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

Optimal. Leaf size=203 \[ \frac{1}{4} \text{PolyLog}\left (2,\frac{-a-b x+1}{-a-b+1}\right )-\frac{1}{4} \text{PolyLog}\left (2,\frac{-a-b x+1}{-a+b+1}\right )+\frac{1}{4} \text{PolyLog}\left (2,\frac{a+b x+1}{a-b+1}\right )-\frac{1}{4} \text{PolyLog}\left (2,\frac{a+b x+1}{a+b+1}\right )+\frac{1}{4} \log \left (-\frac{b (1-x)}{-a-b+1}\right ) \log (-a-b x+1)-\frac{1}{4} \log \left (\frac{b (x+1)}{-a+b+1}\right ) \log (-a-b x+1)-\frac{1}{4} \log \left (\frac{b (1-x)}{a+b+1}\right ) \log (a+b x+1)+\frac{1}{4} \log \left (-\frac{b (x+1)}{a-b+1}\right ) \log (a+b x+1) \]

[Out]

(Log[-((b*(1 - x))/(1 - a - b))]*Log[1 - a - b*x])/4 - (Log[(b*(1 + x))/(1 - a + b)]*Log[1 - a - b*x])/4 - (Lo
g[(b*(1 - x))/(1 + a + b)]*Log[1 + a + b*x])/4 + (Log[-((b*(1 + x))/(1 + a - b))]*Log[1 + a + b*x])/4 + PolyLo
g[2, (1 - a - b*x)/(1 - a - b)]/4 - PolyLog[2, (1 - a - b*x)/(1 - a + b)]/4 + PolyLog[2, (1 + a + b*x)/(1 + a
- b)]/4 - PolyLog[2, (1 + a + b*x)/(1 + a + b)]/4

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Rubi [A]  time = 0.257796, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {6115, 2409, 2394, 2393, 2391} \[ \frac{1}{4} \text{PolyLog}\left (2,\frac{-a-b x+1}{-a-b+1}\right )-\frac{1}{4} \text{PolyLog}\left (2,\frac{-a-b x+1}{-a+b+1}\right )+\frac{1}{4} \text{PolyLog}\left (2,\frac{a+b x+1}{a-b+1}\right )-\frac{1}{4} \text{PolyLog}\left (2,\frac{a+b x+1}{a+b+1}\right )+\frac{1}{4} \log \left (-\frac{b (1-x)}{-a-b+1}\right ) \log (-a-b x+1)-\frac{1}{4} \log \left (\frac{b (x+1)}{-a+b+1}\right ) \log (-a-b x+1)-\frac{1}{4} \log \left (\frac{b (1-x)}{a+b+1}\right ) \log (a+b x+1)+\frac{1}{4} \log \left (-\frac{b (x+1)}{a-b+1}\right ) \log (a+b x+1) \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a + b*x]/(1 - x^2),x]

[Out]

(Log[-((b*(1 - x))/(1 - a - b))]*Log[1 - a - b*x])/4 - (Log[(b*(1 + x))/(1 - a + b)]*Log[1 - a - b*x])/4 - (Lo
g[(b*(1 - x))/(1 + a + b)]*Log[1 + a + b*x])/4 + (Log[-((b*(1 + x))/(1 + a - b))]*Log[1 + a + b*x])/4 + PolyLo
g[2, (1 - a - b*x)/(1 - a - b)]/4 - PolyLog[2, (1 - a - b*x)/(1 - a + b)]/4 + PolyLog[2, (1 + a + b*x)/(1 + a
- b)]/4 - PolyLog[2, (1 + a + b*x)/(1 + a + b)]/4

Rule 6115

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

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}(a+b x)}{1-x^2} \, dx &=-\left (\frac{1}{2} \int \frac{\log (1-a-b x)}{1-x^2} \, dx\right )+\frac{1}{2} \int \frac{\log (1+a+b x)}{1-x^2} \, dx\\ &=-\left (\frac{1}{2} \int \left (\frac{\log (1-a-b x)}{2 (1-x)}+\frac{\log (1-a-b x)}{2 (1+x)}\right ) \, dx\right )+\frac{1}{2} \int \left (\frac{\log (1+a+b x)}{2 (1-x)}+\frac{\log (1+a+b x)}{2 (1+x)}\right ) \, dx\\ &=-\left (\frac{1}{4} \int \frac{\log (1-a-b x)}{1-x} \, dx\right )-\frac{1}{4} \int \frac{\log (1-a-b x)}{1+x} \, dx+\frac{1}{4} \int \frac{\log (1+a+b x)}{1-x} \, dx+\frac{1}{4} \int \frac{\log (1+a+b x)}{1+x} \, dx\\ &=\frac{1}{4} \log \left (-\frac{b (1-x)}{1-a-b}\right ) \log (1-a-b x)-\frac{1}{4} \log \left (\frac{b (1+x)}{1-a+b}\right ) \log (1-a-b x)-\frac{1}{4} \log \left (\frac{b (1-x)}{1+a+b}\right ) \log (1+a+b x)+\frac{1}{4} \log \left (-\frac{b (1+x)}{1+a-b}\right ) \log (1+a+b x)+\frac{1}{4} b \int \frac{\log \left (-\frac{b (1-x)}{1-a-b}\right )}{1-a-b x} \, dx+\frac{1}{4} b \int \frac{\log \left (\frac{b (1-x)}{1+a+b}\right )}{1+a+b x} \, dx-\frac{1}{4} b \int \frac{\log \left (-\frac{b (1+x)}{-1+a-b}\right )}{1-a-b x} \, dx-\frac{1}{4} b \int \frac{\log \left (\frac{b (1+x)}{-1-a+b}\right )}{1+a+b x} \, dx\\ &=\frac{1}{4} \log \left (-\frac{b (1-x)}{1-a-b}\right ) \log (1-a-b x)-\frac{1}{4} \log \left (\frac{b (1+x)}{1-a+b}\right ) \log (1-a-b x)-\frac{1}{4} \log \left (\frac{b (1-x)}{1+a+b}\right ) \log (1+a+b x)+\frac{1}{4} \log \left (-\frac{b (1+x)}{1+a-b}\right ) \log (1+a+b x)-\frac{1}{4} \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{1-a-b}\right )}{x} \, dx,x,1-a-b x\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{x}{-1+a-b}\right )}{x} \, dx,x,1-a-b x\right )-\frac{1}{4} \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{x}{-1-a+b}\right )}{x} \, dx,x,1+a+b x\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{1+a+b}\right )}{x} \, dx,x,1+a+b x\right )\\ &=\frac{1}{4} \log \left (-\frac{b (1-x)}{1-a-b}\right ) \log (1-a-b x)-\frac{1}{4} \log \left (\frac{b (1+x)}{1-a+b}\right ) \log (1-a-b x)-\frac{1}{4} \log \left (\frac{b (1-x)}{1+a+b}\right ) \log (1+a+b x)+\frac{1}{4} \log \left (-\frac{b (1+x)}{1+a-b}\right ) \log (1+a+b x)+\frac{1}{4} \text{Li}_2\left (\frac{1-a-b x}{1-a-b}\right )-\frac{1}{4} \text{Li}_2\left (\frac{1-a-b x}{1-a+b}\right )+\frac{1}{4} \text{Li}_2\left (\frac{1+a+b x}{1+a-b}\right )-\frac{1}{4} \text{Li}_2\left (\frac{1+a+b x}{1+a+b}\right )\\ \end{align*}

Mathematica [A]  time = 0.0392179, size = 203, normalized size = 1. \[ \frac{1}{4} \text{PolyLog}\left (2,\frac{-a-b x+1}{-a-b+1}\right )-\frac{1}{4} \text{PolyLog}\left (2,\frac{-a-b x+1}{-a+b+1}\right )+\frac{1}{4} \text{PolyLog}\left (2,\frac{a+b x+1}{a-b+1}\right )-\frac{1}{4} \text{PolyLog}\left (2,\frac{a+b x+1}{a+b+1}\right )+\frac{1}{4} \log \left (-\frac{b (1-x)}{-a-b+1}\right ) \log (-a-b x+1)-\frac{1}{4} \log \left (\frac{b (x+1)}{-a+b+1}\right ) \log (-a-b x+1)-\frac{1}{4} \log \left (\frac{b (1-x)}{a+b+1}\right ) \log (a+b x+1)+\frac{1}{4} \log \left (-\frac{b (x+1)}{a-b+1}\right ) \log (a+b x+1) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[a + b*x]/(1 - x^2),x]

[Out]

(Log[-((b*(1 - x))/(1 - a - b))]*Log[1 - a - b*x])/4 - (Log[(b*(1 + x))/(1 - a + b)]*Log[1 - a - b*x])/4 - (Lo
g[(b*(1 - x))/(1 + a + b)]*Log[1 + a + b*x])/4 + (Log[-((b*(1 + x))/(1 + a - b))]*Log[1 + a + b*x])/4 + PolyLo
g[2, (1 - a - b*x)/(1 - a - b)]/4 - PolyLog[2, (1 - a - b*x)/(1 - a + b)]/4 + PolyLog[2, (1 + a + b*x)/(1 + a
- b)]/4 - PolyLog[2, (1 + a + b*x)/(1 + a + b)]/4

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Maple [A]  time = 0.161, size = 196, normalized size = 1. \begin{align*} -{\frac{{\it Artanh} \left ( bx+a \right ) \ln \left ( bx-b \right ) }{2}}+{\frac{{\it Artanh} \left ( bx+a \right ) \ln \left ( bx+b \right ) }{2}}-{\frac{1}{4}{\it dilog} \left ({\frac{bx+a+1}{1+a-b}} \right ) }-{\frac{\ln \left ( bx+b \right ) }{4}\ln \left ({\frac{bx+a+1}{1+a-b}} \right ) }+{\frac{1}{4}{\it dilog} \left ({\frac{bx+a-1}{-b-1+a}} \right ) }+{\frac{\ln \left ( bx+b \right ) }{4}\ln \left ({\frac{bx+a-1}{-b-1+a}} \right ) }-{\frac{1}{4}{\it dilog} \left ({\frac{bx+a-1}{b-1+a}} \right ) }-{\frac{\ln \left ( bx-b \right ) }{4}\ln \left ({\frac{bx+a-1}{b-1+a}} \right ) }+{\frac{1}{4}{\it dilog} \left ({\frac{bx+a+1}{1+a+b}} \right ) }+{\frac{\ln \left ( bx-b \right ) }{4}\ln \left ({\frac{bx+a+1}{1+a+b}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.972205, size = 267, normalized size = 1.32 \begin{align*} \frac{1}{4} \, b{\left (\frac{\log \left (x - 1\right ) \log \left (\frac{b x - b}{a + b + 1} + 1\right ) +{\rm Li}_2\left (-\frac{b x - b}{a + b + 1}\right )}{b} - \frac{\log \left (x - 1\right ) \log \left (\frac{b x - b}{a + b - 1} + 1\right ) +{\rm Li}_2\left (-\frac{b x - b}{a + b - 1}\right )}{b} - \frac{\log \left (x + 1\right ) \log \left (\frac{b x + b}{a - b + 1} + 1\right ) +{\rm Li}_2\left (-\frac{b x + b}{a - b + 1}\right )}{b} + \frac{\log \left (x + 1\right ) \log \left (\frac{b x + b}{a - b - 1} + 1\right ) +{\rm Li}_2\left (-\frac{b x + b}{a - 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 + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(atanh(a + 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 + a\right )}{x^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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