3.238 \(\int \frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{11/2}} \, dx\)

Optimal. Leaf size=72 \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{9 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{4 b \tanh ^{-1}(\tanh (a+b x))^{7/2}}{63 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

[Out]

(4*b*ArcTanh[Tanh[a + b*x]]^(7/2))/(63*x^(7/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*ArcTanh[Tanh[a + b*x]]^(
7/2))/(9*x^(9/2)*(b*x - ArcTanh[Tanh[a + b*x]]))

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Rubi [A]  time = 0.0327222, 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} \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{9 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{4 b \tanh ^{-1}(\tanh (a+b x))^{7/2}}{63 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(5/2)/x^(11/2),x]

[Out]

(4*b*ArcTanh[Tanh[a + b*x]]^(7/2))/(63*x^(7/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*ArcTanh[Tanh[a + b*x]]^(
7/2))/(9*x^(9/2)*(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{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{11/2}} \, dx &=\frac{2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{9 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{(2 b) \int \frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{9/2}} \, dx}{9 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{4 b \tanh ^{-1}(\tanh (a+b x))^{7/2}}{63 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{9 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end{align*}

Mathematica [A]  time = 0.0453176, size = 48, normalized size = 0.67 \[ \frac{2 \left (9 b x-7 \tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}{63 x^{9/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(5/2)/x^(11/2),x]

[Out]

(2*(9*b*x - 7*ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(7/2))/(63*x^(9/2)*(-(b*x) + ArcTanh[Tanh[a + b*x
]])^2)

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Maple [A]  time = 0.134, size = 59, normalized size = 0.8 \begin{align*} -{\frac{2}{9\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -9\,bx} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{7}{2}}}{x}^{-{\frac{9}{2}}}}+{\frac{4\,b}{63\, \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{7}{2}}}{x}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(5/2)/x^(11/2),x)

[Out]

-2/9/(arctanh(tanh(b*x+a))-b*x)/x^(9/2)*arctanh(tanh(b*x+a))^(7/2)+4/63*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(7/2)
*arctanh(tanh(b*x+a))^(7/2)

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Maxima [A]  time = 1.48213, size = 46, normalized size = 0.64 \begin{align*} \frac{2 \,{\left (2 \, b^{2} x^{2} - 5 \, a b x - 7 \, a^{2}\right )}{\left (b x + a\right )}^{\frac{5}{2}}}{63 \, a^{2} x^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(11/2),x, algorithm="maxima")

[Out]

2/63*(2*b^2*x^2 - 5*a*b*x - 7*a^2)*(b*x + a)^(5/2)/(a^2*x^(9/2))

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Fricas [A]  time = 2.06608, size = 130, normalized size = 1.81 \begin{align*} \frac{2 \,{\left (2 \, b^{4} x^{4} - a b^{3} x^{3} - 15 \, a^{2} b^{2} x^{2} - 19 \, a^{3} b x - 7 \, a^{4}\right )} \sqrt{b x + a}}{63 \, a^{2} x^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(11/2),x, algorithm="fricas")

[Out]

2/63*(2*b^4*x^4 - a*b^3*x^3 - 15*a^2*b^2*x^2 - 19*a^3*b*x - 7*a^4)*sqrt(b*x + a)/(a^2*x^(9/2))

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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))**(5/2)/x**(11/2),x)

[Out]

Timed out

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Giac [A]  time = 1.27105, size = 80, normalized size = 1.11 \begin{align*} \frac{\sqrt{2}{\left (\frac{2 \, \sqrt{2}{\left (b x + a\right )} b^{9}}{a^{2}} - \frac{9 \, \sqrt{2} b^{9}}{a}\right )}{\left (b x + a\right )}^{\frac{7}{2}} b}{63 \,{\left ({\left (b x + a\right )} b - a b\right )}^{\frac{9}{2}}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(11/2),x, algorithm="giac")

[Out]

1/63*sqrt(2)*(2*sqrt(2)*(b*x + a)*b^9/a^2 - 9*sqrt(2)*b^9/a)*(b*x + a)^(7/2)*b/(((b*x + a)*b - a*b)^(9/2)*abs(
b))