3.350 \(\int x^3 \sin ^{-1}(x^2) \, dx\)

Optimal. Leaf size=38 \[ \frac{1}{8} \sqrt{1-x^4} x^2+\frac{1}{4} x^4 \sin ^{-1}\left (x^2\right )-\frac{1}{8} \sin ^{-1}\left (x^2\right ) \]

[Out]

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

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Rubi [A]  time = 0.0233579, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {4842, 12, 275, 321, 216} \[ \frac{1}{8} \sqrt{1-x^4} x^2+\frac{1}{4} x^4 \sin ^{-1}\left (x^2\right )-\frac{1}{8} \sin ^{-1}\left (x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcSin[x^2],x]

[Out]

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

Rule 4842

Int[((a_.) + ArcSin[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcSin[
u]))/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/Sqrt[1 - u^2], x]
, x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)^(
m + 1), u, x] &&  !FunctionOfExponentialQ[u, 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 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 216

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

Rubi steps

\begin{align*} \int x^3 \sin ^{-1}\left (x^2\right ) \, dx &=\frac{1}{4} x^4 \sin ^{-1}\left (x^2\right )-\frac{1}{4} \int \frac{2 x^5}{\sqrt{1-x^4}} \, dx\\ &=\frac{1}{4} x^4 \sin ^{-1}\left (x^2\right )-\frac{1}{2} \int \frac{x^5}{\sqrt{1-x^4}} \, dx\\ &=\frac{1}{4} x^4 \sin ^{-1}\left (x^2\right )-\frac{1}{4} \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-x^2}} \, dx,x,x^2\right )\\ &=\frac{1}{8} x^2 \sqrt{1-x^4}+\frac{1}{4} x^4 \sin ^{-1}\left (x^2\right )-\frac{1}{8} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2}} \, dx,x,x^2\right )\\ &=\frac{1}{8} x^2 \sqrt{1-x^4}-\frac{1}{8} \sin ^{-1}\left (x^2\right )+\frac{1}{4} x^4 \sin ^{-1}\left (x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.011075, size = 32, normalized size = 0.84 \[ \frac{1}{8} \left (\sqrt{1-x^4} x^2+\left (2 x^4-1\right ) \sin ^{-1}\left (x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcSin[x^2],x]

[Out]

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

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Maple [A]  time = 0.002, size = 31, normalized size = 0.8 \begin{align*} -{\frac{\arcsin \left ({x}^{2} \right ) }{8}}+{\frac{{x}^{4}\arcsin \left ({x}^{2} \right ) }{4}}+{\frac{{x}^{2}}{8}\sqrt{-{x}^{4}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arcsin(x^2),x)

[Out]

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

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Maxima [A]  time = 1.43772, size = 72, normalized size = 1.89 \begin{align*} \frac{1}{4} \, x^{4} \arcsin \left (x^{2}\right ) - \frac{\sqrt{-x^{4} + 1}}{8 \, x^{2}{\left (\frac{x^{4} - 1}{x^{4}} - 1\right )}} + \frac{1}{8} \, \arctan \left (\frac{\sqrt{-x^{4} + 1}}{x^{2}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsin(x^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.18518, size = 73, normalized size = 1.92 \begin{align*} \frac{1}{8} \, \sqrt{-x^{4} + 1} x^{2} + \frac{1}{8} \,{\left (2 \, x^{4} - 1\right )} \arcsin \left (x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsin(x^2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.562222, size = 29, normalized size = 0.76 \begin{align*} \frac{x^{4} \operatorname{asin}{\left (x^{2} \right )}}{4} + \frac{x^{2} \sqrt{1 - x^{4}}}{8} - \frac{\operatorname{asin}{\left (x^{2} \right )}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*asin(x**2),x)

[Out]

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

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Giac [A]  time = 1.07989, size = 43, normalized size = 1.13 \begin{align*} \frac{1}{8} \, \sqrt{-x^{4} + 1} x^{2} + \frac{1}{4} \,{\left (x^{4} - 1\right )} \arcsin \left (x^{2}\right ) + \frac{1}{8} \, \arcsin \left (x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsin(x^2),x, algorithm="giac")

[Out]

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