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

Optimal. Leaf size=35 \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

[Out]

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

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Rubi [A]  time = 0.0141844, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {2167} \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2167

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, -Simp[(u^(m + 1)*v^
(n + 1))/((m + 1)*(b*u - a*v)), x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n}, x] && PiecewiseLinearQ[u, v, x] && E
qQ[m + n + 2, 0] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0311113, size = 34, normalized size = 0.97 \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{3/2} \left (3 b x-3 \tanh ^{-1}(\tanh (a+b x))\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.138, size = 29, normalized size = 0.8 \begin{align*} -{\frac{2}{3\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -3\,bx} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{3}{2}}}{x}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(1/2)/x^(5/2),x)

[Out]

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

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Maxima [A]  time = 1.48269, size = 20, normalized size = 0.57 \begin{align*} -\frac{2 \,{\left (b x + a\right )}^{\frac{3}{2}}}{3 \, a x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^(5/2),x, algorithm="maxima")

[Out]

-2/3*(b*x + a)^(3/2)/(a*x^(3/2))

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Fricas [A]  time = 1.99663, size = 46, normalized size = 1.31 \begin{align*} -\frac{2 \,{\left (b x + a\right )}^{\frac{3}{2}}}{3 \, a x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^(5/2),x, algorithm="fricas")

[Out]

-2/3*(b*x + a)^(3/2)/(a*x^(3/2))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}}{x^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**(1/2)/x**(5/2),x)

[Out]

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

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Giac [B]  time = 1.14085, size = 80, normalized size = 2.29 \begin{align*} \frac{4 \,{\left (3 \, b^{\frac{3}{2}}{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{4} + a^{2} b^{\frac{3}{2}}\right )}}{3 \,{\left ({\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{2} - a\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^(5/2),x, algorithm="giac")

[Out]

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