Optimal. Leaf size=611 \[ -\frac{i c^2 e^{-\frac{i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \text{Gamma}\left (n+1,-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{2 d^3}+\frac{i c^2 e^{\frac{i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \text{Gamma}\left (n+1,\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{2 d^3}-\frac{i e^{-\frac{i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \text{Gamma}\left (n+1,-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{8 d^3}+\frac{c 2^{-n-2} e^{-\frac{2 i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \text{Gamma}\left (n+1,-\frac{2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{d^3}+\frac{i 3^{-n-1} e^{-\frac{3 i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \text{Gamma}\left (n+1,-\frac{3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{8 d^3}+\frac{i e^{\frac{i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \text{Gamma}\left (n+1,\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{8 d^3}+\frac{c 2^{-n-2} e^{\frac{2 i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \text{Gamma}\left (n+1,\frac{2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{d^3}-\frac{i 3^{-n-1} e^{\frac{3 i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{-n} \text{Gamma}\left (n+1,\frac{3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{8 d^3} \]
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Rubi [A] time = 1.11638, antiderivative size = 611, normalized size of antiderivative = 1., 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 4805
Rule 4747
Rule 6741
Rule 12
Rule 6742
Rule 3307
Rule 2181
Rule 4406
Rule 3308
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*}
Mathematica [A] time = 0.503547, size = 419, normalized size = 0.69 \[ \frac{2^{-n-3} 3^{-n-1} e^{-\frac{3 i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^n \left (\frac{\left (a+b \sin ^{-1}(c+d x)\right )^2}{b^2}\right )^{-n} \left (i \left (4 c^2+1\right ) 2^n 3^{n+1} e^{\frac{4 i a}{b}} \left (-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^n \text{Gamma}\left (n+1,\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )-i \left (4 c^2+1\right ) 2^n 3^{n+1} e^{\frac{2 i a}{b}} \left (\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^n \text{Gamma}\left (n+1,-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+2 c 3^{n+1} e^{\frac{5 i a}{b}} \left (-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^n \text{Gamma}\left (n+1,\frac{2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )-i 2^n e^{\frac{6 i a}{b}} \left (-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^n \text{Gamma}\left (n+1,\frac{3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+2 c 3^{n+1} e^{\frac{i a}{b}} \left (\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^n \text{Gamma}\left (n+1,-\frac{2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+i 2^n \left (\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^n \text{Gamma}\left (n+1,-\frac{3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )\right )}{d^3} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.405, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \left ( a+b\arcsin \left ( dx+c \right ) \right ) ^{n}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \arcsin \left (d x + c\right ) + a\right )}^{n} x^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \arcsin \left (d x + c\right ) + a\right )}^{n} x^{2}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \left (a + b \operatorname{asin}{\left (c + d x \right )}\right )^{n}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \arcsin \left (d x + c\right ) + a\right )}^{n} x^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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