3.1 \(\int x^4 \sin ^{-1}(a x) \, dx\)

Optimal. Leaf size=75 \[ \frac{\left (1-a^2 x^2\right )^{5/2}}{25 a^5}-\frac{2 \left (1-a^2 x^2\right )^{3/2}}{15 a^5}+\frac{\sqrt{1-a^2 x^2}}{5 a^5}+\frac{1}{5} x^5 \sin ^{-1}(a x) \]

[Out]

Sqrt[1 - a^2*x^2]/(5*a^5) - (2*(1 - a^2*x^2)^(3/2))/(15*a^5) + (1 - a^2*x^2)^(5/2)/(25*a^5) + (x^5*ArcSin[a*x]
)/5

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Rubi [A]  time = 0.0489039, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {4627, 266, 43} \[ \frac{\left (1-a^2 x^2\right )^{5/2}}{25 a^5}-\frac{2 \left (1-a^2 x^2\right )^{3/2}}{15 a^5}+\frac{\sqrt{1-a^2 x^2}}{5 a^5}+\frac{1}{5} x^5 \sin ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[x^4*ArcSin[a*x],x]

[Out]

Sqrt[1 - a^2*x^2]/(5*a^5) - (2*(1 - a^2*x^2)^(3/2))/(15*a^5) + (1 - a^2*x^2)^(5/2)/(25*a^5) + (x^5*ArcSin[a*x]
)/5

Rule 4627

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int x^4 \sin ^{-1}(a x) \, dx &=\frac{1}{5} x^5 \sin ^{-1}(a x)-\frac{1}{5} a \int \frac{x^5}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{1}{5} x^5 \sin ^{-1}(a x)-\frac{1}{10} a \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-a^2 x}} \, dx,x,x^2\right )\\ &=\frac{1}{5} x^5 \sin ^{-1}(a x)-\frac{1}{10} a \operatorname{Subst}\left (\int \left (\frac{1}{a^4 \sqrt{1-a^2 x}}-\frac{2 \sqrt{1-a^2 x}}{a^4}+\frac{\left (1-a^2 x\right )^{3/2}}{a^4}\right ) \, dx,x,x^2\right )\\ &=\frac{\sqrt{1-a^2 x^2}}{5 a^5}-\frac{2 \left (1-a^2 x^2\right )^{3/2}}{15 a^5}+\frac{\left (1-a^2 x^2\right )^{5/2}}{25 a^5}+\frac{1}{5} x^5 \sin ^{-1}(a x)\\ \end{align*}

Mathematica [A]  time = 0.0320767, size = 51, normalized size = 0.68 \[ \frac{\sqrt{1-a^2 x^2} \left (3 a^4 x^4+4 a^2 x^2+8\right )}{75 a^5}+\frac{1}{5} x^5 \sin ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*ArcSin[a*x],x]

[Out]

(Sqrt[1 - a^2*x^2]*(8 + 4*a^2*x^2 + 3*a^4*x^4))/(75*a^5) + (x^5*ArcSin[a*x])/5

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Maple [A]  time = 0.032, size = 72, normalized size = 1. \begin{align*}{\frac{1}{{a}^{5}} \left ({\frac{{a}^{5}{x}^{5}\arcsin \left ( ax \right ) }{5}}+{\frac{{a}^{4}{x}^{4}}{25}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{4\,{a}^{2}{x}^{2}}{75}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{8}{75}\sqrt{-{a}^{2}{x}^{2}+1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*arcsin(a*x),x)

[Out]

1/a^5*(1/5*a^5*x^5*arcsin(a*x)+1/25*a^4*x^4*(-a^2*x^2+1)^(1/2)+4/75*a^2*x^2*(-a^2*x^2+1)^(1/2)+8/75*(-a^2*x^2+
1)^(1/2))

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Maxima [A]  time = 1.68565, size = 96, normalized size = 1.28 \begin{align*} \frac{1}{5} \, x^{5} \arcsin \left (a x\right ) + \frac{1}{75} \,{\left (\frac{3 \, \sqrt{-a^{2} x^{2} + 1} x^{4}}{a^{2}} + \frac{4 \, \sqrt{-a^{2} x^{2} + 1} x^{2}}{a^{4}} + \frac{8 \, \sqrt{-a^{2} x^{2} + 1}}{a^{6}}\right )} a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsin(a*x),x, algorithm="maxima")

[Out]

1/5*x^5*arcsin(a*x) + 1/75*(3*sqrt(-a^2*x^2 + 1)*x^4/a^2 + 4*sqrt(-a^2*x^2 + 1)*x^2/a^4 + 8*sqrt(-a^2*x^2 + 1)
/a^6)*a

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Fricas [A]  time = 2.13446, size = 113, normalized size = 1.51 \begin{align*} \frac{15 \, a^{5} x^{5} \arcsin \left (a x\right ) +{\left (3 \, a^{4} x^{4} + 4 \, a^{2} x^{2} + 8\right )} \sqrt{-a^{2} x^{2} + 1}}{75 \, a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsin(a*x),x, algorithm="fricas")

[Out]

1/75*(15*a^5*x^5*arcsin(a*x) + (3*a^4*x^4 + 4*a^2*x^2 + 8)*sqrt(-a^2*x^2 + 1))/a^5

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Sympy [A]  time = 7.08271, size = 70, normalized size = 0.93 \begin{align*} \begin{cases} \frac{x^{5} \operatorname{asin}{\left (a x \right )}}{5} + \frac{x^{4} \sqrt{- a^{2} x^{2} + 1}}{25 a} + \frac{4 x^{2} \sqrt{- a^{2} x^{2} + 1}}{75 a^{3}} + \frac{8 \sqrt{- a^{2} x^{2} + 1}}{75 a^{5}} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*asin(a*x),x)

[Out]

Piecewise((x**5*asin(a*x)/5 + x**4*sqrt(-a**2*x**2 + 1)/(25*a) + 4*x**2*sqrt(-a**2*x**2 + 1)/(75*a**3) + 8*sqr
t(-a**2*x**2 + 1)/(75*a**5), Ne(a, 0)), (0, True))

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Giac [A]  time = 1.2218, size = 153, normalized size = 2.04 \begin{align*} \frac{{\left (a^{2} x^{2} - 1\right )}^{2} x \arcsin \left (a x\right )}{5 \, a^{4}} + \frac{2 \,{\left (a^{2} x^{2} - 1\right )} x \arcsin \left (a x\right )}{5 \, a^{4}} + \frac{x \arcsin \left (a x\right )}{5 \, a^{4}} + \frac{{\left (a^{2} x^{2} - 1\right )}^{2} \sqrt{-a^{2} x^{2} + 1}}{25 \, a^{5}} - \frac{2 \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{15 \, a^{5}} + \frac{\sqrt{-a^{2} x^{2} + 1}}{5 \, a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsin(a*x),x, algorithm="giac")

[Out]

1/5*(a^2*x^2 - 1)^2*x*arcsin(a*x)/a^4 + 2/5*(a^2*x^2 - 1)*x*arcsin(a*x)/a^4 + 1/5*x*arcsin(a*x)/a^4 + 1/25*(a^
2*x^2 - 1)^2*sqrt(-a^2*x^2 + 1)/a^5 - 2/15*(-a^2*x^2 + 1)^(3/2)/a^5 + 1/5*sqrt(-a^2*x^2 + 1)/a^5