\(\int \frac {1}{\coth ^{-1}(\tanh (a+b x))^2} \, dx\) [171]

Optimal result

Integrand size = 9, antiderivative size = 14 \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))^2} \, dx=-\frac {1}{b \coth ^{-1}(\tanh (a+b x))} \]

[Out]

-1/b/arccoth(tanh(b*x+a))

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, 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, 30} \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))^2} \, dx=-\frac {1}{b \coth ^{-1}(\tanh (a+b x))} \]

[In]

Int[ArcCoth[Tanh[a + b*x]]^(-2),x]

[Out]

-(1/(b*ArcCoth[Tanh[a + b*x]]))

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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^2} \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{b} \\ & = -\frac {1}{b \coth ^{-1}(\tanh (a+b x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))^2} \, dx=-\frac {1}{b \coth ^{-1}(\tanh (a+b x))} \]

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^(-2),x]

[Out]

-(1/(b*ArcCoth[Tanh[a + b*x]]))

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07

method	result	size
derivativedivides	\(-\frac {1}{b \,\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}\)	\(15\)
default	\(-\frac {1}{b \,\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}\)	\(15\)
parallelrisch	\(-\frac {1}{b \,\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}\)	\(15\)
risch	\(-\frac {4 i}{b \left (\pi \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-2 \pi \,\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )+\pi \operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )+\pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}-\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+2 \pi \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}-2 \pi \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+4 i \ln \left ({\mathrm e}^{b x +a}\right )+2 \pi \right )}\)	\(319\)

[In]

int(1/arccoth(tanh(b*x+a))^2,x,method=_RETURNVERBOSE)

[Out]

-1/b/arccoth(tanh(b*x+a))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.36 \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))^2} \, dx=-\frac {2}{2 \, b^{2} x + i \, \pi b + 2 \, a b} \]

[In]

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

[Out]

-2/(2*b^2*x + I*pi*b + 2*a*b)

Sympy [A] (verification not implemented)

Time = 12.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.43 \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))^2} \, dx=\begin {cases} - \frac {1}{b \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}} & \text {for}\: b \neq 0 \\\frac {x}{\operatorname {acoth}^{2}{\left (\tanh {\left (a \right )} \right )}} & \text {otherwise} \end {cases} \]

[In]

integrate(1/acoth(tanh(b*x+a))**2,x)

[Out]

Piecewise((-1/(b*acoth(tanh(a + b*x))), Ne(b, 0)), (x/acoth(tanh(a))**2, True))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.29 \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))^2} \, dx=\frac {4}{-2 \, {\left (i \, \pi + 2 \, b x + 2 \, a\right )} b} \]

[In]

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

[Out]

4/((-2*I*pi - 4*b*x - 4*a)*b)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.36 \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))^2} \, dx=-\frac {2}{2 \, b^{2} x + i \, \pi b + 2 \, a b} \]

[In]

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

[Out]

-2/(2*b^2*x + I*pi*b + 2*a*b)

Mupad [B] (verification not implemented)

Time = 3.84 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))^2} \, dx=-\frac {1}{b\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )} \]

[In]

int(1/acoth(tanh(a + b*x))^2,x)

[Out]

-1/(b*acoth(tanh(a + b*x)))