3.2.31 \(\int \coth ^{-1}(\tanh (a+b x)) \, dx\) [131]

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

[Out]

1/2*arccoth(tanh(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 = {2188, 30} \begin {gather*} \frac {\coth ^{-1}(\tanh (a+b x))^2}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Tanh[a + b*x]],x]

[Out]

ArcCoth[Tanh[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*} \int \coth ^{-1}(\tanh (a+b x)) \, dx &=\frac {\text {Subst}\left (\int x \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{b}\\ &=\frac {\coth ^{-1}(\tanh (a+b x))^2}{2 b}\\ \end {align*}

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

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[Tanh[a + b*x]],x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(31\) vs. \(2(14)=28\).
time = 0.24, size = 32, normalized size = 2.00

method	result	size
derivativedivides	\(\frac {\arctanh \left (\tanh \left (b x +a \right )\right ) \mathrm {arccoth}\left (\tanh \left (b x +a \right )\right )-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{2}}{b}\)	\(32\)
default	\(\frac {\arctanh \left (\tanh \left (b x +a \right )\right ) \mathrm {arccoth}\left (\tanh \left (b x +a \right )\right )-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{2}}{b}\)	\(32\)
risch	\(x \ln \left ({\mathrm e}^{b x +a}\right )-\frac {i \pi x \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}}{4}+\frac {i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2} x}{2}+\frac {i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2} x}{4}-\frac {i \pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right ) x}{4}-\frac {i \pi x \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )}{4}-\frac {i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3} x}{4}+\frac {i \pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2} x}{4}-\frac {i \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3} x}{2}-\frac {b \,x^{2}}{2}+\frac {i \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2} x}{2}-\frac {i \pi x}{2}\)	\(340\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 16, normalized size = 1.00 \begin {gather*} -\frac {1}{2} \, b x^{2} + x \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*x^2 + x*arccoth(tanh(b*x + a))

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Fricas [A]
time = 0.34, size = 10, normalized size = 0.62 \begin {gather*} \frac {1}{2} \, b x^{2} + a x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*x^2 + a*x

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Sympy [A]
time = 0.06, size = 19, normalized size = 1.19 \begin {gather*} \begin {cases} \frac {\operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{2 b} & \text {for}\: b \neq 0 \\x \operatorname {acoth}{\left (\tanh {\left (a \right )} \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (14) = 28\).
time = 0.40, size = 69, normalized size = 4.31 \begin {gather*} -\frac {1}{2} \, b x^{2} + \frac {1}{2} \, x \log \left (-\frac {\frac {e^{\left (2 \, b x + 2 \, a\right )} + 1}{e^{\left (2 \, b x + 2 \, a\right )} - 1} + 1}{\frac {e^{\left (2 \, b x + 2 \, a\right )} + 1}{e^{\left (2 \, b x + 2 \, a\right )} - 1} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*x^2 + 1/2*x*log(-((e^(2*b*x + 2*a) + 1)/(e^(2*b*x + 2*a) - 1) + 1)/((e^(2*b*x + 2*a) + 1)/(e^(2*b*x + 2
*a) - 1) - 1))

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Mupad [B]
time = 1.12, size = 16, normalized size = 1.00 \begin {gather*} x\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )-\frac {b\,x^2}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(tanh(a + b*x)),x)

[Out]

x*acoth(tanh(a + b*x)) - (b*x^2)/2

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