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

Optimal. Leaf size=89 \[ 2 i b^2 c \text{PolyLog}\left (2,-i e^{i \cos ^{-1}(c x)}\right )-2 i b^2 c \text{PolyLog}\left (2,i e^{i \cos ^{-1}(c x)}\right )-\frac{\left (a+b \cos ^{-1}(c x)\right )^2}{x}-4 i b c \tan ^{-1}\left (e^{i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right ) \]

[Out]

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

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Rubi [A]  time = 0.127538, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {4628, 4710, 4181, 2279, 2391} \[ 2 i b^2 c \text{PolyLog}\left (2,-i e^{i \cos ^{-1}(c x)}\right )-2 i b^2 c \text{PolyLog}\left (2,i e^{i \cos ^{-1}(c x)}\right )-\frac{\left (a+b \cos ^{-1}(c x)\right )^2}{x}-4 i b c \tan ^{-1}\left (e^{i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCos[c*x])^2/x^2,x]

[Out]

-((a + b*ArcCos[c*x])^2/x) - (4*I)*b*c*(a + b*ArcCos[c*x])*ArcTan[E^(I*ArcCos[c*x])] + (2*I)*b^2*c*PolyLog[2,
(-I)*E^(I*ArcCos[c*x])] - (2*I)*b^2*c*PolyLog[2, 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 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{\left (a+b \cos ^{-1}(c x)\right )^2}{x^2} \, dx &=-\frac{\left (a+b \cos ^{-1}(c x)\right )^2}{x}-(2 b c) \int \frac{a+b \cos ^{-1}(c x)}{x \sqrt{1-c^2 x^2}} \, dx\\ &=-\frac{\left (a+b \cos ^{-1}(c x)\right )^2}{x}+(2 b c) \operatorname{Subst}\left (\int (a+b x) \sec (x) \, dx,x,\cos ^{-1}(c x)\right )\\ &=-\frac{\left (a+b \cos ^{-1}(c x)\right )^2}{x}-4 i b c \left (a+b \cos ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \cos ^{-1}(c x)}\right )-\left (2 b^2 c\right ) \operatorname{Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\cos ^{-1}(c x)\right )+\left (2 b^2 c\right ) \operatorname{Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\cos ^{-1}(c x)\right )\\ &=-\frac{\left (a+b \cos ^{-1}(c x)\right )^2}{x}-4 i b c \left (a+b \cos ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \cos ^{-1}(c x)}\right )+\left (2 i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i \cos ^{-1}(c x)}\right )-\left (2 i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i \cos ^{-1}(c x)}\right )\\ &=-\frac{\left (a+b \cos ^{-1}(c x)\right )^2}{x}-4 i b c \left (a+b \cos ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \cos ^{-1}(c x)}\right )+2 i b^2 c \text{Li}_2\left (-i e^{i \cos ^{-1}(c x)}\right )-2 i b^2 c \text{Li}_2\left (i e^{i \cos ^{-1}(c x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.207137, size = 134, normalized size = 1.51 \[ -\frac{b^2 \left (\cos ^{-1}(c x)^2-2 c x \left (i \left (\text{PolyLog}\left (2,-i e^{i \cos ^{-1}(c x)}\right )-\text{PolyLog}\left (2,i e^{i \cos ^{-1}(c x)}\right )\right )+\cos ^{-1}(c x) \left (\log \left (1-i e^{i \cos ^{-1}(c x)}\right )-\log \left (1+i e^{i \cos ^{-1}(c x)}\right )\right )\right )\right )+a^2+2 a b \left (\cos ^{-1}(c x)-c x \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )\right )}{x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((a^2 + 2*a*b*(ArcCos[c*x] - c*x*ArcTanh[Sqrt[1 - c^2*x^2]]) + b^2*(ArcCos[c*x]^2 - 2*c*x*(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])]))))/x)

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Maple [A]  time = 0.053, size = 187, normalized size = 2.1 \begin{align*} -{\frac{{a}^{2}}{x}}-{\frac{{b}^{2} \left ( \arccos \left ( cx \right ) \right ) ^{2}}{x}}-2\,c{b}^{2}\arccos \left ( cx \right ) \ln \left ( 1+i \left ( cx+i\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) +2\,c{b}^{2}\arccos \left ( cx \right ) \ln \left ( 1-i \left ( cx+i\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) +2\,ic{b}^{2}{\it dilog} \left ( 1+i \left ( cx+i\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) -2\,ic{b}^{2}{\it dilog} \left ( 1-i \left ( cx+i\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) -2\,{\frac{ab\arccos \left ( cx \right ) }{x}}+2\,cab{\it Artanh} \left ({\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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