3.157 \(\int \frac{(a+b \cos ^{-1}(c x))^3}{x^2} \, dx\)

Optimal. Leaf size=151 \[ 6 i b^2 c \text{PolyLog}\left (2,-i e^{i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right )-6 i b^2 c \text{PolyLog}\left (2,i e^{i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right )-6 b^3 c \text{PolyLog}\left (3,-i e^{i \cos ^{-1}(c x)}\right )+6 b^3 c \text{PolyLog}\left (3,i e^{i \cos ^{-1}(c x)}\right )-\frac{\left (a+b \cos ^{-1}(c x)\right )^3}{x}-6 i b c \tan ^{-1}\left (e^{i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right )^2 \]

[Out]

-((a + b*ArcCos[c*x])^3/x) - (6*I)*b*c*(a + b*ArcCos[c*x])^2*ArcTan[E^(I*ArcCos[c*x])] + (6*I)*b^2*c*(a + b*Ar
cCos[c*x])*PolyLog[2, (-I)*E^(I*ArcCos[c*x])] - (6*I)*b^2*c*(a + b*ArcCos[c*x])*PolyLog[2, I*E^(I*ArcCos[c*x])
] - 6*b^3*c*PolyLog[3, (-I)*E^(I*ArcCos[c*x])] + 6*b^3*c*PolyLog[3, I*E^(I*ArcCos[c*x])]

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Rubi [A]  time = 0.213895, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {4628, 4710, 4181, 2531, 2282, 6589} \[ 6 i b^2 c \text{PolyLog}\left (2,-i e^{i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right )-6 i b^2 c \text{PolyLog}\left (2,i e^{i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right )-6 b^3 c \text{PolyLog}\left (3,-i e^{i \cos ^{-1}(c x)}\right )+6 b^3 c \text{PolyLog}\left (3,i e^{i \cos ^{-1}(c x)}\right )-\frac{\left (a+b \cos ^{-1}(c x)\right )^3}{x}-6 i b c \tan ^{-1}\left (e^{i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right )^2 \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCos[c*x])^3/x^2,x]

[Out]

-((a + b*ArcCos[c*x])^3/x) - (6*I)*b*c*(a + b*ArcCos[c*x])^2*ArcTan[E^(I*ArcCos[c*x])] + (6*I)*b^2*c*(a + b*Ar
cCos[c*x])*PolyLog[2, (-I)*E^(I*ArcCos[c*x])] - (6*I)*b^2*c*(a + b*ArcCos[c*x])*PolyLog[2, I*E^(I*ArcCos[c*x])
] - 6*b^3*c*PolyLog[3, (-I)*E^(I*ArcCos[c*x])] + 6*b^3*c*PolyLog[3, I*E^(I*ArcCos[c*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 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{\left (a+b \cos ^{-1}(c x)\right )^3}{x^2} \, dx &=-\frac{\left (a+b \cos ^{-1}(c x)\right )^3}{x}-(3 b c) \int \frac{\left (a+b \cos ^{-1}(c x)\right )^2}{x \sqrt{1-c^2 x^2}} \, dx\\ &=-\frac{\left (a+b \cos ^{-1}(c x)\right )^3}{x}+(3 b c) \operatorname{Subst}\left (\int (a+b x)^2 \sec (x) \, dx,x,\cos ^{-1}(c x)\right )\\ &=-\frac{\left (a+b \cos ^{-1}(c x)\right )^3}{x}-6 i b c \left (a+b \cos ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \cos ^{-1}(c x)}\right )-\left (6 b^2 c\right ) \operatorname{Subst}\left (\int (a+b x) \log \left (1-i e^{i x}\right ) \, dx,x,\cos ^{-1}(c x)\right )+\left (6 b^2 c\right ) \operatorname{Subst}\left (\int (a+b x) \log \left (1+i e^{i x}\right ) \, dx,x,\cos ^{-1}(c x)\right )\\ &=-\frac{\left (a+b \cos ^{-1}(c x)\right )^3}{x}-6 i b c \left (a+b \cos ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \cos ^{-1}(c x)}\right )+6 i b^2 c \left (a+b \cos ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \cos ^{-1}(c x)}\right )-6 i b^2 c \left (a+b \cos ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \cos ^{-1}(c x)}\right )-\left (6 i b^3 c\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^{i x}\right ) \, dx,x,\cos ^{-1}(c x)\right )+\left (6 i b^3 c\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^{i x}\right ) \, dx,x,\cos ^{-1}(c x)\right )\\ &=-\frac{\left (a+b \cos ^{-1}(c x)\right )^3}{x}-6 i b c \left (a+b \cos ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \cos ^{-1}(c x)}\right )+6 i b^2 c \left (a+b \cos ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \cos ^{-1}(c x)}\right )-6 i b^2 c \left (a+b \cos ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \cos ^{-1}(c x)}\right )-\left (6 b^3 c\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{i \cos ^{-1}(c x)}\right )+\left (6 b^3 c\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{i \cos ^{-1}(c x)}\right )\\ &=-\frac{\left (a+b \cos ^{-1}(c x)\right )^3}{x}-6 i b c \left (a+b \cos ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \cos ^{-1}(c x)}\right )+6 i b^2 c \left (a+b \cos ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \cos ^{-1}(c x)}\right )-6 i b^2 c \left (a+b \cos ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \cos ^{-1}(c x)}\right )-6 b^3 c \text{Li}_3\left (-i e^{i \cos ^{-1}(c x)}\right )+6 b^3 c \text{Li}_3\left (i e^{i \cos ^{-1}(c x)}\right )\\ \end{align*}

Mathematica [B]  time = 0.291687, size = 308, normalized size = 2.04 \[ 3 a b^2 c \left (-\frac{\cos ^{-1}(c x)^2}{c x}+2 \left (i \left (\text{PolyLog}\left (2,-i e^{i \cos ^{-1}(c x)}\right )-\text{PolyLog}\left (2,i e^{i \cos ^{-1}(c x)}\right )\right )+\cos ^{-1}(c x) \left (\log \left (1-i e^{i \cos ^{-1}(c x)}\right )-\log \left (1+i e^{i \cos ^{-1}(c x)}\right )\right )\right )\right )+b^3 c \left (-\frac{\cos ^{-1}(c x)^3}{c x}+3 \left (2 i \cos ^{-1}(c x) \left (\text{PolyLog}\left (2,-i e^{i \cos ^{-1}(c x)}\right )-\text{PolyLog}\left (2,i e^{i \cos ^{-1}(c x)}\right )\right )-2 \left (\text{PolyLog}\left (3,-i e^{i \cos ^{-1}(c x)}\right )-\text{PolyLog}\left (3,i e^{i \cos ^{-1}(c x)}\right )\right )+\cos ^{-1}(c x)^2 \left (\log \left (1-i e^{i \cos ^{-1}(c x)}\right )-\log \left (1+i e^{i \cos ^{-1}(c x)}\right )\right )\right )\right )+3 a^2 b c \log \left (\sqrt{1-c^2 x^2}+1\right )-3 a^2 b c \log (x)-\frac{3 a^2 b \cos ^{-1}(c x)}{x}-\frac{a^3}{x} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcCos[c*x])^3/x^2,x]

[Out]

-(a^3/x) - (3*a^2*b*ArcCos[c*x])/x - 3*a^2*b*c*Log[x] + 3*a^2*b*c*Log[1 + Sqrt[1 - c^2*x^2]] + 3*a*b^2*c*(-(Ar
cCos[c*x]^2/(c*x)) + 2*(ArcCos[c*x]*(Log[1 - I*E^(I*ArcCos[c*x])] - Log[1 + I*E^(I*ArcCos[c*x])]) + I*(PolyLog
[2, (-I)*E^(I*ArcCos[c*x])] - PolyLog[2, I*E^(I*ArcCos[c*x])]))) + b^3*c*(-(ArcCos[c*x]^3/(c*x)) + 3*(ArcCos[c
*x]^2*(Log[1 - I*E^(I*ArcCos[c*x])] - Log[1 + I*E^(I*ArcCos[c*x])]) + (2*I)*ArcCos[c*x]*(PolyLog[2, (-I)*E^(I*
ArcCos[c*x])] - PolyLog[2, I*E^(I*ArcCos[c*x])]) - 2*(PolyLog[3, (-I)*E^(I*ArcCos[c*x])] - PolyLog[3, I*E^(I*A
rcCos[c*x])])))

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Maple [F]  time = 0.161, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\arccos \left ( cx \right ) \right ) ^{3}}{{x}^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccos(c*x))^3/x^2,x)

[Out]

int((a+b*arccos(c*x))^3/x^2,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccos(c*x))^3/x^2,x, algorithm="maxima")

[Out]

3*(c*log(2*sqrt(-c^2*x^2 + 1)/abs(x) + 2/abs(x)) - arccos(c*x)/x)*a^2*b - a^3/x - (b^3*arctan2(sqrt(c*x + 1)*s
qrt(-c*x + 1), c*x)^3 - x*integrate(3*(sqrt(c*x + 1)*sqrt(-c*x + 1)*b^3*c*x*arctan2(sqrt(c*x + 1)*sqrt(-c*x +
1), c*x)^2 + (a*b^2*c^2*x^2 - a*b^2)*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x)^2)/(c^2*x^4 - x^2), x))/x

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{3} \arccos \left (c x\right )^{3} + 3 \, a b^{2} \arccos \left (c x\right )^{2} + 3 \, a^{2} b \arccos \left (c x\right ) + a^{3}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccos(c*x))^3/x^2,x, algorithm="fricas")

[Out]

integral((b^3*arccos(c*x)^3 + 3*a*b^2*arccos(c*x)^2 + 3*a^2*b*arccos(c*x) + a^3)/x^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{acos}{\left (c x \right )}\right )^{3}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acos(c*x))**3/x**2,x)

[Out]

Integral((a + b*acos(c*x))**3/x**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arccos \left (c x\right ) + a\right )}^{3}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccos(c*x))^3/x^2,x, algorithm="giac")

[Out]

integrate((b*arccos(c*x) + a)^3/x^2, x)