3.1.21 \(\int \frac {x^3 \sin ^{-1}(x)}{\sqrt {1-x^4}} \, dx\) [21]

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

[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 = 0.03, 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 = {267, 4871, 12, 26, 201, 221} \begin {gather*} -\frac {1}{2} \sqrt {1-x^4} \text {ArcSin}(x)+\frac {1}{4} \sqrt {x^2+1} x+\frac {1}{4} \sinh ^{-1}(x) \end {gather*}

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 201

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 221

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 267

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 4871

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] Leaf count is larger than twice the leaf count of optimal. \(85\) vs. \(2(38)=76\).
time = 0.06, size = 85, normalized size = 2.24 \begin {gather*} \frac {1}{4} \left (\frac {x \sqrt {1-x^4}}{\sqrt {1-x^2}}-2 \sqrt {1-x^4} \sin ^{-1}(x)+\log \left (1-x^2\right )-\log \left (-x+x^3+\sqrt {1-x^2} \sqrt {1-x^4}\right )\right ) \end {gather*}

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 = 0.23, size = 0, normalized size = 0.00 \[\int \frac {x^{3} \arcsin \left (x \right )}{\sqrt {-x^{4}+1}}\, dx\]

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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (28) = 56\).
time = 0.50, size = 138, normalized size = 3.63 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.49, size = 38, normalized size = 1.00 \begin {gather*} \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 {gather*}

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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {x^3\,\mathrm {asin}\left (x\right )}{\sqrt {1-x^4}} \,d x \end {gather*}

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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