\(\int \frac {1}{\coth ^{-1}(\tanh (a+b x))} \, dx\) [162]

Optimal result

Integrand size = 9, antiderivative size = 12 \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))} \, dx=\frac {\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b} \]

[Out]

ln(arccoth(tanh(b*x+a)))/b

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2188, 29} \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))} \, dx=\frac {\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b} \]

[In]

Int[ArcCoth[Tanh[a + b*x]]^(-1),x]

[Out]

Log[ArcCoth[Tanh[a + b*x]]]/b

Rule 29

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

Rule 2188

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]

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))} \, dx=\frac {\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b} \]

[In]

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

[Out]

Log[ArcCoth[Tanh[a + b*x]]]/b

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08

method	result	size
derivativedivides	\(\frac {\ln \left (\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )\right )}{b}\)	\(13\)
default	\(\frac {\ln \left (\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )\right )}{b}\)	\(13\)
parallelrisch	\(\frac {\ln \left (\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )\right )}{b}\)	\(13\)
risch	\(\frac {\ln \left (\ln \left ({\mathrm e}^{b x +a}\right )-\frac {i \pi \left (\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )-\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+2 \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}-2 \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )-2 \,\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+2\right )}{4}\right )}{b}\)	\(305\)

[In]

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

[Out]

ln(arccoth(tanh(b*x+a)))/b

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.33 \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))} \, dx=\frac {\log \left (i \, \pi + 2 \, b x + 2 \, a\right )}{b} \]

[In]

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

[Out]

log(I*pi + 2*b*x + 2*a)/b

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.42 \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))} \, dx=\begin {cases} \frac {\log {\left (\operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )} \right )}}{b} & \text {for}\: b \neq 0 \\\frac {x}{\operatorname {acoth}{\left (\tanh {\left (a \right )} \right )}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((log(acoth(tanh(a + b*x)))/b, Ne(b, 0)), (x/acoth(tanh(a)), True))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.33 \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))} \, dx=\frac {\log \left (-\frac {1}{2} i \, \pi - b x - a\right )}{b} \]

[In]

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

[Out]

log(-1/2*I*pi - b*x - a)/b

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17 \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))} \, dx=\frac {\log \left (\pi - 2 i \, b x - 2 i \, a\right )}{b} \]

[In]

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

[Out]

log(pi - 2*I*b*x - 2*I*a)/b

Mupad [B] (verification not implemented)

Time = 3.80 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))} \, dx=\frac {\ln \left (\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )\right )}{b} \]

[In]

int(1/acoth(tanh(a + b*x)),x)

[Out]

log(acoth(tanh(a + b*x)))/b