3.7.0 ∫ x4 arcsin(x) √ 1−x2 dx [660]

Integrand size = 17, antiderivative size = 61 \[ \int \frac {x^4 \arcsin (x)}{\sqrt {1-x^2}} \, dx=\frac {3 x^2}{16}+\frac {x^4}{16}-\frac {3}{8} x \sqrt {1-x^2} \arcsin (x)-\frac {1}{4} x^3 \sqrt {1-x^2} \arcsin (x)+\frac {3 \arcsin (x)^2}{16} \]

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)

Rubi [A] (veriﬁed)

Time = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {4795, 4737, 30} \[ \int \frac {x^4 \arcsin (x)}{\sqrt {1-x^2}} \, dx=-\frac {3}{8} \sqrt {1-x^2} x \arcsin (x)-\frac {1}{4} \sqrt {1-x^2} x^3 \arcsin (x)+\frac {3 \arcsin (x)^2}{16}+\frac {x^4}{16}+\frac {3 x^2}{16} \]

(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[(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

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

Mathematica [A] (veriﬁed)

Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.70 \[ \int \frac {x^4 \arcsin (x)}{\sqrt {1-x^2}} \, dx=\frac {1}{16} \left (x^2 \left (3+x^2\right )-2 x \sqrt {1-x^2} \left (3+2 x^2\right ) \arcsin (x)+3 \arcsin (x)^2\right ) \]

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

Maple [A] (veriﬁed)

Time = 0.45 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89


method	result	size

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

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*(-x^2+1)^(1/2)+3*arcsin(x))-3/16*arcsin(x)^2+1/64*(2*x^2+3)^2

Fricas [A] (veriﬁcation not implemented)

Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.64 \[ \int \frac {x^4 \arcsin (x)}{\sqrt {1-x^2}} \, dx=\frac {1}{16} \, x^{4} - \frac {1}{8} \, {\left (2 \, x^{3} + 3 \, x\right )} \sqrt {-x^{2} + 1} \arcsin \left (x\right ) + \frac {3}{16} \, x^{2} + \frac {3}{16} \, \arcsin \left (x\right )^{2} \]

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

Sympy [A] (veriﬁcation not implemented)

Time = 0.21 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87 \[ \int \frac {x^4 \arcsin (x)}{\sqrt {1-x^2}} \, dx=\frac {x^{4}}{16} - \frac {x^{3} \sqrt {1 - x^{2}} \operatorname {asin}{\left (x \right )}}{4} + \frac {3 x^{2}}{16} - \frac {3 x \sqrt {1 - x^{2}} \operatorname {asin}{\left (x \right )}}{8} + \frac {3 \operatorname {asin}^{2}{\left (x \right )}}{16} \]

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

Maxima [A] (veriﬁcation not implemented)

Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.85 \[ \int \frac {x^4 \arcsin (x)}{\sqrt {1-x^2}} \, dx=\frac {1}{16} \, x^{4} + \frac {3}{16} \, x^{2} - \frac {1}{8} \, {\left (2 \, \sqrt {-x^{2} + 1} x^{3} + 3 \, \sqrt {-x^{2} + 1} x - 3 \, \arcsin \left (x\right )\right )} \arcsin \left (x\right ) - \frac {3}{16} \, \arcsin \left (x\right )^{2} \]

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)

Giac [A] (veriﬁcation not implemented)

Time = 0.33 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.82 \[ \int \frac {x^4 \arcsin (x)}{\sqrt {1-x^2}} \, dx=\frac {1}{4} \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}} x \arcsin \left (x\right ) - \frac {5}{8} \, \sqrt {-x^{2} + 1} x \arcsin \left (x\right ) + \frac {1}{16} \, {\left (x^{2} - 1\right )}^{2} + \frac {5}{16} \, x^{2} + \frac {3}{16} \, \arcsin \left (x\right )^{2} - \frac {23}{128} \]

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(

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 \arcsin (x)}{\sqrt {1-x^2}} \, dx=\int \frac {x^4\,\mathrm {asin}\left (x\right )}{\sqrt {1-x^2}} \,d x \]

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

Optimal result