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

Optimal. Leaf size=102 \[ -\frac{3}{2} i a^2 \text{PolyLog}\left (2,-e^{2 i \cos ^{-1}(a x)}\right )+\frac{3 a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{2 x}-\frac{3}{2} i a^2 \cos ^{-1}(a x)^2+3 a^2 \cos ^{-1}(a x) \log \left (1+e^{2 i \cos ^{-1}(a x)}\right )-\frac{\cos ^{-1}(a x)^3}{2 x^2} \]

[Out]

((-3*I)/2)*a^2*ArcCos[a*x]^2 + (3*a*Sqrt[1 - a^2*x^2]*ArcCos[a*x]^2)/(2*x) - ArcCos[a*x]^3/(2*x^2) + 3*a^2*Arc
Cos[a*x]*Log[1 + E^((2*I)*ArcCos[a*x])] - ((3*I)/2)*a^2*PolyLog[2, -E^((2*I)*ArcCos[a*x])]

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Rubi [A]  time = 0.175343, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {4628, 4682, 4626, 3719, 2190, 2279, 2391} \[ -\frac{3}{2} i a^2 \text{PolyLog}\left (2,-e^{2 i \cos ^{-1}(a x)}\right )+\frac{3 a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{2 x}-\frac{3}{2} i a^2 \cos ^{-1}(a x)^2+3 a^2 \cos ^{-1}(a x) \log \left (1+e^{2 i \cos ^{-1}(a x)}\right )-\frac{\cos ^{-1}(a x)^3}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*I)/2)*a^2*ArcCos[a*x]^2 + (3*a*Sqrt[1 - a^2*x^2]*ArcCos[a*x]^2)/(2*x) - ArcCos[a*x]^3/(2*x^2) + 3*a^2*Arc
Cos[a*x]*Log[1 + E^((2*I)*ArcCos[a*x])] - ((3*I)/2)*a^2*PolyLog[2, -E^((2*I)*ArcCos[a*x])]

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 4682

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(
(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcCos[c*x])^n)/(d*f*(m + 1)), x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x
^2)^FracPart[p])/(f*(m + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcCo
s[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && EqQ[m + 2*p
+ 3, 0] && NeQ[m, -1]

Rule 4626

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

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{\cos ^{-1}(a x)^3}{x^3} \, dx &=-\frac{\cos ^{-1}(a x)^3}{2 x^2}-\frac{1}{2} (3 a) \int \frac{\cos ^{-1}(a x)^2}{x^2 \sqrt{1-a^2 x^2}} \, dx\\ &=\frac{3 a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{2 x}-\frac{\cos ^{-1}(a x)^3}{2 x^2}+\left (3 a^2\right ) \int \frac{\cos ^{-1}(a x)}{x} \, dx\\ &=\frac{3 a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{2 x}-\frac{\cos ^{-1}(a x)^3}{2 x^2}-\left (3 a^2\right ) \operatorname{Subst}\left (\int x \tan (x) \, dx,x,\cos ^{-1}(a x)\right )\\ &=-\frac{3}{2} i a^2 \cos ^{-1}(a x)^2+\frac{3 a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{2 x}-\frac{\cos ^{-1}(a x)^3}{2 x^2}+\left (6 i a^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 i x} x}{1+e^{2 i x}} \, dx,x,\cos ^{-1}(a x)\right )\\ &=-\frac{3}{2} i a^2 \cos ^{-1}(a x)^2+\frac{3 a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{2 x}-\frac{\cos ^{-1}(a x)^3}{2 x^2}+3 a^2 \cos ^{-1}(a x) \log \left (1+e^{2 i \cos ^{-1}(a x)}\right )-\left (3 a^2\right ) \operatorname{Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\cos ^{-1}(a x)\right )\\ &=-\frac{3}{2} i a^2 \cos ^{-1}(a x)^2+\frac{3 a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{2 x}-\frac{\cos ^{-1}(a x)^3}{2 x^2}+3 a^2 \cos ^{-1}(a x) \log \left (1+e^{2 i \cos ^{-1}(a x)}\right )+\frac{1}{2} \left (3 i a^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i \cos ^{-1}(a x)}\right )\\ &=-\frac{3}{2} i a^2 \cos ^{-1}(a x)^2+\frac{3 a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{2 x}-\frac{\cos ^{-1}(a x)^3}{2 x^2}+3 a^2 \cos ^{-1}(a x) \log \left (1+e^{2 i \cos ^{-1}(a x)}\right )-\frac{3}{2} i a^2 \text{Li}_2\left (-e^{2 i \cos ^{-1}(a x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.192443, size = 92, normalized size = 0.9 \[ \frac{1}{2} \left (-3 i a^2 \text{PolyLog}\left (2,-e^{2 i \cos ^{-1}(a x)}\right )+\frac{3 a \left (\sqrt{1-a^2 x^2}-i a x\right ) \cos ^{-1}(a x)^2}{x}+6 a^2 \cos ^{-1}(a x) \log \left (1+e^{2 i \cos ^{-1}(a x)}\right )-\frac{\cos ^{-1}(a x)^3}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a*((-I)*a*x + Sqrt[1 - a^2*x^2])*ArcCos[a*x]^2)/x - ArcCos[a*x]^3/x^2 + 6*a^2*ArcCos[a*x]*Log[1 + E^((2*I)
*ArcCos[a*x])] - (3*I)*a^2*PolyLog[2, -E^((2*I)*ArcCos[a*x])])/2

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Maple [A]  time = 0.139, size = 113, normalized size = 1.1 \begin{align*} -{\frac{3\,i}{2}}{a}^{2} \left ( \arccos \left ( ax \right ) \right ) ^{2}-{\frac{ \left ( \arccos \left ( ax \right ) \right ) ^{3}}{2\,{x}^{2}}}+3\,{a}^{2}\arccos \left ( ax \right ) \ln \left ( 1+ \left ( i\sqrt{-{a}^{2}{x}^{2}+1}+ax \right ) ^{2} \right ) -{\frac{3\,i}{2}}{a}^{2}{\it polylog} \left ( 2,- \left ( i\sqrt{-{a}^{2}{x}^{2}+1}+ax \right ) ^{2} \right ) +{\frac{3\,a \left ( \arccos \left ( ax \right ) \right ) ^{2}}{2\,x}\sqrt{-{a}^{2}{x}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*I*a^2*arccos(a*x)^2-1/2*arccos(a*x)^3/x^2+3*a^2*arccos(a*x)*ln(1+(I*(-a^2*x^2+1)^(1/2)+a*x)^2)-3/2*I*a^2*
polylog(2,-(I*(-a^2*x^2+1)^(1/2)+a*x)^2)+3/2*a*arccos(a*x)^2*(-a^2*x^2+1)^(1/2)/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(6*a*x^2*integrate(1/2*sqrt(a*x + 1)*sqrt(-a*x + 1)*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^2/(a^2*x^4
- x^2), x) - arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^3)/x^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arccos(a*x)^3/x^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acos(a*x)**3/x**3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccos(a*x)^3/x^3, x)