3.3.77 \(\int \tanh ^{-1}(\coth (a+b x)) \, dx\) [277]

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

[Out]

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

________________________________________________________________________________________

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 {\tanh ^{-1}(\coth (a+b x))^2}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Coth[a + b*x]],x]

[Out]

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

________________________________________________________________________________________

Mathematica [A]
time = 0.00, size = 18, normalized size = 1.12 \begin {gather*} -\frac {b x^2}{2}+x \tanh ^{-1}(\coth (a+b x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Coth[a + b*x]],x]

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.10, size = 15, normalized size = 0.94

method	result	size
derivativedivides	\(\frac {\arctanh \left (\coth \left (b x +a \right )\right )^{2}}{2 b}\)	\(15\)
default	\(\frac {\arctanh \left (\coth \left (b x +a \right )\right )^{2}}{2 b}\)	\(15\)
risch	\(x \ln \left ({\mathrm e}^{b x +a}\right )-\frac {i \pi x \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )}{4}+\frac {i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2} x}{2}-\frac {i \pi x \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}}{4}-\frac {i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}-1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}-1}\right ) x}{4}+\frac {i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}-1}\right )^{2} x}{4}-\frac {i \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}-1}\right )^{3} x}{2}+\frac {i \pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}-1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}-1}\right )^{2} x}{4}-\frac {i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}-1}\right )^{3} x}{4}-\frac {b \,x^{2}}{2}+\frac {i \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}-1}\right )^{2} x}{2}-\frac {i \pi x}{2}\)	\(340\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*x^2 + x*arctanh(coth(b*x + a))

________________________________________________________________________________________

Fricas [A]
time = 0.51, 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(arctanh(coth(b*x+a)),x, algorithm="fricas")

[Out]

1/2*b*x^2 + a*x

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (12) = 24\).
time = 2.23, size = 80, normalized size = 5.00 \begin {gather*} \begin {cases} - \frac {\log {\left (- e^{- b x} \right )} \operatorname {atanh}{\left (\coth {\left (b x + \log {\left (- e^{- b x} \right )} \right )} \right )}}{b} & \text {for}\: a = \log {\left (- e^{- b x} \right )} \\x \operatorname {atanh}{\left (\coth {\left (b x + \log {\left (e^{- b x} \right )} \right )} \right )} & \text {for}\: a = \log {\left (e^{- b x} \right )} \\x \operatorname {atanh}{\left (\coth {\left (a \right )} \right )} & \text {for}\: b = 0 \\\frac {\operatorname {atanh}^{2}{\left (\frac {1}{\tanh {\left (a + b x \right )}} \right )}}{2 b} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-log(-exp(-b*x))*atanh(coth(b*x + log(-exp(-b*x))))/b, Eq(a, log(-exp(-b*x)))), (x*atanh(coth(b*x +
log(exp(-b*x)))), Eq(a, log(exp(-b*x)))), (x*atanh(coth(a)), Eq(b, 0)), (atanh(1/tanh(a + b*x))**2/(2*b), Tru
e))

________________________________________________________________________________________

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(arctanh(coth(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))

________________________________________________________________________________________

Mupad [B]
time = 0.02, size = 16, normalized size = 1.00 \begin {gather*} x\,\mathrm {atanh}\left (\mathrm {coth}\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(atanh(coth(a + b*x)),x)

[Out]

x*atanh(coth(a + b*x)) - (b*x^2)/2

________________________________________________________________________________________