Optimal. Leaf size=258 \[ -\frac {\sqrt {c+d x^2} (2 b e-a f) E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{\sqrt {e} f^{3/2} \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac {x \sqrt {c+d x^2} (b e-a f)}{e f \sqrt {e+f x^2}}+\frac {x \sqrt {c+d x^2} (2 b e-a f)}{e f \sqrt {e+f x^2}}+\frac {b \sqrt {e} \sqrt {c+d x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{f^{3/2} \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}} \]
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Rubi [A] time = 0.16, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {526, 531, 418, 492, 411} \[ -\frac {\sqrt {c+d x^2} (2 b e-a f) E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{\sqrt {e} f^{3/2} \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac {x \sqrt {c+d x^2} (b e-a f)}{e f \sqrt {e+f x^2}}+\frac {x \sqrt {c+d x^2} (2 b e-a f)}{e f \sqrt {e+f x^2}}+\frac {b \sqrt {e} \sqrt {c+d x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{f^{3/2} \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}} \]
Antiderivative was successfully verified.
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Rule 411
Rule 418
Rule 492
Rule 526
Rule 531
Rubi steps
\begin {align*} \int \frac {\left (a+b x^2\right ) \sqrt {c+d x^2}}{\left (e+f x^2\right )^{3/2}} \, dx &=-\frac {(b e-a f) x \sqrt {c+d x^2}}{e f \sqrt {e+f x^2}}-\frac {\int \frac {-b c e-d (2 b e-a f) x^2}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{e f}\\ &=-\frac {(b e-a f) x \sqrt {c+d x^2}}{e f \sqrt {e+f x^2}}+\frac {(b c) \int \frac {1}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{f}+\frac {(d (2 b e-a f)) \int \frac {x^2}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{e f}\\ &=-\frac {(b e-a f) x \sqrt {c+d x^2}}{e f \sqrt {e+f x^2}}+\frac {(2 b e-a f) x \sqrt {c+d x^2}}{e f \sqrt {e+f x^2}}+\frac {b \sqrt {e} \sqrt {c+d x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{f^{3/2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}-\frac {(2 b e-a f) \int \frac {\sqrt {c+d x^2}}{\left (e+f x^2\right )^{3/2}} \, dx}{f}\\ &=-\frac {(b e-a f) x \sqrt {c+d x^2}}{e f \sqrt {e+f x^2}}+\frac {(2 b e-a f) x \sqrt {c+d x^2}}{e f \sqrt {e+f x^2}}-\frac {(2 b e-a f) \sqrt {c+d x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{\sqrt {e} f^{3/2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}+\frac {b \sqrt {e} \sqrt {c+d x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{f^{3/2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}\\ \end {align*}
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Mathematica [C] time = 0.41, size = 208, normalized size = 0.81 \[ \frac {f x \sqrt {\frac {d}{c}} \left (c+d x^2\right ) (a f-b e)-i e \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} (a d f+b c f-2 b d e) F\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )-i d e \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} (2 b e-a f) E\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )}{e f^2 \sqrt {\frac {d}{c}} \sqrt {c+d x^2} \sqrt {e+f x^2}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )} \sqrt {d x^{2} + c} \sqrt {f x^{2} + e}}{f^{2} x^{4} + 2 \, e f x^{2} + e^{2}}, x\right ) \]
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 \[ \int \frac {{\left (b x^{2} + a\right )} \sqrt {d x^{2} + c}}{{\left (f x^{2} + e\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.03, size = 393, normalized size = 1.52 \[ \frac {\sqrt {d \,x^{2}+c}\, \sqrt {f \,x^{2}+e}\, \left (\sqrt {-\frac {d}{c}}\, a d \,f^{2} x^{3}-\sqrt {-\frac {d}{c}}\, b d e f \,x^{3}+\sqrt {-\frac {d}{c}}\, a c \,f^{2} x -\sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, a d e f \EllipticE \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )+\sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, a d e f \EllipticF \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )-\sqrt {-\frac {d}{c}}\, b c e f x +\sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, b c e f \EllipticF \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )+2 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, b d \,e^{2} \EllipticE \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )-2 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, b d \,e^{2} \EllipticF \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )\right )}{\left (d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e \right ) \sqrt {-\frac {d}{c}}\, e \,f^{2}} \]
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 \[ \int \frac {{\left (b x^{2} + a\right )} \sqrt {d x^{2} + c}}{{\left (f x^{2} + e\right )}^{\frac {3}{2}}}\,{d x} \]
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 \[ \int \frac {\left (b\,x^2+a\right )\,\sqrt {d\,x^2+c}}{{\left (f\,x^2+e\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x^{2}\right ) \sqrt {c + d x^{2}}}{\left (e + f x^{2}\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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