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

Optimal. Leaf size=83 \[ \frac{\text{Si}\left (2 \cos ^{-1}(a x)\right )}{2 a^4}+\frac{\text{Si}\left (4 \cos ^{-1}(a x)\right )}{a^4}+\frac{x^3 \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)^2}-\frac{3 x^2}{2 a^2 \cos ^{-1}(a x)}+\frac{2 x^4}{\cos ^{-1}(a x)} \]

[Out]

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

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Rubi [A]  time = 0.304897, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {4634, 4720, 4636, 4406, 3299, 12} \[ \frac{\text{Si}\left (2 \cos ^{-1}(a x)\right )}{2 a^4}+\frac{\text{Si}\left (4 \cos ^{-1}(a x)\right )}{a^4}+\frac{x^3 \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)^2}-\frac{3 x^2}{2 a^2 \cos ^{-1}(a x)}+\frac{2 x^4}{\cos ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4634

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> -Simp[(x^m*Sqrt[1 - c^2*x^2]*(a + b*ArcCo
s[c*x])^(n + 1))/(b*c*(n + 1)), x] + (-Dist[(c*(m + 1))/(b*(n + 1)), Int[(x^(m + 1)*(a + b*ArcCos[c*x])^(n + 1
))/Sqrt[1 - c^2*x^2], x], x] + Dist[m/(b*c*(n + 1)), Int[(x^(m - 1)*(a + b*ArcCos[c*x])^(n + 1))/Sqrt[1 - c^2*
x^2], x], x]) /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && LtQ[n, -2]

Rule 4720

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

Rule 4636

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> -Dist[(c^(m + 1))^(-1), Subst[Int[(a + b*
x)^n*Cos[x]^m*Sin[x], x], x, ArcCos[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rubi steps

\begin{align*} \int \frac{x^3}{\cos ^{-1}(a x)^3} \, dx &=\frac{x^3 \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)^2}-\frac{3 \int \frac{x^2}{\sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2} \, dx}{2 a}+(2 a) \int \frac{x^4}{\sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2} \, dx\\ &=\frac{x^3 \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)^2}-\frac{3 x^2}{2 a^2 \cos ^{-1}(a x)}+\frac{2 x^4}{\cos ^{-1}(a x)}-8 \int \frac{x^3}{\cos ^{-1}(a x)} \, dx+\frac{3 \int \frac{x}{\cos ^{-1}(a x)} \, dx}{a^2}\\ &=\frac{x^3 \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)^2}-\frac{3 x^2}{2 a^2 \cos ^{-1}(a x)}+\frac{2 x^4}{\cos ^{-1}(a x)}-\frac{3 \operatorname{Subst}\left (\int \frac{\cos (x) \sin (x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{a^4}+\frac{8 \operatorname{Subst}\left (\int \frac{\cos ^3(x) \sin (x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{a^4}\\ &=\frac{x^3 \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)^2}-\frac{3 x^2}{2 a^2 \cos ^{-1}(a x)}+\frac{2 x^4}{\cos ^{-1}(a x)}-\frac{3 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{2 x} \, dx,x,\cos ^{-1}(a x)\right )}{a^4}+\frac{8 \operatorname{Subst}\left (\int \left (\frac{\sin (2 x)}{4 x}+\frac{\sin (4 x)}{8 x}\right ) \, dx,x,\cos ^{-1}(a x)\right )}{a^4}\\ &=\frac{x^3 \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)^2}-\frac{3 x^2}{2 a^2 \cos ^{-1}(a x)}+\frac{2 x^4}{\cos ^{-1}(a x)}+\frac{\operatorname{Subst}\left (\int \frac{\sin (4 x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{a^4}-\frac{3 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{2 a^4}+\frac{2 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{a^4}\\ &=\frac{x^3 \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)^2}-\frac{3 x^2}{2 a^2 \cos ^{-1}(a x)}+\frac{2 x^4}{\cos ^{-1}(a x)}+\frac{\text{Si}\left (2 \cos ^{-1}(a x)\right )}{2 a^4}+\frac{\text{Si}\left (4 \cos ^{-1}(a x)\right )}{a^4}\\ \end{align*}

Mathematica [A]  time = 0.152084, size = 70, normalized size = 0.84 \[ \frac{\frac{a^2 x^2 \left (a x \sqrt{1-a^2 x^2}+\left (4 a^2 x^2-3\right ) \cos ^{-1}(a x)\right )}{\cos ^{-1}(a x)^2}+\text{Si}\left (2 \cos ^{-1}(a x)\right )+2 \text{Si}\left (4 \cos ^{-1}(a x)\right )}{2 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.055, size = 82, normalized size = 1. \begin{align*}{\frac{1}{{a}^{4}} \left ({\frac{\sin \left ( 2\,\arccos \left ( ax \right ) \right ) }{8\, \left ( \arccos \left ( ax \right ) \right ) ^{2}}}+{\frac{\cos \left ( 2\,\arccos \left ( ax \right ) \right ) }{4\,\arccos \left ( ax \right ) }}+{\frac{{\it Si} \left ( 2\,\arccos \left ( ax \right ) \right ) }{2}}+{\frac{\sin \left ( 4\,\arccos \left ( ax \right ) \right ) }{16\, \left ( \arccos \left ( ax \right ) \right ) ^{2}}}+{\frac{\cos \left ( 4\,\arccos \left ( ax \right ) \right ) }{4\,\arccos \left ( ax \right ) }}+{\it Si} \left ( 4\,\arccos \left ( ax \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^4*(1/8/arccos(a*x)^2*sin(2*arccos(a*x))+1/4/arccos(a*x)*cos(2*arccos(a*x))+1/2*Si(2*arccos(a*x))+1/16*sin(
4*arccos(a*x))/arccos(a*x)^2+1/4/arccos(a*x)*cos(4*arccos(a*x))+Si(4*arccos(a*x)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{a x + 1} \sqrt{-a x + 1} a x^{3} - 2 \, \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )^{2} \int \frac{{\left (8 \, a^{2} x^{2} - 3\right )} x}{\arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )}\,{d x} +{\left (4 \, a^{2} x^{4} - 3 \, x^{2}\right )} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )}{2 \, a^{2} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(sqrt(a*x + 1)*sqrt(-a*x + 1)*a*x^3 - 2*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^2*integrate((8*a^2*x^3
- 3*x)/arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x), x) + (4*a^2*x^4 - 3*x^2)*arctan2(sqrt(a*x + 1)*sqrt(-a*x +
1), a*x))/(a^2*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^2)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{3}}{\arccos \left (a x\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^3/arccos(a*x)^3, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\operatorname{acos}^{3}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3/acos(a*x)**3, x)

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Giac [A]  time = 1.20064, size = 101, normalized size = 1.22 \begin{align*} \frac{2 \, x^{4}}{\arccos \left (a x\right )} + \frac{\sqrt{-a^{2} x^{2} + 1} x^{3}}{2 \, a \arccos \left (a x\right )^{2}} - \frac{3 \, x^{2}}{2 \, a^{2} \arccos \left (a x\right )} + \frac{\operatorname{Si}\left (4 \, \arccos \left (a x\right )\right )}{a^{4}} + \frac{\operatorname{Si}\left (2 \, \arccos \left (a x\right )\right )}{2 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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