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

Optimal. Leaf size=171 \[ \frac {1}{4} \text {Li}_2\left (\frac {1-b x}{1-b}\right )-\frac {1}{4} \text {Li}_2\left (\frac {1-b x}{b+1}\right )+\frac {1}{4} \text {Li}_2\left (\frac {b x+1}{1-b}\right )-\frac {1}{4} \text {Li}_2\left (\frac {b x+1}{b+1}\right )+\frac {1}{4} \log \left (-\frac {b (1-x)}{1-b}\right ) \log (1-b x)-\frac {1}{4} \log \left (\frac {b (x+1)}{b+1}\right ) \log (1-b x)-\frac {1}{4} \log \left (\frac {b (1-x)}{b+1}\right ) \log (b x+1)+\frac {1}{4} \log \left (-\frac {b (x+1)}{1-b}\right ) \log (b x+1) \]

[Out]

1/4*ln(-b*(1-x)/(1-b))*ln(-b*x+1)-1/4*ln(b*(1+x)/(1+b))*ln(-b*x+1)-1/4*ln(b*(1-x)/(1+b))*ln(b*x+1)+1/4*ln(-b*(
1+x)/(1-b))*ln(b*x+1)+1/4*polylog(2,(-b*x+1)/(1-b))-1/4*polylog(2,(-b*x+1)/(1+b))+1/4*polylog(2,(b*x+1)/(1-b))
-1/4*polylog(2,(b*x+1)/(1+b))

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Rubi [A]  time = 0.24, antiderivative size = 171, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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]

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*}

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Mathematica [C]  time = 0.94, size = 576, normalized size = 3.37 \[ -\frac {b \left (i \left (\text {Li}_2\left (\frac {\left (b^2-2 i \sqrt {-b^2}+1\right ) \left (b-i \sqrt {-b^2} x\right )}{\left (b^2-1\right ) \left (b+i \sqrt {-b^2} x\right )}\right )-\text {Li}_2\left (\frac {\left (b^2+2 i \sqrt {-b^2}+1\right ) \left (b-i \sqrt {-b^2} x\right )}{\left (b^2-1\right ) \left (b+i \sqrt {-b^2} x\right )}\right )\right )+2 i \cos ^{-1}\left (\frac {b^2+1}{1-b^2}\right ) \tan ^{-1}\left (\frac {b x}{\sqrt {-b^2}}\right )-4 \tan ^{-1}\left (\frac {\sqrt {-b^2}}{b x}\right ) \tanh ^{-1}(b x)-\log \left (\frac {2 b \left (\sqrt {-b^2}-i\right ) (b x-1)}{\left (b^2-1\right ) \left (\sqrt {-b^2} x-i b\right )}\right ) \left (\cos ^{-1}\left (\frac {b^2+1}{1-b^2}\right )-2 \tan ^{-1}\left (\frac {b x}{\sqrt {-b^2}}\right )\right )-\log \left (\frac {2 b \left (\sqrt {-b^2}+i\right ) (b x+1)}{\left (b^2-1\right ) \left (\sqrt {-b^2} x-i b\right )}\right ) \left (2 \tan ^{-1}\left (\frac {b x}{\sqrt {-b^2}}\right )+\cos ^{-1}\left (\frac {b^2+1}{1-b^2}\right )\right )+\left (\cos ^{-1}\left (\frac {b^2+1}{1-b^2}\right )-2 \left (\tan ^{-1}\left (\frac {\sqrt {-b^2}}{b x}\right )+\tan ^{-1}\left (\frac {b x}{\sqrt {-b^2}}\right )\right )\right ) \log \left (\frac {\sqrt {2} \sqrt {-b^2} e^{-\tanh ^{-1}(b x)}}{\sqrt {b^2-1} \sqrt {\left (b^2-1\right ) \cosh \left (2 \tanh ^{-1}(b x)\right )+b^2+1}}\right )+\left (2 \left (\tan ^{-1}\left (\frac {\sqrt {-b^2}}{b x}\right )+\tan ^{-1}\left (\frac {b x}{\sqrt {-b^2}}\right )\right )+\cos ^{-1}\left (\frac {b^2+1}{1-b^2}\right )\right ) \log \left (\frac {\sqrt {2} \sqrt {-b^2} e^{\tanh ^{-1}(b x)}}{\sqrt {b^2-1} \sqrt {\left (b^2-1\right ) \cosh \left (2 \tanh ^{-1}(b x)\right )+b^2+1}}\right )\right )}{4 \sqrt {-b^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[b*x]/(1 - x^2),x]

[Out]

-1/4*(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]
- (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[(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))] -
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))])))/Sqrt[-b^
2]

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\operatorname {artanh}\left (b x\right )}{x^{2} - 1}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {\operatorname {artanh}\left (b x\right )}{x^{2} - 1}\,{d x} \]

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)

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maple [A]  time = 0.06, size = 176, normalized size = 1.03 \[ \frac {\arctanh \left (b x \right ) \ln \left (b x +b \right )}{2}-\frac {\arctanh \left (b x \right ) \ln \left (b x -b \right )}{2}+\frac {\dilog \left (\frac {b x -1}{-b -1}\right )}{4}+\frac {\ln \left (b x +b \right ) \ln \left (\frac {b x -1}{-b -1}\right )}{4}-\frac {\dilog \left (\frac {b x +1}{1-b}\right )}{4}-\frac {\ln \left (b x +b \right ) \ln \left (\frac {b x +1}{1-b}\right )}{4}-\frac {\dilog \left (\frac {b x -1}{-1+b}\right )}{4}-\frac {\ln \left (b x -b \right ) \ln \left (\frac {b x -1}{-1+b}\right )}{4}+\frac {\dilog \left (\frac {b x +1}{1+b}\right )}{4}+\frac {\ln \left (b x -b \right ) \ln \left (\frac {b x +1}{1+b}\right )}{4} \]

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)/(-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))+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 = 0.31, size = 180, normalized size = 1.05 \[ \frac {1}{4} \, b {\left (\frac {\log \left (x + 1\right ) \log \left (-\frac {b x + b}{b + 1} + 1\right ) + {\rm Li}_2\left (\frac {b x + b}{b + 1}\right )}{b} + \frac {\log \left (x - 1\right ) \log \left (\frac {b x - b}{b + 1} + 1\right ) + {\rm Li}_2\left (-\frac {b x - b}{b + 1}\right )}{b} - \frac {\log \left (x + 1\right ) \log \left (-\frac {b x + b}{b - 1} + 1\right ) + {\rm Li}_2\left (\frac {b x + b}{b - 1}\right )}{b} - \frac {\log \left (x - 1\right ) \log \left (\frac {b x - b}{b - 1} + 1\right ) + {\rm Li}_2\left (-\frac {b x - b}{b - 1}\right )}{b}\right )} + \frac {1}{2} \, {\left (\log \left (x + 1\right ) - \log \left (x - 1\right )\right )} \operatorname {artanh}\left (b x\right ) \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {\mathrm {atanh}\left (b\,x\right )}{x^2-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-atanh(b*x)/(x^2 - 1),x)

[Out]

-int(atanh(b*x)/(x^2 - 1), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {\operatorname {atanh}{\left (b x \right )}}{x^{2} - 1}\, dx \]

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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