3.21 \(\int \frac {x^3 \sin ^{-1}(x)}{\sqrt {1-x^4}} \, dx\)

Optimal. Leaf size=38 \[ -\frac {1}{2} \sqrt {1-x^4} \sin ^{-1}(x)+\frac {1}{4} \sqrt {x^2+1} x+\frac {1}{4} \sinh ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {261, 4787, 12, 26, 195, 215} \[ \frac {1}{4} \sqrt {x^2+1} x-\frac {1}{2} \sqrt {1-x^4} \sin ^{-1}(x)+\frac {1}{4} \sinh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x^3*ArcSin[x])/Sqrt[1 - x^4],x]

[Out]

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

Rule 12

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

Rule 26

Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(m_.)*((c_) + (d_.)*(x_)^(j_))^(p_.), x_Symbol] :> Dist[(-(b^2/d))^m, Int[
u/(a - b*x^n)^m, x], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[j, 2*n] && EqQ[p, -m] && EqQ[b^2*c + a^2*d,
0] && GtQ[a, 0] && LtQ[d, 0]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 4787

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*(u_), x_Symbol] :> With[{v = IntHide[u, x]}, Dist[a + b*ArcSin[c*x], v,
x] - Dist[b*c, Int[SimplifyIntegrand[v/Sqrt[1 - c^2*x^2], x], x], x] /; InverseFunctionFreeQ[v, x]] /; FreeQ[
{a, b, c}, x]

Rubi steps

\begin {align*} \int \frac {x^3 \sin ^{-1}(x)}{\sqrt {1-x^4}} \, dx &=-\frac {1}{2} \sqrt {1-x^4} \sin ^{-1}(x)-\int -\frac {\sqrt {1-x^4}}{2 \sqrt {1-x^2}} \, dx\\ &=-\frac {1}{2} \sqrt {1-x^4} \sin ^{-1}(x)+\frac {1}{2} \int \frac {\sqrt {1-x^4}}{\sqrt {1-x^2}} \, dx\\ &=-\frac {1}{2} \sqrt {1-x^4} \sin ^{-1}(x)+\frac {1}{2} \int \sqrt {1+x^2} \, dx\\ &=\frac {1}{4} x \sqrt {1+x^2}-\frac {1}{2} \sqrt {1-x^4} \sin ^{-1}(x)+\frac {1}{4} \int \frac {1}{\sqrt {1+x^2}} \, dx\\ &=\frac {1}{4} x \sqrt {1+x^2}-\frac {1}{2} \sqrt {1-x^4} \sin ^{-1}(x)+\frac {1}{4} \sinh ^{-1}(x)\\ \end {align*}

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Mathematica [B]  time = 0.09, size = 85, normalized size = 2.24 \[ \frac {1}{4} \left (-2 \sqrt {1-x^4} \sin ^{-1}(x)+\log \left (1-x^2\right )+\frac {\sqrt {1-x^4} x}{\sqrt {1-x^2}}-\log \left (x^3+\sqrt {1-x^2} \sqrt {1-x^4}-x\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*ArcSin[x])/Sqrt[1 - x^4],x]

[Out]

((x*Sqrt[1 - x^4])/Sqrt[1 - x^2] - 2*Sqrt[1 - x^4]*ArcSin[x] + Log[1 - x^2] - Log[-x + x^3 + Sqrt[1 - x^2]*Sqr
t[1 - x^4]])/4

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fricas [B]  time = 0.45, size = 138, normalized size = 3.63 \[ -\frac {4 \, \sqrt {-x^{4} + 1} {\left (x^{2} - 1\right )} \arcsin \relax (x) + 2 \, \sqrt {-x^{4} + 1} \sqrt {-x^{2} + 1} x + {\left (x^{2} - 1\right )} \log \left (\frac {x^{3} + \sqrt {-x^{4} + 1} \sqrt {-x^{2} + 1} - x}{x^{3} - x}\right ) - {\left (x^{2} - 1\right )} \log \left (-\frac {x^{3} - \sqrt {-x^{4} + 1} \sqrt {-x^{2} + 1} - x}{x^{3} - x}\right )}{8 \, {\left (x^{2} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsin(x)/(-x^4+1)^(1/2),x, algorithm="fricas")

[Out]

-1/8*(4*sqrt(-x^4 + 1)*(x^2 - 1)*arcsin(x) + 2*sqrt(-x^4 + 1)*sqrt(-x^2 + 1)*x + (x^2 - 1)*log((x^3 + sqrt(-x^
4 + 1)*sqrt(-x^2 + 1) - x)/(x^3 - x)) - (x^2 - 1)*log(-(x^3 - sqrt(-x^4 + 1)*sqrt(-x^2 + 1) - x)/(x^3 - x)))/(
x^2 - 1)

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giac [A]  time = 1.10, size = 38, normalized size = 1.00 \[ \frac {1}{4} \, \sqrt {x^{2} + 1} x - \frac {1}{2} \, \sqrt {-x^{4} + 1} \arcsin \relax (x) - \frac {1}{4} \, \log \left (-x + \sqrt {x^{2} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsin(x)/(-x^4+1)^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.41, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \arcsin \relax (x )}{\sqrt {-x^{4}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arcsin(x)/(-x^4+1)^(1/2),x)

[Out]

int(x^3*arcsin(x)/(-x^4+1)^(1/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsin(x)/(-x^4+1)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {x^3\,\mathrm {asin}\relax (x)}{\sqrt {1-x^4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*asin(x))/(1 - x^4)^(1/2),x)

[Out]

int((x^3*asin(x))/(1 - x^4)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*asin(x)/(-x**4+1)**(1/2),x)

[Out]

Integral(x**3*asin(x)/sqrt(-(x - 1)*(x + 1)*(x**2 + 1)), x)

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