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

Optimal. Leaf size=44 \[ \frac{\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b x-\coth ^{-1}(\tanh (a+b x))}-\frac{\log (x)}{b x-\coth ^{-1}(\tanh (a+b x))} \]

[Out]

-(Log[x]/(b*x - ArcCoth[Tanh[a + b*x]])) + Log[ArcCoth[Tanh[a + b*x]]]/(b*x - ArcCoth[Tanh[a + b*x]])

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Rubi [A]  time = 0.0284736, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2160, 2157, 29} \[ \frac{\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b x-\coth ^{-1}(\tanh (a+b x))}-\frac{\log (x)}{b x-\coth ^{-1}(\tanh (a+b x))} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*ArcCoth[Tanh[a + b*x]]),x]

[Out]

-(Log[x]/(b*x - ArcCoth[Tanh[a + b*x]])) + Log[ArcCoth[Tanh[a + b*x]]]/(b*x - ArcCoth[Tanh[a + b*x]])

Rule 2160

Int[1/((u_)*(v_)), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Dist[b/(b*u - a*v), Int[1
/v, x], x] - Dist[a/(b*u - a*v), Int[1/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x]

Rule 2157

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rubi steps

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

Mathematica [A]  time = 0.0183468, size = 29, normalized size = 0.66 \[ \frac{\log \left (\coth ^{-1}(\tanh (a+b x))\right )-\log (x)}{b x-\coth ^{-1}(\tanh (a+b x))} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*ArcCoth[Tanh[a + b*x]]),x]

[Out]

(-Log[x] + Log[ArcCoth[Tanh[a + b*x]]])/(b*x - ArcCoth[Tanh[a + b*x]])

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Maple [C]  time = 11.445, size = 972, normalized size = 22.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/arccoth(tanh(b*x+a)),x)

[Out]

-4*I/(2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^3+Pi*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+
2*a)/(exp(2*b*x+2*a)+1))-Pi*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+Pi*csgn(I*e
xp(2*b*x+2*a))^3-2*Pi*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2-Pi*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2
*a)/(exp(2*b*x+2*a)+1))^2+Pi*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))+Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2
*a)+1))^3-4*I*b*x-2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^2+4*I*ln(exp(b*x+a))+2*Pi)*ln(-2*Pi*csgn(I/(exp(2*b*x+2*a)+1
))^2+Pi*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))-Pi*csgn(I/
(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^3+Pi*csgn(I*ex
p(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-2*Pi*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2+Pi*csgn(I*exp(2*b*x+2*a))^
3-Pi*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*
a)+1))^3+4*I*b*x+4*I*(ln(exp(b*x+a))-b*x-a)+4*I*a+2*Pi)+4*I/(2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^3+Pi*csgn(I/(exp(
2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))-Pi*csgn(I/(exp(2*b*x+2*a)+1))*
csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+Pi*csgn(I*exp(2*b*x+2*a))^3-2*Pi*csgn(I*exp(b*x+a))*csgn(I*exp(2*b
*x+2*a))^2-Pi*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+Pi*csgn(I*exp(b*x+a))^2*csgn(
I*exp(2*b*x+2*a))+Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3-4*I*b*x-2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^2+4*I
*ln(exp(b*x+a))+2*Pi)*ln(x)

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Maxima [C]  time = 1.81207, size = 50, normalized size = 1.14 \begin{align*} \frac{2 \, \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{i \, \pi - 2 \, a} - \frac{2 \, \log \left (x\right )}{i \, \pi - 2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/arccoth(tanh(b*x+a)),x, algorithm="maxima")

[Out]

2*log(-I*pi + 2*b*x + 2*a)/(I*pi - 2*a) - 2*log(x)/(I*pi - 2*a)

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Fricas [A]  time = 1.69563, size = 205, normalized size = 4.66 \begin{align*} -\frac{2 \,{\left (2 \, \pi \arctan \left (-\frac{2 \, b x + 2 \, a - \sqrt{4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}}}{\pi }\right ) + a \log \left (4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}\right ) - 2 \, a \log \left (x\right )\right )}}{\pi ^{2} + 4 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/arccoth(tanh(b*x+a)),x, algorithm="fricas")

[Out]

-2*(2*pi*arctan(-(2*b*x + 2*a - sqrt(4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a^2))/pi) + a*log(4*b^2*x^2 + 8*a*b*x + pi
^2 + 4*a^2) - 2*a*log(x))/(pi^2 + 4*a^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \operatorname{acoth}{\left (\tanh{\left (a + b x \right )} \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/acoth(tanh(b*x+a)),x)

[Out]

Integral(1/(x*acoth(tanh(a + b*x))), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/arccoth(tanh(b*x+a)),x, algorithm="giac")

[Out]

integrate(1/(x*arccoth(tanh(b*x + a))), x)