3.115 \(\int \sqrt {\tanh ^{-1}(\tanh (a+b x))} \, dx\)

Optimal. Leaf size=18 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b} \]

[Out]

2/3*arctanh(tanh(b*x+a))^(3/2)/b

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Rubi [A]  time = 0.00, antiderivative size = 18, 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 = {2157, 30} \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[ArcTanh[Tanh[a + b*x]]],x]

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(3/2))/(3*b)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2157

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 \sqrt {\tanh ^{-1}(\tanh (a+b x))} \, dx &=\frac {\operatorname {Subst}\left (\int \sqrt {x} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b}\\ &=\frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 1.00 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(3/2))/(3*b)

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fricas [A]  time = 0.53, size = 12, normalized size = 0.67 \[ \frac {2 \, {\left (b x + a\right )}^{\frac {3}{2}}}{3 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(b*x + a)^(3/2)/b

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giac [A]  time = 0.16, size = 18, normalized size = 1.00 \[ \frac {\sqrt {2} {\left (2 \, b x + 2 \, a\right )}^{\frac {3}{2}}}{6 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(2)*(2*b*x + 2*a)^(3/2)/b

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maple [A]  time = 0.03, size = 15, normalized size = 0.83 \[ \frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*arctanh(tanh(b*x+a))^(3/2)/b

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maxima [A]  time = 0.49, size = 12, normalized size = 0.67 \[ \frac {2 \, {\left (b x + a\right )}^{\frac {3}{2}}}{3 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(b*x + a)^(3/2)/b

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mupad [B]  time = 1.12, size = 95, normalized size = 5.28 \[ -\frac {\left (\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )\,\sqrt {\frac {\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}}{3\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(1/2),x)

[Out]

-((log(1/(exp(2*a)*exp(2*b*x) + 1)) - log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)))*(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))/(3*b)

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sympy [A]  time = 0.58, size = 26, normalized size = 1.44 \[ \begin {cases} \frac {2 \operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{3 b} & \text {for}\: b \neq 0 \\x \sqrt {\operatorname {atanh}{\left (\tanh {\relax (a )} \right )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*atanh(tanh(a + b*x))**(3/2)/(3*b), Ne(b, 0)), (x*sqrt(atanh(tanh(a))), True))

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