3.416 \(\int \sin ^{-1}(1-x^2)^2 \, dx\)

Optimal. Leaf size=44 \[ x \sin ^{-1}\left (1-x^2\right )^2-\frac {4 \sqrt {2 x^2-x^4} \sin ^{-1}\left (1-x^2\right )}{x}-8 x \]

[Out]

-8*x+x*arcsin(x^2-1)^2+4*arcsin(x^2-1)*(-x^4+2*x^2)^(1/2)/x

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Rubi [A]  time = 0.01, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4814, 8} \[ x \sin ^{-1}\left (1-x^2\right )^2-\frac {4 \sqrt {2 x^2-x^4} \sin ^{-1}\left (1-x^2\right )}{x}-8 x \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[1 - x^2]^2,x]

[Out]

-8*x - (4*Sqrt[2*x^2 - x^4]*ArcSin[1 - x^2])/x + x*ArcSin[1 - x^2]^2

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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]

Rubi steps

\begin {align*} \int \sin ^{-1}\left (1-x^2\right )^2 \, dx &=-\frac {4 \sqrt {2 x^2-x^4} \sin ^{-1}\left (1-x^2\right )}{x}+x \sin ^{-1}\left (1-x^2\right )^2-8 \int 1 \, dx\\ &=-8 x-\frac {4 \sqrt {2 x^2-x^4} \sin ^{-1}\left (1-x^2\right )}{x}+x \sin ^{-1}\left (1-x^2\right )^2\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 44, normalized size = 1.00 \[ x \sin ^{-1}\left (1-x^2\right )^2-\frac {4 \sqrt {2 x^2-x^4} \sin ^{-1}\left (1-x^2\right )}{x}-8 x \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[1 - x^2]^2,x]

[Out]

-8*x - (4*Sqrt[2*x^2 - x^4]*ArcSin[1 - x^2])/x + x*ArcSin[1 - x^2]^2

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fricas [A]  time = 0.40, size = 43, normalized size = 0.98 \[ \frac {x^{2} \arcsin \left (x^{2} - 1\right )^{2} - 8 \, x^{2} + 4 \, \sqrt {-x^{4} + 2 \, x^{2}} \arcsin \left (x^{2} - 1\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(x^2-1)^2,x, algorithm="fricas")

[Out]

(x^2*arcsin(x^2 - 1)^2 - 8*x^2 + 4*sqrt(-x^4 + 2*x^2)*arcsin(x^2 - 1))/x

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giac [A]  time = 1.76, size = 58, normalized size = 1.32 \[ x \arcsin \left (x^{2} - 1\right )^{2} + 2 \, {\left (\sqrt {2} \pi - 4 \, \sqrt {2}\right )} \mathrm {sgn}\relax (x) + \frac {4 \, {\left (\sqrt {-x^{2} + 2} \arcsin \left (x^{2} - 1\right ) + 2 \, \sqrt {2} - 2 \, {\left | x \right |}\right )}}{\mathrm {sgn}\relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(x^2-1)^2,x, algorithm="giac")

[Out]

x*arcsin(x^2 - 1)^2 + 2*(sqrt(2)*pi - 4*sqrt(2))*sgn(x) + 4*(sqrt(-x^2 + 2)*arcsin(x^2 - 1) + 2*sqrt(2) - 2*ab
s(x))/sgn(x)

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \arcsin \left (x^{2}-1\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(x^2-1)^2,x)

[Out]

int(arcsin(x^2-1)^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ x \arctan \left (x^{2} - 1, \sqrt {-x^{2} + 2} x\right )^{2} + 4 \, \int \frac {\sqrt {-x^{2} + 2} x \arctan \left (x^{2} - 1, \sqrt {-x^{2} + 2} x\right )}{x^{2} - 2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(x^2-1)^2,x, algorithm="maxima")

[Out]

x*arctan2(x^2 - 1, sqrt(-x^2 + 2)*x)^2 + 4*integrate(sqrt(-x^2 + 2)*x*arctan2(x^2 - 1, sqrt(-x^2 + 2)*x)/(x^2
- 2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\mathrm {asin}\left (x^2-1\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(x^2 - 1)^2,x)

[Out]

int(asin(x^2 - 1)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asin(x**2-1)**2,x)

[Out]

Integral(asin(x**2 - 1)**2, x)

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