3.20 \(\int \frac{\cos ^{-1}(a x)^2}{x^4} \, dx\)

Optimal. Leaf size=124 \[ \frac{1}{3} i a^3 \text{PolyLog}\left (2,-i e^{i \cos ^{-1}(a x)}\right )-\frac{1}{3} i a^3 \text{PolyLog}\left (2,i e^{i \cos ^{-1}(a x)}\right )+\frac{a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{3 x^2}-\frac{a^2}{3 x}-\frac{2}{3} i a^3 \cos ^{-1}(a x) \tan ^{-1}\left (e^{i \cos ^{-1}(a x)}\right )-\frac{\cos ^{-1}(a x)^2}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.176919, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {4628, 4702, 4710, 4181, 2279, 2391, 30} \[ \frac{1}{3} i a^3 \text{PolyLog}\left (2,-i e^{i \cos ^{-1}(a x)}\right )-\frac{1}{3} i a^3 \text{PolyLog}\left (2,i e^{i \cos ^{-1}(a x)}\right )+\frac{a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{3 x^2}-\frac{a^2}{3 x}-\frac{2}{3} i a^3 \cos ^{-1}(a x) \tan ^{-1}\left (e^{i \cos ^{-1}(a x)}\right )-\frac{\cos ^{-1}(a x)^2}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a*x]^2/x^4,x]

[Out]

-a^2/(3*x) + (a*Sqrt[1 - a^2*x^2]*ArcCos[a*x])/(3*x^2) - ArcCos[a*x]^2/(3*x^3) - ((2*I)/3)*a^3*ArcCos[a*x]*Arc
Tan[E^(I*ArcCos[a*x])] + (I/3)*a^3*PolyLog[2, (-I)*E^(I*ArcCos[a*x])] - (I/3)*a^3*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 4702

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[(c^2*(m + 2*p + 3))/(f^2*(m
+ 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcCos[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^F
racPart[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*ArcCos[c*x
])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[m, -1] && Inte
gerQ[m]

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]

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 \frac{\cos ^{-1}(a x)^2}{x^4} \, dx &=-\frac{\cos ^{-1}(a x)^2}{3 x^3}-\frac{1}{3} (2 a) \int \frac{\cos ^{-1}(a x)}{x^3 \sqrt{1-a^2 x^2}} \, dx\\ &=\frac{a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{3 x^2}-\frac{\cos ^{-1}(a x)^2}{3 x^3}+\frac{1}{3} a^2 \int \frac{1}{x^2} \, dx-\frac{1}{3} a^3 \int \frac{\cos ^{-1}(a x)}{x \sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{a^2}{3 x}+\frac{a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{3 x^2}-\frac{\cos ^{-1}(a x)^2}{3 x^3}+\frac{1}{3} a^3 \operatorname{Subst}\left (\int x \sec (x) \, dx,x,\cos ^{-1}(a x)\right )\\ &=-\frac{a^2}{3 x}+\frac{a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{3 x^2}-\frac{\cos ^{-1}(a x)^2}{3 x^3}-\frac{2}{3} i a^3 \cos ^{-1}(a x) \tan ^{-1}\left (e^{i \cos ^{-1}(a x)}\right )-\frac{1}{3} a^3 \operatorname{Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\cos ^{-1}(a x)\right )+\frac{1}{3} a^3 \operatorname{Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\cos ^{-1}(a x)\right )\\ &=-\frac{a^2}{3 x}+\frac{a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{3 x^2}-\frac{\cos ^{-1}(a x)^2}{3 x^3}-\frac{2}{3} i a^3 \cos ^{-1}(a x) \tan ^{-1}\left (e^{i \cos ^{-1}(a x)}\right )+\frac{1}{3} \left (i a^3\right ) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i \cos ^{-1}(a x)}\right )-\frac{1}{3} \left (i a^3\right ) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i \cos ^{-1}(a x)}\right )\\ &=-\frac{a^2}{3 x}+\frac{a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{3 x^2}-\frac{\cos ^{-1}(a x)^2}{3 x^3}-\frac{2}{3} i a^3 \cos ^{-1}(a x) \tan ^{-1}\left (e^{i \cos ^{-1}(a x)}\right )+\frac{1}{3} i a^3 \text{Li}_2\left (-i e^{i \cos ^{-1}(a x)}\right )-\frac{1}{3} i a^3 \text{Li}_2\left (i e^{i \cos ^{-1}(a x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.571418, size = 152, normalized size = 1.23 \[ -\frac{-i a^3 x^3 \text{PolyLog}\left (2,-i e^{i \cos ^{-1}(a x)}\right )+i a^3 x^3 \text{PolyLog}\left (2,i e^{i \cos ^{-1}(a x)}\right )+a^2 x^2-a x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)-a^3 x^3 \cos ^{-1}(a x) \log \left (1-i e^{i \cos ^{-1}(a x)}\right )+a^3 x^3 \cos ^{-1}(a x) \log \left (1+i e^{i \cos ^{-1}(a x)}\right )+\cos ^{-1}(a x)^2}{3 x^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCos[a*x]^2/x^4,x]

[Out]

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

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Maple [A]  time = 0.19, size = 173, normalized size = 1.4 \begin{align*}{\frac{a\arccos \left ( ax \right ) }{3\,{x}^{2}}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{{a}^{2}}{3\,x}}-{\frac{ \left ( \arccos \left ( ax \right ) \right ) ^{2}}{3\,{x}^{3}}}-{\frac{{a}^{3}\arccos \left ( ax \right ) }{3}\ln \left ( 1+i \left ( i\sqrt{-{a}^{2}{x}^{2}+1}+ax \right ) \right ) }+{\frac{{a}^{3}\arccos \left ( ax \right ) }{3}\ln \left ( 1-i \left ( i\sqrt{-{a}^{2}{x}^{2}+1}+ax \right ) \right ) }+{\frac{i}{3}}{a}^{3}{\it dilog} \left ( 1+i \left ( i\sqrt{-{a}^{2}{x}^{2}+1}+ax \right ) \right ) -{\frac{i}{3}}{a}^{3}{\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^4,x)

[Out]

1/3*a*arccos(a*x)*(-a^2*x^2+1)^(1/2)/x^2-1/3*a^2/x-1/3*arccos(a*x)^2/x^3-1/3*a^3*arccos(a*x)*ln(1+I*(I*(-a^2*x
^2+1)^(1/2)+a*x))+1/3*a^3*arccos(a*x)*ln(1-I*(I*(-a^2*x^2+1)^(1/2)+a*x))+1/3*I*a^3*dilog(1+I*(I*(-a^2*x^2+1)^(
1/2)+a*x))-1/3*I*a^3*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^{3} \int \frac{\sqrt{a x + 1} \sqrt{-a x + 1} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )}{a^{2} x^{5} - x^{3}}\,{d x} - \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )^{2}}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x)^2/x^4,x, algorithm="maxima")

[Out]

1/3*(6*a*x^3*integrate(1/3*sqrt(a*x + 1)*sqrt(-a*x + 1)*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)/(a^2*x^5 -
x^3), x) - arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^2)/x^3

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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^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x)^2/x^4,x, algorithm="fricas")

[Out]

integral(arccos(a*x)^2/x^4, 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^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acos(a*x)**2/x**4,x)

[Out]

Integral(acos(a*x)**2/x**4, 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^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x)^2/x^4,x, algorithm="giac")

[Out]

integrate(arccos(a*x)^2/x^4, x)