3.31 \(\int \frac{\cos ^{-1}(a x)^3}{x^5} \, dx\)

Optimal. Leaf size=169 \[ -\frac{1}{2} i a^4 \text{PolyLog}\left (2,-e^{2 i \cos ^{-1}(a x)}\right )+\frac{a^3 \sqrt{1-a^2 x^2}}{4 x}+\frac{a^3 \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{2 x}-\frac{a^2 \cos ^{-1}(a x)}{4 x^2}+\frac{a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{4 x^3}-\frac{1}{2} i a^4 \cos ^{-1}(a x)^2+a^4 \cos ^{-1}(a x) \log \left (1+e^{2 i \cos ^{-1}(a x)}\right )-\frac{\cos ^{-1}(a x)^3}{4 x^4} \]

[Out]

(a^3*Sqrt[1 - a^2*x^2])/(4*x) - (a^2*ArcCos[a*x])/(4*x^2) - (I/2)*a^4*ArcCos[a*x]^2 + (a*Sqrt[1 - a^2*x^2]*Arc
Cos[a*x]^2)/(4*x^3) + (a^3*Sqrt[1 - a^2*x^2]*ArcCos[a*x]^2)/(2*x) - ArcCos[a*x]^3/(4*x^4) + a^4*ArcCos[a*x]*Lo
g[1 + E^((2*I)*ArcCos[a*x])] - (I/2)*a^4*PolyLog[2, -E^((2*I)*ArcCos[a*x])]

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Rubi [A]  time = 0.289569, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.9, Rules used = {4628, 4702, 4682, 4626, 3719, 2190, 2279, 2391, 264} \[ -\frac{1}{2} i a^4 \text{PolyLog}\left (2,-e^{2 i \cos ^{-1}(a x)}\right )+\frac{a^3 \sqrt{1-a^2 x^2}}{4 x}+\frac{a^3 \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{2 x}-\frac{a^2 \cos ^{-1}(a x)}{4 x^2}+\frac{a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{4 x^3}-\frac{1}{2} i a^4 \cos ^{-1}(a x)^2+a^4 \cos ^{-1}(a x) \log \left (1+e^{2 i \cos ^{-1}(a x)}\right )-\frac{\cos ^{-1}(a x)^3}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a*x]^3/x^5,x]

[Out]

(a^3*Sqrt[1 - a^2*x^2])/(4*x) - (a^2*ArcCos[a*x])/(4*x^2) - (I/2)*a^4*ArcCos[a*x]^2 + (a*Sqrt[1 - a^2*x^2]*Arc
Cos[a*x]^2)/(4*x^3) + (a^3*Sqrt[1 - a^2*x^2]*ArcCos[a*x]^2)/(2*x) - ArcCos[a*x]^3/(4*x^4) + a^4*ArcCos[a*x]*Lo
g[1 + E^((2*I)*ArcCos[a*x])] - (I/2)*a^4*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 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 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]

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\cos ^{-1}(a x)^3}{x^5} \, dx &=-\frac{\cos ^{-1}(a x)^3}{4 x^4}-\frac{1}{4} (3 a) \int \frac{\cos ^{-1}(a x)^2}{x^4 \sqrt{1-a^2 x^2}} \, dx\\ &=\frac{a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{4 x^3}-\frac{\cos ^{-1}(a x)^3}{4 x^4}+\frac{1}{2} a^2 \int \frac{\cos ^{-1}(a x)}{x^3} \, dx-\frac{1}{2} a^3 \int \frac{\cos ^{-1}(a x)^2}{x^2 \sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{a^2 \cos ^{-1}(a x)}{4 x^2}+\frac{a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{4 x^3}+\frac{a^3 \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{2 x}-\frac{\cos ^{-1}(a x)^3}{4 x^4}-\frac{1}{4} a^3 \int \frac{1}{x^2 \sqrt{1-a^2 x^2}} \, dx+a^4 \int \frac{\cos ^{-1}(a x)}{x} \, dx\\ &=\frac{a^3 \sqrt{1-a^2 x^2}}{4 x}-\frac{a^2 \cos ^{-1}(a x)}{4 x^2}+\frac{a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{4 x^3}+\frac{a^3 \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{2 x}-\frac{\cos ^{-1}(a x)^3}{4 x^4}-a^4 \operatorname{Subst}\left (\int x \tan (x) \, dx,x,\cos ^{-1}(a x)\right )\\ &=\frac{a^3 \sqrt{1-a^2 x^2}}{4 x}-\frac{a^2 \cos ^{-1}(a x)}{4 x^2}-\frac{1}{2} i a^4 \cos ^{-1}(a x)^2+\frac{a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{4 x^3}+\frac{a^3 \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{2 x}-\frac{\cos ^{-1}(a x)^3}{4 x^4}+\left (2 i a^4\right ) \operatorname{Subst}\left (\int \frac{e^{2 i x} x}{1+e^{2 i x}} \, dx,x,\cos ^{-1}(a x)\right )\\ &=\frac{a^3 \sqrt{1-a^2 x^2}}{4 x}-\frac{a^2 \cos ^{-1}(a x)}{4 x^2}-\frac{1}{2} i a^4 \cos ^{-1}(a x)^2+\frac{a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{4 x^3}+\frac{a^3 \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{2 x}-\frac{\cos ^{-1}(a x)^3}{4 x^4}+a^4 \cos ^{-1}(a x) \log \left (1+e^{2 i \cos ^{-1}(a x)}\right )-a^4 \operatorname{Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\cos ^{-1}(a x)\right )\\ &=\frac{a^3 \sqrt{1-a^2 x^2}}{4 x}-\frac{a^2 \cos ^{-1}(a x)}{4 x^2}-\frac{1}{2} i a^4 \cos ^{-1}(a x)^2+\frac{a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{4 x^3}+\frac{a^3 \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{2 x}-\frac{\cos ^{-1}(a x)^3}{4 x^4}+a^4 \cos ^{-1}(a x) \log \left (1+e^{2 i \cos ^{-1}(a x)}\right )+\frac{1}{2} \left (i a^4\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i \cos ^{-1}(a x)}\right )\\ &=\frac{a^3 \sqrt{1-a^2 x^2}}{4 x}-\frac{a^2 \cos ^{-1}(a x)}{4 x^2}-\frac{1}{2} i a^4 \cos ^{-1}(a x)^2+\frac{a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{4 x^3}+\frac{a^3 \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2}{2 x}-\frac{\cos ^{-1}(a x)^3}{4 x^4}+a^4 \cos ^{-1}(a x) \log \left (1+e^{2 i \cos ^{-1}(a x)}\right )-\frac{1}{2} i a^4 \text{Li}_2\left (-e^{2 i \cos ^{-1}(a x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.470525, size = 149, normalized size = 0.88 \[ -\frac{\cos ^{-1}(a x)^3}{4 x^4}+\frac{1}{4} a^4 \sqrt{1-a^2 x^2} \left (-\frac{2 i \text{PolyLog}\left (2,-e^{2 i \cos ^{-1}(a x)}\right )}{\sqrt{1-a^2 x^2}}+\frac{\frac{\cos ^{-1}(a x)^2}{a^2 x^2}+2 \cos ^{-1}(a x)^2+1}{a x}+\frac{\cos ^{-1}(a x) \left (-\frac{1}{a^2 x^2}-2 i \cos ^{-1}(a x)+4 \log \left (1+e^{2 i \cos ^{-1}(a x)}\right )\right )}{\sqrt{1-a^2 x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[a*x]^3/x^5,x]

[Out]

-ArcCos[a*x]^3/(4*x^4) + (a^4*Sqrt[1 - a^2*x^2]*((1 + 2*ArcCos[a*x]^2 + ArcCos[a*x]^2/(a^2*x^2))/(a*x) + (ArcC
os[a*x]*(-(1/(a^2*x^2)) - (2*I)*ArcCos[a*x] + 4*Log[1 + E^((2*I)*ArcCos[a*x])]))/Sqrt[1 - a^2*x^2] - ((2*I)*Po
lyLog[2, -E^((2*I)*ArcCos[a*x])])/Sqrt[1 - a^2*x^2]))/4

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Maple [A]  time = 0.188, size = 176, normalized size = 1. \begin{align*} -{\frac{i}{2}}{a}^{4} \left ( \arccos \left ( ax \right ) \right ) ^{2}+{\frac{i}{4}}{a}^{4}+{\frac{{a}^{3} \left ( \arccos \left ( ax \right ) \right ) ^{2}}{2\,x}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{{a}^{3}}{4\,x}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{a \left ( \arccos \left ( ax \right ) \right ) ^{2}}{4\,{x}^{3}}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{{a}^{2}\arccos \left ( ax \right ) }{4\,{x}^{2}}}-{\frac{ \left ( \arccos \left ( ax \right ) \right ) ^{3}}{4\,{x}^{4}}}+{a}^{4}\arccos \left ( ax \right ) \ln \left ( 1+ \left ( i\sqrt{-{a}^{2}{x}^{2}+1}+ax \right ) ^{2} \right ) -{\frac{i}{2}}{a}^{4}{\it polylog} \left ( 2,- \left ( i\sqrt{-{a}^{2}{x}^{2}+1}+ax \right ) ^{2} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccos(a*x)^3/x^5,x)

[Out]

-1/2*I*a^4*arccos(a*x)^2+1/4*I*a^4+1/2*a^3*arccos(a*x)^2*(-a^2*x^2+1)^(1/2)/x+1/4*a^3*(-a^2*x^2+1)^(1/2)/x+1/4
*a*arccos(a*x)^2*(-a^2*x^2+1)^(1/2)/x^3-1/4*a^2*arccos(a*x)/x^2-1/4*arccos(a*x)^3/x^4+a^4*arccos(a*x)*ln(1+(I*
(-a^2*x^2+1)^(1/2)+a*x)^2)-1/2*I*a^4*polylog(2,-(I*(-a^2*x^2+1)^(1/2)+a*x)^2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\frac{1}{4} \,{\left ({\left (2 \, a^{2} x^{2} + 1\right )} \sqrt{a x + 1} \sqrt{-a x + 1} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )^{2} + 12 \, x^{3} \int \frac{9 \, \sqrt{a x + 1} \sqrt{-a x + 1} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )^{2} + 2 \,{\left (2 \, a^{5} x^{5} - a^{3} x^{3} - a x\right )} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )}{12 \,{\left (a^{2} x^{6} - x^{4}\right )}}\,{d x}\right )} a x - \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )^{3}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x)^3/x^5,x, algorithm="maxima")

[Out]

1/4*(12*a*x^4*integrate(1/4*sqrt(a*x + 1)*sqrt(-a*x + 1)*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^2/(a^2*x^6
- x^4), x) - arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^3)/x^4

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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^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x)^3/x^5,x, algorithm="fricas")

[Out]

integral(arccos(a*x)^3/x^5, 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^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acos(a*x)**3/x**5,x)

[Out]

Integral(acos(a*x)**3/x**5, 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^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x)^3/x^5,x, algorithm="giac")

[Out]

integrate(arccos(a*x)^3/x^5, x)