3.1.89 \(\int \frac {1}{\tanh ^{-1}(\tanh (a+b x))} \, dx\) [89]

Optimal. Leaf size=12 \[ \frac {\log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b} \]

[Out]

ln(arctanh(tanh(b*x+a)))/b

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Rubi [A]
time = 0.00, antiderivative size = 12, 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, 29} \begin {gather*} \frac {\log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(-1),x]

[Out]

Log[ArcTanh[Tanh[a + b*x]]]/b

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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

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Mathematica [A]
time = 0.04, size = 12, normalized size = 1.00 \begin {gather*} \frac {\log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(-1),x]

[Out]

Log[ArcTanh[Tanh[a + b*x]]]/b

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Maple [A]
time = 0.06, size = 13, normalized size = 1.08

method	result	size
derivativedivides	\(\frac {\ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )}{b}\)	\(13\)
default	\(\frac {\ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )}{b}\)	\(13\)
risch	\(\frac {\ln \left (\ln \left ({\mathrm e}^{b x +a}\right )+\frac {i \pi \left (-\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )+\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}-\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )+2 \,\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}-\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}+\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}-\mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}\right )}{4}\right )}{b}\)	\(266\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/arctanh(tanh(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

ln(arctanh(tanh(b*x+a)))/b

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Maxima [A]
time = 0.47, size = 13, normalized size = 1.08 \begin {gather*} \frac {\log \left (-b x - a\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(-b*x - a)/b

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Fricas [A]
time = 0.34, size = 10, normalized size = 0.83 \begin {gather*} \frac {\log \left (b x + a\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(b*x + a)/b

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Sympy [A]
time = 12.13, size = 17, normalized size = 1.42 \begin {gather*} \begin {cases} \frac {\log {\left (\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )} \right )}}{b} & \text {for}\: b \neq 0 \\\frac {x}{\operatorname {atanh}{\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/atanh(tanh(b*x+a)),x)

[Out]

Piecewise((log(atanh(tanh(a + b*x)))/b, Ne(b, 0)), (x/atanh(tanh(a)), True))

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Giac [A]
time = 0.40, size = 11, normalized size = 0.92 \begin {gather*} \frac {\log \left ({\left | b x + a \right |}\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(abs(b*x + a))/b

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Mupad [B]
time = 1.06, size = 12, normalized size = 1.00 \begin {gather*} \frac {\ln \left (\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/atanh(tanh(a + b*x)),x)

[Out]

log(atanh(tanh(a + b*x)))/b

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