Optimal. Leaf size=57 \[ -\frac {\text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{b}-\frac {\text {Si}\left (4 \sin ^{-1}(a+b x)\right )}{2 b}-\frac {\left (1-(a+b x)^2\right )^2}{b \sin ^{-1}(a+b x)} \]
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Rubi [A] time = 0.15, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {4807, 4659, 4723, 4406, 3299} \[ -\frac {\text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{b}-\frac {\text {Si}\left (4 \sin ^{-1}(a+b x)\right )}{2 b}-\frac {\left (1-(a+b x)^2\right )^2}{b \sin ^{-1}(a+b x)} \]
Antiderivative was successfully verified.
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Rule 3299
Rule 4406
Rule 4659
Rule 4723
Rule 4807
Rubi steps
\begin {align*} \int \frac {\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}}{\sin ^{-1}(a+b x)^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^{3/2}}{\sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{b \sin ^{-1}(a+b x)}-\frac {4 \operatorname {Subst}\left (\int \frac {x \left (1-x^2\right )}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{b \sin ^{-1}(a+b x)}-\frac {4 \operatorname {Subst}\left (\int \frac {\cos ^3(x) \sin (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{b \sin ^{-1}(a+b x)}-\frac {4 \operatorname {Subst}\left (\int \left (\frac {\sin (2 x)}{4 x}+\frac {\sin (4 x)}{8 x}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{b \sin ^{-1}(a+b x)}-\frac {\operatorname {Subst}\left (\int \frac {\sin (4 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b}-\frac {\operatorname {Subst}\left (\int \frac {\sin (2 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{b \sin ^{-1}(a+b x)}-\frac {\text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{b}-\frac {\text {Si}\left (4 \sin ^{-1}(a+b x)\right )}{2 b}\\ \end {align*}
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Mathematica [A] time = 0.33, size = 70, normalized size = 1.23 \[ -\frac {2 \left (a^2+2 a b x+b^2 x^2-1\right )^2+2 \sin ^{-1}(a+b x) \text {Si}\left (2 \sin ^{-1}(a+b x)\right )+\sin ^{-1}(a+b x) \text {Si}\left (4 \sin ^{-1}(a+b x)\right )}{2 b \sin ^{-1}(a+b x)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}}}{\arcsin \left (b x + a\right )^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 2.52, size = 61, normalized size = 1.07 \[ -\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}^{2}}{b \arcsin \left (b x + a\right )} - \frac {\operatorname {Si}\left (4 \, \arcsin \left (b x + a\right )\right )}{2 \, b} - \frac {\operatorname {Si}\left (2 \, \arcsin \left (b x + a\right )\right )}{b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.17, size = 70, normalized size = 1.23 \[ -\frac {8 \Si \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )+4 \Si \left (4 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )+4 \cos \left (2 \arcsin \left (b x +a \right )\right )+\cos \left (4 \arcsin \left (b x +a \right )\right )+3}{8 b \arcsin \left (b x +a \right )} \]
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 \[ -\frac {b^{4} x^{4} + 4 \, a b^{3} x^{3} + 2 \, {\left (3 \, a^{2} - 1\right )} b^{2} x^{2} + a^{4} + 4 \, {\left (a^{3} - a\right )} b x - 4 \, b \arctan \left (b x + a, \sqrt {b x + a + 1} \sqrt {-b x - a + 1}\right ) \int \frac {b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} - 1\right )} b x - a}{\arctan \left (b x + a, \sqrt {b x + a + 1} \sqrt {-b x - a + 1}\right )}\,{d x} - 2 \, a^{2} + 1}{b \arctan \left (b x + a, \sqrt {b x + a + 1} \sqrt {-b x - a + 1}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (-a^2-2\,a\,b\,x-b^2\,x^2+1\right )}^{3/2}}{{\mathrm {asin}\left (a+b\,x\right )}^2} \,d x \]
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 \[ \int \frac {\left (- \left (a + b x - 1\right ) \left (a + b x + 1\right )\right )^{\frac {3}{2}}}{\operatorname {asin}^{2}{\left (a + b x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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