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

Optimal. Leaf size=48 \[ -\frac{2 \tanh ^{-1}(\tanh (a+b x))^2}{3 x^{3/2}}-\frac{8 b \tanh ^{-1}(\tanh (a+b x))}{3 \sqrt{x}}+\frac{16 b^2 \sqrt{x}}{3} \]

[Out]

(16*b^2*Sqrt[x])/3 - (8*b*ArcTanh[Tanh[a + b*x]])/(3*Sqrt[x]) - (2*ArcTanh[Tanh[a + b*x]]^2)/(3*x^(3/2))

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Rubi [A]  time = 0.0230305, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2168, 30} \[ -\frac{2 \tanh ^{-1}(\tanh (a+b x))^2}{3 x^{3/2}}-\frac{8 b \tanh ^{-1}(\tanh (a+b x))}{3 \sqrt{x}}+\frac{16 b^2 \sqrt{x}}{3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*b^2*Sqrt[x])/3 - (8*b*ArcTanh[Tanh[a + b*x]])/(3*Sqrt[x]) - (2*ArcTanh[Tanh[a + b*x]]^2)/(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))^2}{x^{5/2}} \, dx &=-\frac{2 \tanh ^{-1}(\tanh (a+b x))^2}{3 x^{3/2}}+\frac{1}{3} (4 b) \int \frac{\tanh ^{-1}(\tanh (a+b x))}{x^{3/2}} \, dx\\ &=-\frac{8 b \tanh ^{-1}(\tanh (a+b x))}{3 \sqrt{x}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^2}{3 x^{3/2}}+\frac{1}{3} \left (8 b^2\right ) \int \frac{1}{\sqrt{x}} \, dx\\ &=\frac{16 b^2 \sqrt{x}}{3}-\frac{8 b \tanh ^{-1}(\tanh (a+b x))}{3 \sqrt{x}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^2}{3 x^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0442962, size = 40, normalized size = 0.83 \[ \frac{2 \left (-4 b x \tanh ^{-1}(\tanh (a+b x))-\tanh ^{-1}(\tanh (a+b x))^2+8 b^2 x^2\right )}{3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(8*b^2*x^2 - 4*b*x*ArcTanh[Tanh[a + b*x]] - ArcTanh[Tanh[a + b*x]]^2))/(3*x^(3/2))

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Maple [A]  time = 0.041, size = 38, normalized size = 0.8 \begin{align*} -{\frac{2\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{3}{x}^{-{\frac{3}{2}}}}+{\frac{8\,b}{3} \left ( -{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ){\frac{1}{\sqrt{x}}}}+2\,b\sqrt{x} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*arctanh(tanh(b*x+a))^2/x^(3/2)+8/3*b*(-arctanh(tanh(b*x+a))/x^(1/2)+2*b*x^(1/2))

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Maxima [A]  time = 1.04449, size = 49, normalized size = 1.02 \begin{align*} \frac{16}{3} \, b^{2} \sqrt{x} - \frac{8 \, b \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )}{3 \, \sqrt{x}} - \frac{2 \, \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{3 \, x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/3*b^2*sqrt(x) - 8/3*b*arctanh(tanh(b*x + a))/sqrt(x) - 2/3*arctanh(tanh(b*x + a))^2/x^(3/2)

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Fricas [A]  time = 1.9352, size = 55, normalized size = 1.15 \begin{align*} \frac{2 \,{\left (3 \, b^{2} x^{2} - 6 \, a b x - a^{2}\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))^2/x^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 22.5462, size = 48, normalized size = 1. \begin{align*} \frac{16 b^{2} \sqrt{x}}{3} - \frac{8 b \operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}{3 \sqrt{x}} - \frac{2 \operatorname{atanh}^{2}{\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))**2/x**(5/2),x)

[Out]

16*b**2*sqrt(x)/3 - 8*b*atanh(tanh(a + b*x))/(3*sqrt(x)) - 2*atanh(tanh(a + b*x))**2/(3*x**(3/2))

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Giac [A]  time = 1.13166, size = 31, normalized size = 0.65 \begin{align*} 2 \, b^{2} \sqrt{x} - \frac{2 \,{\left (6 \, a b x + a^{2}\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))^2/x^(5/2),x, algorithm="giac")

[Out]

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