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

Optimal. Leaf size=56 \[ \frac {1}{2} \text {Li}_3\left (1+\frac {2}{b x}\right )-\text {Li}_2\left (1+\frac {2}{b x}\right ) \tanh ^{-1}(b x+1)-\log \left (-\frac {2}{b x}\right ) \tanh ^{-1}(b x+1)^2 \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6111, 5918, 5948, 6058, 6610} \[ \frac {1}{2} \text {PolyLog}\left (3,\frac {2}{b x}+1\right )-\tanh ^{-1}(b x+1) \text {PolyLog}\left (2,\frac {2}{b x}+1\right )-\log \left (-\frac {2}{b x}\right ) \tanh ^{-1}(b x+1)^2 \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[1 + b*x]^2/x,x]

[Out]

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

Rule 5918

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

Rule 5948

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

Rule 6058

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[((a + b*ArcT
anh[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] + Dist[(b*p)/2, Int[((a + b*ArcTanh[c*x])^(p - 1)*PolyLog[2, 1 - u]
)/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[(1 - u)^2 - (1 -
2/(1 - c*x))^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]

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 75, normalized size = 1.34 \[ \tanh ^{-1}(b x+1) \text {Li}_2\left (-e^{-2 \tanh ^{-1}(b x+1)}\right )+\frac {1}{2} \text {Li}_3\left (-e^{-2 \tanh ^{-1}(b x+1)}\right )-\frac {2}{3} \tanh ^{-1}(b x+1)^3-\tanh ^{-1}(b x+1)^2 \log \left (e^{-2 \tanh ^{-1}(b x+1)}+1\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTanh[1 + b*x]^2/x,x]

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arctanh(b*x + 1)^2/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arctanh(b*x + 1)^2/x, x)

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maple [C]  time = 0.45, size = 160, normalized size = 2.86 \[ \ln \left (b x \right ) \arctanh \left (b x +1\right )^{2}-\arctanh \left (b x +1\right ) \polylog \left (2, -\frac {\left (b x +2\right )^{2}}{-\left (b x +1\right )^{2}+1}\right )+\frac {\polylog \left (3, -\frac {\left (b x +2\right )^{2}}{-\left (b x +1\right )^{2}+1}\right )}{2}-\left (i \pi \mathrm {csgn}\left (\frac {i}{1+\frac {\left (b x +2\right )^{2}}{-\left (b x +1\right )^{2}+1}}\right )^{3}-i \pi \mathrm {csgn}\left (\frac {i}{1+\frac {\left (b x +2\right )^{2}}{-\left (b x +1\right )^{2}+1}}\right )^{2}+i \pi +\ln \relax (2)\right ) \arctanh \left (b x +1\right )^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{12} \, \log \left (-b x\right )^{3} + \frac {1}{4} \, \log \left (b x + 2\right )^{2} \log \left (-x\right ) - \frac {1}{4} \, \int \frac {2 \, {\left (b x \log \relax (b) + 2 \, {\left (b x + 1\right )} \log \left (-x\right ) + 2 \, \log \relax (b)\right )} \log \left (b x + 2\right )}{b x^{2} + 2 \, x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*log(-b*x)^3 + 1/4*log(b*x + 2)^2*log(-x) - 1/4*integrate(2*(b*x*log(b) + 2*(b*x + 1)*log(-x) + 2*log(b))*
log(b*x + 2)/(b*x^2 + 2*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\mathrm {atanh}\left (b\,x+1\right )}^2}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(b*x + 1)^2/x,x)

[Out]

int(atanh(b*x + 1)^2/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atanh(b*x + 1)**2/x, x)

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