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

Optimal. Leaf size=177 \[ -i \text{PolyLog}\left (2,\frac{e^{i \cos ^{-1}(a+b x)}}{a-i \sqrt{1-a^2}}\right )-i \text{PolyLog}\left (2,\frac{e^{i \cos ^{-1}(a+b x)}}{a+i \sqrt{1-a^2}}\right )+\cos ^{-1}(a+b x) \log \left (1-\frac{e^{i \cos ^{-1}(a+b x)}}{a-i \sqrt{1-a^2}}\right )+\cos ^{-1}(a+b x) \log \left (1-\frac{e^{i \cos ^{-1}(a+b x)}}{a+i \sqrt{1-a^2}}\right )-\frac{1}{2} i \cos ^{-1}(a+b x)^2 \]

[Out]

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

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Rubi [A]  time = 0.276321, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {4806, 4742, 4522, 2190, 2279, 2391} \[ -i \text{PolyLog}\left (2,\frac{e^{i \cos ^{-1}(a+b x)}}{a-i \sqrt{1-a^2}}\right )-i \text{PolyLog}\left (2,\frac{e^{i \cos ^{-1}(a+b x)}}{a+i \sqrt{1-a^2}}\right )+\cos ^{-1}(a+b x) \log \left (1-\frac{e^{i \cos ^{-1}(a+b x)}}{a-i \sqrt{1-a^2}}\right )+\cos ^{-1}(a+b x) \log \left (1-\frac{e^{i \cos ^{-1}(a+b x)}}{a+i \sqrt{1-a^2}}\right )-\frac{1}{2} i \cos ^{-1}(a+b x)^2 \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a + b*x]/x,x]

[Out]

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

Rule 4806

Int[((a_.) + ArcCos[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcCos[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rule 4742

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Subst[Int[((a + b*x)^n*Sin[x])
/(c*d + e*Cos[x]), x], x, ArcCos[c*x]] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[n, 0]

Rule 4522

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)])/(Cos[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :>
Simp[(I*(e + f*x)^(m + 1))/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(I*(c + d*x)))/(I*a - Rt[-a^2 + b^2, 2] + I
*b*E^(I*(c + d*x))), x] + Int[((e + f*x)^m*E^(I*(c + d*x)))/(I*a + Rt[-a^2 + b^2, 2] + I*b*E^(I*(c + d*x))), x
]) /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && NegQ[a^2 - b^2]

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]

Rubi steps

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

Mathematica [A]  time = 0.191965, size = 228, normalized size = 1.29 \[ -i \left (\text{PolyLog}\left (2,\left (a-\sqrt{a^2-1}\right ) e^{i \cos ^{-1}(a+b x)}\right )+\text{PolyLog}\left (2,\left (\sqrt{a^2-1}+a\right ) e^{i \cos ^{-1}(a+b x)}\right )\right )+\log \left (1+\left (\sqrt{a^2-1}-a\right ) e^{i \cos ^{-1}(a+b x)}\right ) \left (\cos ^{-1}(a+b x)-2 \sin ^{-1}\left (\frac{\sqrt{1-a}}{\sqrt{2}}\right )\right )+\log \left (1-\left (\sqrt{a^2-1}+a\right ) e^{i \cos ^{-1}(a+b x)}\right ) \left (\cos ^{-1}(a+b x)+2 \sin ^{-1}\left (\frac{\sqrt{1-a}}{\sqrt{2}}\right )\right )-4 i \sin ^{-1}\left (\frac{\sqrt{1-a}}{\sqrt{2}}\right ) \tan ^{-1}\left (\frac{(a+1) \tan \left (\frac{1}{2} \cos ^{-1}(a+b x)\right )}{\sqrt{a^2-1}}\right )-\frac{1}{2} i \cos ^{-1}(a+b x)^2 \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCos[a + b*x]/x,x]

[Out]

(-I/2)*ArcCos[a + b*x]^2 - (4*I)*ArcSin[Sqrt[1 - a]/Sqrt[2]]*ArcTan[((1 + a)*Tan[ArcCos[a + b*x]/2])/Sqrt[-1 +
a^2]] + (ArcCos[a + b*x] - 2*ArcSin[Sqrt[1 - a]/Sqrt[2]])*Log[1 + (-a + Sqrt[-1 + a^2])*E^(I*ArcCos[a + b*x])
] + (ArcCos[a + b*x] + 2*ArcSin[Sqrt[1 - a]/Sqrt[2]])*Log[1 - (a + Sqrt[-1 + a^2])*E^(I*ArcCos[a + b*x])] - I*
(PolyLog[2, (a - Sqrt[-1 + a^2])*E^(I*ArcCos[a + b*x])] + PolyLog[2, (a + Sqrt[-1 + a^2])*E^(I*ArcCos[a + b*x]
)])

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Maple [A]  time = 0.104, size = 199, normalized size = 1.1 \begin{align*} -{\frac{i}{2}} \left ( \arccos \left ( bx+a \right ) \right ) ^{2}+\arccos \left ( bx+a \right ) \ln \left ({ \left ( \sqrt{{a}^{2}-1}+bx+i\sqrt{1- \left ( bx+a \right ) ^{2}} \right ) \left ( -a+\sqrt{{a}^{2}-1} \right ) ^{-1}} \right ) +\arccos \left ( bx+a \right ) \ln \left ({ \left ( \sqrt{{a}^{2}-1}-bx-i\sqrt{1- \left ( bx+a \right ) ^{2}} \right ) \left ( a+\sqrt{{a}^{2}-1} \right ) ^{-1}} \right ) -i{\it dilog} \left ({ \left ( \sqrt{{a}^{2}-1}+bx+i\sqrt{1- \left ( bx+a \right ) ^{2}} \right ) \left ( -a+\sqrt{{a}^{2}-1} \right ) ^{-1}} \right ) -i{\it dilog} \left ({ \left ( \sqrt{{a}^{2}-1}-bx-i\sqrt{1- \left ( bx+a \right ) ^{2}} \right ) \left ( a+\sqrt{{a}^{2}-1} \right ) ^{-1}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccos(b*x+a)/x,x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arccos(b*x + a)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acos(a + b*x)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccos(b*x + a)/x, x)