3.2.71 \(\int \frac {1}{\coth ^{-1}(\tanh (a+b x))^2} \, dx\) [171]

Optimal. Leaf size=14 \[ -\frac {1}{b \coth ^{-1}(\tanh (a+b x))} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2188, 30} \begin {gather*} -\frac {1}{b \coth ^{-1}(\tanh (a+b x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Tanh[a + b*x]]^(-2),x]

[Out]

-(1/(b*ArcCoth[Tanh[a + b*x]]))

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

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Mathematica [A]
time = 0.01, size = 14, normalized size = 1.00 \begin {gather*} -\frac {1}{b \coth ^{-1}(\tanh (a+b x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^(-2),x]

[Out]

-(1/(b*ArcCoth[Tanh[a + b*x]]))

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Maple [A]
time = 0.21, size = 15, normalized size = 1.07

method	result	size
derivativedivides	\(-\frac {1}{b \,\mathrm {arccoth}\left (\tanh \left (b x +a \right )\right )}\)	\(15\)
default	\(-\frac {1}{b \,\mathrm {arccoth}\left (\tanh \left (b x +a \right )\right )}\)	\(15\)
risch	\(-\frac {4 i}{b \left (\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}-2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+\pi \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )-2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+2 \pi +4 i \ln \left ({\mathrm e}^{b x +a}\right )\right )}\)	\(319\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/arccoth(tanh(b*x+a))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.47, size = 18, normalized size = 1.29 \begin {gather*} \frac {4}{-2 \, {\left (i \, \pi + 2 \, b x + 2 \, a\right )} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccoth(tanh(b*x+a))^2,x, algorithm="maxima")

[Out]

4/((-2*I*pi - 4*b*x - 4*a)*b)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 36 vs. \(2 (14) = 28\).
time = 0.42, size = 36, normalized size = 2.57 \begin {gather*} -\frac {4 \, {\left (b x + a\right )}}{4 \, b^{3} x^{2} + 8 \, a b^{2} x + \pi ^{2} b + 4 \, a^{2} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccoth(tanh(b*x+a))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 21.87, size = 20, normalized size = 1.43 \begin {gather*} \begin {cases} - \frac {1}{b \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}} & \text {for}\: b \neq 0 \\\frac {x}{\operatorname {acoth}^{2}{\left (\tanh {\left (a \right )} \right )}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/acoth(tanh(b*x+a))**2,x)

[Out]

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

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Giac [C] Result contains complex when optimal does not.
time = 0.40, size = 19, normalized size = 1.36 \begin {gather*} -\frac {2}{2 \, b^{2} x + i \, \pi b + 2 \, a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccoth(tanh(b*x+a))^2,x, algorithm="giac")

[Out]

-2/(2*b^2*x + I*pi*b + 2*a*b)

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Mupad [B]
time = 1.14, size = 14, normalized size = 1.00 \begin {gather*} -\frac {1}{b\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/acoth(tanh(a + b*x))^2,x)

[Out]

-1/(b*acoth(tanh(a + b*x)))

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