3.66 \(\int \frac {\coth ^{-1}(a+b x)}{x} \, dx\)

Optimal. Leaf size=92 \[ \frac {1}{2} \text {Li}_2\left (1-\frac {2}{a+b x+1}\right )-\frac {1}{2} \text {Li}_2\left (1-\frac {2 b x}{(1-a) (a+b x+1)}\right )+\log \left (\frac {2}{a+b x+1}\right ) \left (-\coth ^{-1}(a+b x)\right )+\log \left (\frac {2 b x}{(1-a) (a+b x+1)}\right ) \coth ^{-1}(a+b x) \]

[Out]

-arccoth(b*x+a)*ln(2/(b*x+a+1))+arccoth(b*x+a)*ln(2*b*x/(1-a)/(b*x+a+1))+1/2*polylog(2,1-2/(b*x+a+1))-1/2*poly
log(2,1-2*b*x/(1-a)/(b*x+a+1))

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Rubi [A]  time = 0.10, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6112, 5921, 2402, 2315, 2447} \[ \frac {1}{2} \text {PolyLog}\left (2,1-\frac {2}{a+b x+1}\right )-\frac {1}{2} \text {PolyLog}\left (2,1-\frac {2 b x}{(1-a) (a+b x+1)}\right )+\log \left (\frac {2}{a+b x+1}\right ) \left (-\coth ^{-1}(a+b x)\right )+\log \left (\frac {2 b x}{(1-a) (a+b x+1)}\right ) \coth ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a + b*x]/x,x]

[Out]

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

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 5921

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

Rule 6112

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

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Mathematica [C]  time = 0.17, size = 259, normalized size = 2.82 \[ \frac {1}{8} \left (-4 \text {Li}_2\left (e^{2 \tanh ^{-1}(a)-2 \tanh ^{-1}(a+b x)}\right )-4 \text {Li}_2\left (-e^{2 \tanh ^{-1}(a+b x)}\right )+4 \left (\tanh ^{-1}(a)-\tanh ^{-1}(a+b x)\right )^2-\left (\pi -2 i \tanh ^{-1}(a+b x)\right )^2-8 \left (\tanh ^{-1}(a)-\tanh ^{-1}(a+b x)\right ) \log \left (1-e^{2 \tanh ^{-1}(a)-2 \tanh ^{-1}(a+b x)}\right )-4 i \left (\pi -2 i \tanh ^{-1}(a+b x)\right ) \log \left (e^{2 \tanh ^{-1}(a+b x)}+1\right )+4 \log \left (\frac {2}{\sqrt {1-(a+b x)^2}}\right ) \left (2 \tanh ^{-1}(a+b x)+i \pi \right )+8 \left (\tanh ^{-1}(a)-\tanh ^{-1}(a+b x)\right ) \log \left (-2 i \sinh \left (\tanh ^{-1}(a)-\tanh ^{-1}(a+b x)\right )\right )\right )+\tanh ^{-1}(a+b x) \left (-\log \left (\frac {1}{\sqrt {1-(a+b x)^2}}\right )+\log \left (-i \sinh \left (\tanh ^{-1}(a)-\tanh ^{-1}(a+b x)\right )\right )\right )+\log (x) \left (\coth ^{-1}(a+b x)-\tanh ^{-1}(a+b x)\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCoth[a + b*x]/x,x]

[Out]

(ArcCoth[a + b*x] - ArcTanh[a + b*x])*Log[x] + ArcTanh[a + b*x]*(-Log[1/Sqrt[1 - (a + b*x)^2]] + Log[(-I)*Sinh
[ArcTanh[a] - ArcTanh[a + b*x]]]) + (4*(ArcTanh[a] - ArcTanh[a + b*x])^2 - (Pi - (2*I)*ArcTanh[a + b*x])^2 - 8
*(ArcTanh[a] - ArcTanh[a + b*x])*Log[1 - E^(2*ArcTanh[a] - 2*ArcTanh[a + b*x])] - (4*I)*(Pi - (2*I)*ArcTanh[a
+ b*x])*Log[1 + E^(2*ArcTanh[a + b*x])] + 4*(I*Pi + 2*ArcTanh[a + b*x])*Log[2/Sqrt[1 - (a + b*x)^2]] + 8*(ArcT
anh[a] - ArcTanh[a + b*x])*Log[(-2*I)*Sinh[ArcTanh[a] - ArcTanh[a + b*x]]] - 4*PolyLog[2, E^(2*ArcTanh[a] - 2*
ArcTanh[a + b*x])] - 4*PolyLog[2, -E^(2*ArcTanh[a + b*x])])/8

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arcoth}\left (b x + a\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(b*x+a)/x,x, algorithm="fricas")

[Out]

integral(arccoth(b*x + a)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(b*x+a)/x,x, algorithm="giac")

[Out]

integrate(arccoth(b*x + a)/x, x)

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maple [A]  time = 0.06, size = 81, normalized size = 0.88 \[ \ln \left (b x \right ) \mathrm {arccoth}\left (b x +a \right )-\frac {\dilog \left (\frac {b x +a +1}{1+a}\right )}{2}-\frac {\ln \left (b x \right ) \ln \left (\frac {b x +a +1}{1+a}\right )}{2}+\frac {\dilog \left (\frac {b x +a -1}{a -1}\right )}{2}+\frac {\ln \left (b x \right ) \ln \left (\frac {b x +a -1}{a -1}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(b*x+a)/x,x)

[Out]

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

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maxima [A]  time = 0.31, size = 128, normalized size = 1.39 \[ -\frac {1}{2} \, b {\left (\frac {\log \left (b x + a + 1\right )}{b} - \frac {\log \left (b x + a - 1\right )}{b}\right )} \log \relax (x) + \frac {1}{2} \, b {\left (\frac {\log \left (b x + a + 1\right ) \log \left (-\frac {b x + a + 1}{a + 1} + 1\right ) + {\rm Li}_2\left (\frac {b x + a + 1}{a + 1}\right )}{b} - \frac {\log \left (b x + a - 1\right ) \log \left (-\frac {b x + a - 1}{a - 1} + 1\right ) + {\rm Li}_2\left (\frac {b x + a - 1}{a - 1}\right )}{b}\right )} + \operatorname {arcoth}\left (b x + a\right ) \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(b*x+a)/x,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(a + b*x)/x,x)

[Out]

int(acoth(a + b*x)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(b*x+a)/x,x)

[Out]

Integral(acoth(a + b*x)/x, x)

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