3.3 \(\int x^2 \cos ^{-1}(a x) \, dx\)

Optimal. Leaf size=54 \[ \frac {\left (1-a^2 x^2\right )^{3/2}}{9 a^3}-\frac {\sqrt {1-a^2 x^2}}{3 a^3}+\frac {1}{3} x^3 \cos ^{-1}(a x) \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 54, normalized size of antiderivative = 1.00, 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 = {4628, 266, 43} \[ \frac {\left (1-a^2 x^2\right )^{3/2}}{9 a^3}-\frac {\sqrt {1-a^2 x^2}}{3 a^3}+\frac {1}{3} x^3 \cos ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcCos[a*x],x]

[Out]

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

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])

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 42, normalized size = 0.78 \[ \frac {1}{3} x^3 \cos ^{-1}(a x)-\frac {\sqrt {1-a^2 x^2} \left (a^2 x^2+2\right )}{9 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcCos[a*x],x]

[Out]

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

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fricas [A]  time = 1.31, size = 41, normalized size = 0.76 \[ \frac {3 \, a^{3} x^{3} \arccos \left (a x\right ) - {\left (a^{2} x^{2} + 2\right )} \sqrt {-a^{2} x^{2} + 1}}{9 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccos(a*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.00, size = 47, normalized size = 0.87 \[ \frac {1}{3} \, x^{3} \arccos \left (a x\right ) - \frac {\sqrt {-a^{2} x^{2} + 1} x^{2}}{9 \, a} - \frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{9 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 52, normalized size = 0.96 \[ \frac {\frac {a^{3} x^{3} \arccos \left (a x \right )}{3}-\frac {a^{2} x^{2} \sqrt {-a^{2} x^{2}+1}}{9}-\frac {2 \sqrt {-a^{2} x^{2}+1}}{9}}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arccos(a*x),x)

[Out]

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

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maxima [A]  time = 0.42, size = 50, normalized size = 0.93 \[ \frac {1}{3} \, x^{3} \arccos \left (a x\right ) - \frac {1}{9} \, a {\left (\frac {\sqrt {-a^{2} x^{2} + 1} x^{2}}{a^{2}} + \frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{a^{4}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccos(a*x),x, algorithm="maxima")

[Out]

1/3*x^3*arccos(a*x) - 1/9*a*(sqrt(-a^2*x^2 + 1)*x^2/a^2 + 2*sqrt(-a^2*x^2 + 1)/a^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \left \{\begin {array}{cl} \frac {x^3\,\mathrm {acos}\left (a\,x\right )}{3}-\frac {\sqrt {\frac {1}{a^2}-x^2}\,\left (\frac {2}{a^2}+x^2\right )}{9} & \text {\ if\ \ }0<a\\ \int x^2\,\mathrm {acos}\left (a\,x\right ) \,d x & \text {\ if\ \ }\neg 0<a \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*acos(a*x),x)

[Out]

piecewise(0 < a, - ((1/a^2 - x^2)^(1/2)*(2/a^2 + x^2))/9 + (x^3*acos(a*x))/3, ~0 < a, int(x^2*acos(a*x), x))

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sympy [A]  time = 0.43, size = 53, normalized size = 0.98 \[ \begin {cases} \frac {x^{3} \operatorname {acos}{\left (a x \right )}}{3} - \frac {x^{2} \sqrt {- a^{2} x^{2} + 1}}{9 a} - \frac {2 \sqrt {- a^{2} x^{2} + 1}}{9 a^{3}} & \text {for}\: a \neq 0 \\\frac {\pi x^{3}}{6} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*acos(a*x),x)

[Out]

Piecewise((x**3*acos(a*x)/3 - x**2*sqrt(-a**2*x**2 + 1)/(9*a) - 2*sqrt(-a**2*x**2 + 1)/(9*a**3), Ne(a, 0)), (p
i*x**3/6, True))

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