Optimal. Leaf size=611 \[ -\frac {i c^2 e^{-\frac {i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \Gamma \left (n+1,-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{2 d^3}+\frac {i c^2 e^{\frac {i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \Gamma \left (n+1,\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{2 d^3}-\frac {i e^{-\frac {i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \Gamma \left (n+1,-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{8 d^3}+\frac {c 2^{-n-2} e^{-\frac {2 i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \Gamma \left (n+1,-\frac {2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{d^3}+\frac {i 3^{-n-1} e^{-\frac {3 i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \Gamma \left (n+1,-\frac {3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{8 d^3}+\frac {i e^{\frac {i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \Gamma \left (n+1,\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{8 d^3}+\frac {c 2^{-n-2} e^{\frac {2 i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \Gamma \left (n+1,\frac {2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{d^3}-\frac {i 3^{-n-1} e^{\frac {3 i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \Gamma \left (n+1,\frac {3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{8 d^3} \]
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Rubi [A] time = 1.12, antiderivative size = 611, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 9, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {4805, 4747, 6741, 12, 6742, 3307, 2181, 4406, 3308} \[ -\frac {i c^2 e^{-\frac {i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \text {Gamma}\left (n+1,-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{2 d^3}+\frac {i c^2 e^{\frac {i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \text {Gamma}\left (n+1,\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{2 d^3}-\frac {i e^{-\frac {i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \text {Gamma}\left (n+1,-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{8 d^3}+\frac {c 2^{-n-2} e^{-\frac {2 i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \text {Gamma}\left (n+1,-\frac {2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{d^3}+\frac {i 3^{-n-1} e^{-\frac {3 i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \text {Gamma}\left (n+1,-\frac {3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{8 d^3}+\frac {i e^{\frac {i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \text {Gamma}\left (n+1,\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{8 d^3}+\frac {c 2^{-n-2} e^{\frac {2 i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \text {Gamma}\left (n+1,\frac {2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{d^3}-\frac {i 3^{-n-1} e^{\frac {3 i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \text {Gamma}\left (n+1,\frac {3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{8 d^3} \]
Antiderivative was successfully verified.
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Rule 12
Rule 2181
Rule 3307
Rule 3308
Rule 4406
Rule 4747
Rule 4805
Rule 6741
Rule 6742
Rubi steps
\begin {align*} \int x^2 \left (a+b \sin ^{-1}(c+d x)\right )^n \, dx &=\frac {\operatorname {Subst}\left (\int \left (-\frac {c}{d}+\frac {x}{d}\right )^2 \left (a+b \sin ^{-1}(x)\right )^n \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int (a+b x)^n \cos (x) \left (-\frac {c}{d}+\frac {\sin (x)}{d}\right )^2 \, dx,x,\sin ^{-1}(c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {(a+b x)^n \cos (x) (c-\sin (x))^2}{d^2} \, dx,x,\sin ^{-1}(c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int (a+b x)^n \cos (x) (c-\sin (x))^2 \, dx,x,\sin ^{-1}(c+d x)\right )}{d^3}\\ &=\frac {\operatorname {Subst}\left (\int \left (c^2 (a+b x)^n \cos (x)-2 c (a+b x)^n \cos (x) \sin (x)+(a+b x)^n \cos (x) \sin ^2(x)\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d^3}\\ &=\frac {\operatorname {Subst}\left (\int (a+b x)^n \cos (x) \sin ^2(x) \, dx,x,\sin ^{-1}(c+d x)\right )}{d^3}-\frac {(2 c) \operatorname {Subst}\left (\int (a+b x)^n \cos (x) \sin (x) \, dx,x,\sin ^{-1}(c+d x)\right )}{d^3}+\frac {c^2 \operatorname {Subst}\left (\int (a+b x)^n \cos (x) \, dx,x,\sin ^{-1}(c+d x)\right )}{d^3}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{4} (a+b x)^n \cos (x)-\frac {1}{4} (a+b x)^n \cos (3 x)\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d^3}-\frac {(2 c) \operatorname {Subst}\left (\int \frac {1}{2} (a+b x)^n \sin (2 x) \, dx,x,\sin ^{-1}(c+d x)\right )}{d^3}+\frac {c^2 \operatorname {Subst}\left (\int e^{-i x} (a+b x)^n \, dx,x,\sin ^{-1}(c+d x)\right )}{2 d^3}+\frac {c^2 \operatorname {Subst}\left (\int e^{i x} (a+b x)^n \, dx,x,\sin ^{-1}(c+d x)\right )}{2 d^3}\\ &=-\frac {i c^2 e^{-\frac {i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \Gamma \left (1+n,-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{2 d^3}+\frac {i c^2 e^{\frac {i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \Gamma \left (1+n,\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{2 d^3}+\frac {\operatorname {Subst}\left (\int (a+b x)^n \cos (x) \, dx,x,\sin ^{-1}(c+d x)\right )}{4 d^3}-\frac {\operatorname {Subst}\left (\int (a+b x)^n \cos (3 x) \, dx,x,\sin ^{-1}(c+d x)\right )}{4 d^3}-\frac {c \operatorname {Subst}\left (\int (a+b x)^n \sin (2 x) \, dx,x,\sin ^{-1}(c+d x)\right )}{d^3}\\ &=-\frac {i c^2 e^{-\frac {i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \Gamma \left (1+n,-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{2 d^3}+\frac {i c^2 e^{\frac {i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \Gamma \left (1+n,\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{2 d^3}+\frac {\operatorname {Subst}\left (\int e^{-i x} (a+b x)^n \, dx,x,\sin ^{-1}(c+d x)\right )}{8 d^3}+\frac {\operatorname {Subst}\left (\int e^{i x} (a+b x)^n \, dx,x,\sin ^{-1}(c+d x)\right )}{8 d^3}-\frac {\operatorname {Subst}\left (\int e^{-3 i x} (a+b x)^n \, dx,x,\sin ^{-1}(c+d x)\right )}{8 d^3}-\frac {\operatorname {Subst}\left (\int e^{3 i x} (a+b x)^n \, dx,x,\sin ^{-1}(c+d x)\right )}{8 d^3}-\frac {(i c) \operatorname {Subst}\left (\int e^{-2 i x} (a+b x)^n \, dx,x,\sin ^{-1}(c+d x)\right )}{2 d^3}+\frac {(i c) \operatorname {Subst}\left (\int e^{2 i x} (a+b x)^n \, dx,x,\sin ^{-1}(c+d x)\right )}{2 d^3}\\ &=-\frac {i e^{-\frac {i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \Gamma \left (1+n,-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{8 d^3}-\frac {i c^2 e^{-\frac {i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \Gamma \left (1+n,-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{2 d^3}+\frac {i e^{\frac {i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \Gamma \left (1+n,\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{8 d^3}+\frac {i c^2 e^{\frac {i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \Gamma \left (1+n,\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{2 d^3}+\frac {2^{-2-n} c e^{-\frac {2 i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \Gamma \left (1+n,-\frac {2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{d^3}+\frac {2^{-2-n} c e^{\frac {2 i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \Gamma \left (1+n,\frac {2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{d^3}+\frac {i 3^{-1-n} e^{-\frac {3 i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \Gamma \left (1+n,-\frac {3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{8 d^3}-\frac {i 3^{-1-n} e^{\frac {3 i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \Gamma \left (1+n,\frac {3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{8 d^3}\\ \end {align*}
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Mathematica [A] time = 0.57, size = 419, normalized size = 0.69 \[ \frac {2^{-n-3} 3^{-n-1} e^{-\frac {3 i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2}{b^2}\right )^{-n} \left (i \left (4 c^2+1\right ) 2^n 3^{n+1} e^{\frac {4 i a}{b}} \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^n \Gamma \left (n+1,\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )-i \left (4 c^2+1\right ) 2^n 3^{n+1} e^{\frac {2 i a}{b}} \left (\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^n \Gamma \left (n+1,-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+2 c 3^{n+1} e^{\frac {5 i a}{b}} \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^n \Gamma \left (n+1,\frac {2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )-i 2^n e^{\frac {6 i a}{b}} \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^n \Gamma \left (n+1,\frac {3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+2 c 3^{n+1} e^{\frac {i a}{b}} \left (\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^n \Gamma \left (n+1,-\frac {2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+i 2^n \left (\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^n \Gamma \left (n+1,-\frac {3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )\right )}{d^3} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \arcsin \left (d x + c\right ) + a\right )}^{n} x^{2}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \arcsin \left (d x + c\right ) + a\right )}^{n} x^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.74, size = 0, normalized size = 0.00 \[ \int x^{2} \left (a +b \arcsin \left (d x +c \right )\right )^{n}\, dx \]
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 \[ \int {\left (b \arcsin \left (d x + c\right ) + a\right )}^{n} x^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int x^2\,{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^n \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \left (a + b \operatorname {asin}{\left (c + d x \right )}\right )^{n}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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