3.173 \(\int \frac{\tanh ^{-1}(\tanh (a+b x))}{x^{5/2}} \, dx\)

Optimal. Leaf size=27 \[ -\frac{2 \tanh ^{-1}(\tanh (a+b x))}{3 x^{3/2}}-\frac{4 b}{3 \sqrt{x}} \]

[Out]

(-4*b)/(3*Sqrt[x]) - (2*ArcTanh[Tanh[a + b*x]])/(3*x^(3/2))

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Rubi [A]  time = 0.0086571, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2168, 30} \[ -\frac{2 \tanh ^{-1}(\tanh (a+b x))}{3 x^{3/2}}-\frac{4 b}{3 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]/x^(5/2),x]

[Out]

(-4*b)/(3*Sqrt[x]) - (2*ArcTanh[Tanh[a + b*x]])/(3*x^(3/2))

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))}{x^{5/2}} \, dx &=-\frac{2 \tanh ^{-1}(\tanh (a+b x))}{3 x^{3/2}}+\frac{1}{3} (2 b) \int \frac{1}{x^{3/2}} \, dx\\ &=-\frac{4 b}{3 \sqrt{x}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))}{3 x^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0181129, size = 21, normalized size = 0.78 \[ -\frac{2 \left (\tanh ^{-1}(\tanh (a+b x))+2 b x\right )}{3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.037, size = 20, normalized size = 0.7 \begin{align*} -{\frac{2\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{3}{x}^{-{\frac{3}{2}}}}-{\frac{4\,b}{3}{\frac{1}{\sqrt{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*arctanh(tanh(b*x+a))/x^(3/2)-4/3*b/x^(1/2)

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Maxima [A]  time = 0.986474, size = 26, normalized size = 0.96 \begin{align*} -\frac{4 \, b}{3 \, \sqrt{x}} - \frac{2 \, \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )}{3 \, x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))/x^(5/2),x, algorithm="maxima")

[Out]

-4/3*b/sqrt(x) - 2/3*arctanh(tanh(b*x + a))/x^(3/2)

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Fricas [A]  time = 2.02828, size = 35, normalized size = 1.3 \begin{align*} -\frac{2 \,{\left (3 \, b x + a\right )}}{3 \, x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))/x^(5/2),x, algorithm="fricas")

[Out]

-2/3*(3*b*x + a)/x^(3/2)

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Sympy [A]  time = 21.2364, size = 27, normalized size = 1. \begin{align*} - \frac{4 b}{3 \sqrt{x}} - \frac{2 \operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}{3 x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))/x**(5/2),x)

[Out]

-4*b/(3*sqrt(x)) - 2*atanh(tanh(a + b*x))/(3*x**(3/2))

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Giac [A]  time = 1.15382, size = 15, normalized size = 0.56 \begin{align*} -\frac{2 \,{\left (3 \, b x + a\right )}}{3 \, x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))/x^(5/2),x, algorithm="giac")

[Out]

-2/3*(3*b*x + a)/x^(3/2)