3.230 \(\int \frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{11/2}} \, dx\)

Optimal. Leaf size=110 \[ \frac{16 b^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{315 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{9 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{8 b \tanh ^{-1}(\tanh (a+b x))^{5/2}}{63 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

[Out]

(16*b^2*ArcTanh[Tanh[a + b*x]]^(5/2))/(315*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3) + (8*b*ArcTanh[Tanh[a + b
*x]]^(5/2))/(63*x^(7/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*ArcTanh[Tanh[a + b*x]]^(5/2))/(9*x^(9/2)*(b*x -
ArcTanh[Tanh[a + b*x]]))

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Rubi [A]  time = 0.0522912, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2171, 2167} \[ \frac{16 b^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{315 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{9 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{8 b \tanh ^{-1}(\tanh (a+b x))^{5/2}}{63 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(3/2)/x^(11/2),x]

[Out]

(16*b^2*ArcTanh[Tanh[a + b*x]]^(5/2))/(315*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3) + (8*b*ArcTanh[Tanh[a + b
*x]]^(5/2))/(63*x^(7/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*ArcTanh[Tanh[a + b*x]]^(5/2))/(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))^{3/2}}{x^{11/2}} \, dx &=\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{9 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{(4 b) \int \frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{9/2}} \, dx}{9 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{8 b \tanh ^{-1}(\tanh (a+b x))^{5/2}}{63 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{9 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{\left (8 b^2\right ) \int \frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{7/2}} \, dx}{63 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac{16 b^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{315 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{8 b \tanh ^{-1}(\tanh (a+b x))^{5/2}}{63 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{9 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end{align*}

Mathematica [A]  time = 0.0447709, size = 66, normalized size = 0.6 \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2} \left (-90 b x \tanh ^{-1}(\tanh (a+b x))+35 \tanh ^{-1}(\tanh (a+b x))^2+63 b^2 x^2\right )}{315 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(3/2)/x^(11/2),x]

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(5/2)*(63*b^2*x^2 - 90*b*x*ArcTanh[Tanh[a + b*x]] + 35*ArcTanh[Tanh[a + b*x]]^2))/(3
15*x^(9/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3)

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Maple [A]  time = 0.136, size = 105, normalized size = 1. \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{5}{2}}}{x}^{-{\frac{9}{2}}}}-{\frac{8\,b}{9\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -9\,bx} \left ( -{\frac{1}{7\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -7\,bx} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{5}{2}}}{x}^{-{\frac{7}{2}}}}+{\frac{2\,b}{35\, \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{5}{2}}}{x}^{-{\frac{5}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(3/2)/x^(11/2),x)

[Out]

-2/9/(arctanh(tanh(b*x+a))-b*x)/x^(9/2)*arctanh(tanh(b*x+a))^(5/2)-8/9*b/(arctanh(tanh(b*x+a))-b*x)*(-1/7/(arc
tanh(tanh(b*x+a))-b*x)/x^(7/2)*arctanh(tanh(b*x+a))^(5/2)+2/35*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(5/2)*arctanh(
tanh(b*x+a))^(5/2))

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Maxima [A]  time = 1.48465, size = 61, normalized size = 0.55 \begin{align*} -\frac{2 \,{\left (8 \, b^{3} x^{3} - 12 \, a b^{2} x^{2} + 15 \, a^{2} b x + 35 \, a^{3}\right )}{\left (b x + a\right )}^{\frac{3}{2}}}{315 \, a^{3} x^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(3/2)/x^(11/2),x, algorithm="maxima")

[Out]

-2/315*(8*b^3*x^3 - 12*a*b^2*x^2 + 15*a^2*b*x + 35*a^3)*(b*x + a)^(3/2)/(a^3*x^(9/2))

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Fricas [A]  time = 2.02936, size = 135, normalized size = 1.23 \begin{align*} -\frac{2 \,{\left (8 \, b^{4} x^{4} - 4 \, a b^{3} x^{3} + 3 \, a^{2} b^{2} x^{2} + 50 \, a^{3} b x + 35 \, a^{4}\right )} \sqrt{b x + a}}{315 \, a^{3} x^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(3/2)/x^(11/2),x, algorithm="fricas")

[Out]

-2/315*(8*b^4*x^4 - 4*a*b^3*x^3 + 3*a^2*b^2*x^2 + 50*a^3*b*x + 35*a^4)*sqrt(b*x + a)/(a^3*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))**(3/2)/x**(11/2),x)

[Out]

Timed out

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Giac [A]  time = 1.33636, size = 105, normalized size = 0.95 \begin{align*} -\frac{\sqrt{2}{\left (\frac{63 \, \sqrt{2} b^{9}}{a} + 4 \,{\left (\frac{2 \, \sqrt{2}{\left (b x + a\right )} b^{9}}{a^{3}} - \frac{9 \, \sqrt{2} b^{9}}{a^{2}}\right )}{\left (b x + a\right )}\right )}{\left (b x + a\right )}^{\frac{5}{2}} b}{315 \,{\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))^(3/2)/x^(11/2),x, algorithm="giac")

[Out]

-1/315*sqrt(2)*(63*sqrt(2)*b^9/a + 4*(2*sqrt(2)*(b*x + a)*b^9/a^3 - 9*sqrt(2)*b^9/a^2)*(b*x + a))*(b*x + a)^(5
/2)*b/(((b*x + a)*b - a*b)^(9/2)*abs(b))