3.141 \(\int x^2 (a+b \sin ^{-1}(c x)) \, dx\)

Optimal. Leaf size=60 \[ \frac{1}{3} x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{b \left (1-c^2 x^2\right )^{3/2}}{9 c^3}+\frac{b \sqrt{1-c^2 x^2}}{3 c^3} \]

[Out]

(b*Sqrt[1 - c^2*x^2])/(3*c^3) - (b*(1 - c^2*x^2)^(3/2))/(9*c^3) + (x^3*(a + b*ArcSin[c*x]))/3

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Rubi [A]  time = 0.0391678, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4627, 266, 43} \[ \frac{1}{3} x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{b \left (1-c^2 x^2\right )^{3/2}}{9 c^3}+\frac{b \sqrt{1-c^2 x^2}}{3 c^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcSin[c*x]),x]

[Out]

(b*Sqrt[1 - c^2*x^2])/(3*c^3) - (b*(1 - c^2*x^2)^(3/2))/(9*c^3) + (x^3*(a + b*ArcSin[c*x]))/3

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^2 \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac{1}{3} x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{3} (b c) \int \frac{x^3}{\sqrt{1-c^2 x^2}} \, dx\\ &=\frac{1}{3} x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{6} (b c) \operatorname{Subst}\left (\int \frac{x}{\sqrt{1-c^2 x}} \, dx,x,x^2\right )\\ &=\frac{1}{3} x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{6} (b c) \operatorname{Subst}\left (\int \left (\frac{1}{c^2 \sqrt{1-c^2 x}}-\frac{\sqrt{1-c^2 x}}{c^2}\right ) \, dx,x,x^2\right )\\ &=\frac{b \sqrt{1-c^2 x^2}}{3 c^3}-\frac{b \left (1-c^2 x^2\right )^{3/2}}{9 c^3}+\frac{1}{3} x^3 \left (a+b \sin ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.0391751, size = 49, normalized size = 0.82 \[ \frac{1}{9} \left (3 a x^3+\frac{b \sqrt{1-c^2 x^2} \left (c^2 x^2+2\right )}{c^3}+3 b x^3 \sin ^{-1}(c x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*ArcSin[c*x]),x]

[Out]

(3*a*x^3 + (b*Sqrt[1 - c^2*x^2]*(2 + c^2*x^2))/c^3 + 3*b*x^3*ArcSin[c*x])/9

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Maple [A]  time = 0.003, size = 64, normalized size = 1.1 \begin{align*}{\frac{1}{{c}^{3}} \left ({\frac{{c}^{3}{x}^{3}a}{3}}+b \left ({\frac{{c}^{3}{x}^{3}\arcsin \left ( cx \right ) }{3}}+{\frac{{c}^{2}{x}^{2}}{9}\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{2}{9}\sqrt{-{c}^{2}{x}^{2}+1}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arcsin(c*x)),x)

[Out]

1/c^3*(1/3*c^3*x^3*a+b*(1/3*c^3*x^3*arcsin(c*x)+1/9*c^2*x^2*(-c^2*x^2+1)^(1/2)+2/9*(-c^2*x^2+1)^(1/2)))

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Maxima [A]  time = 1.6864, size = 80, normalized size = 1.33 \begin{align*} \frac{1}{3} \, a x^{3} + \frac{1}{9} \,{\left (3 \, x^{3} \arcsin \left (c x\right ) + c{\left (\frac{\sqrt{-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac{2 \, \sqrt{-c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsin(c*x)),x, algorithm="maxima")

[Out]

1/3*a*x^3 + 1/9*(3*x^3*arcsin(c*x) + c*(sqrt(-c^2*x^2 + 1)*x^2/c^2 + 2*sqrt(-c^2*x^2 + 1)/c^4))*b

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Fricas [A]  time = 1.52999, size = 119, normalized size = 1.98 \begin{align*} \frac{3 \, b c^{3} x^{3} \arcsin \left (c x\right ) + 3 \, a c^{3} x^{3} +{\left (b c^{2} x^{2} + 2 \, b\right )} \sqrt{-c^{2} x^{2} + 1}}{9 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

1/9*(3*b*c^3*x^3*arcsin(c*x) + 3*a*c^3*x^3 + (b*c^2*x^2 + 2*b)*sqrt(-c^2*x^2 + 1))/c^3

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Sympy [A]  time = 0.921519, size = 65, normalized size = 1.08 \begin{align*} \begin{cases} \frac{a x^{3}}{3} + \frac{b x^{3} \operatorname{asin}{\left (c x \right )}}{3} + \frac{b x^{2} \sqrt{- c^{2} x^{2} + 1}}{9 c} + \frac{2 b \sqrt{- c^{2} x^{2} + 1}}{9 c^{3}} & \text{for}\: c \neq 0 \\\frac{a x^{3}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*asin(c*x)),x)

[Out]

Piecewise((a*x**3/3 + b*x**3*asin(c*x)/3 + b*x**2*sqrt(-c**2*x**2 + 1)/(9*c) + 2*b*sqrt(-c**2*x**2 + 1)/(9*c**
3), Ne(c, 0)), (a*x**3/3, True))

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Giac [A]  time = 1.34687, size = 100, normalized size = 1.67 \begin{align*} \frac{1}{3} \, a x^{3} + \frac{{\left (c^{2} x^{2} - 1\right )} b x \arcsin \left (c x\right )}{3 \, c^{2}} + \frac{b x \arcsin \left (c x\right )}{3 \, c^{2}} - \frac{{\left (-c^{2} x^{2} + 1\right )}^{\frac{3}{2}} b}{9 \, c^{3}} + \frac{\sqrt{-c^{2} x^{2} + 1} b}{3 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsin(c*x)),x, algorithm="giac")

[Out]

1/3*a*x^3 + 1/3*(c^2*x^2 - 1)*b*x*arcsin(c*x)/c^2 + 1/3*b*x*arcsin(c*x)/c^2 - 1/9*(-c^2*x^2 + 1)^(3/2)*b/c^3 +
1/3*sqrt(-c^2*x^2 + 1)*b/c^3