Optimal. Leaf size=89 \[ \frac {2 \sqrt {1-b^2 x^4} F\left (\left .\sin ^{-1}\left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b} \sqrt {b^2 x^4-1}}-\frac {\sqrt {1-b^2 x^4} E\left (\left .\sin ^{-1}\left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b} \sqrt {b^2 x^4-1}} \]
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Rubi [A] time = 0.05, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1200, 1199, 423, 424, 248, 221} \[ \frac {2 \sqrt {1-b^2 x^4} F\left (\left .\sin ^{-1}\left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b} \sqrt {b^2 x^4-1}}-\frac {\sqrt {1-b^2 x^4} E\left (\left .\sin ^{-1}\left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b} \sqrt {b^2 x^4-1}} \]
Antiderivative was successfully verified.
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Rule 221
Rule 248
Rule 423
Rule 424
Rule 1199
Rule 1200
Rubi steps
\begin {align*} \int \frac {1-b x^2}{\sqrt {-1+b^2 x^4}} \, dx &=\frac {\sqrt {1-b^2 x^4} \int \frac {1-b x^2}{\sqrt {1-b^2 x^4}} \, dx}{\sqrt {-1+b^2 x^4}}\\ &=\frac {\sqrt {1-b^2 x^4} \int \frac {\sqrt {1-b x^2}}{\sqrt {1+b x^2}} \, dx}{\sqrt {-1+b^2 x^4}}\\ &=-\frac {\sqrt {1-b^2 x^4} \int \frac {\sqrt {1+b x^2}}{\sqrt {1-b x^2}} \, dx}{\sqrt {-1+b^2 x^4}}+\frac {\left (2 \sqrt {1-b^2 x^4}\right ) \int \frac {1}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx}{\sqrt {-1+b^2 x^4}}\\ &=-\frac {\sqrt {1-b^2 x^4} E\left (\left .\sin ^{-1}\left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b} \sqrt {-1+b^2 x^4}}+\frac {\left (2 \sqrt {1-b^2 x^4}\right ) \int \frac {1}{\sqrt {1-b^2 x^4}} \, dx}{\sqrt {-1+b^2 x^4}}\\ &=-\frac {\sqrt {1-b^2 x^4} E\left (\left .\sin ^{-1}\left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b} \sqrt {-1+b^2 x^4}}+\frac {2 \sqrt {1-b^2 x^4} F\left (\left .\sin ^{-1}\left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b} \sqrt {-1+b^2 x^4}}\\ \end {align*}
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Mathematica [C] time = 0.02, size = 74, normalized size = 0.83 \[ -\frac {\sqrt {1-b^2 x^4} \left (b x^3 \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};b^2 x^4\right )-3 x \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};b^2 x^4\right )\right )}{3 \sqrt {b^2 x^4-1}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {b^{2} x^{4} - 1}}{b x^{2} + 1}, x\right ) \]
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 \[ \int -\frac {b x^{2} - 1}{\sqrt {b^{2} x^{4} - 1}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.01, size = 108, normalized size = 1.21 \[ \frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \EllipticF \left (\sqrt {-b}\, x , i\right )}{\sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}-\frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \left (-\EllipticE \left (\sqrt {-b}\, x , i\right )+\EllipticF \left (\sqrt {-b}\, x , i\right )\right )}{\sqrt {-b}\, \sqrt {b^{2} x^{4}-1}} \]
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 \[ -\int \frac {b x^{2} - 1}{\sqrt {b^{2} x^{4} - 1}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {b\,x^2-1}{\sqrt {b^2\,x^4-1}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 2.19, size = 60, normalized size = 0.67 \[ \frac {i b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} - \frac {i x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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