∫ x4 sin−1(x)_______

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

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Rubi [A] time = 0.11, 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 = {4707, 4641, 30} \[ \frac {x^4}{16}+\frac {3 x^2}{16}-\frac {1}{4} \sqrt {1-x^2} x^3 \sin ^{-1}(x)-\frac {3}{8} \sqrt {1-x^2} x \sin ^{-1}(x)+\frac {3}{16} \sin ^{-1}(x)^2 \]

(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

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

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,

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] &&

\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 \[ \frac {1}{16} \left (\left (x^2+3\right ) x^2-2 \sqrt {1-x^2} \left (2 x^2+3\right ) x \sin ^{-1}(x)+3 \sin ^{-1}(x)^2\right ) \]

(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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IntegrateAlgebraic [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^4 \sin ^{-1}(x)}{\sqrt {1-x^2}} \, dx \]

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

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fricas [A] time = 1.04, size = 39, normalized size = 0.64 \[ \frac {1}{16} \, x^{4} - \frac {1}{8} \, {\left (2 \, x^{3} + 3 \, x\right )} \sqrt {-x^{2} + 1} \arcsin \relax (x) + \frac {3}{16} \, x^{2} + \frac {3}{16} \, \arcsin \relax (x)^{2} \]

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

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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giac [A] time = 1.41, size = 50, normalized size = 0.82 \[ \frac {1}{4} \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}} x \arcsin \relax (x) - \frac {5}{8} \, \sqrt {-x^{2} + 1} x \arcsin \relax (x) + \frac {1}{16} \, {\left (x^{2} - 1\right )}^{2} + \frac {5}{16} \, x^{2} + \frac {3}{16} \, \arcsin \relax (x)^{2} - \frac {23}{128} \]

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

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(

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method	result	size

default	\(\frac {\arcsin \relax (x ) \left (-2 \sqrt {-x^{2}+1}\, x^{3}-3 \sqrt {-x^{2}+1}\, x +3 \arcsin \relax (x )\right )}{8}-\frac {3 \arcsin \relax (x )^{2}}{16}+\frac {x^{4}}{16}+\frac {3 x^{2}}{16}\)	\(53\)

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

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

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maxima [A] time = 0.97, size = 52, normalized size = 0.85 \[ \frac {1}{16} \, x^{4} + \frac {3}{16} \, x^{2} - \frac {1}{8} \, {\left (2 \, \sqrt {-x^{2} + 1} x^{3} + 3 \, \sqrt {-x^{2} + 1} x - 3 \, \arcsin \relax (x)\right )} \arcsin \relax (x) - \frac {3}{16} \, \arcsin \relax (x)^{2} \]

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

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)

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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^4\,\mathrm {asin}\relax (x)}{\sqrt {1-x^2}} \,d x \]

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sympy [A] time = 1.32, size = 53, normalized size = 0.87 \[ \frac {x^{4}}{16} - \frac {x^{3} \sqrt {1 - x^{2}} \operatorname {asin}{\relax (x )}}{4} + \frac {3 x^{2}}{16} - \frac {3 x \sqrt {1 - x^{2}} \operatorname {asin}{\relax (x )}}{8} + \frac {3 \operatorname {asin}^{2}{\relax (x )}}{16} \]

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