3.6.6 \(\int \frac {\tanh ^{-1}(b x)}{1-x^2} \, dx\) [506]

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

[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 = 0.17, 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 = {6119, 2456, 2441, 2440, 2438} \begin {gather*} \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) \end {gather*}

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 2438

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 2440

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 2441

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 2456

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 6119

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} \text {Subst}\left (\int \frac {\log \left (1+\frac {x}{-1-b}\right )}{x} \, dx,x,1-b x\right )-\frac {1}{4} \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{1-b}\right )}{x} \, dx,x,1-b x\right )-\frac {1}{4} \text {Subst}\left (\int \frac {\log \left (1+\frac {x}{-1+b}\right )}{x} \, dx,x,1+b x\right )+\frac {1}{4} \text {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] Result contains complex when optimal does not.
time = 0.65, size = 576, normalized size = 3.37 \begin {gather*} -\frac {b \left (2 i \text {ArcCos}\left (\frac {1+b^2}{1-b^2}\right ) \text {ArcTan}\left (\frac {b x}{\sqrt {-b^2}}\right )-4 \text {ArcTan}\left (\frac {\sqrt {-b^2}}{b x}\right ) \tanh ^{-1}(b x)-\left (\text {ArcCos}\left (\frac {1+b^2}{1-b^2}\right )-2 \text {ArcTan}\left (\frac {b x}{\sqrt {-b^2}}\right )\right ) \log \left (\frac {2 b \left (-i+\sqrt {-b^2}\right ) (-1+b x)}{\left (-1+b^2\right ) \left (-i b+\sqrt {-b^2} x\right )}\right )-\left (\text {ArcCos}\left (\frac {1+b^2}{1-b^2}\right )+2 \text {ArcTan}\left (\frac {b x}{\sqrt {-b^2}}\right )\right ) \log \left (\frac {2 b \left (i+\sqrt {-b^2}\right ) (1+b x)}{\left (-1+b^2\right ) \left (-i b+\sqrt {-b^2} x\right )}\right )+\left (\text {ArcCos}\left (\frac {1+b^2}{1-b^2}\right )-2 \left (\text {ArcTan}\left (\frac {\sqrt {-b^2}}{b x}\right )+\text {ArcTan}\left (\frac {b x}{\sqrt {-b^2}}\right )\right )\right ) \log \left (\frac {\sqrt {2} \sqrt {-b^2} e^{-\tanh ^{-1}(b x)}}{\sqrt {-1+b^2} \sqrt {1+b^2+\left (-1+b^2\right ) \cosh \left (2 \tanh ^{-1}(b x)\right )}}\right )+\left (\text {ArcCos}\left (\frac {1+b^2}{1-b^2}\right )+2 \left (\text {ArcTan}\left (\frac {\sqrt {-b^2}}{b x}\right )+\text {ArcTan}\left (\frac {b x}{\sqrt {-b^2}}\right )\right )\right ) \log \left (\frac {\sqrt {2} \sqrt {-b^2} e^{\tanh ^{-1}(b x)}}{\sqrt {-1+b^2} \sqrt {1+b^2+\left (-1+b^2\right ) \cosh \left (2 \tanh ^{-1}(b x)\right )}}\right )+i \left (\text {PolyLog}\left (2,\frac {\left (1+b^2-2 i \sqrt {-b^2}\right ) \left (b-i \sqrt {-b^2} x\right )}{\left (-1+b^2\right ) \left (b+i \sqrt {-b^2} x\right )}\right )-\text {PolyLog}\left (2,\frac {\left (1+b^2+2 i \sqrt {-b^2}\right ) \left (b-i \sqrt {-b^2} x\right )}{\left (-1+b^2\right ) \left (b+i \sqrt {-b^2} x\right )}\right )\right )\right )}{4 \sqrt {-b^2}} \end {gather*}

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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Maple [A]
time = 6.54, size = 221, normalized size = 1.29

method	result	size
risch	\(\frac {\ln \left (-b x +1\right ) \ln \left (\frac {-b x +b}{-1+b}\right )}{4}+\frac {\dilog \left (\frac {-b x +b}{-1+b}\right )}{4}-\frac {\ln \left (-b x +1\right ) \ln \left (\frac {-b x -b}{-b -1}\right )}{4}-\frac {\dilog \left (\frac {-b x -b}{-b -1}\right )}{4}+\frac {\ln \left (b x +1\right ) \ln \left (\frac {b x +b}{-1+b}\right )}{4}+\frac {\dilog \left (\frac {b x +b}{-1+b}\right )}{4}-\frac {\ln \left (b x +1\right ) \ln \left (\frac {b x -b}{-b -1}\right )}{4}-\frac {\dilog \left (\frac {b x -b}{-b -1}\right )}{4}\)	\(160\)
derivativedivides	\(\frac {-\frac {\arctanh \left (b x \right ) b \ln \left (-b x +b \right )}{2}+\frac {\arctanh \left (b x \right ) b \ln \left (b x +b \right )}{2}-\frac {b^{2} \left (\frac {\dilog \left (\frac {-b x +1}{1-b}\right )}{2 b}+\frac {\ln \left (-b x +b \right ) \ln \left (\frac {-b x +1}{1-b}\right )}{2 b}-\frac {\dilog \left (\frac {-b x -1}{-b -1}\right )}{2 b}-\frac {\ln \left (-b x +b \right ) \ln \left (\frac {-b x -1}{-b -1}\right )}{2 b}-\frac {\dilog \left (\frac {b x -1}{-b -1}\right )}{2 b}-\frac {\ln \left (b x +b \right ) \ln \left (\frac {b x -1}{-b -1}\right )}{2 b}+\frac {\dilog \left (\frac {b x +1}{1-b}\right )}{2 b}+\frac {\ln \left (b x +b \right ) \ln \left (\frac {b x +1}{1-b}\right )}{2 b}\right )}{2}}{b}\)	\(221\)
default	\(\frac {-\frac {\arctanh \left (b x \right ) b \ln \left (-b x +b \right )}{2}+\frac {\arctanh \left (b x \right ) b \ln \left (b x +b \right )}{2}-\frac {b^{2} \left (\frac {\dilog \left (\frac {-b x +1}{1-b}\right )}{2 b}+\frac {\ln \left (-b x +b \right ) \ln \left (\frac {-b x +1}{1-b}\right )}{2 b}-\frac {\dilog \left (\frac {-b x -1}{-b -1}\right )}{2 b}-\frac {\ln \left (-b x +b \right ) \ln \left (\frac {-b x -1}{-b -1}\right )}{2 b}-\frac {\dilog \left (\frac {b x -1}{-b -1}\right )}{2 b}-\frac {\ln \left (b x +b \right ) \ln \left (\frac {b x -1}{-b -1}\right )}{2 b}+\frac {\dilog \left (\frac {b x +1}{1-b}\right )}{2 b}+\frac {\ln \left (b x +b \right ) \ln \left (\frac {b x +1}{1-b}\right )}{2 b}\right )}{2}}{b}\)	\(221\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 180, normalized size = 1.05 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {\operatorname {atanh}{\left (b x \right )}}{x^{2} - 1}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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