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

Optimal. Leaf size=63 \[ -16 b^2 x^{3/2} \tanh ^{-1}(\tanh (a+b x))+12 b \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^2-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{\sqrt{x}}+\frac{32}{5} b^3 x^{5/2} \]

[Out]

(32*b^3*x^(5/2))/5 - 16*b^2*x^(3/2)*ArcTanh[Tanh[a + b*x]] + 12*b*Sqrt[x]*ArcTanh[Tanh[a + b*x]]^2 - (2*ArcTan
h[Tanh[a + b*x]]^3)/Sqrt[x]

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Rubi [A]  time = 0.0359327, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2168, 30} \[ -16 b^2 x^{3/2} \tanh ^{-1}(\tanh (a+b x))+12 b \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^2-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{\sqrt{x}}+\frac{32}{5} b^3 x^{5/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*b^3*x^(5/2))/5 - 16*b^2*x^(3/2)*ArcTanh[Tanh[a + b*x]] + 12*b*Sqrt[x]*ArcTanh[Tanh[a + b*x]]^2 - (2*ArcTan
h[Tanh[a + b*x]]^3)/Sqrt[x]

Rule 2168

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(u^(m + 1)*v^
n)/(a*(m + 1)), x] - Dist[(b*n)/(a*(m + 1)), Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m
, n}, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0] &&  !(ILtQ[m + n, -2] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

Rule 30

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

Rubi steps

\begin{align*} \int \frac{\tanh ^{-1}(\tanh (a+b x))^3}{x^{3/2}} \, dx &=-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{\sqrt{x}}+(6 b) \int \frac{\tanh ^{-1}(\tanh (a+b x))^2}{\sqrt{x}} \, dx\\ &=12 b \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^2-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{\sqrt{x}}-\left (24 b^2\right ) \int \sqrt{x} \tanh ^{-1}(\tanh (a+b x)) \, dx\\ &=-16 b^2 x^{3/2} \tanh ^{-1}(\tanh (a+b x))+12 b \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^2-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{\sqrt{x}}+\left (16 b^3\right ) \int x^{3/2} \, dx\\ &=\frac{32}{5} b^3 x^{5/2}-16 b^2 x^{3/2} \tanh ^{-1}(\tanh (a+b x))+12 b \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^2-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{\sqrt{x}}\\ \end{align*}

Mathematica [A]  time = 0.0309749, size = 57, normalized size = 0.9 \[ \frac{2 \left (-40 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+30 b x \tanh ^{-1}(\tanh (a+b x))^2-5 \tanh ^{-1}(\tanh (a+b x))^3+16 b^3 x^3\right )}{5 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(16*b^3*x^3 - 40*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 30*b*x*ArcTanh[Tanh[a + b*x]]^2 - 5*ArcTanh[Tanh[a + b*x]
]^3))/(5*Sqrt[x])

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Maple [A]  time = 0.043, size = 64, normalized size = 1. \begin{align*} -2\,{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3}}{\sqrt{x}}}+12\,b \left ( 1/5\,{b}^{2}{x}^{5/2}+2/3\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) b{x}^{3/2}+ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}\sqrt{x} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*arctanh(tanh(b*x+a))^3/x^(1/2)+12*b*(1/5*b^2*x^(5/2)+2/3*(arctanh(tanh(b*x+a))-b*x)*b*x^(3/2)+(arctanh(tanh
(b*x+a))-b*x)^2*x^(1/2))

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Maxima [A]  time = 1.04928, size = 74, normalized size = 1.17 \begin{align*} 12 \, b \sqrt{x} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2} - \frac{2 \, \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{\sqrt{x}} + \frac{16}{5} \,{\left (2 \, b^{2} x^{\frac{5}{2}} - 5 \, b x^{\frac{3}{2}} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

12*b*sqrt(x)*arctanh(tanh(b*x + a))^2 - 2*arctanh(tanh(b*x + a))^3/sqrt(x) + 16/5*(2*b^2*x^(5/2) - 5*b*x^(3/2)
*arctanh(tanh(b*x + a)))*b

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Fricas [A]  time = 1.92586, size = 78, normalized size = 1.24 \begin{align*} \frac{2 \,{\left (b^{3} x^{3} + 5 \, a b^{2} x^{2} + 15 \, a^{2} b x - 5 \, a^{3}\right )}}{5 \, \sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{atanh}^{3}{\left (\tanh{\left (a + b x \right )} \right )}}{x^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.11838, size = 47, normalized size = 0.75 \begin{align*} \frac{2}{5} \, b^{3} x^{\frac{5}{2}} + 2 \, a b^{2} x^{\frac{3}{2}} + 6 \, a^{2} b \sqrt{x} - \frac{2 \, a^{3}}{\sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*b^3*x^(5/2) + 2*a*b^2*x^(3/2) + 6*a^2*b*sqrt(x) - 2*a^3/sqrt(x)