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

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

[Out]

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

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Rubi [A]  time = 0.0524984, antiderivative size = 38, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [B]  time = 0.0811957, size = 85, normalized size = 2.24 \[ \frac{1}{4} \left (\frac{\sqrt{1-x^4} x}{\sqrt{1-x^2}}+\log \left (1-x^2\right )-\log \left (x^3+\sqrt{1-x^2} \sqrt{1-x^4}-x\right )-2 \sqrt{1-x^4} \sin ^{-1}(x)\right ) \]

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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Maple [F]  time = 0.263, size = 0, normalized size = 0. \begin{align*} \int{{x}^{3}\arcsin \left ( x \right ){\frac{1}{\sqrt{-{x}^{4}+1}}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} -\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} \end{align*}

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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Fricas [B]  time = 2.3, size = 311, normalized size = 8.18 \begin{align*} -\frac{4 \, \sqrt{-x^{4} + 1}{\left (x^{2} - 1\right )} \arcsin \left (x\right ) + 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 )}} \end{align*}

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

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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Giac [A]  time = 1.11245, size = 51, normalized size = 1.34 \begin{align*} \frac{1}{4} \, \sqrt{x^{2} + 1} x - \frac{1}{2} \, \sqrt{-x^{4} + 1} \arcsin \left (x\right ) - \frac{1}{4} \, \log \left (-x + \sqrt{x^{2} + 1}\right ) \end{align*}

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))