3.409 \(\int (a-b \sin ^{-1}(1-d x^2))^3 \, dx\)

Optimal. Leaf size=115 \[ -24 a b^2 x+\frac{6 b \sqrt{2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2}{d x}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^3-\frac{48 b^3 \sqrt{2 d x^2-d^2 x^4}}{d x}+24 b^3 x \sin ^{-1}\left (1-d x^2\right ) \]

[Out]

-24*a*b^2*x - (48*b^3*Sqrt[2*d*x^2 - d^2*x^4])/(d*x) + 24*b^3*x*ArcSin[1 - d*x^2] + (6*b*Sqrt[2*d*x^2 - d^2*x^
4]*(a - b*ArcSin[1 - d*x^2])^2)/(d*x) + x*(a - b*ArcSin[1 - d*x^2])^3

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Rubi [A]  time = 0.0607515, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4814, 4840, 12, 1588} \[ -24 a b^2 x+\frac{6 b \sqrt{2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2}{d x}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^3-\frac{48 b^3 \sqrt{2 d x^2-d^2 x^4}}{d x}+24 b^3 x \sin ^{-1}\left (1-d x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a - b*ArcSin[1 - d*x^2])^3,x]

[Out]

-24*a*b^2*x - (48*b^3*Sqrt[2*d*x^2 - d^2*x^4])/(d*x) + 24*b^3*x*ArcSin[1 - d*x^2] + (6*b*Sqrt[2*d*x^2 - d^2*x^
4]*(a - b*ArcSin[1 - d*x^2])^2)/(d*x) + x*(a - b*ArcSin[1 - d*x^2])^3

Rule 4814

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(n_), x_Symbol] :> Simp[x*(a + b*ArcSin[c + d*x^2])^n, x] + (-
Dist[4*b^2*n*(n - 1), Int[(a + b*ArcSin[c + d*x^2])^(n - 2), x], x] + Simp[(2*b*n*Sqrt[-2*c*d*x^2 - d^2*x^4]*(
a + b*ArcSin[c + d*x^2])^(n - 1))/(d*x), x]) /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1] && GtQ[n, 1]

Rule 4840

Int[ArcSin[u_], x_Symbol] :> Simp[x*ArcSin[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/Sqrt[1 - u^2], x], x] /;
InverseFunctionFreeQ[u, x] &&  !FunctionOfExponentialQ[u, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1588

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*x^(p - q
+ 1)*Qq^(m + 1))/((p + m*q + 1)*Coeff[Qq, x, q]), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^3 \, dx &=\frac{6 b \sqrt{2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2}{d x}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^3-\left (24 b^2\right ) \int \left (a-b \sin ^{-1}\left (1-d x^2\right )\right ) \, dx\\ &=-24 a b^2 x+\frac{6 b \sqrt{2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2}{d x}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^3+\left (24 b^3\right ) \int \sin ^{-1}\left (1-d x^2\right ) \, dx\\ &=-24 a b^2 x+24 b^3 x \sin ^{-1}\left (1-d x^2\right )+\frac{6 b \sqrt{2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2}{d x}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^3-\left (24 b^3\right ) \int -\frac{2 d x^2}{\sqrt{2 d x^2-d^2 x^4}} \, dx\\ &=-24 a b^2 x+24 b^3 x \sin ^{-1}\left (1-d x^2\right )+\frac{6 b \sqrt{2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2}{d x}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^3+\left (48 b^3 d\right ) \int \frac{x^2}{\sqrt{2 d x^2-d^2 x^4}} \, dx\\ &=-24 a b^2 x-\frac{48 b^3 \sqrt{2 d x^2-d^2 x^4}}{d x}+24 b^3 x \sin ^{-1}\left (1-d x^2\right )+\frac{6 b \sqrt{2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2}{d x}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^3\\ \end{align*}

Mathematica [A]  time = 0.116136, size = 166, normalized size = 1.44 \[ \frac{a d x^2 \left (a^2-24 b^2\right )+6 b \left (a^2-8 b^2\right ) \sqrt{d x^2 \left (2-d x^2\right )}-3 b \sin ^{-1}\left (1-d x^2\right ) \left (a^2 d x^2+4 a b \sqrt{-d x^2 \left (d x^2-2\right )}-8 b^2 d x^2\right )+3 b^2 \sin ^{-1}\left (1-d x^2\right )^2 \left (a d x^2+2 b \sqrt{-d x^2 \left (d x^2-2\right )}\right )-b^3 d x^2 \sin ^{-1}\left (1-d x^2\right )^3}{d x} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*ArcSin[1 - d*x^2])^3,x]

[Out]

(a*(a^2 - 24*b^2)*d*x^2 + 6*b*(a^2 - 8*b^2)*Sqrt[d*x^2*(2 - d*x^2)] - 3*b*(a^2*d*x^2 - 8*b^2*d*x^2 + 4*a*b*Sqr
t[-(d*x^2*(-2 + d*x^2))])*ArcSin[1 - d*x^2] + 3*b^2*(a*d*x^2 + 2*b*Sqrt[-(d*x^2*(-2 + d*x^2))])*ArcSin[1 - d*x
^2]^2 - b^3*d*x^2*ArcSin[1 - d*x^2]^3)/(d*x)

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Maple [F]  time = 0.103, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\arcsin \left ( d{x}^{2}-1 \right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(d*x^2-1))^3,x)

[Out]

int((a+b*arcsin(d*x^2-1))^3,x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x^2-1))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.28059, size = 328, normalized size = 2.85 \begin{align*} \frac{b^{3} d x^{2} \arcsin \left (d x^{2} - 1\right )^{3} + 3 \, a b^{2} d x^{2} \arcsin \left (d x^{2} - 1\right )^{2} + 3 \,{\left (a^{2} b - 8 \, b^{3}\right )} d x^{2} \arcsin \left (d x^{2} - 1\right ) +{\left (a^{3} - 24 \, a b^{2}\right )} d x^{2} + 6 \, \sqrt{-d^{2} x^{4} + 2 \, d x^{2}}{\left (b^{3} \arcsin \left (d x^{2} - 1\right )^{2} + 2 \, a b^{2} \arcsin \left (d x^{2} - 1\right ) + a^{2} b - 8 \, b^{3}\right )}}{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x^2-1))^3,x, algorithm="fricas")

[Out]

(b^3*d*x^2*arcsin(d*x^2 - 1)^3 + 3*a*b^2*d*x^2*arcsin(d*x^2 - 1)^2 + 3*(a^2*b - 8*b^3)*d*x^2*arcsin(d*x^2 - 1)
+ (a^3 - 24*a*b^2)*d*x^2 + 6*sqrt(-d^2*x^4 + 2*d*x^2)*(b^3*arcsin(d*x^2 - 1)^2 + 2*a*b^2*arcsin(d*x^2 - 1) +
a^2*b - 8*b^3))/(d*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \operatorname{asin}{\left (d x^{2} - 1 \right )}\right )^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(d*x**2-1))**3,x)

[Out]

Integral((a + b*asin(d*x**2 - 1))**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x^2-1))^3,x, algorithm="giac")

[Out]

integrate((b*arcsin(d*x^2 - 1) + a)^3, x)