Optimal. Leaf size=39 \[ -\frac {\text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{b}-\frac {1-(a+b x)^2}{b \sin ^{-1}(a+b x)} \]
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Rubi [A] time = 0.11, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4807, 4659, 4635, 4406, 12, 3299} \[ -\frac {\text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{b}-\frac {1-(a+b x)^2}{b \sin ^{-1}(a+b x)} \]
Antiderivative was successfully verified.
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Rule 12
Rule 3299
Rule 4406
Rule 4635
Rule 4659
Rule 4807
Rubi steps
\begin {align*} \int \frac {\sqrt {1-a^2-2 a b x-b^2 x^2}}{\sin ^{-1}(a+b x)^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sqrt {1-x^2}}{\sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{b}\\ &=-\frac {1-(a+b x)^2}{b \sin ^{-1}(a+b x)}-\frac {2 \operatorname {Subst}\left (\int \frac {x}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{b}\\ &=-\frac {1-(a+b x)^2}{b \sin ^{-1}(a+b x)}-\frac {2 \operatorname {Subst}\left (\int \frac {\cos (x) \sin (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac {1-(a+b x)^2}{b \sin ^{-1}(a+b x)}-\frac {2 \operatorname {Subst}\left (\int \frac {\sin (2 x)}{2 x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac {1-(a+b x)^2}{b \sin ^{-1}(a+b x)}-\frac {\operatorname {Subst}\left (\int \frac {\sin (2 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac {1-(a+b x)^2}{b \sin ^{-1}(a+b x)}-\frac {\text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{b}\\ \end {align*}
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Mathematica [A] time = 0.07, size = 46, normalized size = 1.18 \[ \frac {a^2-\sin ^{-1}(a+b x) \text {Si}\left (2 \sin ^{-1}(a+b x)\right )+2 a b x+b^2 x^2-1}{b \sin ^{-1}(a+b x)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{\arcsin \left (b x + a\right )^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.45, size = 44, normalized size = 1.13 \[ -\frac {\operatorname {Si}\left (2 \, \arcsin \left (b x + a\right )\right )}{b} + \frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b \arcsin \left (b x + a\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.18, size = 42, normalized size = 1.08 \[ -\frac {2 \Si \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )+\cos \left (2 \arcsin \left (b x +a \right )\right )+1}{2 b \arcsin \left (b x +a \right )} \]
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 \[ \frac {b^{2} x^{2} + 2 \, a b x - 2 \, b \arctan \left (b x + a, \sqrt {b x + a + 1} \sqrt {-b x - a + 1}\right ) \int \frac {b x + a}{\arctan \left (b x + a, \sqrt {b x + a + 1} \sqrt {-b x - a + 1}\right )}\,{d x} + a^{2} - 1}{b \arctan \left (b x + a, \sqrt {b x + a + 1} \sqrt {-b x - a + 1}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}}{{\mathrm {asin}\left (a+b\,x\right )}^2} \,d x \]
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 \[ \int \frac {\sqrt {- \left (a + b x - 1\right ) \left (a + b x + 1\right )}}{\operatorname {asin}^{2}{\left (a + b x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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