Optimal. Leaf size=199 \[ -\frac {(a+b x) \left (1-(a+b x)^2\right )^{3/2}}{32 b}-\frac {15 (a+b x) \sqrt {1-(a+b x)^2}}{64 b}+\frac {\sin ^{-1}(a+b x)^3}{8 b}+\frac {(a+b x) \left (1-(a+b x)^2\right )^{3/2} \sin ^{-1}(a+b x)^2}{4 b}+\frac {3 (a+b x) \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{8 b}-\frac {3 (a+b x)^2 \sin ^{-1}(a+b x)}{8 b}+\frac {\left (1-(a+b x)^2\right )^2 \sin ^{-1}(a+b x)}{8 b}+\frac {9 \sin ^{-1}(a+b x)}{64 b} \]
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Rubi [A] time = 0.20, antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4807, 4649, 4647, 4641, 4627, 321, 216, 4677, 195} \[ -\frac {(a+b x) \left (1-(a+b x)^2\right )^{3/2}}{32 b}-\frac {15 (a+b x) \sqrt {1-(a+b x)^2}}{64 b}+\frac {\sin ^{-1}(a+b x)^3}{8 b}+\frac {(a+b x) \left (1-(a+b x)^2\right )^{3/2} \sin ^{-1}(a+b x)^2}{4 b}+\frac {3 (a+b x) \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{8 b}-\frac {3 (a+b x)^2 \sin ^{-1}(a+b x)}{8 b}+\frac {\left (1-(a+b x)^2\right )^2 \sin ^{-1}(a+b x)}{8 b}+\frac {9 \sin ^{-1}(a+b x)}{64 b} \]
Antiderivative was successfully verified.
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Rule 195
Rule 216
Rule 321
Rule 4627
Rule 4641
Rule 4647
Rule 4649
Rule 4677
Rule 4807
Rubi steps
\begin {align*} \int \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 \left (1-x^2\right )^{3/2} \sin ^{-1}(x)^2 \, dx,x,a+b x\right )}{b}\\ &=\frac {(a+b x) \left (1-(a+b x)^2\right )^{3/2} \sin ^{-1}(a+b x)^2}{4 b}-\frac {\operatorname {Subst}\left (\int x \left (1-x^2\right ) \sin ^{-1}(x) \, dx,x,a+b x\right )}{2 b}+\frac {3 \operatorname {Subst}\left (\int \sqrt {1-x^2} \sin ^{-1}(x)^2 \, dx,x,a+b x\right )}{4 b}\\ &=\frac {\left (1-(a+b x)^2\right )^2 \sin ^{-1}(a+b x)}{8 b}+\frac {3 (a+b x) \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{8 b}+\frac {(a+b x) \left (1-(a+b x)^2\right )^{3/2} \sin ^{-1}(a+b x)^2}{4 b}-\frac {\operatorname {Subst}\left (\int \left (1-x^2\right )^{3/2} \, dx,x,a+b x\right )}{8 b}+\frac {3 \operatorname {Subst}\left (\int \frac {\sin ^{-1}(x)^2}{\sqrt {1-x^2}} \, dx,x,a+b x\right )}{8 b}-\frac {3 \operatorname {Subst}\left (\int x \sin ^{-1}(x) \, dx,x,a+b x\right )}{4 b}\\ &=-\frac {(a+b x) \left (1-(a+b x)^2\right )^{3/2}}{32 b}-\frac {3 (a+b x)^2 \sin ^{-1}(a+b x)}{8 b}+\frac {\left (1-(a+b x)^2\right )^2 \sin ^{-1}(a+b x)}{8 b}+\frac {3 (a+b x) \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{8 b}+\frac {(a+b x) \left (1-(a+b x)^2\right )^{3/2} \sin ^{-1}(a+b x)^2}{4 b}+\frac {\sin ^{-1}(a+b x)^3}{8 b}-\frac {3 \operatorname {Subst}\left (\int \sqrt {1-x^2} \, dx,x,a+b x\right )}{32 b}+\frac {3 \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-x^2}} \, dx,x,a+b x\right )}{8 b}\\ &=-\frac {15 (a+b x) \sqrt {1-(a+b x)^2}}{64 b}-\frac {(a+b x) \left (1-(a+b x)^2\right )^{3/2}}{32 b}-\frac {3 (a+b x)^2 \sin ^{-1}(a+b x)}{8 b}+\frac {\left (1-(a+b x)^2\right )^2 \sin ^{-1}(a+b x)}{8 b}+\frac {3 (a+b x) \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{8 b}+\frac {(a+b x) \left (1-(a+b x)^2\right )^{3/2} \sin ^{-1}(a+b x)^2}{4 b}+\frac {\sin ^{-1}(a+b x)^3}{8 b}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,a+b x\right )}{64 b}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,a+b x\right )}{16 b}\\ &=-\frac {15 (a+b x) \sqrt {1-(a+b x)^2}}{64 b}-\frac {(a+b x) \left (1-(a+b x)^2\right )^{3/2}}{32 b}+\frac {9 \sin ^{-1}(a+b x)}{64 b}-\frac {3 (a+b x)^2 \sin ^{-1}(a+b x)}{8 b}+\frac {\left (1-(a+b x)^2\right )^2 \sin ^{-1}(a+b x)}{8 b}+\frac {3 (a+b x) \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{8 b}+\frac {(a+b x) \left (1-(a+b x)^2\right )^{3/2} \sin ^{-1}(a+b x)^2}{4 b}+\frac {\sin ^{-1}(a+b x)^3}{8 b}\\ \end {align*}
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Mathematica [A] time = 0.20, size = 216, normalized size = 1.09 \[ \frac {\left (8 a^4-40 a^2+17\right ) \sin ^{-1}(a+b x)+\sqrt {-a^2-2 a b x-b^2 x^2+1} \left (2 a^3+6 a^2 b x+6 a b^2 x^2-17 a+2 b^3 x^3-17 b x\right )-8 \sqrt {-a^2-2 a b x-b^2 x^2+1} \left (2 a^3+6 a^2 b x+6 a b^2 x^2-5 a+2 b^3 x^3-5 b x\right ) \sin ^{-1}(a+b x)^2+8 b x \left (4 a^3+6 a^2 b x+4 a b^2 x^2-10 a+b^3 x^3-5 b x\right ) \sin ^{-1}(a+b x)+8 \sin ^{-1}(a+b x)^3}{64 b} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.58, size = 185, normalized size = 0.93 \[ \frac {8 \, \arcsin \left (b x + a\right )^{3} + {\left (8 \, b^{4} x^{4} + 32 \, a b^{3} x^{3} + 8 \, {\left (6 \, a^{2} - 5\right )} b^{2} x^{2} + 8 \, a^{4} + 16 \, {\left (2 \, a^{3} - 5 \, a\right )} b x - 40 \, a^{2} + 17\right )} \arcsin \left (b x + a\right ) + {\left (2 \, b^{3} x^{3} + 6 \, a b^{2} x^{2} + 2 \, a^{3} + {\left (6 \, a^{2} - 17\right )} b x - 8 \, {\left (2 \, b^{3} x^{3} + 6 \, a b^{2} x^{2} + 2 \, a^{3} + {\left (6 \, a^{2} - 5\right )} b x - 5 \, a\right )} \arcsin \left (b x + a\right )^{2} - 17 \, a\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{64 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.48, size = 227, normalized size = 1.14 \[ \frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}} {\left (b x + a\right )} \arcsin \left (b x + a\right )^{2}}{4 \, b} + \frac {3 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )} \arcsin \left (b x + a\right )^{2}}{8 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}^{2} \arcsin \left (b x + a\right )}{8 \, b} + \frac {\arcsin \left (b x + a\right )^{3}}{8 \, b} - \frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}} {\left (b x + a\right )}}{32 \, b} - \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )} \arcsin \left (b x + a\right )}{8 \, b} - \frac {15 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{64 \, b} - \frac {15 \, \arcsin \left (b x + a\right )}{64 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.22, size = 515, normalized size = 2.59 \[ \frac {-16 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, \arcsin \left (b x +a \right )^{2} x^{3} b^{3}+8 \arcsin \left (b x +a \right ) x^{4} b^{4}-48 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, \arcsin \left (b x +a \right )^{2} x^{2} a \,b^{2}+32 \arcsin \left (b x +a \right ) x^{3} a \,b^{3}-48 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, \arcsin \left (b x +a \right )^{2} x \,a^{2} b +48 \arcsin \left (b x +a \right ) x^{2} a^{2} b^{2}+2 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, x^{3} b^{3}-16 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, \arcsin \left (b x +a \right )^{2} a^{3}+32 \arcsin \left (b x +a \right ) x \,a^{3} b +6 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, x^{2} a \,b^{2}+40 \arcsin \left (b x +a \right )^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, x b -40 \arcsin \left (b x +a \right ) x^{2} b^{2}+8 \arcsin \left (b x +a \right ) a^{4}+6 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, x \,a^{2} b +40 \arcsin \left (b x +a \right )^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a -80 \arcsin \left (b x +a \right ) x a b +2 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a^{3}+8 \arcsin \left (b x +a \right )^{3}-40 \arcsin \left (b x +a \right ) a^{2}-17 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, x b -17 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a +17 \arcsin \left (b x +a \right )}{64 b} \]
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 \[ \int {\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}} \arcsin \left (b x + a\right )^{2}\,{d x} \]
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 \[ \int {\mathrm {asin}\left (a+b\,x\right )}^2\,{\left (-a^2-2\,a\,b\,x-b^2\,x^2+1\right )}^{3/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 8.42, size = 568, normalized size = 2.85 \[ \begin {cases} \frac {a^{4} \operatorname {asin}{\left (a + b x \right )}}{8 b} + \frac {a^{3} x \operatorname {asin}{\left (a + b x \right )}}{2} - \frac {a^{3} \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname {asin}^{2}{\left (a + b x \right )}}{4 b} + \frac {a^{3} \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{32 b} + \frac {3 a^{2} b x^{2} \operatorname {asin}{\left (a + b x \right )}}{4} - \frac {3 a^{2} x \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname {asin}^{2}{\left (a + b x \right )}}{4} + \frac {3 a^{2} x \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{32} - \frac {5 a^{2} \operatorname {asin}{\left (a + b x \right )}}{8 b} + \frac {a b^{2} x^{3} \operatorname {asin}{\left (a + b x \right )}}{2} - \frac {3 a b x^{2} \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname {asin}^{2}{\left (a + b x \right )}}{4} + \frac {3 a b x^{2} \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{32} - \frac {5 a x \operatorname {asin}{\left (a + b x \right )}}{4} + \frac {5 a \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname {asin}^{2}{\left (a + b x \right )}}{8 b} - \frac {17 a \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{64 b} + \frac {b^{3} x^{4} \operatorname {asin}{\left (a + b x \right )}}{8} - \frac {b^{2} x^{3} \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname {asin}^{2}{\left (a + b x \right )}}{4} + \frac {b^{2} x^{3} \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{32} - \frac {5 b x^{2} \operatorname {asin}{\left (a + b x \right )}}{8} + \frac {5 x \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname {asin}^{2}{\left (a + b x \right )}}{8} - \frac {17 x \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{64} + \frac {\operatorname {asin}^{3}{\left (a + b x \right )}}{8 b} + \frac {17 \operatorname {asin}{\left (a + b x \right )}}{64 b} & \text {for}\: b \neq 0 \\x \left (1 - a^{2}\right )^{\frac {3}{2}} \operatorname {asin}^{2}{\relax (a )} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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