Optimal. Leaf size=140 \[ \frac {e^2 (c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )^2}{3 d}+\frac {2 b e^2 \sqrt {1-(c+d x)^2} (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )}{9 d}+\frac {4 b e^2 \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{9 d}-\frac {2 b^2 e^2 (c+d x)^3}{27 d}-\frac {4}{9} b^2 e^2 x \]
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Rubi [A] time = 0.21, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4805, 12, 4627, 4707, 4677, 8, 30} \[ \frac {e^2 (c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )^2}{3 d}+\frac {2 b e^2 \sqrt {1-(c+d x)^2} (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )}{9 d}+\frac {4 b e^2 \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{9 d}-\frac {2 b^2 e^2 (c+d x)^3}{27 d}-\frac {4}{9} b^2 e^2 x \]
Antiderivative was successfully verified.
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Rule 8
Rule 12
Rule 30
Rule 4627
Rule 4677
Rule 4707
Rule 4805
Rubi steps
\begin {align*} \int (c e+d e x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \, dx &=\frac {\operatorname {Subst}\left (\int e^2 x^2 \left (a+b \sin ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {e^2 \operatorname {Subst}\left (\int x^2 \left (a+b \sin ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {e^2 (c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )^2}{3 d}-\frac {\left (2 b e^2\right ) \operatorname {Subst}\left (\int \frac {x^3 \left (a+b \sin ^{-1}(x)\right )}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{3 d}\\ &=\frac {2 b e^2 (c+d x)^2 \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{9 d}+\frac {e^2 (c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )^2}{3 d}-\frac {\left (4 b e^2\right ) \operatorname {Subst}\left (\int \frac {x \left (a+b \sin ^{-1}(x)\right )}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{9 d}-\frac {\left (2 b^2 e^2\right ) \operatorname {Subst}\left (\int x^2 \, dx,x,c+d x\right )}{9 d}\\ &=-\frac {2 b^2 e^2 (c+d x)^3}{27 d}+\frac {4 b e^2 \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{9 d}+\frac {2 b e^2 (c+d x)^2 \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{9 d}+\frac {e^2 (c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )^2}{3 d}-\frac {\left (4 b^2 e^2\right ) \operatorname {Subst}(\int 1 \, dx,x,c+d x)}{9 d}\\ &=-\frac {4}{9} b^2 e^2 x-\frac {2 b^2 e^2 (c+d x)^3}{27 d}+\frac {4 b e^2 \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{9 d}+\frac {2 b e^2 (c+d x)^2 \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{9 d}+\frac {e^2 (c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )^2}{3 d}\\ \end {align*}
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Mathematica [A] time = 0.25, size = 112, normalized size = 0.80 \[ \frac {e^2 \left ((c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )^2-\frac {2}{9} b \left (-3 \sqrt {1-(c+d x)^2} (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )-6 \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )+b (c+d x)^3+6 b d x\right )\right )}{3 d} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.47, size = 307, normalized size = 2.19 \[ \frac {{\left (9 \, a^{2} - 2 \, b^{2}\right )} d^{3} e^{2} x^{3} + 3 \, {\left (9 \, a^{2} - 2 \, b^{2}\right )} c d^{2} e^{2} x^{2} + 3 \, {\left ({\left (9 \, a^{2} - 2 \, b^{2}\right )} c^{2} - 4 \, b^{2}\right )} d e^{2} x + 9 \, {\left (b^{2} d^{3} e^{2} x^{3} + 3 \, b^{2} c d^{2} e^{2} x^{2} + 3 \, b^{2} c^{2} d e^{2} x + b^{2} c^{3} e^{2}\right )} \arcsin \left (d x + c\right )^{2} + 18 \, {\left (a b d^{3} e^{2} x^{3} + 3 \, a b c d^{2} e^{2} x^{2} + 3 \, a b c^{2} d e^{2} x + a b c^{3} e^{2}\right )} \arcsin \left (d x + c\right ) + 6 \, {\left (a b d^{2} e^{2} x^{2} + 2 \, a b c d e^{2} x + {\left (a b c^{2} + 2 \, a b\right )} e^{2} + {\left (b^{2} d^{2} e^{2} x^{2} + 2 \, b^{2} c d e^{2} x + {\left (b^{2} c^{2} + 2 \, b^{2}\right )} e^{2}\right )} \arcsin \left (d x + c\right )\right )} \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}}{27 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.26, size = 263, normalized size = 1.88 \[ \frac {{\left ({\left (d x + c\right )}^{2} - 1\right )} {\left (d x + c\right )} b^{2} \arcsin \left (d x + c\right )^{2} e^{2}}{3 \, d} + \frac {{\left (d x + c\right )}^{3} a^{2} e^{2}}{3 \, d} + \frac {2 \, {\left ({\left (d x + c\right )}^{2} - 1\right )} {\left (d x + c\right )} a b \arcsin \left (d x + c\right ) e^{2}}{3 \, d} + \frac {{\left (d x + c\right )} b^{2} \arcsin \left (d x + c\right )^{2} e^{2}}{3 \, d} - \frac {2 \, {\left (-{\left (d x + c\right )}^{2} + 1\right )}^{\frac {3}{2}} b^{2} \arcsin \left (d x + c\right ) e^{2}}{9 \, d} - \frac {2 \, {\left ({\left (d x + c\right )}^{2} - 1\right )} {\left (d x + c\right )} b^{2} e^{2}}{27 \, d} + \frac {2 \, {\left (d x + c\right )} a b \arcsin \left (d x + c\right ) e^{2}}{3 \, d} - \frac {2 \, {\left (-{\left (d x + c\right )}^{2} + 1\right )}^{\frac {3}{2}} a b e^{2}}{9 \, d} + \frac {2 \, \sqrt {-{\left (d x + c\right )}^{2} + 1} b^{2} \arcsin \left (d x + c\right ) e^{2}}{3 \, d} - \frac {14 \, {\left (d x + c\right )} b^{2} e^{2}}{27 \, d} + \frac {2 \, \sqrt {-{\left (d x + c\right )}^{2} + 1} a b e^{2}}{3 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 152, normalized size = 1.09 \[ \frac {\frac {e^{2} \left (d x +c \right )^{3} a^{2}}{3}+e^{2} b^{2} \left (\frac {\arcsin \left (d x +c \right )^{2} \left (d x +c \right )^{3}}{3}+\frac {2 \arcsin \left (d x +c \right ) \left (\left (d x +c \right )^{2}+2\right ) \sqrt {1-\left (d x +c \right )^{2}}}{9}-\frac {2 \left (d x +c \right )^{3}}{27}-\frac {4 d x}{9}-\frac {4 c}{9}\right )+2 e^{2} a b \left (\frac {\arcsin \left (d x +c \right ) \left (d x +c \right )^{3}}{3}+\frac {\left (d x +c \right )^{2} \sqrt {1-\left (d x +c \right )^{2}}}{9}+\frac {2 \sqrt {1-\left (d x +c \right )^{2}}}{9}\right )}{d} \]
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 \[ \frac {1}{3} \, a^{2} d^{2} e^{2} x^{3} + a^{2} c d e^{2} x^{2} + {\left (2 \, x^{2} \arcsin \left (d x + c\right ) + d {\left (\frac {3 \, c^{2} \arcsin \left (-\frac {d^{2} x + c d}{\sqrt {c^{2} d^{2} - {\left (c^{2} - 1\right )} d^{2}}}\right )}{d^{3}} + \frac {\sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} x}{d^{2}} - \frac {{\left (c^{2} - 1\right )} \arcsin \left (-\frac {d^{2} x + c d}{\sqrt {c^{2} d^{2} - {\left (c^{2} - 1\right )} d^{2}}}\right )}{d^{3}} - \frac {3 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} c}{d^{3}}\right )}\right )} a b c d e^{2} + \frac {1}{9} \, {\left (6 \, x^{3} \arcsin \left (d x + c\right ) + d {\left (\frac {2 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} x^{2}}{d^{2}} - \frac {15 \, c^{3} \arcsin \left (-\frac {d^{2} x + c d}{\sqrt {c^{2} d^{2} - {\left (c^{2} - 1\right )} d^{2}}}\right )}{d^{4}} - \frac {5 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} c x}{d^{3}} + \frac {9 \, {\left (c^{2} - 1\right )} c \arcsin \left (-\frac {d^{2} x + c d}{\sqrt {c^{2} d^{2} - {\left (c^{2} - 1\right )} d^{2}}}\right )}{d^{4}} + \frac {15 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} c^{2}}{d^{4}} - \frac {4 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} {\left (c^{2} - 1\right )}}{d^{4}}\right )}\right )} a b d^{2} e^{2} + a^{2} c^{2} e^{2} x + \frac {2 \, {\left ({\left (d x + c\right )} \arcsin \left (d x + c\right ) + \sqrt {-{\left (d x + c\right )}^{2} + 1}\right )} a b c^{2} e^{2}}{d} + \frac {1}{3} \, {\left (b^{2} d^{2} e^{2} x^{3} + 3 \, b^{2} c d e^{2} x^{2} + 3 \, b^{2} c^{2} e^{2} x\right )} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{2} + \int \frac {2 \, {\left (b^{2} d^{3} e^{2} x^{3} + 3 \, b^{2} c d^{2} e^{2} x^{2} + 3 \, b^{2} c^{2} d e^{2} x\right )} \sqrt {d x + c + 1} \sqrt {-d x - c + 1} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )}{3 \, {\left (d^{2} x^{2} + 2 \, c d x + c^{2} - 1\right )}}\,{d x} \]
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 \[ \int {\left (c\,e+d\,e\,x\right )}^2\,{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.78, size = 610, normalized size = 4.36 \[ \begin {cases} a^{2} c^{2} e^{2} x + a^{2} c d e^{2} x^{2} + \frac {a^{2} d^{2} e^{2} x^{3}}{3} + \frac {2 a b c^{3} e^{2} \operatorname {asin}{\left (c + d x \right )}}{3 d} + 2 a b c^{2} e^{2} x \operatorname {asin}{\left (c + d x \right )} + \frac {2 a b c^{2} e^{2} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{9 d} + 2 a b c d e^{2} x^{2} \operatorname {asin}{\left (c + d x \right )} + \frac {4 a b c e^{2} x \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{9} + \frac {2 a b d^{2} e^{2} x^{3} \operatorname {asin}{\left (c + d x \right )}}{3} + \frac {2 a b d e^{2} x^{2} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{9} + \frac {4 a b e^{2} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{9 d} + \frac {b^{2} c^{3} e^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}{3 d} + b^{2} c^{2} e^{2} x \operatorname {asin}^{2}{\left (c + d x \right )} - \frac {2 b^{2} c^{2} e^{2} x}{9} + \frac {2 b^{2} c^{2} e^{2} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1} \operatorname {asin}{\left (c + d x \right )}}{9 d} + b^{2} c d e^{2} x^{2} \operatorname {asin}^{2}{\left (c + d x \right )} - \frac {2 b^{2} c d e^{2} x^{2}}{9} + \frac {4 b^{2} c e^{2} x \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1} \operatorname {asin}{\left (c + d x \right )}}{9} + \frac {b^{2} d^{2} e^{2} x^{3} \operatorname {asin}^{2}{\left (c + d x \right )}}{3} - \frac {2 b^{2} d^{2} e^{2} x^{3}}{27} + \frac {2 b^{2} d e^{2} x^{2} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1} \operatorname {asin}{\left (c + d x \right )}}{9} - \frac {4 b^{2} e^{2} x}{9} + \frac {4 b^{2} e^{2} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1} \operatorname {asin}{\left (c + d x \right )}}{9 d} & \text {for}\: d \neq 0 \\c^{2} e^{2} x \left (a + b \operatorname {asin}{\relax (c )}\right )^{2} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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