3.40 \(\int \frac{\cos ^{-1}(a x)^4}{x^3} \, dx\)

Optimal. Leaf size=121 \[ -6 i a^2 \cos ^{-1}(a x) \text{PolyLog}\left (2,-e^{2 i \cos ^{-1}(a x)}\right )+3 a^2 \text{PolyLog}\left (3,-e^{2 i \cos ^{-1}(a x)}\right )+\frac{2 a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^3}{x}-2 i a^2 \cos ^{-1}(a x)^3+6 a^2 \cos ^{-1}(a x)^2 \log \left (1+e^{2 i \cos ^{-1}(a x)}\right )-\frac{\cos ^{-1}(a x)^4}{2 x^2} \]

[Out]

(-2*I)*a^2*ArcCos[a*x]^3 + (2*a*Sqrt[1 - a^2*x^2]*ArcCos[a*x]^3)/x - ArcCos[a*x]^4/(2*x^2) + 6*a^2*ArcCos[a*x]
^2*Log[1 + E^((2*I)*ArcCos[a*x])] - (6*I)*a^2*ArcCos[a*x]*PolyLog[2, -E^((2*I)*ArcCos[a*x])] + 3*a^2*PolyLog[3
, -E^((2*I)*ArcCos[a*x])]

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Rubi [A]  time = 0.213345, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.8, Rules used = {4628, 4682, 4626, 3719, 2190, 2531, 2282, 6589} \[ -6 i a^2 \cos ^{-1}(a x) \text{PolyLog}\left (2,-e^{2 i \cos ^{-1}(a x)}\right )+3 a^2 \text{PolyLog}\left (3,-e^{2 i \cos ^{-1}(a x)}\right )+\frac{2 a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^3}{x}-2 i a^2 \cos ^{-1}(a x)^3+6 a^2 \cos ^{-1}(a x)^2 \log \left (1+e^{2 i \cos ^{-1}(a x)}\right )-\frac{\cos ^{-1}(a x)^4}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a*x]^4/x^3,x]

[Out]

(-2*I)*a^2*ArcCos[a*x]^3 + (2*a*Sqrt[1 - a^2*x^2]*ArcCos[a*x]^3)/x - ArcCos[a*x]^4/(2*x^2) + 6*a^2*ArcCos[a*x]
^2*Log[1 + E^((2*I)*ArcCos[a*x])] - (6*I)*a^2*ArcCos[a*x]*PolyLog[2, -E^((2*I)*ArcCos[a*x])] + 3*a^2*PolyLog[3
, -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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int \frac{\cos ^{-1}(a x)^4}{x^3} \, dx &=-\frac{\cos ^{-1}(a x)^4}{2 x^2}-(2 a) \int \frac{\cos ^{-1}(a x)^3}{x^2 \sqrt{1-a^2 x^2}} \, dx\\ &=\frac{2 a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^3}{x}-\frac{\cos ^{-1}(a x)^4}{2 x^2}+\left (6 a^2\right ) \int \frac{\cos ^{-1}(a x)^2}{x} \, dx\\ &=\frac{2 a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^3}{x}-\frac{\cos ^{-1}(a x)^4}{2 x^2}-\left (6 a^2\right ) \operatorname{Subst}\left (\int x^2 \tan (x) \, dx,x,\cos ^{-1}(a x)\right )\\ &=-2 i a^2 \cos ^{-1}(a x)^3+\frac{2 a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^3}{x}-\frac{\cos ^{-1}(a x)^4}{2 x^2}+\left (12 i a^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 i x} x^2}{1+e^{2 i x}} \, dx,x,\cos ^{-1}(a x)\right )\\ &=-2 i a^2 \cos ^{-1}(a x)^3+\frac{2 a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^3}{x}-\frac{\cos ^{-1}(a x)^4}{2 x^2}+6 a^2 \cos ^{-1}(a x)^2 \log \left (1+e^{2 i \cos ^{-1}(a x)}\right )-\left (12 a^2\right ) \operatorname{Subst}\left (\int x \log \left (1+e^{2 i x}\right ) \, dx,x,\cos ^{-1}(a x)\right )\\ &=-2 i a^2 \cos ^{-1}(a x)^3+\frac{2 a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^3}{x}-\frac{\cos ^{-1}(a x)^4}{2 x^2}+6 a^2 \cos ^{-1}(a x)^2 \log \left (1+e^{2 i \cos ^{-1}(a x)}\right )-6 i a^2 \cos ^{-1}(a x) \text{Li}_2\left (-e^{2 i \cos ^{-1}(a x)}\right )+\left (6 i a^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-e^{2 i x}\right ) \, dx,x,\cos ^{-1}(a x)\right )\\ &=-2 i a^2 \cos ^{-1}(a x)^3+\frac{2 a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^3}{x}-\frac{\cos ^{-1}(a x)^4}{2 x^2}+6 a^2 \cos ^{-1}(a x)^2 \log \left (1+e^{2 i \cos ^{-1}(a x)}\right )-6 i a^2 \cos ^{-1}(a x) \text{Li}_2\left (-e^{2 i \cos ^{-1}(a x)}\right )+\left (3 a^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 i \cos ^{-1}(a x)}\right )\\ &=-2 i a^2 \cos ^{-1}(a x)^3+\frac{2 a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^3}{x}-\frac{\cos ^{-1}(a x)^4}{2 x^2}+6 a^2 \cos ^{-1}(a x)^2 \log \left (1+e^{2 i \cos ^{-1}(a x)}\right )-6 i a^2 \cos ^{-1}(a x) \text{Li}_2\left (-e^{2 i \cos ^{-1}(a x)}\right )+3 a^2 \text{Li}_3\left (-e^{2 i \cos ^{-1}(a x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.35785, size = 115, normalized size = 0.95 \[ -\frac{\cos ^{-1}(a x)^4}{2 x^2}-a^2 \left (6 i \cos ^{-1}(a x) \text{PolyLog}\left (2,-e^{2 i \cos ^{-1}(a x)}\right )-3 \text{PolyLog}\left (3,-e^{2 i \cos ^{-1}(a x)}\right )-2 \cos ^{-1}(a x)^2 \left (\frac{\sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{a x}-i \cos ^{-1}(a x)+3 \log \left (1+e^{2 i \cos ^{-1}(a x)}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[a*x]^4/x^3,x]

[Out]

-ArcCos[a*x]^4/(2*x^2) - a^2*(-2*ArcCos[a*x]^2*((-I)*ArcCos[a*x] + (Sqrt[1 - a^2*x^2]*ArcCos[a*x])/(a*x) + 3*L
og[1 + E^((2*I)*ArcCos[a*x])]) + (6*I)*ArcCos[a*x]*PolyLog[2, -E^((2*I)*ArcCos[a*x])] - 3*PolyLog[3, -E^((2*I)
*ArcCos[a*x])])

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Maple [A]  time = 0.124, size = 149, normalized size = 1.2 \begin{align*} -2\,i{a}^{2} \left ( \arccos \left ( ax \right ) \right ) ^{3}-{\frac{ \left ( \arccos \left ( ax \right ) \right ) ^{4}}{2\,{x}^{2}}}+6\,{a}^{2} \left ( \arccos \left ( ax \right ) \right ) ^{2}\ln \left ( 1+ \left ( i\sqrt{-{a}^{2}{x}^{2}+1}+ax \right ) ^{2} \right ) -6\,i{a}^{2}\arccos \left ( ax \right ){\it polylog} \left ( 2,- \left ( i\sqrt{-{a}^{2}{x}^{2}+1}+ax \right ) ^{2} \right ) +3\,{a}^{2}{\it polylog} \left ( 3,- \left ( i\sqrt{-{a}^{2}{x}^{2}+1}+ax \right ) ^{2} \right ) +2\,{\frac{a \left ( \arccos \left ( ax \right ) \right ) ^{3}\sqrt{-{a}^{2}{x}^{2}+1}}{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccos(a*x)^4/x^3,x)

[Out]

-2*I*a^2*arccos(a*x)^3-1/2*arccos(a*x)^4/x^2+6*a^2*arccos(a*x)^2*ln(1+(I*(-a^2*x^2+1)^(1/2)+a*x)^2)-6*I*a^2*ar
ccos(a*x)*polylog(2,-(I*(-a^2*x^2+1)^(1/2)+a*x)^2)+3*a^2*polylog(3,-(I*(-a^2*x^2+1)^(1/2)+a*x)^2)+2*a*arccos(a
*x)^3*(-a^2*x^2+1)^(1/2)/x

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{\arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )^{4} - \frac{1}{2} \,{\left (\sqrt{a x + 1} \sqrt{-a x + 1} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )^{3} + 8 \, x \int \frac{7 \, \sqrt{a x + 1} \sqrt{-a x + 1} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )^{3} + 3 \,{\left (a^{3} x^{3} - a x\right )} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )^{2}}{8 \,{\left (a^{2} x^{4} - x^{2}\right )}}\,{d x}\right )} a x}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x)^4/x^3,x, algorithm="maxima")

[Out]

-1/2*(arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^4 - 4*a*x^2*integrate(sqrt(a*x + 1)*sqrt(-a*x + 1)*arctan2(sq
rt(a*x + 1)*sqrt(-a*x + 1), a*x)^3/(a^2*x^4 - x^2), x))/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 )^{4}}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x)^4/x^3,x, algorithm="fricas")

[Out]

integral(arccos(a*x)^4/x^3, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{acos}^{4}{\left (a x \right )}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acos(a*x)**4/x**3,x)

[Out]

Integral(acos(a*x)**4/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 )^{4}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x)^4/x^3,x, algorithm="giac")

[Out]

integrate(arccos(a*x)^4/x^3, x)