3.506 \(\int \frac{\tanh ^{-1}(b x)}{1-x^2} \, dx\)

Optimal. Leaf size=171 \[ \frac{1}{4} \text{PolyLog}\left (2,\frac{1-b x}{1-b}\right )-\frac{1}{4} \text{PolyLog}\left (2,\frac{1-b x}{b+1}\right )+\frac{1}{4} \text{PolyLog}\left (2,\frac{b x+1}{1-b}\right )-\frac{1}{4} \text{PolyLog}\left (2,\frac{b x+1}{b+1}\right )+\frac{1}{4} \log \left (-\frac{b (1-x)}{1-b}\right ) \log (1-b x)-\frac{1}{4} \log \left (\frac{b (x+1)}{b+1}\right ) \log (1-b x)-\frac{1}{4} \log \left (\frac{b (1-x)}{b+1}\right ) \log (b x+1)+\frac{1}{4} \log \left (-\frac{b (x+1)}{1-b}\right ) \log (b x+1) \]

[Out]

(Log[-((b*(1 - x))/(1 - b))]*Log[1 - b*x])/4 - (Log[(b*(1 + x))/(1 + b)]*Log[1 - b*x])/4 - (Log[(b*(1 - x))/(1
+ b)]*Log[1 + b*x])/4 + (Log[-((b*(1 + x))/(1 - b))]*Log[1 + b*x])/4 + PolyLog[2, (1 - b*x)/(1 - b)]/4 - Poly
Log[2, (1 - b*x)/(1 + b)]/4 + PolyLog[2, (1 + b*x)/(1 - b)]/4 - PolyLog[2, (1 + b*x)/(1 + b)]/4

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Rubi [A]  time = 0.240118, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {5972, 2409, 2394, 2393, 2391} \[ \frac{1}{4} \text{PolyLog}\left (2,\frac{1-b x}{1-b}\right )-\frac{1}{4} \text{PolyLog}\left (2,\frac{1-b x}{b+1}\right )+\frac{1}{4} \text{PolyLog}\left (2,\frac{b x+1}{1-b}\right )-\frac{1}{4} \text{PolyLog}\left (2,\frac{b x+1}{b+1}\right )+\frac{1}{4} \log \left (-\frac{b (1-x)}{1-b}\right ) \log (1-b x)-\frac{1}{4} \log \left (\frac{b (x+1)}{b+1}\right ) \log (1-b x)-\frac{1}{4} \log \left (\frac{b (1-x)}{b+1}\right ) \log (b x+1)+\frac{1}{4} \log \left (-\frac{b (x+1)}{1-b}\right ) \log (b x+1) \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[b*x]/(1 - x^2),x]

[Out]

(Log[-((b*(1 - x))/(1 - b))]*Log[1 - b*x])/4 - (Log[(b*(1 + x))/(1 + b)]*Log[1 - b*x])/4 - (Log[(b*(1 - x))/(1
+ b)]*Log[1 + b*x])/4 + (Log[-((b*(1 + x))/(1 - b))]*Log[1 + b*x])/4 + PolyLog[2, (1 - b*x)/(1 - b)]/4 - Poly
Log[2, (1 - b*x)/(1 + b)]/4 + PolyLog[2, (1 + b*x)/(1 - b)]/4 - PolyLog[2, (1 + b*x)/(1 + b)]/4

Rule 5972

Int[ArcTanh[(c_.)*(x_)]/((d_.) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/2, Int[Log[1 + c*x]/(d + e*x^2), x], x] -
Dist[1/2, Int[Log[1 - c*x]/(d + e*x^2), x], x] /; FreeQ[{c, d, e}, x]

Rule 2409

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_)^(r_))^(q_.), x_Symbol] :> In
t[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (f + g*x^r)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, r}, x]
&& IGtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[r] && NeQ[r, 1]))

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 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{\tanh ^{-1}(b x)}{1-x^2} \, dx &=-\left (\frac{1}{2} \int \frac{\log (1-b x)}{1-x^2} \, dx\right )+\frac{1}{2} \int \frac{\log (1+b x)}{1-x^2} \, dx\\ &=-\left (\frac{1}{2} \int \left (\frac{\log (1-b x)}{2 (1-x)}+\frac{\log (1-b x)}{2 (1+x)}\right ) \, dx\right )+\frac{1}{2} \int \left (\frac{\log (1+b x)}{2 (1-x)}+\frac{\log (1+b x)}{2 (1+x)}\right ) \, dx\\ &=-\left (\frac{1}{4} \int \frac{\log (1-b x)}{1-x} \, dx\right )-\frac{1}{4} \int \frac{\log (1-b x)}{1+x} \, dx+\frac{1}{4} \int \frac{\log (1+b x)}{1-x} \, dx+\frac{1}{4} \int \frac{\log (1+b x)}{1+x} \, dx\\ &=\frac{1}{4} \log \left (-\frac{b (1-x)}{1-b}\right ) \log (1-b x)-\frac{1}{4} \log \left (\frac{b (1+x)}{1+b}\right ) \log (1-b x)-\frac{1}{4} \log \left (\frac{b (1-x)}{1+b}\right ) \log (1+b x)+\frac{1}{4} \log \left (-\frac{b (1+x)}{1-b}\right ) \log (1+b x)+\frac{1}{4} b \int \frac{\log \left (-\frac{b (1-x)}{1-b}\right )}{1-b x} \, dx+\frac{1}{4} b \int \frac{\log \left (\frac{b (1-x)}{1+b}\right )}{1+b x} \, dx-\frac{1}{4} b \int \frac{\log \left (-\frac{b (1+x)}{-1-b}\right )}{1-b x} \, dx-\frac{1}{4} b \int \frac{\log \left (\frac{b (1+x)}{-1+b}\right )}{1+b x} \, dx\\ &=\frac{1}{4} \log \left (-\frac{b (1-x)}{1-b}\right ) \log (1-b x)-\frac{1}{4} \log \left (\frac{b (1+x)}{1+b}\right ) \log (1-b x)-\frac{1}{4} \log \left (\frac{b (1-x)}{1+b}\right ) \log (1+b x)+\frac{1}{4} \log \left (-\frac{b (1+x)}{1-b}\right ) \log (1+b x)+\frac{1}{4} \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{x}{-1-b}\right )}{x} \, dx,x,1-b x\right )-\frac{1}{4} \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{1-b}\right )}{x} \, dx,x,1-b x\right )-\frac{1}{4} \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{x}{-1+b}\right )}{x} \, dx,x,1+b x\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{1+b}\right )}{x} \, dx,x,1+b x\right )\\ &=\frac{1}{4} \log \left (-\frac{b (1-x)}{1-b}\right ) \log (1-b x)-\frac{1}{4} \log \left (\frac{b (1+x)}{1+b}\right ) \log (1-b x)-\frac{1}{4} \log \left (\frac{b (1-x)}{1+b}\right ) \log (1+b x)+\frac{1}{4} \log \left (-\frac{b (1+x)}{1-b}\right ) \log (1+b x)+\frac{1}{4} \text{Li}_2\left (\frac{1-b x}{1-b}\right )-\frac{1}{4} \text{Li}_2\left (\frac{1-b x}{1+b}\right )+\frac{1}{4} \text{Li}_2\left (\frac{1+b x}{1-b}\right )-\frac{1}{4} \text{Li}_2\left (\frac{1+b x}{1+b}\right )\\ \end{align*}

Mathematica [C]  time = 0.885032, size = 576, normalized size = 3.37 \[ -\frac{b \left (i \left (\text{PolyLog}\left (2,\frac{\left (b^2-2 i \sqrt{-b^2}+1\right ) \left (b-i \sqrt{-b^2} x\right )}{\left (b^2-1\right ) \left (b+i \sqrt{-b^2} x\right )}\right )-\text{PolyLog}\left (2,\frac{\left (b^2+2 i \sqrt{-b^2}+1\right ) \left (b-i \sqrt{-b^2} x\right )}{\left (b^2-1\right ) \left (b+i \sqrt{-b^2} x\right )}\right )\right )+2 i \cos ^{-1}\left (\frac{b^2+1}{1-b^2}\right ) \tan ^{-1}\left (\frac{b x}{\sqrt{-b^2}}\right )-4 \tan ^{-1}\left (\frac{\sqrt{-b^2}}{b x}\right ) \tanh ^{-1}(b x)-\log \left (\frac{2 b \left (\sqrt{-b^2}-i\right ) (b x-1)}{\left (b^2-1\right ) \left (\sqrt{-b^2} x-i b\right )}\right ) \left (\cos ^{-1}\left (\frac{b^2+1}{1-b^2}\right )-2 \tan ^{-1}\left (\frac{b x}{\sqrt{-b^2}}\right )\right )-\log \left (\frac{2 b \left (\sqrt{-b^2}+i\right ) (b x+1)}{\left (b^2-1\right ) \left (\sqrt{-b^2} x-i b\right )}\right ) \left (2 \tan ^{-1}\left (\frac{b x}{\sqrt{-b^2}}\right )+\cos ^{-1}\left (\frac{b^2+1}{1-b^2}\right )\right )+\left (\cos ^{-1}\left (\frac{b^2+1}{1-b^2}\right )-2 \left (\tan ^{-1}\left (\frac{\sqrt{-b^2}}{b x}\right )+\tan ^{-1}\left (\frac{b x}{\sqrt{-b^2}}\right )\right )\right ) \log \left (\frac{\sqrt{2} \sqrt{-b^2} e^{-\tanh ^{-1}(b x)}}{\sqrt{b^2-1} \sqrt{\left (b^2-1\right ) \cosh \left (2 \tanh ^{-1}(b x)\right )+b^2+1}}\right )+\left (2 \left (\tan ^{-1}\left (\frac{\sqrt{-b^2}}{b x}\right )+\tan ^{-1}\left (\frac{b x}{\sqrt{-b^2}}\right )\right )+\cos ^{-1}\left (\frac{b^2+1}{1-b^2}\right )\right ) \log \left (\frac{\sqrt{2} \sqrt{-b^2} e^{\tanh ^{-1}(b x)}}{\sqrt{b^2-1} \sqrt{\left (b^2-1\right ) \cosh \left (2 \tanh ^{-1}(b x)\right )+b^2+1}}\right )\right )}{4 \sqrt{-b^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[b*x]/(1 - x^2),x]

[Out]

-(b*((2*I)*ArcCos[(1 + b^2)/(1 - b^2)]*ArcTan[(b*x)/Sqrt[-b^2]] - 4*ArcTan[Sqrt[-b^2]/(b*x)]*ArcTanh[b*x] - (A
rcCos[(1 + b^2)/(1 - b^2)] - 2*ArcTan[(b*x)/Sqrt[-b^2]])*Log[(2*b*(-I + Sqrt[-b^2])*(-1 + b*x))/((-1 + b^2)*((
-I)*b + Sqrt[-b^2]*x))] - (ArcCos[(1 + b^2)/(1 - b^2)] + 2*ArcTan[(b*x)/Sqrt[-b^2]])*Log[(2*b*(I + Sqrt[-b^2])
*(1 + b*x))/((-1 + b^2)*((-I)*b + Sqrt[-b^2]*x))] + (ArcCos[(1 + b^2)/(1 - b^2)] - 2*(ArcTan[Sqrt[-b^2]/(b*x)]
+ ArcTan[(b*x)/Sqrt[-b^2]]))*Log[(Sqrt[2]*Sqrt[-b^2])/(Sqrt[-1 + b^2]*E^ArcTanh[b*x]*Sqrt[1 + b^2 + (-1 + b^2
)*Cosh[2*ArcTanh[b*x]]])] + (ArcCos[(1 + b^2)/(1 - b^2)] + 2*(ArcTan[Sqrt[-b^2]/(b*x)] + ArcTan[(b*x)/Sqrt[-b^
2]]))*Log[(Sqrt[2]*Sqrt[-b^2]*E^ArcTanh[b*x])/(Sqrt[-1 + b^2]*Sqrt[1 + b^2 + (-1 + b^2)*Cosh[2*ArcTanh[b*x]]])
] + I*(PolyLog[2, ((1 + b^2 - (2*I)*Sqrt[-b^2])*(b - I*Sqrt[-b^2]*x))/((-1 + b^2)*(b + I*Sqrt[-b^2]*x))] - Pol
yLog[2, ((1 + b^2 + (2*I)*Sqrt[-b^2])*(b - I*Sqrt[-b^2]*x))/((-1 + b^2)*(b + I*Sqrt[-b^2]*x))])))/(4*Sqrt[-b^2
])

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Maple [A]  time = 0.106, size = 176, normalized size = 1. \begin{align*}{\frac{{\it Artanh} \left ( bx \right ) \ln \left ( bx+b \right ) }{2}}-{\frac{{\it Artanh} \left ( bx \right ) \ln \left ( bx-b \right ) }{2}}-{\frac{1}{4}{\it dilog} \left ({\frac{bx+1}{1-b}} \right ) }-{\frac{\ln \left ( bx+b \right ) }{4}\ln \left ({\frac{bx+1}{1-b}} \right ) }+{\frac{1}{4}{\it dilog} \left ({\frac{bx-1}{-b-1}} \right ) }+{\frac{\ln \left ( bx+b \right ) }{4}\ln \left ({\frac{bx-1}{-b-1}} \right ) }+{\frac{1}{4}{\it dilog} \left ({\frac{bx+1}{1+b}} \right ) }+{\frac{\ln \left ( bx-b \right ) }{4}\ln \left ({\frac{bx+1}{1+b}} \right ) }-{\frac{1}{4}{\it dilog} \left ({\frac{bx-1}{-1+b}} \right ) }-{\frac{\ln \left ( bx-b \right ) }{4}\ln \left ({\frac{bx-1}{-1+b}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(b*x)/(-x^2+1),x)

[Out]

1/2*arctanh(b*x)*ln(b*x+b)-1/2*arctanh(b*x)*ln(b*x-b)-1/4*dilog((b*x+1)/(1-b))-1/4*ln(b*x+b)*ln((b*x+1)/(1-b))
+1/4*dilog((b*x-1)/(-b-1))+1/4*ln(b*x+b)*ln((b*x-1)/(-b-1))+1/4*dilog((b*x+1)/(1+b))+1/4*ln(b*x-b)*ln((b*x+1)/
(1+b))-1/4*dilog((b*x-1)/(-1+b))-1/4*ln(b*x-b)*ln((b*x-1)/(-1+b))

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Maxima [A]  time = 0.959717, size = 243, normalized size = 1.42 \begin{align*} \frac{1}{4} \, b{\left (\frac{\log \left (x + 1\right ) \log \left (-\frac{b x + b}{b + 1} + 1\right ) +{\rm Li}_2\left (\frac{b x + b}{b + 1}\right )}{b} + \frac{\log \left (x - 1\right ) \log \left (\frac{b x - b}{b + 1} + 1\right ) +{\rm Li}_2\left (-\frac{b x - b}{b + 1}\right )}{b} - \frac{\log \left (x + 1\right ) \log \left (-\frac{b x + b}{b - 1} + 1\right ) +{\rm Li}_2\left (\frac{b x + b}{b - 1}\right )}{b} - \frac{\log \left (x - 1\right ) \log \left (\frac{b x - b}{b - 1} + 1\right ) +{\rm Li}_2\left (-\frac{b x - b}{b - 1}\right )}{b}\right )} + \frac{1}{2} \,{\left (\log \left (x + 1\right ) - \log \left (x - 1\right )\right )} \operatorname{artanh}\left (b x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(b*x)/(-x^2+1),x, algorithm="maxima")

[Out]

1/4*b*((log(x + 1)*log(-(b*x + b)/(b + 1) + 1) + dilog((b*x + b)/(b + 1)))/b + (log(x - 1)*log((b*x - b)/(b +
1) + 1) + dilog(-(b*x - b)/(b + 1)))/b - (log(x + 1)*log(-(b*x + b)/(b - 1) + 1) + dilog((b*x + b)/(b - 1)))/b
- (log(x - 1)*log((b*x - b)/(b - 1) + 1) + dilog(-(b*x - b)/(b - 1)))/b) + 1/2*(log(x + 1) - log(x - 1))*arct
anh(b*x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\operatorname{artanh}\left (b x\right )}{x^{2} - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(b*x)/(-x^2+1),x, algorithm="fricas")

[Out]

integral(-arctanh(b*x)/(x^2 - 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{\operatorname{atanh}{\left (b x \right )}}{x^{2} - 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(b*x)/(-x**2+1),x)

[Out]

-Integral(atanh(b*x)/(x**2 - 1), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{\operatorname{artanh}\left (b x\right )}{x^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(b*x)/(-x^2+1),x, algorithm="giac")

[Out]

integrate(-arctanh(b*x)/(x^2 - 1), x)