Optimal. Leaf size=466 \[ -\frac{i d 2^{-n-3} e^{-\frac{2 i a}{b}} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (-\frac{i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \text{Gamma}\left (n+1,-\frac{2 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{c \sqrt{1-c^2 x^2}}-\frac{i d 2^{-2 (n+3)} e^{-\frac{4 i a}{b}} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (-\frac{i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \text{Gamma}\left (n+1,-\frac{4 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{c \sqrt{1-c^2 x^2}}+\frac{i d 2^{-n-3} e^{\frac{2 i a}{b}} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (\frac{i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \text{Gamma}\left (n+1,\frac{2 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{c \sqrt{1-c^2 x^2}}+\frac{i d 2^{-2 (n+3)} e^{\frac{4 i a}{b}} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (\frac{i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \text{Gamma}\left (n+1,\frac{4 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{c \sqrt{1-c^2 x^2}}+\frac{3 d \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^{n+1}}{8 b c (n+1) \sqrt{1-c^2 x^2}} \]
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Rubi [A] time = 0.406926, antiderivative size = 466, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {4663, 4661, 3312, 3307, 2181} \[ -\frac{i d 2^{-n-3} e^{-\frac{2 i a}{b}} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (-\frac{i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \text{Gamma}\left (n+1,-\frac{2 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{c \sqrt{1-c^2 x^2}}-\frac{i d 2^{-2 (n+3)} e^{-\frac{4 i a}{b}} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (-\frac{i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \text{Gamma}\left (n+1,-\frac{4 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{c \sqrt{1-c^2 x^2}}+\frac{i d 2^{-n-3} e^{\frac{2 i a}{b}} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (\frac{i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \text{Gamma}\left (n+1,\frac{2 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{c \sqrt{1-c^2 x^2}}+\frac{i d 2^{-2 (n+3)} e^{\frac{4 i a}{b}} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (\frac{i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \text{Gamma}\left (n+1,\frac{4 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{c \sqrt{1-c^2 x^2}}+\frac{3 d \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^{n+1}}{8 b c (n+1) \sqrt{1-c^2 x^2}} \]
Antiderivative was successfully verified.
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Rule 4663
Rule 4661
Rule 3312
Rule 3307
Rule 2181
Rubi steps
\begin{align*} \int \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^n \, dx &=\frac{\left (d \sqrt{d-c^2 d x^2}\right ) \int \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^n \, dx}{\sqrt{1-c^2 x^2}}\\ &=\frac{\left (d \sqrt{d-c^2 d x^2}\right ) \operatorname{Subst}\left (\int (a+b x)^n \cos ^4(x) \, dx,x,\sin ^{-1}(c x)\right )}{c \sqrt{1-c^2 x^2}}\\ &=\frac{\left (d \sqrt{d-c^2 d x^2}\right ) \operatorname{Subst}\left (\int \left (\frac{3}{8} (a+b x)^n+\frac{1}{2} (a+b x)^n \cos (2 x)+\frac{1}{8} (a+b x)^n \cos (4 x)\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c \sqrt{1-c^2 x^2}}\\ &=\frac{3 d \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^{1+n}}{8 b c (1+n) \sqrt{1-c^2 x^2}}+\frac{\left (d \sqrt{d-c^2 d x^2}\right ) \operatorname{Subst}\left (\int (a+b x)^n \cos (4 x) \, dx,x,\sin ^{-1}(c x)\right )}{8 c \sqrt{1-c^2 x^2}}+\frac{\left (d \sqrt{d-c^2 d x^2}\right ) \operatorname{Subst}\left (\int (a+b x)^n \cos (2 x) \, dx,x,\sin ^{-1}(c x)\right )}{2 c \sqrt{1-c^2 x^2}}\\ &=\frac{3 d \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^{1+n}}{8 b c (1+n) \sqrt{1-c^2 x^2}}+\frac{\left (d \sqrt{d-c^2 d x^2}\right ) \operatorname{Subst}\left (\int e^{-4 i x} (a+b x)^n \, dx,x,\sin ^{-1}(c x)\right )}{16 c \sqrt{1-c^2 x^2}}+\frac{\left (d \sqrt{d-c^2 d x^2}\right ) \operatorname{Subst}\left (\int e^{4 i x} (a+b x)^n \, dx,x,\sin ^{-1}(c x)\right )}{16 c \sqrt{1-c^2 x^2}}+\frac{\left (d \sqrt{d-c^2 d x^2}\right ) \operatorname{Subst}\left (\int e^{-2 i x} (a+b x)^n \, dx,x,\sin ^{-1}(c x)\right )}{4 c \sqrt{1-c^2 x^2}}+\frac{\left (d \sqrt{d-c^2 d x^2}\right ) \operatorname{Subst}\left (\int e^{2 i x} (a+b x)^n \, dx,x,\sin ^{-1}(c x)\right )}{4 c \sqrt{1-c^2 x^2}}\\ &=\frac{3 d \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^{1+n}}{8 b c (1+n) \sqrt{1-c^2 x^2}}-\frac{i 2^{-3-n} d e^{-\frac{2 i a}{b}} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (-\frac{i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \Gamma \left (1+n,-\frac{2 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{c \sqrt{1-c^2 x^2}}+\frac{i 2^{-3-n} d e^{\frac{2 i a}{b}} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (\frac{i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \Gamma \left (1+n,\frac{2 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{c \sqrt{1-c^2 x^2}}-\frac{i 4^{-3-n} d e^{-\frac{4 i a}{b}} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (-\frac{i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \Gamma \left (1+n,-\frac{4 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{c \sqrt{1-c^2 x^2}}+\frac{i 4^{-3-n} d e^{\frac{4 i a}{b}} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (\frac{i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \Gamma \left (1+n,\frac{4 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{c \sqrt{1-c^2 x^2}}\\ \end{align*}
Mathematica [A] time = 1.65971, size = 326, normalized size = 0.7 \[ \frac{d^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (i 4^{-n} e^{-\frac{4 i a}{b}} \left (\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{b^2}\right )^{-n} \left (e^{\frac{8 i a}{b}} \left (-\frac{i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^n \text{Gamma}\left (n+1,\frac{4 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )-\left (\frac{i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^n \text{Gamma}\left (n+1,-\frac{4 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )\right )+8 \left (-i 2^{-n} e^{-\frac{2 i a}{b}} \left (-\frac{i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \text{Gamma}\left (n+1,-\frac{2 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )+i 2^{-n} e^{\frac{2 i a}{b}} \left (\frac{i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \text{Gamma}\left (n+1,\frac{2 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )+\frac{4 a+4 b \sin ^{-1}(c x)}{b n+b}\right )-\frac{8 \left (a+b \sin ^{-1}(c x)\right )}{b n+b}\right )}{64 c \sqrt{d-c^2 d x^2}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.133, size = 0, normalized size = 0. \begin{align*} \int \left ( -{c}^{2}d{x}^{2}+d \right ) ^{{\frac{3}{2}}} \left ( a+b\arcsin \left ( cx \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 (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}}{\left (b \arcsin \left (c x\right ) + a\right )}^{n}\,{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 (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}}{\left (b \arcsin \left (c x\right ) + a\right )}^{n}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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 (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}}{\left (b \arcsin \left (c x\right ) + a\right )}^{n}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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