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

Optimal. Leaf size=16 \[ \frac {\coth ^{-1}(\coth (a+b x))^2}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.00, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2157, 30} \[ \frac {\coth ^{-1}(\coth (a+b x))^2}{2 b} \]

Antiderivative was successfully verified.

[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 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]

Rubi steps

\begin {align*} \int \coth ^{-1}(\coth (a+b x)) \, dx &=\frac {\operatorname {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*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 1.12 \[ x \coth ^{-1}(\coth (a+b x))-\frac {b x^2}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[Coth[a + b*x]],x]

[Out]

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

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fricas [A]  time = 0.51, size = 10, normalized size = 0.62 \[ \frac {1}{2} x^{2} b + x a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*b + x*a

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giac [A]  time = 0.12, size = 10, normalized size = 0.62 \[ \frac {1}{2} \, b x^{2} + a x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*x^2 + a*x

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maple [B]  time = 0.06, size = 32, normalized size = 2.00 \[ \frac {\arctanh \left (\coth \left (b x +a \right )\right ) \mathrm {arccoth}\left (\coth \left (b x +a \right )\right )-\frac {\arctanh \left (\coth \left (b x +a \right )\right )^{2}}{2}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(coth(b*x+a)),x)

[Out]

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

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maxima [A]  time = 0.31, size = 10, normalized size = 0.62 \[ \frac {1}{2} \, b x^{2} + a x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*x^2 + a*x

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mupad [B]  time = 1.21, size = 16, normalized size = 1.00 \[ x\,\mathrm {acoth}\left (\mathrm {coth}\left (a+b\,x\right )\right )-\frac {b\,x^2}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(coth(a + b*x)),x)

[Out]

x*acoth(coth(a + b*x)) - (b*x^2)/2

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sympy [A]  time = 2.93, size = 37, normalized size = 2.31 \[ \begin {cases} x \operatorname {acoth}{\left (\coth {\relax (a )} \right )} & \text {for}\: b = 0 \\0 & \text {for}\: a = \log {\left (- e^{- b x} \right )} \vee a = \log {\left (e^{- b x} \right )} \\\frac {\operatorname {acoth}^{2}{\left (\frac {1}{\tanh {\left (a + b x \right )}} \right )}}{2 b} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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