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

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

[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*ArcCos[c*x]))/3

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Rubi [A]  time = 0.0418091, 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 = {4628, 266, 43} \[ \frac{1}{3} x^3 \left (a+b \cos ^{-1}(c x)\right )+\frac{b \left (1-c^2 x^2\right )^{3/2}}{9 c^3}-\frac{b \sqrt{1-c^2 x^2}}{3 c^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcCos[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*ArcCos[c*x]))/3

Rule 4628

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcCo
s[c*x])^n)/(d*(m + 1)), x] + Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCos[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 \cos ^{-1}(c x)\right ) \, dx &=\frac{1}{3} x^3 \left (a+b \cos ^{-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 \cos ^{-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 \cos ^{-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 \cos ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.0417632, size = 55, normalized size = 0.92 \[ \frac{a x^3}{3}+b \left (-\frac{2}{9 c^3}-\frac{x^2}{9 c}\right ) \sqrt{1-c^2 x^2}+\frac{1}{3} b x^3 \cos ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*ArcCos[c*x]),x]

[Out]

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

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Maple [A]  time = 0.005, 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}\arccos \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*arccos(c*x)),x)

[Out]

1/c^3*(1/3*c^3*x^3*a+b*(1/3*c^3*x^3*arccos(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 = 1.4429, size = 81, normalized size = 1.35 \begin{align*} \frac{1}{3} \, a x^{3} + \frac{1}{9} \,{\left (3 \, x^{3} \arccos \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*arccos(c*x)),x, algorithm="maxima")

[Out]

1/3*a*x^3 + 1/9*(3*x^3*arccos(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 = 2.59659, size = 119, normalized size = 1.98 \begin{align*} \frac{3 \, b c^{3} x^{3} \arccos \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*arccos(c*x)),x, algorithm="fricas")

[Out]

1/9*(3*b*c^3*x^3*arccos(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 = 0.642933, size = 70, normalized size = 1.17 \begin{align*} \begin{cases} \frac{a x^{3}}{3} + \frac{b x^{3} \operatorname{acos}{\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{x^{3} \left (a + \frac{\pi b}{2}\right )}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*x**3/3 + b*x**3*acos(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)), (x**3*(a + pi*b/2)/3, True))

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Giac [A]  time = 1.1333, size = 76, normalized size = 1.27 \begin{align*} \frac{1}{3} \, b x^{3} \arccos \left (c x\right ) + \frac{1}{3} \, a x^{3} - \frac{\sqrt{-c^{2} x^{2} + 1} b x^{2}}{9 \, c} - \frac{2 \, \sqrt{-c^{2} x^{2} + 1} b}{9 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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