\(\int \frac {1}{\coth ^{-1}(\tanh (a+b x))^3} \, dx\) [180]

Optimal result

Integrand size = 9, antiderivative size = 16 \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=-\frac {1}{2 b \coth ^{-1}(\tanh (a+b x))^2} \]

[Out]

-1/2/b/arccoth(tanh(b*x+a))^2

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, 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} \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=-\frac {1}{2 b \coth ^{-1}(\tanh (a+b x))^2} \]

[In]

Int[ArcCoth[Tanh[a + b*x]]^(-3),x]

[Out]

-1/2*1/(b*ArcCoth[Tanh[a + b*x]]^2)

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^3} \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{b} \\ & = -\frac {1}{2 b \coth ^{-1}(\tanh (a+b x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=-\frac {1}{2 b \coth ^{-1}(\tanh (a+b x))^2} \]

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^(-3),x]

[Out]

-1/2*1/(b*ArcCoth[Tanh[a + b*x]]^2)

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(-\frac {1}{2 b \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2}}\)	\(15\)
default	\(-\frac {1}{2 b \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2}}\)	\(15\)
parallelrisch	\(-\frac {1}{2 b \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2}}\)	\(15\)
risch	\(\frac {8}{b {\left (\pi \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-2 \pi \,\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )+\pi \operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )+\pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}-\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+2 \pi \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}-2 \pi \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+4 i \ln \left ({\mathrm e}^{b x +a}\right )+2 \pi \right )}^{2}}\)	\(318\)

[In]

int(1/arccoth(tanh(b*x+a))^3,x,method=_RETURNVERBOSE)

[Out]

-1/2/b/arccoth(tanh(b*x+a))^2

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.75 \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=-\frac {2}{4 \, b^{3} x^{2} + 8 \, a b^{2} x - \pi ^{2} b + 4 \, a^{2} b + 4 i \, \pi {\left (b^{2} x + a b\right )}} \]

[In]

integrate(1/arccoth(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

-2/(4*b^3*x^2 + 8*a*b^2*x - pi^2*b + 4*a^2*b + 4*I*pi*(b^2*x + a*b))

Sympy [A] (verification not implemented)

Time = 24.85 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=\begin {cases} - \frac {1}{2 b \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}} & \text {for}\: b \neq 0 \\\frac {x}{\operatorname {acoth}^{3}{\left (\tanh {\left (a \right )} \right )}} & \text {otherwise} \end {cases} \]

[In]

integrate(1/acoth(tanh(b*x+a))**3,x)

[Out]

Piecewise((-1/(2*b*acoth(tanh(a + b*x))**2), Ne(b, 0)), (x/acoth(tanh(a))**3, True))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.75 \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=\frac {2}{{\left (\pi ^{2} - 4 i \, \pi {\left (b x + a\right )} - 4 \, {\left (b x + a\right )}^{2}\right )} b} \]

[In]

integrate(1/arccoth(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

2/((pi^2 - 4*I*pi*(b*x + a) - 4*(b*x + a)^2)*b)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.75 \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=-\frac {2 i}{4 i \, b^{3} x^{2} - 4 \, \pi b^{2} x + 8 i \, a b^{2} x - i \, \pi ^{2} b - 4 \, \pi a b + 4 i \, a^{2} b} \]

[In]

integrate(1/arccoth(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

-2*I/(4*I*b^3*x^2 - 4*pi*b^2*x + 8*I*a*b^2*x - I*pi^2*b - 4*pi*a*b + 4*I*a^2*b)

Mupad [B] (verification not implemented)

Time = 3.80 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=-\frac {1}{2\,b\,{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2} \]

[In]

int(1/acoth(tanh(a + b*x))^3,x)

[Out]

-1/(2*b*acoth(tanh(a + b*x))^2)