3.255 \(\int \frac{1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0540588, 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} \[ -\frac{16 b^2 \sqrt{x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{2}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{8 b}{3 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*ArcTanh[Tanh[a + b*x]]^(3/2)),x]

[Out]

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

Mathematica [A]  time = 0.0491826, size = 64, normalized size = 0.58 \[ \frac{2 \left (-6 b x \tanh ^{-1}(\tanh (a+b x))+\tanh ^{-1}(\tanh (a+b x))^2-3 b^2 x^2\right )}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*ArcTanh[Tanh[a + b*x]]^(3/2)),x]

[Out]

(2*(-3*b^2*x^2 - 6*b*x*ArcTanh[Tanh[a + b*x]] + ArcTanh[Tanh[a + b*x]]^2))/(3*x^(3/2)*(b*x - ArcTanh[Tanh[a +
b*x]])^3*Sqrt[ArcTanh[Tanh[a + b*x]]])

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Maple [A]  time = 0.148, size = 105, normalized size = 1. \begin{align*} -{\frac{2}{3\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -3\,bx}{x}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}}}-{\frac{8\,b}{3\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -3\,bx} \left ( -{\frac{1}{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx}{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}}}-2\,{\frac{b\sqrt{x}}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(5/2)/arctanh(tanh(b*x+a))^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.50479, size = 61, normalized size = 0.55 \begin{align*} \frac{2 \,{\left (8 \, b^{3} x^{3} + 12 \, a b^{2} x^{2} + 3 \, a^{2} b x - a^{3}\right )}}{3 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{3} x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/arctanh(tanh(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

2/3*(8*b^3*x^3 + 12*a*b^2*x^2 + 3*a^2*b*x - a^3)/((b*x + a)^(3/2)*a^3*x^(3/2))

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Fricas [A]  time = 2.05685, size = 104, normalized size = 0.95 \begin{align*} \frac{2 \,{\left (8 \, b^{2} x^{2} + 4 \, a b x - a^{2}\right )} \sqrt{b x + a} \sqrt{x}}{3 \,{\left (a^{3} b x^{3} + a^{4} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/arctanh(tanh(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

2/3*(8*b^2*x^2 + 4*a*b*x - a^2)*sqrt(b*x + a)*sqrt(x)/(a^3*b*x^3 + a^4*x^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(1/x**(5/2)/atanh(tanh(b*x+a))**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.27649, size = 144, normalized size = 1.31 \begin{align*} \frac{2 \, b^{2} \sqrt{x}}{\sqrt{b x + a} a^{3}} - \frac{4 \,{\left (3 \, b^{\frac{3}{2}}{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{4} - 12 \, a b^{\frac{3}{2}}{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{2} + 5 \, a^{2} b^{\frac{3}{2}}\right )}}{3 \,{\left ({\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{2} - a\right )}^{3} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/arctanh(tanh(b*x+a))^(3/2),x, algorithm="giac")

[Out]

2*b^2*sqrt(x)/(sqrt(b*x + a)*a^3) - 4/3*(3*b^(3/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^4 - 12*a*b^(3/2)*(sqrt(b)
*sqrt(x) - sqrt(b*x + a))^2 + 5*a^2*b^(3/2))/(((sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a)^3*a^2)