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

Optimal. Leaf size=51 \[ \frac{\sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)^2}+\frac{\text{Si}\left (\cos ^{-1}(a x)\right )}{2 a}+\frac{x}{2 \cos ^{-1}(a x)} \]

[Out]

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

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Rubi [A]  time = 0.0828363, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {4622, 4720, 4624, 3299} \[ \frac{\sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)^2}+\frac{\text{Si}\left (\cos ^{-1}(a x)\right )}{2 a}+\frac{x}{2 \cos ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a*x]^(-3),x]

[Out]

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

Rule 4622

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

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 4624

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

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]

Rubi steps

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

Mathematica [A]  time = 0.0306565, size = 47, normalized size = 0.92 \[ \frac{\sqrt{1-a^2 x^2}+\cos ^{-1}(a x)^2 \text{Si}\left (\cos ^{-1}(a x)\right )+a x \cos ^{-1}(a x)}{2 a \cos ^{-1}(a x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[a*x]^(-3),x]

[Out]

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

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Maple [A]  time = 0.045, size = 43, normalized size = 0.8 \begin{align*}{\frac{1}{a} \left ({\frac{1}{2\, \left ( \arccos \left ( ax \right ) \right ) ^{2}}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{ax}{2\,\arccos \left ( ax \right ) }}+{\frac{{\it Si} \left ( \arccos \left ( ax \right ) \right ) }{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/arccos(a*x)^3,x)

[Out]

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

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

[Out]

-1/2*(a*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^2*integrate(1/arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x), x
) - a*x*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x) - sqrt(a*x + 1)*sqrt(-a*x + 1))/(a*arctan2(sqrt(a*x + 1)*sq
rt(-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{1}{\arccos \left (a x\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccos(a*x)^3,x, algorithm="fricas")

[Out]

integral(arccos(a*x)^(-3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\operatorname{acos}^{3}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/acos(a*x)**3,x)

[Out]

Integral(acos(a*x)**(-3), x)

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Giac [A]  time = 1.15819, size = 58, normalized size = 1.14 \begin{align*} \frac{x}{2 \, \arccos \left (a x\right )} + \frac{\operatorname{Si}\left (\arccos \left (a x\right )\right )}{2 \, a} + \frac{\sqrt{-a^{2} x^{2} + 1}}{2 \, a \arccos \left (a x\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccos(a*x)^3,x, algorithm="giac")

[Out]

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