3.13 \(\int x^3 \cos ^{-1}(a x)^2 \, dx\)

Optimal. Leaf size=98 \[ -\frac{3 x^2}{32 a^2}-\frac{x^3 \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{8 a}-\frac{3 x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{16 a^3}-\frac{3 \cos ^{-1}(a x)^2}{32 a^4}+\frac{1}{4} x^4 \cos ^{-1}(a x)^2-\frac{x^4}{32} \]

[Out]

(-3*x^2)/(32*a^2) - x^4/32 - (3*x*Sqrt[1 - a^2*x^2]*ArcCos[a*x])/(16*a^3) - (x^3*Sqrt[1 - a^2*x^2]*ArcCos[a*x]
)/(8*a) - (3*ArcCos[a*x]^2)/(32*a^4) + (x^4*ArcCos[a*x]^2)/4

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Rubi [A]  time = 0.169663, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4628, 4708, 4642, 30} \[ -\frac{3 x^2}{32 a^2}-\frac{x^3 \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{8 a}-\frac{3 x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{16 a^3}-\frac{3 \cos ^{-1}(a x)^2}{32 a^4}+\frac{1}{4} x^4 \cos ^{-1}(a x)^2-\frac{x^4}{32} \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcCos[a*x]^2,x]

[Out]

(-3*x^2)/(32*a^2) - x^4/32 - (3*x*Sqrt[1 - a^2*x^2]*ArcCos[a*x])/(16*a^3) - (x^3*Sqrt[1 - a^2*x^2]*ArcCos[a*x]
)/(8*a) - (3*ArcCos[a*x]^2)/(32*a^4) + (x^4*ArcCos[a*x]^2)/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 4708

Int[(((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[
(f*(f*x)^(m - 1)*Sqrt[d + e*x^2]*(a + b*ArcCos[c*x])^n)/(e*m), x] + (Dist[(f^2*(m - 1))/(c^2*m), Int[((f*x)^(m
- 2)*(a + b*ArcCos[c*x])^n)/Sqrt[d + e*x^2], x], x] - Dist[(b*f*n*Sqrt[1 - c^2*x^2])/(c*m*Sqrt[d + e*x^2]), I
nt[(f*x)^(m - 1)*(a + b*ArcCos[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] &&
GtQ[n, 0] && GtQ[m, 1] && IntegerQ[m]

Rule 4642

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> -Simp[(a + b*ArcCos[c*x])
^(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0] && NeQ[n,
-1]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int x^3 \cos ^{-1}(a x)^2 \, dx &=\frac{1}{4} x^4 \cos ^{-1}(a x)^2+\frac{1}{2} a \int \frac{x^4 \cos ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{x^3 \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{8 a}+\frac{1}{4} x^4 \cos ^{-1}(a x)^2-\frac{\int x^3 \, dx}{8}+\frac{3 \int \frac{x^2 \cos ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{8 a}\\ &=-\frac{x^4}{32}-\frac{3 x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{16 a^3}-\frac{x^3 \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{8 a}+\frac{1}{4} x^4 \cos ^{-1}(a x)^2+\frac{3 \int \frac{\cos ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{16 a^3}-\frac{3 \int x \, dx}{16 a^2}\\ &=-\frac{3 x^2}{32 a^2}-\frac{x^4}{32}-\frac{3 x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{16 a^3}-\frac{x^3 \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{8 a}-\frac{3 \cos ^{-1}(a x)^2}{32 a^4}+\frac{1}{4} x^4 \cos ^{-1}(a x)^2\\ \end{align*}

Mathematica [A]  time = 0.0376655, size = 74, normalized size = 0.76 \[ \frac{-a^2 x^2 \left (a^2 x^2+3\right )-2 a x \sqrt{1-a^2 x^2} \left (2 a^2 x^2+3\right ) \cos ^{-1}(a x)+\left (8 a^4 x^4-3\right ) \cos ^{-1}(a x)^2}{32 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcCos[a*x]^2,x]

[Out]

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

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Maple [A]  time = 0.054, size = 93, normalized size = 1. \begin{align*}{\frac{1}{{a}^{4}} \left ({\frac{{a}^{4}{x}^{4} \left ( \arccos \left ( ax \right ) \right ) ^{2}}{4}}-{\frac{\arccos \left ( ax \right ) }{16} \left ( 2\,{a}^{3}{x}^{3}\sqrt{-{a}^{2}{x}^{2}+1}+3\,ax\sqrt{-{a}^{2}{x}^{2}+1}+3\,\arccos \left ( ax \right ) \right ) }+{\frac{3\, \left ( \arccos \left ( ax \right ) \right ) ^{2}}{32}}-{\frac{{a}^{4}{x}^{4}}{32}}-{\frac{3\,{a}^{2}{x}^{2}}{32}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arccos(a*x)^2,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{4} \, x^{4} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )^{2} - a \int \frac{\sqrt{a x + 1} \sqrt{-a x + 1} x^{4} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )}{2 \,{\left (a^{2} x^{2} - 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccos(a*x)^2,x, algorithm="maxima")

[Out]

1/4*x^4*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^2 - a*integrate(1/2*sqrt(a*x + 1)*sqrt(-a*x + 1)*x^4*arctan
2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)/(a^2*x^2 - 1), x)

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Fricas [A]  time = 1.95436, size = 162, normalized size = 1.65 \begin{align*} -\frac{a^{4} x^{4} + 3 \, a^{2} x^{2} -{\left (8 \, a^{4} x^{4} - 3\right )} \arccos \left (a x\right )^{2} + 2 \,{\left (2 \, a^{3} x^{3} + 3 \, a x\right )} \sqrt{-a^{2} x^{2} + 1} \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)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.38094, size = 97, normalized size = 0.99 \begin{align*} \begin{cases} \frac{x^{4} \operatorname{acos}^{2}{\left (a x \right )}}{4} - \frac{x^{4}}{32} - \frac{x^{3} \sqrt{- a^{2} x^{2} + 1} \operatorname{acos}{\left (a x \right )}}{8 a} - \frac{3 x^{2}}{32 a^{2}} - \frac{3 x \sqrt{- a^{2} x^{2} + 1} \operatorname{acos}{\left (a x \right )}}{16 a^{3}} - \frac{3 \operatorname{acos}^{2}{\left (a x \right )}}{32 a^{4}} & \text{for}\: a \neq 0 \\\frac{\pi ^{2} x^{4}}{16} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*acos(a*x)**2,x)

[Out]

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

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Giac [A]  time = 1.1796, size = 117, normalized size = 1.19 \begin{align*} \frac{1}{4} \, x^{4} \arccos \left (a x\right )^{2} - \frac{1}{32} \, x^{4} - \frac{\sqrt{-a^{2} x^{2} + 1} x^{3} \arccos \left (a x\right )}{8 \, a} - \frac{3 \, x^{2}}{32 \, a^{2}} - \frac{3 \, \sqrt{-a^{2} x^{2} + 1} x \arccos \left (a x\right )}{16 \, a^{3}} - \frac{3 \, \arccos \left (a x\right )^{2}}{32 \, a^{4}} + \frac{15}{256 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccos(a*x)^2,x, algorithm="giac")

[Out]

1/4*x^4*arccos(a*x)^2 - 1/32*x^4 - 1/8*sqrt(-a^2*x^2 + 1)*x^3*arccos(a*x)/a - 3/32*x^2/a^2 - 3/16*sqrt(-a^2*x^
2 + 1)*x*arccos(a*x)/a^3 - 3/32*arccos(a*x)^2/a^4 + 15/256/a^4