\(\int \coth ^{-1}(\coth (a+b x)) \, dx\) [196]

Optimal result

Integrand size = 7, antiderivative size = 16 \[ \int \coth ^{-1}(\coth (a+b x)) \, dx=\frac {\coth ^{-1}(\coth (a+b x))^2}{2 b} \]

[Out]

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

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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2188, 30} \[ \int \coth ^{-1}(\coth (a+b x)) \, dx=\frac {\coth ^{-1}(\coth (a+b x))^2}{2 b} \]

[In]

Int[ArcCoth[Coth[a + b*x]],x]

[Out]

ArcCoth[Coth[a + b*x]]^2/(2*b)

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12 \[ \int \coth ^{-1}(\coth (a+b x)) \, dx=-\frac {b x^2}{2}+x \coth ^{-1}(\coth (a+b x)) \]

[In]

Integrate[ArcCoth[Coth[a + b*x]],x]

[Out]

-1/2*(b*x^2) + x*ArcCoth[Coth[a + b*x]]

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19

method	result	size
parallelrisch	\(-\frac {x \left (b x -2 \,\operatorname {arccoth}\left (\frac {1}{\tanh \left (b x +a \right )}\right )\right )}{2}\)	\(19\)
derivativedivides	\(\frac {\operatorname {arctanh}\left (\coth \left (b x +a \right )\right ) \operatorname {arccoth}\left (\coth \left (b x +a \right )\right )-\frac {\operatorname {arctanh}\left (\coth \left (b x +a \right )\right )^{2}}{2}}{b}\)	\(32\)
default	\(\frac {\operatorname {arctanh}\left (\coth \left (b x +a \right )\right ) \operatorname {arccoth}\left (\coth \left (b x +a \right )\right )-\frac {\operatorname {arctanh}\left (\coth \left (b x +a \right )\right )^{2}}{2}}{b}\)	\(32\)
parts	\(x \,\operatorname {arccoth}\left (\coth \left (b x +a \right )\right )+\frac {-\frac {\left (b x +a \right )^{2}}{2}+\left (b x +a \right ) a}{b}\)	\(32\)
risch	\(x \ln \left ({\mathrm e}^{b x +a}\right )-\frac {i \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}-1}\right )^{3} x}{4}+\frac {i \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}-1}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) x}{4}+\frac {i \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}-1}\right )^{2} \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}-1}\right ) x}{4}-\frac {i \pi \,\operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}-1}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}-1}\right ) x}{4}-\frac {i \pi \operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) x}{4}+\frac {i \pi \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right ) x}{2}-\frac {i \pi \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3} x}{4}-\frac {b \,x^{2}}{2}\)	\(287\)

[In]

int(arccoth(coth(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \coth ^{-1}(\coth (a+b x)) \, dx=\frac {1}{2} x^{2} b + x a \]

[In]

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

[Out]

1/2*x^2*b + x*a

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (12) = 24\).

Time = 1.47 (sec) , antiderivative size = 78, normalized size of antiderivative = 4.88 \[ \int \coth ^{-1}(\coth (a+b x)) \, dx=\begin {cases} x \operatorname {acoth}{\left (\coth {\left (a \right )} \right )} & \text {for}\: b = 0 \\x \operatorname {acoth}{\left (\coth {\left (b x + \log {\left (- e^{- b x} \right )} \right )} \right )} & \text {for}\: a = \log {\left (- e^{- b x} \right )} \\- \frac {\log {\left (e^{- b x} \right )} \operatorname {acoth}{\left (\coth {\left (b x + \log {\left (e^{- b x} \right )} \right )} \right )}}{b} & \text {for}\: a = \log {\left (e^{- b x} \right )} \\\frac {\operatorname {acoth}^{2}{\left (\frac {1}{\tanh {\left (a + b x \right )}} \right )}}{2 b} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x*acoth(coth(a)), Eq(b, 0)), (x*acoth(coth(b*x + log(-exp(-b*x)))), Eq(a, log(-exp(-b*x)))), (-log(
exp(-b*x))*acoth(coth(b*x + log(exp(-b*x))))/b, Eq(a, log(exp(-b*x)))), (acoth(1/tanh(a + b*x))**2/(2*b), True
))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \coth ^{-1}(\coth (a+b x)) \, dx=\frac {1}{2} \, b x^{2} + a x \]

[In]

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

[Out]

1/2*b*x^2 + a*x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \coth ^{-1}(\coth (a+b x)) \, dx=\frac {1}{2} \, b x^{2} + a x \]

[In]

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

[Out]

1/2*b*x^2 + a*x

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \coth ^{-1}(\coth (a+b x)) \, dx=x\,\mathrm {acoth}\left (\mathrm {coth}\left (a+b\,x\right )\right )-\frac {b\,x^2}{2} \]

[In]

int(acoth(coth(a + b*x)),x)

[Out]

x*acoth(coth(a + b*x)) - (b*x^2)/2