3.124 \(\int \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx\)

Optimal. Leaf size=18 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b} \]

[Out]

2/5*arctanh(tanh(b*x+a))^(5/2)/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} \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(3/2),x]

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(5/2))/(5*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 \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx &=\frac {\operatorname {Subst}\left (\int x^{3/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b}\\ &=\frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 1.00 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(5/2))/(5*b)

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fricas [A]  time = 0.52, size = 28, normalized size = 1.56 \[ \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {b x + a}}{5 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

2/5*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(b*x + a)/b

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giac [B]  time = 0.20, size = 84, normalized size = 4.67 \[ \frac {\sqrt {2} {\left (15 \, \sqrt {2} \sqrt {b x + a} a^{2} + 10 \, \sqrt {2} {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} a + \sqrt {2} {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )}\right )}}{15 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(3/2),x, algorithm="giac")

[Out]

1/15*sqrt(2)*(15*sqrt(2)*sqrt(b*x + a)*a^2 + 10*sqrt(2)*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*a + sqrt(2)*(3*(
b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2))/b

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maple [A]  time = 0.03, size = 15, normalized size = 0.83 \[ \frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*arctanh(tanh(b*x+a))^(5/2)/b

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maxima [A]  time = 0.49, size = 12, normalized size = 0.67 \[ \frac {2 \, {\left (b x + a\right )}^{\frac {5}{2}}}{5 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

2/5*(b*x + a)^(5/2)/b

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mupad [B]  time = 1.16, size = 97, normalized size = 5.39 \[ \frac {{\left (\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )}^2\,\sqrt {\frac {\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}}{10\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(3/2),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)))^2*(log((exp(2*a)*ex
p(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))/(10*b)

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sympy [A]  time = 11.53, size = 26, normalized size = 1.44 \[ \begin {cases} \frac {2 \operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{5 b} & \text {for}\: b \neq 0 \\x \operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\relax (a )} \right )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**(3/2),x)

[Out]

Piecewise((2*atanh(tanh(a + b*x))**(5/2)/(5*b), Ne(b, 0)), (x*atanh(tanh(a))**(3/2), True))

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