Optimal. Leaf size=240 \[ -\frac {x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \text {Ci}\left (-\frac {a-b \sin ^{-1}\left (1-d x^2\right )}{2 b}\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}+\frac {x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a}{2 b}-\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}+\frac {x}{8 b^2 \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )}-\frac {\sqrt {2 d x^2-d^2 x^4}}{4 b d x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2} \]
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Rubi [A] time = 0.05, antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4828, 4816} \[ -\frac {x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \text {CosIntegral}\left (-\frac {a-b \sin ^{-1}\left (1-d x^2\right )}{2 b}\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}+\frac {x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a}{2 b}-\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}+\frac {x}{8 b^2 \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )}-\frac {\sqrt {2 d x^2-d^2 x^4}}{4 b d x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2} \]
Antiderivative was successfully verified.
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Rule 4816
Rule 4828
Rubi steps
\begin {align*} \int \frac {1}{\left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^3} \, dx &=-\frac {\sqrt {2 d x^2-d^2 x^4}}{4 b d x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2}+\frac {x}{8 b^2 \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )}-\frac {\int \frac {1}{a-b \sin ^{-1}\left (1-d x^2\right )} \, dx}{8 b^2}\\ &=-\frac {\sqrt {2 d x^2-d^2 x^4}}{4 b d x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2}+\frac {x}{8 b^2 \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )}-\frac {x \text {Ci}\left (-\frac {a-b \sin ^{-1}\left (1-d x^2\right )}{2 b}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}+\frac {x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a}{2 b}-\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}\\ \end {align*}
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Mathematica [A] time = 0.57, size = 195, normalized size = 0.81 \[ -\frac {\frac {4 b^2 \sqrt {-d x^2 \left (d x^2-2\right )}}{d \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2}+\frac {\left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right ) \left (\left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \text {Ci}\left (\frac {1}{2} \left (\sin ^{-1}\left (1-d x^2\right )-\frac {a}{b}\right )\right )+\left (\sin \left (\frac {a}{2 b}\right )-\cos \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a-b \sin ^{-1}\left (1-d x^2\right )}{2 b}\right )\right )}{d}-\frac {2 b x^2}{a-b \sin ^{-1}\left (1-d x^2\right )}}{16 b^3 x} \]
Antiderivative was successfully verified.
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fricas [F] time = 1.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{b^{3} \arcsin \left (d x^{2} - 1\right )^{3} + 3 \, a b^{2} \arcsin \left (d x^{2} - 1\right )^{2} + 3 \, a^{2} b \arcsin \left (d x^{2} - 1\right ) + a^{3}}, x\right ) \]
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 \[ \int \frac {1}{{\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a +b \arcsin \left (d \,x^{2}-1\right )\right )^{3}}\, dx \]
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 \[ \frac {b d x \arctan \left (d x^{2} - 1, \sqrt {-d x^{2} + 2} \sqrt {d} x\right ) + a d x - 2 \, \sqrt {-d x^{2} + 2} b \sqrt {d} - {\left (b^{4} d \arctan \left (d x^{2} - 1, \sqrt {-d x^{2} + 2} \sqrt {d} x\right )^{2} + 2 \, a b^{3} d \arctan \left (d x^{2} - 1, \sqrt {-d x^{2} + 2} \sqrt {d} x\right ) + a^{2} b^{2} d\right )} \int \frac {1}{b^{3} \arctan \left (d x^{2} - 1, \sqrt {-d x^{2} + 2} \sqrt {d} x\right ) + a b^{2}}\,{d x}}{8 \, {\left (b^{4} d \arctan \left (d x^{2} - 1, \sqrt {-d x^{2} + 2} \sqrt {d} x\right )^{2} + 2 \, a b^{3} d \arctan \left (d x^{2} - 1, \sqrt {-d x^{2} + 2} \sqrt {d} x\right ) + a^{2} b^{2} d\right )}} \]
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 \[ \int \frac {1}{{\left (a+b\,\mathrm {asin}\left (d\,x^2-1\right )\right )}^3} \,d x \]
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 \[ \int \frac {1}{\left (a + b \operatorname {asin}{\left (d x^{2} - 1 \right )}\right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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