3.220 \(\int \frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x^{7/2}} \, dx\)

Optimal. Leaf size=72 \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{4 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

[Out]

(4*b*ArcTanh[Tanh[a + b*x]]^(3/2))/(15*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*ArcTanh[Tanh[a + b*x]]^(
3/2))/(5*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]]))

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Rubi [A]  time = 0.033482, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2171, 2167} \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{4 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[ArcTanh[Tanh[a + b*x]]]/x^(7/2),x]

[Out]

(4*b*ArcTanh[Tanh[a + b*x]]^(3/2))/(15*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*ArcTanh[Tanh[a + b*x]]^(
3/2))/(5*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]]))

Rule 2171

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] + Dist[(b*(m + n + 2))/((m + 1)*(b*u - a*v)), Int[u^(m + 1)*v^n, x], x] /;
NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && LtQ[m, -1]

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^{7/2}} \, dx &=\frac{2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{(2 b) \int \frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x^{5/2}} \, dx}{5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{4 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end{align*}

Mathematica [A]  time = 0.0343497, size = 48, normalized size = 0.67 \[ \frac{2 \left (5 b x-3 \tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}{15 x^{5/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[ArcTanh[Tanh[a + b*x]]]/x^(7/2),x]

[Out]

(2*(5*b*x - 3*ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(3/2))/(15*x^(5/2)*(-(b*x) + ArcTanh[Tanh[a + b*x
]])^2)

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Maple [A]  time = 0.154, size = 59, normalized size = 0.8 \begin{align*} -{\frac{2}{5\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -5\,bx} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{3}{2}}}{x}^{-{\frac{5}{2}}}}+{\frac{4\,b}{15\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}} \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^(7/2),x)

[Out]

-2/5/(arctanh(tanh(b*x+a))-b*x)/x^(5/2)*arctanh(tanh(b*x+a))^(3/2)+4/15*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(3/2)
*arctanh(tanh(b*x+a))^(3/2)

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Maxima [A]  time = 1.48247, size = 46, normalized size = 0.64 \begin{align*} \frac{2 \,{\left (2 \, b^{2} x^{2} - a b x - 3 \, a^{2}\right )} \sqrt{b x + a}}{15 \, a^{2} x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^(7/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.15894, size = 84, normalized size = 1.17 \begin{align*} \frac{2 \,{\left (2 \, b^{2} x^{2} - a b x - 3 \, a^{2}\right )} \sqrt{b x + a}}{15 \, a^{2} x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^(7/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**(1/2)/x**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 1.1564, size = 151, normalized size = 2.1 \begin{align*} \frac{8 \,{\left (15 \, b^{\frac{5}{2}}{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{6} + 5 \, a b^{\frac{5}{2}}{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{4} + 5 \, a^{2} b^{\frac{5}{2}}{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{2} - a^{3} b^{\frac{5}{2}}\right )}}{15 \,{\left ({\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{2} - a\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^(7/2),x, algorithm="giac")

[Out]

8/15*(15*b^(5/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^6 + 5*a*b^(5/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^4 + 5*a^2
*b^(5/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a^3*b^(5/2))/((sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a)^5