3.660 \(\int \frac{x^4 \sin ^{-1}(x)}{\sqrt{1-x^2}} \, dx\)

Optimal. Leaf size=61 \[ \frac{x^4}{16}+\frac{3 x^2}{16}-\frac{1}{4} \sqrt{1-x^2} x^3 \sin ^{-1}(x)-\frac{3}{8} \sqrt{1-x^2} x \sin ^{-1}(x)+\frac{3}{16} \sin ^{-1}(x)^2 \]

[Out]

(3*x^2)/16 + x^4/16 - (3*x*Sqrt[1 - x^2]*ArcSin[x])/8 - (x^3*Sqrt[1 - x^2]*ArcSin[x])/4 + (3*ArcSin[x]^2)/16

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Rubi [A]  time = 0.106788, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {4707, 4641, 30} \[ \frac{x^4}{16}+\frac{3 x^2}{16}-\frac{1}{4} \sqrt{1-x^2} x^3 \sin ^{-1}(x)-\frac{3}{8} \sqrt{1-x^2} x \sin ^{-1}(x)+\frac{3}{16} \sin ^{-1}(x)^2 \]

Antiderivative was successfully verified.

[In]

Int[(x^4*ArcSin[x])/Sqrt[1 - x^2],x]

[Out]

(3*x^2)/16 + x^4/16 - (3*x*Sqrt[1 - x^2]*ArcSin[x])/8 - (x^3*Sqrt[1 - x^2]*ArcSin[x])/4 + (3*ArcSin[x]^2)/16

Rule 4707

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[
(f*(f*x)^(m - 1)*Sqrt[d + e*x^2]*(a + b*ArcSin[c*x])^n)/(e*m), x] + (Dist[(f^2*(m - 1))/(c^2*m), Int[((f*x)^(m
- 2)*(a + b*ArcSin[c*x])^n)/Sqrt[d + e*x^2], x], x] + Dist[(b*f*n*Sqrt[1 - c^2*x^2])/(c*m*Sqrt[d + e*x^2]), I
nt[(f*x)^(m - 1)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] &&
GtQ[n, 0] && GtQ[m, 1] && IntegerQ[m]

Rule 4641

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSin[c*x])^
(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0] && NeQ[n,
-1]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{x^4 \sin ^{-1}(x)}{\sqrt{1-x^2}} \, dx &=-\frac{1}{4} x^3 \sqrt{1-x^2} \sin ^{-1}(x)+\frac{\int x^3 \, dx}{4}+\frac{3}{4} \int \frac{x^2 \sin ^{-1}(x)}{\sqrt{1-x^2}} \, dx\\ &=\frac{x^4}{16}-\frac{3}{8} x \sqrt{1-x^2} \sin ^{-1}(x)-\frac{1}{4} x^3 \sqrt{1-x^2} \sin ^{-1}(x)+\frac{3 \int x \, dx}{8}+\frac{3}{8} \int \frac{\sin ^{-1}(x)}{\sqrt{1-x^2}} \, dx\\ &=\frac{3 x^2}{16}+\frac{x^4}{16}-\frac{3}{8} x \sqrt{1-x^2} \sin ^{-1}(x)-\frac{1}{4} x^3 \sqrt{1-x^2} \sin ^{-1}(x)+\frac{3}{16} \sin ^{-1}(x)^2\\ \end{align*}

Mathematica [A]  time = 0.0250347, size = 43, normalized size = 0.7 \[ \frac{1}{16} \left (\left (x^2+3\right ) x^2-2 \sqrt{1-x^2} \left (2 x^2+3\right ) x \sin ^{-1}(x)+3 \sin ^{-1}(x)^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*ArcSin[x])/Sqrt[1 - x^2],x]

[Out]

(x^2*(3 + x^2) - 2*x*Sqrt[1 - x^2]*(3 + 2*x^2)*ArcSin[x] + 3*ArcSin[x]^2)/16

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Maple [A]  time = 0.044, size = 53, normalized size = 0.9 \begin{align*}{\frac{\arcsin \left ( x \right ) }{8} \left ( -2\,\sqrt{-{x}^{2}+1}{x}^{3}-3\,x\sqrt{-{x}^{2}+1}+3\,\arcsin \left ( x \right ) \right ) }-{\frac{3\, \left ( \arcsin \left ( x \right ) \right ) ^{2}}{16}}+{\frac{{x}^{4}}{16}}+{\frac{3\,{x}^{2}}{16}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*arcsin(x)*(-2*(-x^2+1)^(1/2)*x^3-3*x*(-x^2+1)^(1/2)+3*arcsin(x))-3/16*arcsin(x)^2+1/16*x^4+3/16*x^2

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Maxima [A]  time = 1.42568, size = 70, normalized size = 1.15 \begin{align*} \frac{1}{16} \, x^{4} + \frac{3}{16} \, x^{2} - \frac{1}{8} \,{\left (2 \, \sqrt{-x^{2} + 1} x^{3} + 3 \, \sqrt{-x^{2} + 1} x - 3 \, \arcsin \left (x\right )\right )} \arcsin \left (x\right ) - \frac{3}{16} \, \arcsin \left (x\right )^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*x^4 + 3/16*x^2 - 1/8*(2*sqrt(-x^2 + 1)*x^3 + 3*sqrt(-x^2 + 1)*x - 3*arcsin(x))*arcsin(x) - 3/16*arcsin(x)
^2

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Fricas [A]  time = 2.44136, size = 115, normalized size = 1.89 \begin{align*} \frac{1}{16} \, x^{4} - \frac{1}{8} \,{\left (2 \, x^{3} + 3 \, x\right )} \sqrt{-x^{2} + 1} \arcsin \left (x\right ) + \frac{3}{16} \, x^{2} + \frac{3}{16} \, \arcsin \left (x\right )^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*x^4 - 1/8*(2*x^3 + 3*x)*sqrt(-x^2 + 1)*arcsin(x) + 3/16*x^2 + 3/16*arcsin(x)^2

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Sympy [A]  time = 1.29486, size = 53, normalized size = 0.87 \begin{align*} \frac{x^{4}}{16} - \frac{x^{3} \sqrt{1 - x^{2}} \operatorname{asin}{\left (x \right )}}{4} + \frac{3 x^{2}}{16} - \frac{3 x \sqrt{1 - x^{2}} \operatorname{asin}{\left (x \right )}}{8} + \frac{3 \operatorname{asin}^{2}{\left (x \right )}}{16} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**4/16 - x**3*sqrt(1 - x**2)*asin(x)/4 + 3*x**2/16 - 3*x*sqrt(1 - x**2)*asin(x)/8 + 3*asin(x)**2/16

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Giac [A]  time = 1.09011, size = 68, normalized size = 1.11 \begin{align*} \frac{1}{4} \,{\left (-x^{2} + 1\right )}^{\frac{3}{2}} x \arcsin \left (x\right ) - \frac{5}{8} \, \sqrt{-x^{2} + 1} x \arcsin \left (x\right ) + \frac{1}{16} \,{\left (x^{2} - 1\right )}^{2} + \frac{5}{16} \, x^{2} + \frac{3}{16} \, \arcsin \left (x\right )^{2} - \frac{23}{128} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(-x^2 + 1)^(3/2)*x*arcsin(x) - 5/8*sqrt(-x^2 + 1)*x*arcsin(x) + 1/16*(x^2 - 1)^2 + 5/16*x^2 + 3/16*arcsin(
x)^2 - 23/128