Optimal. Leaf size=137 \[ -\frac {\left (4 a \left (19 a^2+16\right )-\left (26 a^2+9\right ) (a+b x)\right ) \sqrt {1-(a+b x)^2}}{96 b^4}-\frac {\left (8 a^4+24 a^2+3\right ) \sin ^{-1}(a+b x)}{32 b^4}-\frac {7 a x^2 \sqrt {1-(a+b x)^2}}{48 b^2}+\frac {1}{4} x^4 \sin ^{-1}(a+b x)+\frac {x^3 \sqrt {1-(a+b x)^2}}{16 b} \]
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Rubi [A] time = 0.20, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4805, 4743, 743, 833, 780, 216} \[ -\frac {\left (4 a \left (19 a^2+16\right )-\left (26 a^2+9\right ) (a+b x)\right ) \sqrt {1-(a+b x)^2}}{96 b^4}-\frac {\left (8 a^4+24 a^2+3\right ) \sin ^{-1}(a+b x)}{32 b^4}-\frac {7 a x^2 \sqrt {1-(a+b x)^2}}{48 b^2}+\frac {x^3 \sqrt {1-(a+b x)^2}}{16 b}+\frac {1}{4} x^4 \sin ^{-1}(a+b x) \]
Antiderivative was successfully verified.
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Rule 216
Rule 743
Rule 780
Rule 833
Rule 4743
Rule 4805
Rubi steps
\begin {align*} \int x^3 \sin ^{-1}(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \left (-\frac {a}{b}+\frac {x}{b}\right )^3 \sin ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac {1}{4} x^4 \sin ^{-1}(a+b x)-\frac {1}{4} \operatorname {Subst}\left (\int \frac {\left (-\frac {a}{b}+\frac {x}{b}\right )^4}{\sqrt {1-x^2}} \, dx,x,a+b x\right )\\ &=\frac {x^3 \sqrt {1-(a+b x)^2}}{16 b}+\frac {1}{4} x^4 \sin ^{-1}(a+b x)+\frac {1}{16} \operatorname {Subst}\left (\int \frac {\left (-\frac {3+4 a^2}{b^2}+\frac {7 a x}{b^2}\right ) \left (-\frac {a}{b}+\frac {x}{b}\right )^2}{\sqrt {1-x^2}} \, dx,x,a+b x\right )\\ &=-\frac {7 a x^2 \sqrt {1-(a+b x)^2}}{48 b^2}+\frac {x^3 \sqrt {1-(a+b x)^2}}{16 b}+\frac {1}{4} x^4 \sin ^{-1}(a+b x)-\frac {1}{48} \operatorname {Subst}\left (\int \frac {\left (-\frac {a \left (23+12 a^2\right )}{b^3}+\frac {\left (9+26 a^2\right ) x}{b^3}\right ) \left (-\frac {a}{b}+\frac {x}{b}\right )}{\sqrt {1-x^2}} \, dx,x,a+b x\right )\\ &=-\frac {7 a x^2 \sqrt {1-(a+b x)^2}}{48 b^2}+\frac {x^3 \sqrt {1-(a+b x)^2}}{16 b}-\frac {\left (4 a \left (16+19 a^2\right )-\left (9+26 a^2\right ) (a+b x)\right ) \sqrt {1-(a+b x)^2}}{96 b^4}+\frac {1}{4} x^4 \sin ^{-1}(a+b x)-\frac {\left (3+24 a^2+8 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,a+b x\right )}{32 b^4}\\ &=-\frac {7 a x^2 \sqrt {1-(a+b x)^2}}{48 b^2}+\frac {x^3 \sqrt {1-(a+b x)^2}}{16 b}-\frac {\left (4 a \left (16+19 a^2\right )-\left (9+26 a^2\right ) (a+b x)\right ) \sqrt {1-(a+b x)^2}}{96 b^4}-\frac {\left (3+24 a^2+8 a^4\right ) \sin ^{-1}(a+b x)}{32 b^4}+\frac {1}{4} x^4 \sin ^{-1}(a+b x)\\ \end {align*}
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Mathematica [A] time = 0.11, size = 99, normalized size = 0.72 \[ \frac {\sqrt {-a^2-2 a b x-b^2 x^2+1} \left (-50 a^3+26 a^2 b x-a \left (14 b^2 x^2+55\right )+6 b^3 x^3+9 b x\right )-3 \left (8 a^4+24 a^2-8 b^4 x^4+3\right ) \sin ^{-1}(a+b x)}{96 b^4} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.46, size = 93, normalized size = 0.68 \[ \frac {3 \, {\left (8 \, b^{4} x^{4} - 8 \, a^{4} - 24 \, a^{2} - 3\right )} \arcsin \left (b x + a\right ) + {\left (6 \, b^{3} x^{3} - 14 \, a b^{2} x^{2} - 50 \, a^{3} + {\left (26 \, a^{2} + 9\right )} b x - 55 \, a\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{96 \, b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.28, size = 284, normalized size = 2.07 \[ -\frac {{\left (b x + a\right )} a^{3} \arcsin \left (b x + a\right )}{b^{4}} - \frac {{\left ({\left (b x + a\right )}^{2} - 1\right )} {\left (b x + a\right )} a \arcsin \left (b x + a\right )}{b^{4}} + \frac {3 \, {\left ({\left (b x + a\right )}^{2} - 1\right )} a^{2} \arcsin \left (b x + a\right )}{2 \, b^{4}} + \frac {3 \, \sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (b x + a\right )} a^{2}}{4 \, b^{4}} - \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} a^{3}}{b^{4}} + \frac {{\left ({\left (b x + a\right )}^{2} - 1\right )}^{2} \arcsin \left (b x + a\right )}{4 \, b^{4}} - \frac {{\left (b x + a\right )} a \arcsin \left (b x + a\right )}{b^{4}} + \frac {3 \, a^{2} \arcsin \left (b x + a\right )}{4 \, b^{4}} - \frac {{\left (-{\left (b x + a\right )}^{2} + 1\right )}^{\frac {3}{2}} {\left (b x + a\right )}}{16 \, b^{4}} + \frac {{\left (-{\left (b x + a\right )}^{2} + 1\right )}^{\frac {3}{2}} a}{3 \, b^{4}} + \frac {{\left ({\left (b x + a\right )}^{2} - 1\right )} \arcsin \left (b x + a\right )}{2 \, b^{4}} + \frac {5 \, \sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (b x + a\right )}}{32 \, b^{4}} - \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} a}{b^{4}} + \frac {5 \, \arcsin \left (b x + a\right )}{32 \, b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.02, size = 213, normalized size = 1.55 \[ \frac {\frac {\arcsin \left (b x +a \right ) \left (b x +a \right )^{4}}{4}-\arcsin \left (b x +a \right ) \left (b x +a \right )^{3} a +\frac {3 \arcsin \left (b x +a \right ) \left (b x +a \right )^{2} a^{2}}{2}-\arcsin \left (b x +a \right ) \left (b x +a \right ) a^{3}+\frac {\left (b x +a \right )^{3} \sqrt {1-\left (b x +a \right )^{2}}}{16}+\frac {3 \left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}}{32}-\frac {3 \arcsin \left (b x +a \right )}{32}+a \left (-\frac {\left (b x +a \right )^{2} \sqrt {1-\left (b x +a \right )^{2}}}{3}-\frac {2 \sqrt {1-\left (b x +a \right )^{2}}}{3}\right )-\frac {3 a^{2} \left (-\frac {\left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}}{2}+\frac {\arcsin \left (b x +a \right )}{2}\right )}{2}-a^{3} \sqrt {1-\left (b x +a \right )^{2}}}{b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.46, size = 333, normalized size = 2.43 \[ \frac {1}{4} \, x^{4} \arcsin \left (b x + a\right ) + \frac {1}{96} \, {\left (\frac {6 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} x^{3}}{b^{2}} - \frac {14 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a x^{2}}{b^{3}} + \frac {105 \, a^{4} \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{b^{5}} + \frac {35 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a^{2} x}{b^{4}} - \frac {90 \, {\left (a^{2} - 1\right )} a^{2} \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{b^{5}} - \frac {105 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a^{3}}{b^{5}} - \frac {9 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{2} - 1\right )} x}{b^{4}} + \frac {9 \, {\left (a^{2} - 1\right )}^{2} \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{b^{5}} + \frac {55 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{2} - 1\right )} a}{b^{5}}\right )} b \]
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 \[ \int x^3\,\mathrm {asin}\left (a+b\,x\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.33, size = 255, normalized size = 1.86 \[ \begin {cases} - \frac {a^{4} \operatorname {asin}{\left (a + b x \right )}}{4 b^{4}} - \frac {25 a^{3} \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{48 b^{4}} + \frac {13 a^{2} x \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{48 b^{3}} - \frac {3 a^{2} \operatorname {asin}{\left (a + b x \right )}}{4 b^{4}} - \frac {7 a x^{2} \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{48 b^{2}} - \frac {55 a \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{96 b^{4}} + \frac {x^{4} \operatorname {asin}{\left (a + b x \right )}}{4} + \frac {x^{3} \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{16 b} + \frac {3 x \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{32 b^{3}} - \frac {3 \operatorname {asin}{\left (a + b x \right )}}{32 b^{4}} & \text {for}\: b \neq 0 \\\frac {x^{4} \operatorname {asin}{\relax (a )}}{4} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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