3.4.50 \(\int x^3 \sin ^{-1}(x^2) \, dx\) [350]

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

[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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Rubi [A]
time = 0.02, antiderivative size = 38, normalized size of antiderivative = 1.00, 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 = {4926, 12, 281, 327, 222} \begin {gather*} -\frac {\text {ArcSin}\left (x^2\right )}{8}+\frac {1}{4} x^4 \text {ArcSin}\left (x^2\right )+\frac {1}{8} \sqrt {1-x^4} x^2 \end {gather*}

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 12

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

Rule 222

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]

Rule 281

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 327

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 4926

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]

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} \text {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} \text {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*}

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Mathematica [A]
time = 0.01, size = 32, normalized size = 0.84 \begin {gather*} \frac {1}{8} \left (x^2 \sqrt {1-x^4}+\left (-1+2 x^4\right ) \sin ^{-1}\left (x^2\right )\right ) \end {gather*}

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 = 0.00, size = 31, normalized size = 0.82

method	result	size
derivativedivides	\(-\frac {\arcsin \left (x^{2}\right )}{8}+\frac {x^{4} \arcsin \left (x^{2}\right )}{4}+\frac {x^{2} \sqrt {-x^{4}+1}}{8}\)	\(31\)
default	\(-\frac {\arcsin \left (x^{2}\right )}{8}+\frac {x^{4} \arcsin \left (x^{2}\right )}{4}+\frac {x^{2} \sqrt {-x^{4}+1}}{8}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arcsin(x^2),x,method=_RETURNVERBOSE)

[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 = 2.38, size = 53, normalized size = 1.39 \begin {gather*} \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 {gather*}

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 = 1.19, size = 28, normalized size = 0.74 \begin {gather*} \frac {1}{8} \, \sqrt {-x^{4} + 1} x^{2} + \frac {1}{8} \, {\left (2 \, x^{4} - 1\right )} \arcsin \left (x^{2}\right ) \end {gather*}

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 = 0.14, size = 29, normalized size = 0.76 \begin {gather*} \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 {gather*}

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 = 1.04, size = 32, normalized size = 0.84 \begin {gather*} \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 {gather*}

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)

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Mupad [B]
time = 0.22, size = 28, normalized size = 0.74 \begin {gather*} \frac {x^2\,\sqrt {1-x^4}}{8}+\frac {\mathrm {asin}\left (x^2\right )\,\left (2\,x^4-1\right )}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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