3.7.60 \(\int \frac {x^4 \sin ^{-1}(x)}{\sqrt {1-x^2}} \, dx\) [660]

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

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 61, normalized size of antiderivative = 1.00, 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 = {4795, 4737, 30} \begin {gather*} -\frac {3}{8} \sqrt {1-x^2} x \text {ArcSin}(x)-\frac {1}{4} \sqrt {1-x^2} x^3 \text {ArcSin}(x)+\frac {3 \text {ArcSin}(x)^2}{16}+\frac {x^4}{16}+\frac {3 x^2}{16} \end {gather*}

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 30

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

Rule 4737

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

Rule 4795

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

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*}

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Mathematica [A]
time = 0.02, size = 43, normalized size = 0.70 \begin {gather*} \frac {1}{16} \left (x^2 \left (3+x^2\right )-2 x \sqrt {1-x^2} \left (3+2 x^2\right ) \sin ^{-1}(x)+3 \sin ^{-1}(x)^2\right ) \end {gather*}

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 = 0.14, size = 54, normalized size = 0.89

method	result	size
default	\(\frac {\arcsin \left (x \right ) \left (-2 \sqrt {-x^{2}+1}\, x^{3}-3 x \sqrt {-x^{2}+1}+3 \arcsin \left (x \right )\right )}{8}-\frac {3 \arcsin \left (x \right )^{2}}{16}+\frac {\left (2 x^{2}+3\right )^{2}}{64}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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+1/64*(2*x^2+3)^2

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Maxima [A]
time = 1.10, size = 52, normalized size = 0.85 \begin {gather*} \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 {gather*}

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 = 0.52, size = 39, normalized size = 0.64 \begin {gather*} \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 {gather*}

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 = 0.24, size = 53, normalized size = 0.87 \begin {gather*} \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 {gather*}

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 = 0.96, size = 50, normalized size = 0.82 \begin {gather*} \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 {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^4\,\mathrm {asin}\left (x\right )}{\sqrt {1-x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*asin(x))/(1 - x^2)^(1/2),x)

[Out]

int((x^4*asin(x))/(1 - x^2)^(1/2), x)

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