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

Optimal. Leaf size=148 \[ \frac {1}{2} \text {Li}_3\left (1-\frac {2}{a+b x+1}\right )-\frac {1}{2} \text {Li}_3\left (1-\frac {2 b x}{(1-a) (a+b x+1)}\right )+\text {Li}_2\left (1-\frac {2}{a+b x+1}\right ) \tanh ^{-1}(a+b x)-\text {Li}_2\left (1-\frac {2 b x}{(1-a) (a+b x+1)}\right ) \tanh ^{-1}(a+b x)-\log \left (\frac {2}{a+b x+1}\right ) \tanh ^{-1}(a+b x)^2+\log \left (\frac {2 b x}{(1-a) (a+b x+1)}\right ) \tanh ^{-1}(a+b x)^2 \]

[Out]

-arctanh(b*x+a)^2*ln(2/(b*x+a+1))+arctanh(b*x+a)^2*ln(2*b*x/(1-a)/(b*x+a+1))+arctanh(b*x+a)*polylog(2,1-2/(b*x
+a+1))-arctanh(b*x+a)*polylog(2,1-2*b*x/(1-a)/(b*x+a+1))+1/2*polylog(3,1-2/(b*x+a+1))-1/2*polylog(3,1-2*b*x/(1
-a)/(b*x+a+1))

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Rubi [A]  time = 0.09, antiderivative size = 148, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [C]  time = 2.64, size = 472, normalized size = 3.19 \[ \frac {2 \sqrt {1-a^2} e^{\tanh ^{-1}(a)} \tanh ^{-1}(a+b x)^3}{3 a}+\tanh ^{-1}(a+b x) \text {Li}_2\left (-e^{-2 \tanh ^{-1}(a+b x)}\right )+2 \tanh ^{-1}(a+b x) \text {Li}_2\left (-e^{\tanh ^{-1}(a+b x)-\tanh ^{-1}(a)}\right )+2 \tanh ^{-1}(a+b x) \text {Li}_2\left (e^{\tanh ^{-1}(a+b x)-\tanh ^{-1}(a)}\right )+\frac {1}{2} \text {Li}_3\left (-e^{-2 \tanh ^{-1}(a+b x)}\right )-2 \text {Li}_3\left (-e^{\tanh ^{-1}(a+b x)-\tanh ^{-1}(a)}\right )-2 \text {Li}_3\left (e^{\tanh ^{-1}(a+b x)-\tanh ^{-1}(a)}\right )-\frac {2 \tanh ^{-1}(a+b x)^3}{3 a}-\frac {4}{3} \tanh ^{-1}(a+b x)^3-\tanh ^{-1}(a+b x)^2 \log \left (e^{-2 \tanh ^{-1}(a+b x)}+1\right )+\tanh ^{-1}(a+b x)^2 \log \left (\frac {1}{2} e^{-\tanh ^{-1}(a+b x)} \left (a e^{2 \tanh ^{-1}(a+b x)}-e^{2 \tanh ^{-1}(a+b x)}+a+1\right )\right )+\tanh ^{-1}(a+b x)^2 \log \left (1-e^{\tanh ^{-1}(a+b x)-\tanh ^{-1}(a)}\right )+\tanh ^{-1}(a+b x)^2 \log \left (e^{\tanh ^{-1}(a+b x)-\tanh ^{-1}(a)}+1\right )-\log \left (-\frac {b x}{\sqrt {1-(a+b x)^2}}\right ) \tanh ^{-1}(a+b x)^2-i \pi \tanh ^{-1}(a+b x) \log \left (\frac {1}{2} \left (e^{-\tanh ^{-1}(a+b x)}+e^{\tanh ^{-1}(a+b x)}\right )\right )-2 \tanh ^{-1}(a) \tanh ^{-1}(a+b x) \log \left (\frac {1}{2} i \left (e^{\tanh ^{-1}(a+b x)-\tanh ^{-1}(a)}-e^{\tanh ^{-1}(a)-\tanh ^{-1}(a+b x)}\right )\right )+i \pi \log \left (\frac {1}{\sqrt {1-(a+b x)^2}}\right ) \tanh ^{-1}(a+b x)+2 \tanh ^{-1}(a) \tanh ^{-1}(a+b x) \log \left (-i \sinh \left (\tanh ^{-1}(a)-\tanh ^{-1}(a+b x)\right )\right ) \]

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 - 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] - ArcTan
h[a + b*x]) + E^(-ArcTanh[a] + ArcTanh[a + b*x]))] + I*Pi*ArcTanh[a + b*x]*Log[1/Sqrt[1 - (a + b*x)^2]] - ArcT
anh[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])] + 2*ArcTanh[a + b*x]*PolyLog[2, -E^
(-ArcTanh[a] + ArcTanh[a + b*x])] + 2*ArcTanh[a + b*x]*PolyLog[2, E^(-ArcTanh[a] + ArcTanh[a + b*x])] + PolyLo
g[3, -E^(-2*ArcTanh[a + b*x])]/2 - 2*PolyLog[3, -E^(-ArcTanh[a] + ArcTanh[a + b*x])] - 2*PolyLog[3, E^(-ArcTan
h[a] + ArcTanh[a + b*x])]

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {artanh}\left (b x + a\right )^{2}}{x}, x\right ) \]

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

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)

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maple [C]  time = 0.85, size = 1022, normalized size = 6.91 \[ \text {result too large to display} \]

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*(1+(b*x+a+1)^2/(1-(b*x+a)^2))-(b*x+a+1)^2/(1-(b*x+a)^2)+1)+1/2*
I*Pi*(csgn(I/(1+(b*x+a+1)^2/(1-(b*x+a)^2)))*csgn(I*(a*(1+(b*x+a+1)^2/(1-(b*x+a)^2))-(b*x+a+1)^2/(1-(b*x+a)^2)+
1))*csgn(I*(a*(1+(b*x+a+1)^2/(1-(b*x+a)^2))-(b*x+a+1)^2/(1-(b*x+a)^2)+1)/(1+(b*x+a+1)^2/(1-(b*x+a)^2)))-csgn(I
/(1+(b*x+a+1)^2/(1-(b*x+a)^2)))*csgn(I*(a*(1+(b*x+a+1)^2/(1-(b*x+a)^2))-(b*x+a+1)^2/(1-(b*x+a)^2)+1)/(1+(b*x+a
+1)^2/(1-(b*x+a)^2)))^2+2*csgn(I*(a*(1+(b*x+a+1)^2/(1-(b*x+a)^2))-(b*x+a+1)^2/(1-(b*x+a)^2)+1)/(1+(b*x+a+1)^2/
(1-(b*x+a)^2)))^2-csgn(I*(a*(1+(b*x+a+1)^2/(1-(b*x+a)^2))-(b*x+a+1)^2/(1-(b*x+a)^2)+1))*csgn(I*(a*(1+(b*x+a+1)
^2/(1-(b*x+a)^2))-(b*x+a+1)^2/(1-(b*x+a)^2)+1)/(1+(b*x+a+1)^2/(1-(b*x+a)^2)))^2-csgn(I*(a*(1+(b*x+a+1)^2/(1-(b
*x+a)^2))-(b*x+a+1)^2/(1-(b*x+a)^2)+1)/(1+(b*x+a+1)^2/(1-(b*x+a)^2)))^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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {artanh}\left (b x + a\right )^{2}}{x}\,{d x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a + b*x)^2/x,x)

[Out]

int(atanh(a + b*x)^2/x, x)

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

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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