3.18 \(\int \frac{\cos ^{-1}(a x)^2}{x^2} \, dx\)

Optimal. Leaf size=74 \[ 2 i a \text{PolyLog}\left (2,-i e^{i \cos ^{-1}(a x)}\right )-2 i a \text{PolyLog}\left (2,i e^{i \cos ^{-1}(a x)}\right )-\frac{\cos ^{-1}(a x)^2}{x}-4 i a \cos ^{-1}(a x) \tan ^{-1}\left (e^{i \cos ^{-1}(a x)}\right ) \]

[Out]

-(ArcCos[a*x]^2/x) - (4*I)*a*ArcCos[a*x]*ArcTan[E^(I*ArcCos[a*x])] + (2*I)*a*PolyLog[2, (-I)*E^(I*ArcCos[a*x])
] - (2*I)*a*PolyLog[2, I*E^(I*ArcCos[a*x])]

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Rubi [A]  time = 0.106937, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4628, 4710, 4181, 2279, 2391} \[ 2 i a \text{PolyLog}\left (2,-i e^{i \cos ^{-1}(a x)}\right )-2 i a \text{PolyLog}\left (2,i e^{i \cos ^{-1}(a x)}\right )-\frac{\cos ^{-1}(a x)^2}{x}-4 i a \cos ^{-1}(a x) \tan ^{-1}\left (e^{i \cos ^{-1}(a x)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcCos[a*x]^2/x) - (4*I)*a*ArcCos[a*x]*ArcTan[E^(I*ArcCos[a*x])] + (2*I)*a*PolyLog[2, (-I)*E^(I*ArcCos[a*x])
] - (2*I)*a*PolyLog[2, I*E^(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 4710

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

Rule 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && 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)^2}{x^2} \, dx &=-\frac{\cos ^{-1}(a x)^2}{x}-(2 a) \int \frac{\cos ^{-1}(a x)}{x \sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{\cos ^{-1}(a x)^2}{x}+(2 a) \operatorname{Subst}\left (\int x \sec (x) \, dx,x,\cos ^{-1}(a x)\right )\\ &=-\frac{\cos ^{-1}(a x)^2}{x}-4 i a \cos ^{-1}(a x) \tan ^{-1}\left (e^{i \cos ^{-1}(a x)}\right )-(2 a) \operatorname{Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\cos ^{-1}(a x)\right )+(2 a) \operatorname{Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\cos ^{-1}(a x)\right )\\ &=-\frac{\cos ^{-1}(a x)^2}{x}-4 i a \cos ^{-1}(a x) \tan ^{-1}\left (e^{i \cos ^{-1}(a x)}\right )+(2 i a) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i \cos ^{-1}(a x)}\right )-(2 i a) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i \cos ^{-1}(a x)}\right )\\ &=-\frac{\cos ^{-1}(a x)^2}{x}-4 i a \cos ^{-1}(a x) \tan ^{-1}\left (e^{i \cos ^{-1}(a x)}\right )+2 i a \text{Li}_2\left (-i e^{i \cos ^{-1}(a x)}\right )-2 i a \text{Li}_2\left (i e^{i \cos ^{-1}(a x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.131581, size = 98, normalized size = 1.32 \[ 2 i a \text{PolyLog}\left (2,-i e^{i \cos ^{-1}(a x)}\right )-2 i a \text{PolyLog}\left (2,i e^{i \cos ^{-1}(a x)}\right )-\frac{\cos ^{-1}(a x) \left (\cos ^{-1}(a x)+2 a x \left (\log \left (1+i e^{i \cos ^{-1}(a x)}\right )-\log \left (1-i e^{i \cos ^{-1}(a x)}\right )\right )\right )}{x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.105, size = 135, normalized size = 1.8 \begin{align*} -{\frac{ \left ( \arccos \left ( ax \right ) \right ) ^{2}}{x}}-2\,a\arccos \left ( ax \right ) \ln \left ( 1+i \left ( i\sqrt{-{a}^{2}{x}^{2}+1}+ax \right ) \right ) +2\,a\arccos \left ( ax \right ) \ln \left ( 1-i \left ( i\sqrt{-{a}^{2}{x}^{2}+1}+ax \right ) \right ) +2\,ia{\it dilog} \left ( 1+i \left ( i\sqrt{-{a}^{2}{x}^{2}+1}+ax \right ) \right ) -2\,ia{\it dilog} \left ( 1-i \left ( i\sqrt{-{a}^{2}{x}^{2}+1}+ax \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arccos(a*x)^2/x-2*a*arccos(a*x)*ln(1+I*(I*(-a^2*x^2+1)^(1/2)+a*x))+2*a*arccos(a*x)*ln(1-I*(I*(-a^2*x^2+1)^(1/
2)+a*x))+2*I*a*dilog(1+I*(I*(-a^2*x^2+1)^(1/2)+a*x))-2*I*a*dilog(1-I*(I*(-a^2*x^2+1)^(1/2)+a*x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*x*integrate(sqrt(a*x + 1)*sqrt(-a*x + 1)*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)/(a^2*x^3 - x), x) - a
rctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^2)/x

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\arccos \left (a x\right )^{2}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x)^2/x^2,x, algorithm="fricas")

[Out]

integral(arccos(a*x)^2/x^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{acos}^{2}{\left (a x \right )}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acos(a*x)**2/x**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arccos \left (a x\right )^{2}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccos(a*x)^2/x^2, x)