Optimal. Leaf size=115 \[ \frac {2 \text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{3 b}-\frac {2 (a+b x)^2}{3 b \sin ^{-1}(a+b x)}+\frac {\sqrt {1-(a+b x)^2} (a+b x)}{3 b \sin ^{-1}(a+b x)^2}+\frac {1}{3 b \sin ^{-1}(a+b x)}-\frac {1-(a+b x)^2}{3 b \sin ^{-1}(a+b x)^3} \]
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Rubi [A] time = 0.22, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4807, 4659, 4633, 4719, 4635, 4406, 12, 3299, 4641} \[ \frac {2 \text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{3 b}-\frac {2 (a+b x)^2}{3 b \sin ^{-1}(a+b x)}+\frac {\sqrt {1-(a+b x)^2} (a+b x)}{3 b \sin ^{-1}(a+b x)^2}+\frac {1}{3 b \sin ^{-1}(a+b x)}-\frac {1-(a+b x)^2}{3 b \sin ^{-1}(a+b x)^3} \]
Antiderivative was successfully verified.
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Rule 12
Rule 3299
Rule 4406
Rule 4633
Rule 4635
Rule 4641
Rule 4659
Rule 4719
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)^4} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sqrt {1-x^2}}{\sin ^{-1}(x)^4} \, dx,x,a+b x\right )}{b}\\ &=-\frac {1-(a+b x)^2}{3 b \sin ^{-1}(a+b x)^3}-\frac {2 \operatorname {Subst}\left (\int \frac {x}{\sin ^{-1}(x)^3} \, dx,x,a+b x\right )}{3 b}\\ &=-\frac {1-(a+b x)^2}{3 b \sin ^{-1}(a+b x)^3}+\frac {(a+b x) \sqrt {1-(a+b x)^2}}{3 b \sin ^{-1}(a+b x)^2}-\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{3 b}+\frac {2 \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-x^2} \sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{3 b}\\ &=-\frac {1-(a+b x)^2}{3 b \sin ^{-1}(a+b x)^3}+\frac {(a+b x) \sqrt {1-(a+b x)^2}}{3 b \sin ^{-1}(a+b x)^2}+\frac {1}{3 b \sin ^{-1}(a+b x)}-\frac {2 (a+b x)^2}{3 b \sin ^{-1}(a+b x)}+\frac {4 \operatorname {Subst}\left (\int \frac {x}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{3 b}\\ &=-\frac {1-(a+b x)^2}{3 b \sin ^{-1}(a+b x)^3}+\frac {(a+b x) \sqrt {1-(a+b x)^2}}{3 b \sin ^{-1}(a+b x)^2}+\frac {1}{3 b \sin ^{-1}(a+b x)}-\frac {2 (a+b x)^2}{3 b \sin ^{-1}(a+b x)}+\frac {4 \operatorname {Subst}\left (\int \frac {\cos (x) \sin (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}\\ &=-\frac {1-(a+b x)^2}{3 b \sin ^{-1}(a+b x)^3}+\frac {(a+b x) \sqrt {1-(a+b x)^2}}{3 b \sin ^{-1}(a+b x)^2}+\frac {1}{3 b \sin ^{-1}(a+b x)}-\frac {2 (a+b x)^2}{3 b \sin ^{-1}(a+b x)}+\frac {4 \operatorname {Subst}\left (\int \frac {\sin (2 x)}{2 x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}\\ &=-\frac {1-(a+b x)^2}{3 b \sin ^{-1}(a+b x)^3}+\frac {(a+b x) \sqrt {1-(a+b x)^2}}{3 b \sin ^{-1}(a+b x)^2}+\frac {1}{3 b \sin ^{-1}(a+b x)}-\frac {2 (a+b x)^2}{3 b \sin ^{-1}(a+b x)}+\frac {2 \operatorname {Subst}\left (\int \frac {\sin (2 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}\\ &=-\frac {1-(a+b x)^2}{3 b \sin ^{-1}(a+b x)^3}+\frac {(a+b x) \sqrt {1-(a+b x)^2}}{3 b \sin ^{-1}(a+b x)^2}+\frac {1}{3 b \sin ^{-1}(a+b x)}-\frac {2 (a+b x)^2}{3 b \sin ^{-1}(a+b x)}+\frac {2 \text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{3 b}\\ \end {align*}
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Mathematica [A] time = 0.12, size = 117, normalized size = 1.02 \[ \frac {-\left (2 a^2+4 a b x+2 b^2 x^2-1\right ) \sin ^{-1}(a+b x)^2+(a+b x) \sqrt {-a^2-2 a b x-b^2 x^2+1} \sin ^{-1}(a+b x)+a^2+2 \sin ^{-1}(a+b x)^3 \text {Si}\left (2 \sin ^{-1}(a+b x)\right )+2 a b x+b^2 x^2-1}{3 b \sin ^{-1}(a+b x)^3} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{\arcsin \left (b x + a\right )^{4}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.52, size = 128, normalized size = 1.11 \[ \frac {2 \, \operatorname {Si}\left (2 \, \arcsin \left (b x + a\right )\right )}{3 \, b} - \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}}{3 \, b \arcsin \left (b x + a\right )} + \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{3 \, b \arcsin \left (b x + a\right )^{2}} - \frac {1}{3 \, b \arcsin \left (b x + a\right )} + \frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{3 \, b \arcsin \left (b x + a\right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.16, size = 81, normalized size = 0.70 \[ \frac {4 \Si \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{3}+2 \cos \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{2}+\sin \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )-\cos \left (2 \arcsin \left (b x +a \right )\right )-1}{6 b \arcsin \left (b x +a \right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
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 \[ \int \frac {\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}}{{\mathrm {asin}\left (a+b\,x\right )}^4} \,d x \]
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 \[ \int \frac {\sqrt {- \left (a + b x - 1\right ) \left (a + b x + 1\right )}}{\operatorname {asin}^{4}{\left (a + b x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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