3.2 \(\int x^3 \cos ^{-1}(a x) \, dx\)

Optimal. Leaf size=69 \[ -\frac{x^3 \sqrt{1-a^2 x^2}}{16 a}-\frac{3 x \sqrt{1-a^2 x^2}}{32 a^3}+\frac{3 \sin ^{-1}(a x)}{32 a^4}+\frac{1}{4} x^4 \cos ^{-1}(a x) \]

[Out]

(-3*x*Sqrt[1 - a^2*x^2])/(32*a^3) - (x^3*Sqrt[1 - a^2*x^2])/(16*a) + (x^4*ArcCos[a*x])/4 + (3*ArcSin[a*x])/(32
*a^4)

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Rubi [A]  time = 0.0308727, antiderivative size = 69, 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 = {4628, 321, 216} \[ -\frac{x^3 \sqrt{1-a^2 x^2}}{16 a}-\frac{3 x \sqrt{1-a^2 x^2}}{32 a^3}+\frac{3 \sin ^{-1}(a x)}{32 a^4}+\frac{1}{4} x^4 \cos ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcCos[a*x],x]

[Out]

(-3*x*Sqrt[1 - a^2*x^2])/(32*a^3) - (x^3*Sqrt[1 - a^2*x^2])/(16*a) + (x^4*ArcCos[a*x])/4 + (3*ArcSin[a*x])/(32
*a^4)

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int x^3 \cos ^{-1}(a x) \, dx &=\frac{1}{4} x^4 \cos ^{-1}(a x)+\frac{1}{4} a \int \frac{x^4}{\sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{x^3 \sqrt{1-a^2 x^2}}{16 a}+\frac{1}{4} x^4 \cos ^{-1}(a x)+\frac{3 \int \frac{x^2}{\sqrt{1-a^2 x^2}} \, dx}{16 a}\\ &=-\frac{3 x \sqrt{1-a^2 x^2}}{32 a^3}-\frac{x^3 \sqrt{1-a^2 x^2}}{16 a}+\frac{1}{4} x^4 \cos ^{-1}(a x)+\frac{3 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{32 a^3}\\ &=-\frac{3 x \sqrt{1-a^2 x^2}}{32 a^3}-\frac{x^3 \sqrt{1-a^2 x^2}}{16 a}+\frac{1}{4} x^4 \cos ^{-1}(a x)+\frac{3 \sin ^{-1}(a x)}{32 a^4}\\ \end{align*}

Mathematica [A]  time = 0.0320315, size = 54, normalized size = 0.78 \[ \frac{-a x \sqrt{1-a^2 x^2} \left (2 a^2 x^2+3\right )+8 a^4 x^4 \cos ^{-1}(a x)+3 \sin ^{-1}(a x)}{32 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcCos[a*x],x]

[Out]

(-(a*x*Sqrt[1 - a^2*x^2]*(3 + 2*a^2*x^2)) + 8*a^4*x^4*ArcCos[a*x] + 3*ArcSin[a*x])/(32*a^4)

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Maple [A]  time = 0.004, size = 60, normalized size = 0.9 \begin{align*}{\frac{1}{{a}^{4}} \left ({\frac{{a}^{4}{x}^{4}\arccos \left ( ax \right ) }{4}}-{\frac{{a}^{3}{x}^{3}}{16}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{3\,ax}{32}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{3\,\arcsin \left ( ax \right ) }{32}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arccos(a*x),x)

[Out]

1/a^4*(1/4*a^4*x^4*arccos(a*x)-1/16*a^3*x^3*(-a^2*x^2+1)^(1/2)-3/32*a*x*(-a^2*x^2+1)^(1/2)+3/32*arcsin(a*x))

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Maxima [A]  time = 1.4617, size = 99, normalized size = 1.43 \begin{align*} \frac{1}{4} \, x^{4} \arccos \left (a x\right ) - \frac{1}{32} \,{\left (\frac{2 \, \sqrt{-a^{2} x^{2} + 1} x^{3}}{a^{2}} + \frac{3 \, \sqrt{-a^{2} x^{2} + 1} x}{a^{4}} - \frac{3 \, \arcsin \left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{\sqrt{a^{2}} a^{4}}\right )} a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccos(a*x),x, algorithm="maxima")

[Out]

1/4*x^4*arccos(a*x) - 1/32*(2*sqrt(-a^2*x^2 + 1)*x^3/a^2 + 3*sqrt(-a^2*x^2 + 1)*x/a^4 - 3*arcsin(a^2*x/sqrt(a^
2))/(sqrt(a^2)*a^4))*a

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Fricas [A]  time = 1.95799, size = 109, normalized size = 1.58 \begin{align*} \frac{{\left (8 \, a^{4} x^{4} - 3\right )} \arccos \left (a x\right ) -{\left (2 \, a^{3} x^{3} + 3 \, a x\right )} \sqrt{-a^{2} x^{2} + 1}}{32 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccos(a*x),x, algorithm="fricas")

[Out]

1/32*((8*a^4*x^4 - 3)*arccos(a*x) - (2*a^3*x^3 + 3*a*x)*sqrt(-a^2*x^2 + 1))/a^4

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Sympy [A]  time = 1.08707, size = 66, normalized size = 0.96 \begin{align*} \begin{cases} \frac{x^{4} \operatorname{acos}{\left (a x \right )}}{4} - \frac{x^{3} \sqrt{- a^{2} x^{2} + 1}}{16 a} - \frac{3 x \sqrt{- a^{2} x^{2} + 1}}{32 a^{3}} - \frac{3 \operatorname{acos}{\left (a x \right )}}{32 a^{4}} & \text{for}\: a \neq 0 \\\frac{\pi x^{4}}{8} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*acos(a*x),x)

[Out]

Piecewise((x**4*acos(a*x)/4 - x**3*sqrt(-a**2*x**2 + 1)/(16*a) - 3*x*sqrt(-a**2*x**2 + 1)/(32*a**3) - 3*acos(a
*x)/(32*a**4), Ne(a, 0)), (pi*x**4/8, True))

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Giac [A]  time = 1.17655, size = 77, normalized size = 1.12 \begin{align*} \frac{1}{4} \, x^{4} \arccos \left (a x\right ) - \frac{\sqrt{-a^{2} x^{2} + 1} x^{3}}{16 \, a} - \frac{3 \, \sqrt{-a^{2} x^{2} + 1} x}{32 \, a^{3}} - \frac{3 \, \arccos \left (a x\right )}{32 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccos(a*x),x, algorithm="giac")

[Out]

1/4*x^4*arccos(a*x) - 1/16*sqrt(-a^2*x^2 + 1)*x^3/a - 3/32*sqrt(-a^2*x^2 + 1)*x/a^3 - 3/32*arccos(a*x)/a^4