3.482 \(\int x^2 \sqrt{d-c^2 d x^2} (a+b \sin ^{-1}(c x))^n \, dx\)

Optimal. Leaf size=259 \[ \frac{i 2^{-2 (n+3)} e^{-\frac{4 i a}{b}} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (-\frac{i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \text{Gamma}\left (n+1,-\frac{4 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{c^3 \sqrt{1-c^2 x^2}}-\frac{i 2^{-2 (n+3)} e^{\frac{4 i a}{b}} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (\frac{i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \text{Gamma}\left (n+1,\frac{4 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{c^3 \sqrt{1-c^2 x^2}}+\frac{\sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^{n+1}}{8 b c^3 (n+1) \sqrt{1-c^2 x^2}} \]

[Out]

(Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^(1 + n))/(8*b*c^3*(1 + n)*Sqrt[1 - c^2*x^2]) + (I*Sqrt[d - c^2*d*x^2]
*(a + b*ArcSin[c*x])^n*Gamma[1 + n, ((-4*I)*(a + b*ArcSin[c*x]))/b])/(2^(2*(3 + n))*c^3*E^(((4*I)*a)/b)*Sqrt[1
- c^2*x^2]*(((-I)*(a + b*ArcSin[c*x]))/b)^n) - (I*E^(((4*I)*a)/b)*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^n*G
amma[1 + n, ((4*I)*(a + b*ArcSin[c*x]))/b])/(2^(2*(3 + n))*c^3*Sqrt[1 - c^2*x^2]*((I*(a + b*ArcSin[c*x]))/b)^n
)

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Rubi [A]  time = 0.455666, antiderivative size = 259, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {4725, 4723, 4406, 3307, 2181} \[ \frac{i 2^{-2 (n+3)} e^{-\frac{4 i a}{b}} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (-\frac{i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \text{Gamma}\left (n+1,-\frac{4 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{c^3 \sqrt{1-c^2 x^2}}-\frac{i 2^{-2 (n+3)} e^{\frac{4 i a}{b}} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (\frac{i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \text{Gamma}\left (n+1,\frac{4 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{c^3 \sqrt{1-c^2 x^2}}+\frac{\sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^{n+1}}{8 b c^3 (n+1) \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^n,x]

[Out]

(Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^(1 + n))/(8*b*c^3*(1 + n)*Sqrt[1 - c^2*x^2]) + (I*Sqrt[d - c^2*d*x^2]
*(a + b*ArcSin[c*x])^n*Gamma[1 + n, ((-4*I)*(a + b*ArcSin[c*x]))/b])/(2^(2*(3 + n))*c^3*E^(((4*I)*a)/b)*Sqrt[1
- c^2*x^2]*(((-I)*(a + b*ArcSin[c*x]))/b)^n) - (I*E^(((4*I)*a)/b)*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^n*G
amma[1 + n, ((4*I)*(a + b*ArcSin[c*x]))/b])/(2^(2*(3 + n))*c^3*Sqrt[1 - c^2*x^2]*((I*(a + b*ArcSin[c*x]))/b)^n
)

Rule 4725

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d^IntPar
t[p]*(d + e*x^2)^FracPart[p])/(1 - c^2*x^2)^FracPart[p], Int[x^m*(1 - c^2*x^2)^p*(a + b*ArcSin[c*x])^n, x], x]
/; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] &&  !(Integ
erQ[p] || GtQ[d, 0])

Rule 4723

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^(
m + 1), Subst[Int[(a + b*x)^n*Sin[x]^m*Cos[x]^(2*p + 1), x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e, n},
x] && EqQ[c^2*d + e, 0] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rule 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 3307

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/(E^(
I*k*Pi)*E^(I*(e + f*x))), x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*k*Pi)*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d
, e, f, m}, x] && IntegerQ[2*k]

Rule 2181

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(F^(g*(e - (c*f)/d))*(c +
d*x)^FracPart[m]*Gamma[m + 1, (-((f*g*Log[F])/d))*(c + d*x)])/(d*(-((f*g*Log[F])/d))^(IntPart[m] + 1)*(-((f*g*
Log[F]*(c + d*x))/d))^FracPart[m]), x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rubi steps

\begin{align*} \int x^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \, dx &=\frac{\sqrt{d-c^2 d x^2} \int x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^n \, dx}{\sqrt{1-c^2 x^2}}\\ &=\frac{\sqrt{d-c^2 d x^2} \operatorname{Subst}\left (\int (a+b x)^n \cos ^2(x) \sin ^2(x) \, dx,x,\sin ^{-1}(c x)\right )}{c^3 \sqrt{1-c^2 x^2}}\\ &=\frac{\sqrt{d-c^2 d x^2} \operatorname{Subst}\left (\int \left (\frac{1}{8} (a+b x)^n-\frac{1}{8} (a+b x)^n \cos (4 x)\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^3 \sqrt{1-c^2 x^2}}\\ &=\frac{\sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^{1+n}}{8 b c^3 (1+n) \sqrt{1-c^2 x^2}}-\frac{\sqrt{d-c^2 d x^2} \operatorname{Subst}\left (\int (a+b x)^n \cos (4 x) \, dx,x,\sin ^{-1}(c x)\right )}{8 c^3 \sqrt{1-c^2 x^2}}\\ &=\frac{\sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^{1+n}}{8 b c^3 (1+n) \sqrt{1-c^2 x^2}}-\frac{\sqrt{d-c^2 d x^2} \operatorname{Subst}\left (\int e^{-4 i x} (a+b x)^n \, dx,x,\sin ^{-1}(c x)\right )}{16 c^3 \sqrt{1-c^2 x^2}}-\frac{\sqrt{d-c^2 d x^2} \operatorname{Subst}\left (\int e^{4 i x} (a+b x)^n \, dx,x,\sin ^{-1}(c x)\right )}{16 c^3 \sqrt{1-c^2 x^2}}\\ &=\frac{\sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^{1+n}}{8 b c^3 (1+n) \sqrt{1-c^2 x^2}}+\frac{i 4^{-3-n} 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^3 \sqrt{1-c^2 x^2}}-\frac{i 4^{-3-n} 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^3 \sqrt{1-c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.832689, size = 189, normalized size = 0.73 \[ \frac{d \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (\frac{8 \left (a+b \sin ^{-1}(c x)\right )}{b n+b}+i 4^{-n} e^{-\frac{4 i a}{b}} \left (\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{b^2}\right )^{-n} \left (\left (\frac{i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^n \text{Gamma}\left (n+1,-\frac{4 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )-e^{\frac{8 i a}{b}} \left (-\frac{i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^n \text{Gamma}\left (n+1,\frac{4 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )\right )\right )}{64 c^3 \sqrt{d \left (1-c^2 x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^n,x]

[Out]

(d*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^n*((8*(a + b*ArcSin[c*x]))/(b + b*n) + (I*(((I*(a + b*ArcSin[c*x]))/b
)^n*Gamma[1 + n, ((-4*I)*(a + b*ArcSin[c*x]))/b] - E^(((8*I)*a)/b)*(((-I)*(a + b*ArcSin[c*x]))/b)^n*Gamma[1 +
n, ((4*I)*(a + b*ArcSin[c*x]))/b]))/(4^n*E^(((4*I)*a)/b)*((a + b*ArcSin[c*x])^2/b^2)^n)))/(64*c^3*Sqrt[d*(1 -
c^2*x^2)])

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Maple [F]  time = 0.36, size = 0, normalized size = 0. \begin{align*} \int{x}^{2}\sqrt{-{c}^{2}d{x}^{2}+d} \left ( a+b\arcsin \left ( cx \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))^n,x)

[Out]

int(x^2*(-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))^n,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{-c^{2} d x^{2} + d}{\left (b \arcsin \left (c x\right ) + a\right )}^{n} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))^n,x, algorithm="maxima")

[Out]

integrate(sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a)^n*x^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{-c^{2} d x^{2} + d}{\left (b \arcsin \left (c x\right ) + a\right )}^{n} x^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))^n,x, algorithm="fricas")

[Out]

integral(sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a)^n*x^2, x)

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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.

[In]

integrate(x**2*(-c**2*d*x**2+d)**(1/2)*(a+b*asin(c*x))**n,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{-c^{2} d x^{2} + d}{\left (b \arcsin \left (c x\right ) + a\right )}^{n} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))^n,x, algorithm="giac")

[Out]

integrate(sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a)^n*x^2, x)