 A c d+b B d) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{\sqrt {b^2-4 a c} \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}+\frac {2 \sqrt {2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (b B-2 A c) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{c \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {a+b x+c x^2}} \]
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Rubi [A] time = 0.40, antiderivative size = 528, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {822, 843, 718, 424, 419} \[ \frac {2 \sqrt {d+e x} \left (-A \left (2 a c e+b^2 (-e)+b c d\right )+c x (-2 a B e+A b e-2 A c d+b B d)+a B (2 c d-b e)\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}-\frac {\sqrt {2} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (-2 a B e+A b e-2 A c d+b B d) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{\sqrt {b^2-4 a c} \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}+\frac {2 \sqrt {2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (b B-2 A c) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{c \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {a+b x+c x^2}} \]
Antiderivative was successfully verified.
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Rule 419
Rule 424
Rule 718
Rule 822
Rule 843
Rubi steps
\begin {align*} \int \frac {A+B x}{\sqrt {d+e x} \left (a+b x+c x^2\right )^{3/2}} \, dx &=\frac {2 \sqrt {d+e x} \left (a B (2 c d-b e)-A \left (b c d-b^2 e+2 a c e\right )+c (b B d-2 A c d+A b e-2 a B e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \sqrt {a+b x+c x^2}}-\frac {2 \int \frac {\frac {1}{2} e \left (b^2 B d-A b c d-2 a B c d-a b B e+2 a A c e\right )+\frac {1}{2} c e (b B d-2 A c d+A b e-2 a B e) x}{\sqrt {d+e x} \sqrt {a+b x+c x^2}} \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )}\\ &=\frac {2 \sqrt {d+e x} \left (a B (2 c d-b e)-A \left (b c d-b^2 e+2 a c e\right )+c (b B d-2 A c d+A b e-2 a B e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \sqrt {a+b x+c x^2}}+\frac {(b B-2 A c) \int \frac {1}{\sqrt {d+e x} \sqrt {a+b x+c x^2}} \, dx}{b^2-4 a c}-\frac {(c (b B d-2 A c d+A b e-2 a B e)) \int \frac {\sqrt {d+e x}}{\sqrt {a+b x+c x^2}} \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )}\\ &=\frac {2 \sqrt {d+e x} \left (a B (2 c d-b e)-A \left (b c d-b^2 e+2 a c e\right )+c (b B d-2 A c d+A b e-2 a B e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \sqrt {a+b x+c x^2}}-\frac {\left (\sqrt {2} (b B d-2 A c d+A b e-2 a B e) \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1+\frac {2 \sqrt {b^2-4 a c} e x^2}{2 c d-b e-\sqrt {b^2-4 a c} e}}}{\sqrt {1-x^2}} \, dx,x,\frac {\sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )}{\sqrt {b^2-4 a c} \left (c d^2-b d e+a e^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-b e-\sqrt {b^2-4 a c} e}} \sqrt {a+b x+c x^2}}+\frac {\left (2 \sqrt {2} (b B-2 A c) \sqrt {\frac {c (d+e x)}{2 c d-b e-\sqrt {b^2-4 a c} e}} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+\frac {2 \sqrt {b^2-4 a c} e x^2}{2 c d-b e-\sqrt {b^2-4 a c} e}}} \, dx,x,\frac {\sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )}{c \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {a+b x+c x^2}}\\ &=\frac {2 \sqrt {d+e x} \left (a B (2 c d-b e)-A \left (b c d-b^2 e+2 a c e\right )+c (b B d-2 A c d+A b e-2 a B e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \sqrt {a+b x+c x^2}}-\frac {\sqrt {2} (b B d-2 A c d+A b e-2 a B e) \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{\sqrt {b^2-4 a c} \left (c d^2-b d e+a e^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}} \sqrt {a+b x+c x^2}}+\frac {2 \sqrt {2} (b B-2 A c) \sqrt {\frac {c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{c \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {a+b x+c x^2}}\\ \end {align*}
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Mathematica [C] time = 9.00, size = 893, normalized size = 1.69 \[ \frac {2 \sqrt {d+e x} \left (-A e b^2+A c d b+a B e b-B c d x b-A c e x b-2 a B c d+2 a A c e+2 A c^2 d x+2 a B c e x\right ) \left (c x^2+b x+a\right )}{\left (4 a c-b^2\right ) \left (c d^2-b e d+a e^2\right ) (a+x (b+c x))^{3/2}}-\frac {2 (d+e x)^{3/2} \left (c x^2+b x+a\right )^{3/2} \left (\frac {i \sqrt {1-\frac {2 \left (c d^2+e (a e-b d)\right )}{\left (2 c d-b e+\sqrt {\left (b^2-4 a c\right ) e^2}\right ) (d+e x)}} \sqrt {\frac {2 \left (c d^2+e (a e-b d)\right )}{\left (-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}\right ) (d+e x)}+1} \left ((b B d-2 A c d+A b e-2 a B e) \left (2 c d-b e+\sqrt {\left (b^2-4 a c\right ) e^2}\right ) E\left (i \sinh ^{-1}\left (\frac {\sqrt {2} \sqrt {\frac {c d^2-b e d+a e^2}{-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}}}{\sqrt {d+e x}}\right )|-\frac {-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}{2 c d-b e+\sqrt {\left (b^2-4 a c\right ) e^2}}\right )-\left (e (B d-A e) b^2+\sqrt {\left (b^2-4 a c\right ) e^2} (B d+A e) b-2 a e \left (2 c d B+\sqrt {\left (b^2-4 a c\right ) e^2} B-2 A c e\right )-2 A c d \sqrt {\left (b^2-4 a c\right ) e^2}\right ) F\left (i \sinh ^{-1}\left (\frac {\sqrt {2} \sqrt {\frac {c d^2-b e d+a e^2}{-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}}}{\sqrt {d+e x}}\right )|-\frac {-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}{2 c d-b e+\sqrt {\left (b^2-4 a c\right ) e^2}}\right )\right )}{2 \sqrt {2} \sqrt {\frac {c d^2+e (a e-b d)}{-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}} \sqrt {d+e x}}-(b B d-2 A c d+A b e-2 a B e) \left (c \left (\frac {d}{d+e x}-1\right )^2+\frac {e \left (-\frac {d b}{d+e x}+b+\frac {a e}{d+e x}\right )}{d+e x}\right )\right )}{\left (4 a c-b^2\right ) e \left (c d^2-b e d+a e^2\right ) (a+x (b+c x))^{3/2} \sqrt {\frac {(d+e x)^2 \left (c \left (\frac {d}{d+e x}-1\right )^2+\frac {e \left (-\frac {d b}{d+e x}+b+\frac {a e}{d+e x}\right )}{d+e x}\right )}{e^2}}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + b x + a} {\left (B x + A\right )} \sqrt {e x + d}}{c^{2} e x^{5} + {\left (c^{2} d + 2 \, b c e\right )} x^{4} + {\left (2 \, b c d + {\left (b^{2} + 2 \, a c\right )} e\right )} x^{3} + a^{2} d + {\left (2 \, a b e + {\left (b^{2} + 2 \, a c\right )} d\right )} x^{2} + {\left (2 \, a b d + a^{2} e\right )} x}, x\right ) \]
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 \[ \int \frac {B x + A}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} \sqrt {e x + d}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.10, size = 5059, normalized size = 9.58 \[ \text {output too large to display} \]
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 \[ \int \frac {B x + A}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} \sqrt {e x + d}}\,{d x} \]
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 \[ \int \frac {A+B\,x}{\sqrt {d+e\,x}\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]
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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