Optimal. Leaf size=165 \[ -\frac {3 b^2 e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )}{4 d}+\frac {3 b e (c+d x) \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{4 d}+\frac {e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^3}{2 d}-\frac {e \left (a+b \sin ^{-1}(c+d x)\right )^3}{4 d}-\frac {3 b^3 e (c+d x) \sqrt {1-(c+d x)^2}}{8 d}+\frac {3 b^3 e \sin ^{-1}(c+d x)}{8 d} \]
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Rubi [A] time = 0.21, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4805, 12, 4627, 4707, 4641, 321, 216} \[ -\frac {3 b^2 e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )}{4 d}+\frac {3 b e (c+d x) \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{4 d}+\frac {e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^3}{2 d}-\frac {e \left (a+b \sin ^{-1}(c+d x)\right )^3}{4 d}-\frac {3 b^3 e (c+d x) \sqrt {1-(c+d x)^2}}{8 d}+\frac {3 b^3 e \sin ^{-1}(c+d x)}{8 d} \]
Antiderivative was successfully verified.
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Rule 12
Rule 216
Rule 321
Rule 4627
Rule 4641
Rule 4707
Rule 4805
Rubi steps
\begin {align*} \int (c e+d e x) \left (a+b \sin ^{-1}(c+d x)\right )^3 \, dx &=\frac {\operatorname {Subst}\left (\int e x \left (a+b \sin ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac {e \operatorname {Subst}\left (\int x \left (a+b \sin ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac {e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^3}{2 d}-\frac {(3 b e) \operatorname {Subst}\left (\int \frac {x^2 \left (a+b \sin ^{-1}(x)\right )^2}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{2 d}\\ &=\frac {3 b e (c+d x) \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{4 d}+\frac {e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^3}{2 d}-\frac {(3 b e) \operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^2}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{4 d}-\frac {\left (3 b^2 e\right ) \operatorname {Subst}\left (\int x \left (a+b \sin ^{-1}(x)\right ) \, dx,x,c+d x\right )}{2 d}\\ &=-\frac {3 b^2 e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )}{4 d}+\frac {3 b e (c+d x) \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{4 d}-\frac {e \left (a+b \sin ^{-1}(c+d x)\right )^3}{4 d}+\frac {e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^3}{2 d}+\frac {\left (3 b^3 e\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{4 d}\\ &=-\frac {3 b^3 e (c+d x) \sqrt {1-(c+d x)^2}}{8 d}-\frac {3 b^2 e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )}{4 d}+\frac {3 b e (c+d x) \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{4 d}-\frac {e \left (a+b \sin ^{-1}(c+d x)\right )^3}{4 d}+\frac {e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^3}{2 d}+\frac {\left (3 b^3 e\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{8 d}\\ &=-\frac {3 b^3 e (c+d x) \sqrt {1-(c+d x)^2}}{8 d}+\frac {3 b^3 e \sin ^{-1}(c+d x)}{8 d}-\frac {3 b^2 e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )}{4 d}+\frac {3 b e (c+d x) \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{4 d}-\frac {e \left (a+b \sin ^{-1}(c+d x)\right )^3}{4 d}+\frac {e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^3}{2 d}\\ \end {align*}
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Mathematica [A] time = 0.24, size = 137, normalized size = 0.83 \[ \frac {e \left (-3 b^2 (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )+3 b (c+d x) \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2+2 (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^3-\left (a+b \sin ^{-1}(c+d x)\right )^3+\frac {3}{2} b^3 \left (\sin ^{-1}(c+d x)-(c+d x) \sqrt {1-(c+d x)^2}\right )\right )}{4 d} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.44, size = 329, normalized size = 1.99 \[ \frac {2 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} d^{2} e x^{2} + 4 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} c d e x + 2 \, {\left (2 \, b^{3} d^{2} e x^{2} + 4 \, b^{3} c d e x + {\left (2 \, b^{3} c^{2} - b^{3}\right )} e\right )} \arcsin \left (d x + c\right )^{3} + 6 \, {\left (2 \, a b^{2} d^{2} e x^{2} + 4 \, a b^{2} c d e x + {\left (2 \, a b^{2} c^{2} - a b^{2}\right )} e\right )} \arcsin \left (d x + c\right )^{2} + 3 \, {\left (2 \, {\left (2 \, a^{2} b - b^{3}\right )} d^{2} e x^{2} + 4 \, {\left (2 \, a^{2} b - b^{3}\right )} c d e x - {\left (2 \, a^{2} b - b^{3} - 2 \, {\left (2 \, a^{2} b - b^{3}\right )} c^{2}\right )} e\right )} \arcsin \left (d x + c\right ) + 3 \, {\left ({\left (2 \, a^{2} b - b^{3}\right )} d e x + {\left (2 \, a^{2} b - b^{3}\right )} c e + 2 \, {\left (b^{3} d e x + b^{3} c e\right )} \arcsin \left (d x + c\right )^{2} + 4 \, {\left (a b^{2} d e x + a b^{2} c e\right )} \arcsin \left (d x + c\right )\right )} \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}}{8 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.34, size = 355, normalized size = 2.15 \[ \frac {{\left ({\left (d x + c\right )}^{2} - 1\right )} b^{3} \arcsin \left (d x + c\right )^{3} e}{2 \, d} + \frac {3 \, \sqrt {-{\left (d x + c\right )}^{2} + 1} {\left (d x + c\right )} b^{3} \arcsin \left (d x + c\right )^{2} e}{4 \, d} + \frac {3 \, {\left ({\left (d x + c\right )}^{2} - 1\right )} a b^{2} \arcsin \left (d x + c\right )^{2} e}{2 \, d} + \frac {b^{3} \arcsin \left (d x + c\right )^{3} e}{4 \, d} + \frac {3 \, \sqrt {-{\left (d x + c\right )}^{2} + 1} {\left (d x + c\right )} a b^{2} \arcsin \left (d x + c\right ) e}{2 \, d} + \frac {3 \, {\left ({\left (d x + c\right )}^{2} - 1\right )} a^{2} b \arcsin \left (d x + c\right ) e}{2 \, d} - \frac {3 \, {\left ({\left (d x + c\right )}^{2} - 1\right )} b^{3} \arcsin \left (d x + c\right ) e}{4 \, d} + \frac {3 \, a b^{2} \arcsin \left (d x + c\right )^{2} e}{4 \, d} + \frac {3 \, \sqrt {-{\left (d x + c\right )}^{2} + 1} {\left (d x + c\right )} a^{2} b e}{4 \, d} - \frac {3 \, \sqrt {-{\left (d x + c\right )}^{2} + 1} {\left (d x + c\right )} b^{3} e}{8 \, d} + \frac {{\left ({\left (d x + c\right )}^{2} - 1\right )} a^{3} e}{2 \, d} - \frac {3 \, {\left ({\left (d x + c\right )}^{2} - 1\right )} a b^{2} e}{4 \, d} + \frac {3 \, a^{2} b \arcsin \left (d x + c\right ) e}{4 \, d} - \frac {3 \, b^{3} \arcsin \left (d x + c\right ) e}{8 \, d} - \frac {3 \, a b^{2} e}{8 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 266, normalized size = 1.61 \[ \frac {\frac {e \left (d x +c \right )^{2} a^{3}}{2}+e \,b^{3} \left (\frac {\arcsin \left (d x +c \right )^{3} \left (\left (d x +c \right )^{2}-1\right )}{2}+\frac {3 \arcsin \left (d x +c \right )^{2} \left (\left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}+\arcsin \left (d x +c \right )\right )}{4}-\frac {3 \arcsin \left (d x +c \right ) \left (\left (d x +c \right )^{2}-1\right )}{4}-\frac {3 \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}}{8}-\frac {3 \arcsin \left (d x +c \right )}{8}-\frac {\arcsin \left (d x +c \right )^{3}}{2}\right )+3 e a \,b^{2} \left (\frac {\arcsin \left (d x +c \right )^{2} \left (\left (d x +c \right )^{2}-1\right )}{2}+\frac {\arcsin \left (d x +c \right ) \left (\left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}+\arcsin \left (d x +c \right )\right )}{2}-\frac {\arcsin \left (d x +c \right )^{2}}{4}-\frac {\left (d x +c \right )^{2}}{4}\right )+3 e \,a^{2} b \left (\frac {\arcsin \left (d x +c \right ) \left (d x +c \right )^{2}}{2}+\frac {\left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}}{4}-\frac {\arcsin \left (d x +c \right )}{4}\right )}{d} \]
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 \[ \frac {1}{2} \, a^{3} d e x^{2} + \frac {3}{4} \, {\left (2 \, x^{2} \arcsin \left (d x + c\right ) + d {\left (\frac {3 \, c^{2} \arcsin \left (-\frac {d^{2} x + c d}{\sqrt {c^{2} d^{2} - {\left (c^{2} - 1\right )} d^{2}}}\right )}{d^{3}} + \frac {\sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} x}{d^{2}} - \frac {{\left (c^{2} - 1\right )} \arcsin \left (-\frac {d^{2} x + c d}{\sqrt {c^{2} d^{2} - {\left (c^{2} - 1\right )} d^{2}}}\right )}{d^{3}} - \frac {3 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} c}{d^{3}}\right )}\right )} a^{2} b d e + a^{3} c e x + \frac {3 \, {\left ({\left (d x + c\right )} \arcsin \left (d x + c\right ) + \sqrt {-{\left (d x + c\right )}^{2} + 1}\right )} a^{2} b c e}{d} + \frac {1}{2} \, {\left (b^{3} d e x^{2} + 2 \, b^{3} c e x\right )} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{3} + \int \frac {3 \, {\left ({\left (b^{3} d^{2} e x^{2} + 2 \, b^{3} c d e x\right )} \sqrt {d x + c + 1} \sqrt {-d x - c + 1} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{2} + 2 \, {\left (a b^{2} d^{3} e x^{3} + 3 \, a b^{2} c d^{2} e x^{2} + {\left (3 \, a b^{2} c^{2} - a b^{2}\right )} d e x + {\left (a b^{2} c^{3} - a b^{2} c\right )} e\right )} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{2}\right )}}{2 \, {\left (d^{2} x^{2} + 2 \, c d x + c^{2} - 1\right )}}\,{d x} \]
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 \[ \int \left (c\,e+d\,e\,x\right )\,{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^3 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.96, size = 685, normalized size = 4.15 \[ \begin {cases} a^{3} c e x + \frac {a^{3} d e x^{2}}{2} + \frac {3 a^{2} b c^{2} e \operatorname {asin}{\left (c + d x \right )}}{2 d} + 3 a^{2} b c e x \operatorname {asin}{\left (c + d x \right )} + \frac {3 a^{2} b c e \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{4 d} + \frac {3 a^{2} b d e x^{2} \operatorname {asin}{\left (c + d x \right )}}{2} + \frac {3 a^{2} b e x \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{4} - \frac {3 a^{2} b e \operatorname {asin}{\left (c + d x \right )}}{4 d} + \frac {3 a b^{2} c^{2} e \operatorname {asin}^{2}{\left (c + d x \right )}}{2 d} + 3 a b^{2} c e x \operatorname {asin}^{2}{\left (c + d x \right )} - \frac {3 a b^{2} c e x}{2} + \frac {3 a b^{2} c e \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1} \operatorname {asin}{\left (c + d x \right )}}{2 d} + \frac {3 a b^{2} d e x^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}{2} - \frac {3 a b^{2} d e x^{2}}{4} + \frac {3 a b^{2} e x \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1} \operatorname {asin}{\left (c + d x \right )}}{2} - \frac {3 a b^{2} e \operatorname {asin}^{2}{\left (c + d x \right )}}{4 d} + \frac {b^{3} c^{2} e \operatorname {asin}^{3}{\left (c + d x \right )}}{2 d} - \frac {3 b^{3} c^{2} e \operatorname {asin}{\left (c + d x \right )}}{4 d} + b^{3} c e x \operatorname {asin}^{3}{\left (c + d x \right )} - \frac {3 b^{3} c e x \operatorname {asin}{\left (c + d x \right )}}{2} + \frac {3 b^{3} c e \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1} \operatorname {asin}^{2}{\left (c + d x \right )}}{4 d} - \frac {3 b^{3} c e \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{8 d} + \frac {b^{3} d e x^{2} \operatorname {asin}^{3}{\left (c + d x \right )}}{2} - \frac {3 b^{3} d e x^{2} \operatorname {asin}{\left (c + d x \right )}}{4} + \frac {3 b^{3} e x \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1} \operatorname {asin}^{2}{\left (c + d x \right )}}{4} - \frac {3 b^{3} e x \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{8} - \frac {b^{3} e \operatorname {asin}^{3}{\left (c + d x \right )}}{4 d} + \frac {3 b^{3} e \operatorname {asin}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\c e x \left (a + b \operatorname {asin}{\relax (c )}\right )^{3} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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