Optimal. Leaf size=198 \[ -\frac {3 b^3 e (c+d x) \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{2 d}-\frac {3 b^2 e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d}+\frac {3 b^2 e \left (a+b \sin ^{-1}(c+d x)\right )^2}{4 d}+\frac {b e (c+d x) \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d}+\frac {e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d}-\frac {e \left (a+b \sin ^{-1}(c+d x)\right )^4}{4 d}+\frac {3 b^4 e (c+d x)^2}{4 d} \]
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Rubi [A] time = 0.31, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4805, 12, 4627, 4707, 4641, 30} \[ -\frac {3 b^3 e (c+d x) \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{2 d}-\frac {3 b^2 e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d}+\frac {3 b^2 e \left (a+b \sin ^{-1}(c+d x)\right )^2}{4 d}+\frac {b e (c+d x) \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d}+\frac {e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d}-\frac {e \left (a+b \sin ^{-1}(c+d x)\right )^4}{4 d}+\frac {3 b^4 e (c+d x)^2}{4 d} \]
Antiderivative was successfully verified.
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Rule 12
Rule 30
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 )^4 \, dx &=\frac {\operatorname {Subst}\left (\int e x \left (a+b \sin ^{-1}(x)\right )^4 \, dx,x,c+d x\right )}{d}\\ &=\frac {e \operatorname {Subst}\left (\int x \left (a+b \sin ^{-1}(x)\right )^4 \, dx,x,c+d x\right )}{d}\\ &=\frac {e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d}-\frac {(2 b e) \operatorname {Subst}\left (\int \frac {x^2 \left (a+b \sin ^{-1}(x)\right )^3}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{d}\\ &=\frac {b e (c+d x) \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d}+\frac {e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d}-\frac {(b e) \operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^3}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{d}-\frac {\left (3 b^2 e\right ) \operatorname {Subst}\left (\int x \left (a+b \sin ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=-\frac {3 b^2 e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d}+\frac {b e (c+d x) \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d}-\frac {e \left (a+b \sin ^{-1}(c+d x)\right )^4}{4 d}+\frac {e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d}+\frac {\left (3 b^3 e\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (a+b \sin ^{-1}(x)\right )}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {3 b^3 e (c+d x) \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{2 d}-\frac {3 b^2 e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d}+\frac {b e (c+d x) \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d}-\frac {e \left (a+b \sin ^{-1}(c+d x)\right )^4}{4 d}+\frac {e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d}+\frac {\left (3 b^3 e\right ) \operatorname {Subst}\left (\int \frac {a+b \sin ^{-1}(x)}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{2 d}+\frac {\left (3 b^4 e\right ) \operatorname {Subst}(\int x \, dx,x,c+d x)}{2 d}\\ &=\frac {3 b^4 e (c+d x)^2}{4 d}-\frac {3 b^3 e (c+d x) \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{2 d}+\frac {3 b^2 e \left (a+b \sin ^{-1}(c+d x)\right )^2}{4 d}-\frac {3 b^2 e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d}+\frac {b e (c+d x) \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d}-\frac {e \left (a+b \sin ^{-1}(c+d x)\right )^4}{4 d}+\frac {e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d}\\ \end {align*}
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Mathematica [A] time = 0.32, size = 163, normalized size = 0.82 \[ -\frac {e \left (3 b^2 \left (2 (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^2+2 b \sqrt {1-(c+d x)^2} (c+d x) \left (a+b \sin ^{-1}(c+d x)\right )-\left (a+b \sin ^{-1}(c+d x)\right )^2-b^2 (c+d x)^2\right )-2 (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^4+\left (a+b \sin ^{-1}(c+d x)\right )^4-4 b (c+d x) \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3\right )}{4 d} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.46, size = 483, normalized size = 2.44 \[ \frac {{\left (2 \, a^{4} - 6 \, a^{2} b^{2} + 3 \, b^{4}\right )} d^{2} e x^{2} + 2 \, {\left (2 \, a^{4} - 6 \, a^{2} b^{2} + 3 \, b^{4}\right )} c d e x + {\left (2 \, b^{4} d^{2} e x^{2} + 4 \, b^{4} c d e x + {\left (2 \, b^{4} c^{2} - b^{4}\right )} e\right )} \arcsin \left (d x + c\right )^{4} + 4 \, {\left (2 \, a b^{3} d^{2} e x^{2} + 4 \, a b^{3} c d e x + {\left (2 \, a b^{3} c^{2} - a b^{3}\right )} e\right )} \arcsin \left (d x + c\right )^{3} + 3 \, {\left (2 \, {\left (2 \, a^{2} b^{2} - b^{4}\right )} d^{2} e x^{2} + 4 \, {\left (2 \, a^{2} b^{2} - b^{4}\right )} c d e x - {\left (2 \, a^{2} b^{2} - b^{4} - 2 \, {\left (2 \, a^{2} b^{2} - b^{4}\right )} c^{2}\right )} e\right )} \arcsin \left (d x + c\right )^{2} + 2 \, {\left (2 \, {\left (2 \, a^{3} b - 3 \, a b^{3}\right )} d^{2} e x^{2} + 4 \, {\left (2 \, a^{3} b - 3 \, a b^{3}\right )} c d e x - {\left (2 \, a^{3} b - 3 \, a b^{3} - 2 \, {\left (2 \, a^{3} b - 3 \, a b^{3}\right )} c^{2}\right )} e\right )} \arcsin \left (d x + c\right ) + 2 \, {\left ({\left (2 \, a^{3} b - 3 \, a b^{3}\right )} d e x + 2 \, {\left (b^{4} d e x + b^{4} c e\right )} \arcsin \left (d x + c\right )^{3} + {\left (2 \, a^{3} b - 3 \, a b^{3}\right )} c e + 6 \, {\left (a b^{3} d e x + a b^{3} c e\right )} \arcsin \left (d x + c\right )^{2} + 3 \, {\left ({\left (2 \, a^{2} b^{2} - b^{4}\right )} d e x + {\left (2 \, a^{2} b^{2} - b^{4}\right )} c e\right )} \arcsin \left (d x + c\right )\right )} \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}}{4 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.35, size = 556, normalized size = 2.81 \[ \frac {{\left ({\left (d x + c\right )}^{2} - 1\right )} b^{4} \arcsin \left (d x + c\right )^{4} e}{2 \, d} + \frac {\sqrt {-{\left (d x + c\right )}^{2} + 1} {\left (d x + c\right )} b^{4} \arcsin \left (d x + c\right )^{3} e}{d} + \frac {2 \, {\left ({\left (d x + c\right )}^{2} - 1\right )} a b^{3} \arcsin \left (d x + c\right )^{3} e}{d} + \frac {b^{4} \arcsin \left (d x + c\right )^{4} e}{4 \, d} + \frac {3 \, \sqrt {-{\left (d x + c\right )}^{2} + 1} {\left (d x + c\right )} a b^{3} \arcsin \left (d x + c\right )^{2} e}{d} + \frac {3 \, {\left ({\left (d x + c\right )}^{2} - 1\right )} a^{2} b^{2} \arcsin \left (d x + c\right )^{2} e}{d} - \frac {3 \, {\left ({\left (d x + c\right )}^{2} - 1\right )} b^{4} \arcsin \left (d x + c\right )^{2} e}{2 \, d} + \frac {a b^{3} \arcsin \left (d x + c\right )^{3} e}{d} + \frac {3 \, \sqrt {-{\left (d x + c\right )}^{2} + 1} {\left (d x + c\right )} a^{2} b^{2} \arcsin \left (d x + c\right ) e}{d} - \frac {3 \, \sqrt {-{\left (d x + c\right )}^{2} + 1} {\left (d x + c\right )} b^{4} \arcsin \left (d x + c\right ) e}{2 \, d} + \frac {2 \, {\left ({\left (d x + c\right )}^{2} - 1\right )} a^{3} b \arcsin \left (d x + c\right ) e}{d} - \frac {3 \, {\left ({\left (d x + c\right )}^{2} - 1\right )} a b^{3} \arcsin \left (d x + c\right ) e}{d} + \frac {3 \, a^{2} b^{2} \arcsin \left (d x + c\right )^{2} e}{2 \, d} - \frac {3 \, b^{4} \arcsin \left (d x + c\right )^{2} e}{4 \, d} + \frac {\sqrt {-{\left (d x + c\right )}^{2} + 1} {\left (d x + c\right )} a^{3} b e}{d} - \frac {3 \, \sqrt {-{\left (d x + c\right )}^{2} + 1} {\left (d x + c\right )} a b^{3} e}{2 \, d} + \frac {{\left ({\left (d x + c\right )}^{2} - 1\right )} a^{4} e}{2 \, d} - \frac {3 \, {\left ({\left (d x + c\right )}^{2} - 1\right )} a^{2} b^{2} e}{2 \, d} + \frac {3 \, {\left ({\left (d x + c\right )}^{2} - 1\right )} b^{4} e}{4 \, d} + \frac {a^{3} b \arcsin \left (d x + c\right ) e}{d} - \frac {3 \, a b^{3} \arcsin \left (d x + c\right ) e}{2 \, d} - \frac {3 \, a^{2} b^{2} e}{4 \, d} + \frac {3 \, b^{4} e}{8 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.06, size = 412, normalized size = 2.08 \[ \frac {\frac {e \left (d x +c \right )^{2} a^{4}}{2}+e \,b^{4} \left (\frac {\left (\left (d x +c \right )^{2}-1\right ) \arcsin \left (d x +c \right )^{4}}{2}+\arcsin \left (d x +c \right )^{3} \left (\left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}+\arcsin \left (d x +c \right )\right )-\frac {3 \arcsin \left (d x +c \right )^{2} \left (\left (d x +c \right )^{2}-1\right )}{2}-\frac {3 \arcsin \left (d x +c \right ) \left (\left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}+\arcsin \left (d x +c \right )\right )}{2}+\frac {3 \arcsin \left (d x +c \right )^{2}}{4}+\frac {3 \left (d x +c \right )^{2}}{4}-\frac {3 \arcsin \left (d x +c \right )^{4}}{4}\right )+4 e a \,b^{3} \left (\frac {\arcsin \left (d x +c \right )^{3} \left (\left (d x +c \right )^{2}-1\right )}{2}+\frac {3 \arcsin \left (d x +c \right )^{2} \left (\left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}+\arcsin \left (d x +c \right )\right )}{4}-\frac {3 \arcsin \left (d x +c \right ) \left (\left (d x +c \right )^{2}-1\right )}{4}-\frac {3 \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}}{8}-\frac {3 \arcsin \left (d x +c \right )}{8}-\frac {\arcsin \left (d x +c \right )^{3}}{2}\right )+6 e \,a^{2} b^{2} \left (\frac {\arcsin \left (d x +c \right )^{2} \left (\left (d x +c \right )^{2}-1\right )}{2}+\frac {\arcsin \left (d x +c \right ) \left (\left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}+\arcsin \left (d x +c \right )\right )}{2}-\frac {\arcsin \left (d x +c \right )^{2}}{4}-\frac {\left (d x +c \right )^{2}}{4}\right )+4 e \,a^{3} b \left (\frac {\arcsin \left (d x +c \right ) \left (d x +c \right )^{2}}{2}+\frac {\left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}}{4}-\frac {\arcsin \left (d x +c \right )}{4}\right )}{d} \]
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 \[ \frac {1}{2} \, a^{4} d e x^{2} + {\left (2 \, x^{2} \arcsin \left (d x + c\right ) + d {\left (\frac {3 \, c^{2} \arcsin \left (-\frac {d^{2} x + c d}{\sqrt {c^{2} d^{2} - {\left (c^{2} - 1\right )} d^{2}}}\right )}{d^{3}} + \frac {\sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} x}{d^{2}} - \frac {{\left (c^{2} - 1\right )} \arcsin \left (-\frac {d^{2} x + c d}{\sqrt {c^{2} d^{2} - {\left (c^{2} - 1\right )} d^{2}}}\right )}{d^{3}} - \frac {3 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} c}{d^{3}}\right )}\right )} a^{3} b d e + a^{4} c e x + \frac {4 \, {\left ({\left (d x + c\right )} \arcsin \left (d x + c\right ) + \sqrt {-{\left (d x + c\right )}^{2} + 1}\right )} a^{3} b c e}{d} + \frac {1}{2} \, {\left (b^{4} d e x^{2} + 2 \, b^{4} c e x\right )} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{4} + \int \frac {2 \, {\left ({\left (b^{4} d^{2} e x^{2} + 2 \, b^{4} c d e x\right )} \sqrt {d x + c + 1} \sqrt {-d x - c + 1} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{3} + 2 \, {\left (a b^{3} d^{3} e x^{3} + 3 \, a b^{3} c d^{2} e x^{2} + {\left (3 \, a b^{3} c^{2} - a b^{3}\right )} d e x + {\left (a b^{3} c^{3} - a b^{3} c\right )} e\right )} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{3} + 3 \, {\left (a^{2} b^{2} d^{3} e x^{3} + 3 \, a^{2} b^{2} c d^{2} e x^{2} + {\left (3 \, a^{2} b^{2} c^{2} - a^{2} b^{2}\right )} d e x + {\left (a^{2} b^{2} c^{3} - a^{2} b^{2} c\right )} e\right )} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{2}\right )}}{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\,{d x} \]
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 \[ \int \left (c\,e+d\,e\,x\right )\,{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^4 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 4.36, size = 1027, normalized size = 5.19 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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