3.2.44 \(\int \frac {1}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx\) [144]

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

[Out]

2*arctanh(tanh(b*x+a))^(1/2)/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 = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2188, 30} \begin {gather*} \frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[ArcTanh[Tanh[a + b*x]]],x]

[Out]

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

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Mathematica [A]
time = 0.01, size = 16, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[ArcTanh[Tanh[a + b*x]]],x]

[Out]

(2*Sqrt[ArcTanh[Tanh[a + b*x]]])/b

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Maple [A]
time = 0.07, size = 15, normalized size = 0.94

method	result	size
derivativedivides	\(\frac {2 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{b}\)	\(15\)
default	\(\frac {2 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{b}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.52, size = 12, normalized size = 0.75 \begin {gather*} \frac {2 \, \sqrt {b x + a}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(b*x + a)/b

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Fricas [A]
time = 0.38, size = 12, normalized size = 0.75 \begin {gather*} \frac {2 \, \sqrt {b x + a}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(b*x + a)/b

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Sympy [A]
time = 23.21, size = 24, normalized size = 1.50 \begin {gather*} \begin {cases} \frac {2 \sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}}{b} & \text {for}\: b \neq 0 \\\frac {x}{\sqrt {\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))**(1/2),x)

[Out]

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

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Giac [A]
time = 0.38, size = 12, normalized size = 0.75 \begin {gather*} \frac {2 \, \sqrt {b x + a}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(b*x + a)/b

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Mupad [B]
time = 1.18, size = 52, normalized size = 3.25 \begin {gather*} \frac {2\,\sqrt {\frac {\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/atanh(tanh(a + b*x))^(1/2),x)

[Out]

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

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