3.1.5 \(\int \frac {\tanh ^{-1}(a+b x)^2}{x} \, dx\) [5]

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

[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 = 0.06, 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 = {6246, 6059} \begin {gather*} \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 \end {gather*}

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 6059

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^2/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTanh[c*x])^2)*(Lo
g[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*(PolyL
og[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 6246

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 {\text {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] Result contains complex when optimal does not.
time = 2.17, size = 634, normalized size = 4.28 \begin {gather*} -\frac {4}{3} \tanh ^{-1}(a+b x)^3-\frac {2 \tanh ^{-1}(a+b x)^3}{3 a}+\frac {2 \sqrt {1-a^2} e^{\tanh ^{-1}(a)} \tanh ^{-1}(a+b x)^3}{3 a}-\tanh ^{-1}(a+b x)^2 \log \left (1+e^{-2 \tanh ^{-1}(a+b x)}\right )-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 )+\tanh ^{-1}(a+b x)^2 \log \left (\frac {1}{2} e^{-\tanh ^{-1}(a+b x)} \left (1+a-e^{2 \tanh ^{-1}(a+b x)}+a e^{2 \tanh ^{-1}(a+b x)}\right )\right )-\tanh ^{-1}(a+b x)^2 \log \left (1+\frac {(-1+a) e^{2 \tanh ^{-1}(a+b x)}}{1+a}\right )+\tanh ^{-1}(a+b x)^2 \log \left (1-e^{-\tanh ^{-1}(a)+\tanh ^{-1}(a+b x)}\right )+\tanh ^{-1}(a+b x)^2 \log \left (1+e^{-\tanh ^{-1}(a)+\tanh ^{-1}(a+b x)}\right )-2 \tanh ^{-1}(a) \tanh ^{-1}(a+b x) \log \left (\frac {1}{2} i \left (-e^{\tanh ^{-1}(a)-\tanh ^{-1}(a+b x)}+e^{-\tanh ^{-1}(a)+\tanh ^{-1}(a+b x)}\right )\right )+\tanh ^{-1}(a+b x)^2 \log \left (1-e^{-2 \tanh ^{-1}(a)+2 \tanh ^{-1}(a+b x)}\right )+i \pi \tanh ^{-1}(a+b x) \log \left (\frac {1}{\sqrt {1-(a+b x)^2}}\right )-\tanh ^{-1}(a+b x)^2 \log \left (-\frac {b x}{\sqrt {1-(a+b x)^2}}\right )+2 \tanh ^{-1}(a) \tanh ^{-1}(a+b x) \log \left (-i \sinh \left (\tanh ^{-1}(a)-\tanh ^{-1}(a+b x)\right )\right )+\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 {(-1+a) e^{2 \tanh ^{-1}(a+b x)}}{1+a}\right )+2 \tanh ^{-1}(a+b x) \text {PolyLog}\left (2,-e^{-\tanh ^{-1}(a)+\tanh ^{-1}(a+b x)}\right )+2 \tanh ^{-1}(a+b x) \text {PolyLog}\left (2,e^{-\tanh ^{-1}(a)+\tanh ^{-1}(a+b x)}\right )+\tanh ^{-1}(a+b x) \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}(a)+2 \tanh ^{-1}(a+b x)}\right )+\frac {1}{2} \text {PolyLog}\left (3,-e^{-2 \tanh ^{-1}(a+b x)}\right )+\frac {1}{2} \text {PolyLog}\left (3,-\frac {(-1+a) e^{2 \tanh ^{-1}(a+b x)}}{1+a}\right )-2 \text {PolyLog}\left (3,-e^{-\tanh ^{-1}(a)+\tanh ^{-1}(a+b x)}\right )-2 \text {PolyLog}\left (3,e^{-\tanh ^{-1}(a)+\tanh ^{-1}(a+b x)}\right )-\frac {1}{2} \text {PolyLog}\left (3,e^{-2 \tanh ^{-1}(a)+2 \tanh ^{-1}(a+b x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[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] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 20.38, size = 931, normalized size = 6.29

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(931\)
default	\(\text {Expression too large to display}\)	\(931\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, 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+a)^2/x,x, algorithm="fricas")

[Out]

integral(arctanh(b*x + a)^2/x, x)

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

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 = 0.00, 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+a)^2/x,x, algorithm="giac")

[Out]

integrate(arctanh(b*x + a)^2/x, x)

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

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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