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

Optimal. Leaf size=148 \[ \frac{1}{2} \text{PolyLog}\left (3,1-\frac{2}{a+b x+1}\right )-\frac{1}{2} \text{PolyLog}\left (3,1-\frac{2 b x}{(1-a) (a+b x+1)}\right )+\tanh ^{-1}(a+b x) \text{PolyLog}\left (2,1-\frac{2}{a+b x+1}\right )-\tanh ^{-1}(a+b x) \text{PolyLog}\left (2,1-\frac{2 b x}{(1-a) (a+b x+1)}\right )-\log \left (\frac{2}{a+b x+1}\right ) \tanh ^{-1}(a+b x)^2+\log \left (\frac{2 b x}{(1-a) (a+b x+1)}\right ) \tanh ^{-1}(a+b x)^2 \]

[Out]

-(ArcTanh[a + b*x]^2*Log[2/(1 + a + b*x)]) + ArcTanh[a + b*x]^2*Log[(2*b*x)/((1 - a)*(1 + a + b*x))] + ArcTanh
[a + b*x]*PolyLog[2, 1 - 2/(1 + a + b*x)] - ArcTanh[a + b*x]*PolyLog[2, 1 - (2*b*x)/((1 - a)*(1 + a + b*x))] +
PolyLog[3, 1 - 2/(1 + a + b*x)]/2 - PolyLog[3, 1 - (2*b*x)/((1 - a)*(1 + a + b*x))]/2

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Rubi [A]  time = 0.0901863, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6111, 5922} \[ \frac{1}{2} \text{PolyLog}\left (3,1-\frac{2}{a+b x+1}\right )-\frac{1}{2} \text{PolyLog}\left (3,1-\frac{2 b x}{(1-a) (a+b x+1)}\right )+\tanh ^{-1}(a+b x) \text{PolyLog}\left (2,1-\frac{2}{a+b x+1}\right )-\tanh ^{-1}(a+b x) \text{PolyLog}\left (2,1-\frac{2 b x}{(1-a) (a+b x+1)}\right )-\log \left (\frac{2}{a+b x+1}\right ) \tanh ^{-1}(a+b x)^2+\log \left (\frac{2 b x}{(1-a) (a+b x+1)}\right ) \tanh ^{-1}(a+b x)^2 \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a + b*x]^2/x,x]

[Out]

-(ArcTanh[a + b*x]^2*Log[2/(1 + a + b*x)]) + ArcTanh[a + b*x]^2*Log[(2*b*x)/((1 - a)*(1 + a + b*x))] + ArcTanh
[a + b*x]*PolyLog[2, 1 - 2/(1 + a + b*x)] - ArcTanh[a + b*x]*PolyLog[2, 1 - (2*b*x)/((1 - a)*(1 + a + b*x))] +
PolyLog[3, 1 - 2/(1 + a + b*x)]/2 - PolyLog[3, 1 - (2*b*x)/((1 - a)*(1 + a + b*x))]/2

Rule 6111

Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcTanh[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &
& IGtQ[p, 0]

Rule 5922

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^2/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^2*Log[
2/(1 + c*x)])/e, x] + (Simp[((a + b*ArcTanh[c*x])^2*Log[(2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/e, x] + Simp[(
b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/e, x] - Simp[(b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - (2*c*(
d + e*x))/((c*d + e)*(1 + c*x))])/e, x] + Simp[(b^2*PolyLog[3, 1 - 2/(1 + c*x)])/(2*e), x] - Simp[(b^2*PolyLog
[3, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*e), x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 - e^2,
0]

Rubi steps

\begin{align*} \int \frac{\tanh ^{-1}(a+b x)^2}{x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\tanh ^{-1}(x)^2}{-\frac{a}{b}+\frac{x}{b}} \, dx,x,a+b x\right )}{b}\\ &=-\tanh ^{-1}(a+b x)^2 \log \left (\frac{2}{1+a+b x}\right )+\tanh ^{-1}(a+b x)^2 \log \left (\frac{2 b x}{(1-a) (1+a+b x)}\right )+\tanh ^{-1}(a+b x) \text{Li}_2\left (1-\frac{2}{1+a+b x}\right )-\tanh ^{-1}(a+b x) \text{Li}_2\left (1-\frac{2 b x}{(1-a) (1+a+b x)}\right )+\frac{1}{2} \text{Li}_3\left (1-\frac{2}{1+a+b x}\right )-\frac{1}{2} \text{Li}_3\left (1-\frac{2 b x}{(1-a) (1+a+b x)}\right )\\ \end{align*}

Mathematica [C]  time = 2.96676, size = 634, normalized size = 4.28 \[ \tanh ^{-1}(a+b x) \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a+b x)}\right )-\tanh ^{-1}(a+b x) \text{PolyLog}\left (2,-\frac{(a-1) e^{2 \tanh ^{-1}(a+b x)}}{a+1}\right )+2 \tanh ^{-1}(a+b x) \text{PolyLog}\left (2,-e^{\tanh ^{-1}(a+b x)-\tanh ^{-1}(a)}\right )+2 \tanh ^{-1}(a+b x) \text{PolyLog}\left (2,e^{\tanh ^{-1}(a+b x)-\tanh ^{-1}(a)}\right )+\tanh ^{-1}(a+b x) \text{PolyLog}\left (2,e^{2 \tanh ^{-1}(a+b x)-2 \tanh ^{-1}(a)}\right )+\frac{1}{2} \text{PolyLog}\left (3,-e^{-2 \tanh ^{-1}(a+b x)}\right )+\frac{1}{2} \text{PolyLog}\left (3,-\frac{(a-1) e^{2 \tanh ^{-1}(a+b x)}}{a+1}\right )-2 \text{PolyLog}\left (3,-e^{\tanh ^{-1}(a+b x)-\tanh ^{-1}(a)}\right )-2 \text{PolyLog}\left (3,e^{\tanh ^{-1}(a+b x)-\tanh ^{-1}(a)}\right )-\frac{1}{2} \text{PolyLog}\left (3,e^{2 \tanh ^{-1}(a+b x)-2 \tanh ^{-1}(a)}\right )+\frac{2 \sqrt{1-a^2} e^{\tanh ^{-1}(a)} \tanh ^{-1}(a+b x)^3}{3 a}-\frac{2 \tanh ^{-1}(a+b x)^3}{3 a}-\frac{4}{3} \tanh ^{-1}(a+b x)^3-\tanh ^{-1}(a+b x)^2 \log \left (e^{-2 \tanh ^{-1}(a+b x)}+1\right )+\tanh ^{-1}(a+b x)^2 \log \left (\frac{1}{2} e^{-\tanh ^{-1}(a+b x)} \left (a e^{2 \tanh ^{-1}(a+b x)}-e^{2 \tanh ^{-1}(a+b x)}+a+1\right )\right )-\tanh ^{-1}(a+b x)^2 \log \left (\frac{(a-1) e^{2 \tanh ^{-1}(a+b x)}}{a+1}+1\right )+\tanh ^{-1}(a+b x)^2 \log \left (1-e^{\tanh ^{-1}(a+b x)-\tanh ^{-1}(a)}\right )+\tanh ^{-1}(a+b x)^2 \log \left (e^{\tanh ^{-1}(a+b x)-\tanh ^{-1}(a)}+1\right )+\tanh ^{-1}(a+b x)^2 \log \left (1-e^{2 \tanh ^{-1}(a+b x)-2 \tanh ^{-1}(a)}\right )-\log \left (-\frac{b x}{\sqrt{1-(a+b x)^2}}\right ) \tanh ^{-1}(a+b x)^2-i \pi \tanh ^{-1}(a+b x) \log \left (\frac{1}{2} \left (e^{-\tanh ^{-1}(a+b x)}+e^{\tanh ^{-1}(a+b x)}\right )\right )-2 \tanh ^{-1}(a) \tanh ^{-1}(a+b x) \log \left (\frac{1}{2} i \left (e^{\tanh ^{-1}(a+b x)-\tanh ^{-1}(a)}-e^{\tanh ^{-1}(a)-\tanh ^{-1}(a+b x)}\right )\right )+i \pi \log \left (\frac{1}{\sqrt{1-(a+b x)^2}}\right ) \tanh ^{-1}(a+b x)+2 \tanh ^{-1}(a) \tanh ^{-1}(a+b x) \log \left (-i \sinh \left (\tanh ^{-1}(a)-\tanh ^{-1}(a+b x)\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTanh[a + b*x]^2/x,x]

[Out]

(-4*ArcTanh[a + b*x]^3)/3 - (2*ArcTanh[a + b*x]^3)/(3*a) + (2*Sqrt[1 - a^2]*E^ArcTanh[a]*ArcTanh[a + b*x]^3)/(
3*a) - ArcTanh[a + b*x]^2*Log[1 + E^(-2*ArcTanh[a + b*x])] - I*Pi*ArcTanh[a + b*x]*Log[(E^(-ArcTanh[a + b*x])
+ E^ArcTanh[a + b*x])/2] + ArcTanh[a + b*x]^2*Log[(1 + a - E^(2*ArcTanh[a + b*x]) + a*E^(2*ArcTanh[a + b*x]))/
(2*E^ArcTanh[a + b*x])] - ArcTanh[a + b*x]^2*Log[1 + ((-1 + a)*E^(2*ArcTanh[a + b*x]))/(1 + a)] + ArcTanh[a +
b*x]^2*Log[1 - E^(-ArcTanh[a] + ArcTanh[a + b*x])] + ArcTanh[a + b*x]^2*Log[1 + E^(-ArcTanh[a] + ArcTanh[a + b
*x])] - 2*ArcTanh[a]*ArcTanh[a + b*x]*Log[(I/2)*(-E^(ArcTanh[a] - ArcTanh[a + b*x]) + E^(-ArcTanh[a] + ArcTanh
[a + b*x]))] + ArcTanh[a + b*x]^2*Log[1 - E^(-2*ArcTanh[a] + 2*ArcTanh[a + b*x])] + I*Pi*ArcTanh[a + b*x]*Log[
1/Sqrt[1 - (a + b*x)^2]] - ArcTanh[a + b*x]^2*Log[-((b*x)/Sqrt[1 - (a + b*x)^2])] + 2*ArcTanh[a]*ArcTanh[a + b
*x]*Log[(-I)*Sinh[ArcTanh[a] - ArcTanh[a + b*x]]] + ArcTanh[a + b*x]*PolyLog[2, -E^(-2*ArcTanh[a + b*x])] - Ar
cTanh[a + b*x]*PolyLog[2, -(((-1 + a)*E^(2*ArcTanh[a + b*x]))/(1 + a))] + 2*ArcTanh[a + b*x]*PolyLog[2, -E^(-A
rcTanh[a] + ArcTanh[a + b*x])] + 2*ArcTanh[a + b*x]*PolyLog[2, E^(-ArcTanh[a] + ArcTanh[a + b*x])] + ArcTanh[a
+ b*x]*PolyLog[2, E^(-2*ArcTanh[a] + 2*ArcTanh[a + b*x])] + PolyLog[3, -E^(-2*ArcTanh[a + b*x])]/2 + PolyLog[
3, -(((-1 + a)*E^(2*ArcTanh[a + b*x]))/(1 + a))]/2 - 2*PolyLog[3, -E^(-ArcTanh[a] + ArcTanh[a + b*x])] - 2*Pol
yLog[3, E^(-ArcTanh[a] + ArcTanh[a + b*x])] - PolyLog[3, E^(-2*ArcTanh[a] + 2*ArcTanh[a + b*x])]/2

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Maple [C]  time = 0.537, size = 1022, normalized size = 6.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(b*x)*arctanh(b*x+a)^2-arctanh(b*x+a)^2*ln(a*((b*x+a+1)^2/(1-(b*x+a)^2)+1)-(b*x+a+1)^2/(1-(b*x+a)^2)+1)+1/2*
I*Pi*(csgn(I/((b*x+a+1)^2/(1-(b*x+a)^2)+1))*csgn(I*(a*((b*x+a+1)^2/(1-(b*x+a)^2)+1)-(b*x+a+1)^2/(1-(b*x+a)^2)+
1))*csgn(I*(a*((b*x+a+1)^2/(1-(b*x+a)^2)+1)-(b*x+a+1)^2/(1-(b*x+a)^2)+1)/((b*x+a+1)^2/(1-(b*x+a)^2)+1))-csgn(I
/((b*x+a+1)^2/(1-(b*x+a)^2)+1))*csgn(I*(a*((b*x+a+1)^2/(1-(b*x+a)^2)+1)-(b*x+a+1)^2/(1-(b*x+a)^2)+1)/((b*x+a+1
)^2/(1-(b*x+a)^2)+1))^2+2*csgn(I*(a*((b*x+a+1)^2/(1-(b*x+a)^2)+1)-(b*x+a+1)^2/(1-(b*x+a)^2)+1)/((b*x+a+1)^2/(1
-(b*x+a)^2)+1))^2-csgn(I*(a*((b*x+a+1)^2/(1-(b*x+a)^2)+1)-(b*x+a+1)^2/(1-(b*x+a)^2)+1))*csgn(I*(a*((b*x+a+1)^2
/(1-(b*x+a)^2)+1)-(b*x+a+1)^2/(1-(b*x+a)^2)+1)/((b*x+a+1)^2/(1-(b*x+a)^2)+1))^2-csgn(I*(a*((b*x+a+1)^2/(1-(b*x
+a)^2)+1)-(b*x+a+1)^2/(1-(b*x+a)^2)+1)/((b*x+a+1)^2/(1-(b*x+a)^2)+1))^3-2)*arctanh(b*x+a)^2-arctanh(b*x+a)*pol
ylog(2,-(b*x+a+1)^2/(1-(b*x+a)^2))+1/2*polylog(3,-(b*x+a+1)^2/(1-(b*x+a)^2))+a/(a-1)*arctanh(b*x+a)^2*ln(1-(a-
1)*(b*x+a+1)^2/(1-(b*x+a)^2)/(-1-a))+a/(a-1)*arctanh(b*x+a)*polylog(2,(a-1)*(b*x+a+1)^2/(1-(b*x+a)^2)/(-1-a))-
1/2*a/(a-1)*polylog(3,(a-1)*(b*x+a+1)^2/(1-(b*x+a)^2)/(-1-a))-1/(a-1)*arctanh(b*x+a)^2*ln(1-(a-1)*(b*x+a+1)^2/
(1-(b*x+a)^2)/(-1-a))-1/(a-1)*arctanh(b*x+a)*polylog(2,(a-1)*(b*x+a+1)^2/(1-(b*x+a)^2)/(-1-a))+1/2/(a-1)*polyl
og(3,(a-1)*(b*x+a+1)^2/(1-(b*x+a)^2)/(-1-a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arctanh(b*x + a)^2/x, 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 + a\right )^{2}}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arctanh(b*x + a)^2/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atanh(a + b*x)**2/x, 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 )^{2}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arctanh(b*x + a)^2/x, x)