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-24*b^3*x*arcsin(d*x^2-1)+x*(a+b*arcsin(d*x^2-1))^3-48*b^3*(-d^2*x^4+2*d*x^2)^(1/2)/d/x+6*b*(a+b*ar
csin(d*x^2-1))^2*(-d^2*x^4+2*d*x^2)^(1/2)/d/x

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Rubi [A]  time = 0.06, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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]

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]

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*}

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Mathematica [A]  time = 0.13, 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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fricas [A]  time = 1.38, size = 144, normalized size = 1.25 \[ \frac {b^{3} d x^{2} \arcsin \left (d x^{2} - 1\right )^{3} + 3 \, a b^{2} d x^{2} \arcsin \left (d x^{2} - 1\right )^{2} + 3 \, {\left (a^{2} b - 8 \, b^{3}\right )} d x^{2} \arcsin \left (d x^{2} - 1\right ) + {\left (a^{3} - 24 \, a b^{2}\right )} d x^{2} + 6 \, \sqrt {-d^{2} x^{4} + 2 \, d x^{2}} {\left (b^{3} \arcsin \left (d x^{2} - 1\right )^{2} + 2 \, a b^{2} \arcsin \left (d x^{2} - 1\right ) + a^{2} b - 8 \, b^{3}\right )}}{d x} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{3}\,{d x} \]

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)

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int \left (a +b \arcsin \left (d \,x^{2}-1\right )\right )^{3}\, dx \]

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]  time = 0.00, size = 0, normalized size = 0.00 \[ b^{3} x \arctan \left (d x^{2} - 1, \sqrt {-d x^{2} + 2} \sqrt {d} x\right )^{3} + 3 \, {\left (x \arcsin \left (d x^{2} - 1\right ) - \frac {2 \, {\left (d^{\frac {3}{2}} x^{2} - 2 \, \sqrt {d}\right )}}{\sqrt {-d x^{2} + 2} d}\right )} a^{2} b + a^{3} x + \int \frac {3 \, {\left (2 \, \sqrt {-d x^{2} + 2} b^{3} \sqrt {d} x \arctan \left (d x^{2} - 1, \sqrt {-d x^{2} + 2} \sqrt {d} x\right )^{2} + {\left (a b^{2} d x^{2} - 2 \, a b^{2}\right )} \arctan \left (d x^{2} - 1, \sqrt {-d x^{2} + 2} \sqrt {d} x\right )^{2}\right )}}{d x^{2} - 2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^3*x*arctan2(d*x^2 - 1, sqrt(-d*x^2 + 2)*sqrt(d)*x)^3 + 3*(x*arcsin(d*x^2 - 1) - 2*(d^(3/2)*x^2 - 2*sqrt(d))/
(sqrt(-d*x^2 + 2)*d))*a^2*b + a^3*x + integrate(3*(2*sqrt(-d*x^2 + 2)*b^3*sqrt(d)*x*arctan2(d*x^2 - 1, sqrt(-d
*x^2 + 2)*sqrt(d)*x)^2 + (a*b^2*d*x^2 - 2*a*b^2)*arctan2(d*x^2 - 1, sqrt(-d*x^2 + 2)*sqrt(d)*x)^2)/(d*x^2 - 2)
, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+b\,\mathrm {asin}\left (d\,x^2-1\right )\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(d*x^2 - 1))^3,x)

[Out]

int((a + b*asin(d*x^2 - 1))^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \operatorname {asin}{\left (d x^{2} - 1 \right )}\right )^{3}\, dx \]

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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