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

Optimal. Leaf size=68 \[ \frac{2}{\sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{4 b \sqrt{x}}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \]

[Out]

(-4*b*Sqrt[x])/((b*x - ArcTanh[Tanh[a + b*x]])^2*Sqrt[ArcTanh[Tanh[a + b*x]]]) + 2/(Sqrt[x]*(b*x - ArcTanh[Tan
h[a + b*x]])*Sqrt[ArcTanh[Tanh[a + b*x]]])

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Rubi [A]  time = 0.032766, antiderivative size = 68, 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} \[ \frac{2}{\sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{4 b \sqrt{x}}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0467405, size = 43, normalized size = 0.63 \[ -\frac{2 \left (\tanh ^{-1}(\tanh (a+b x))+b x\right )}{\sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.146, size = 59, normalized size = 0.9 \begin{align*} -2\,{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) \sqrt{x}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}}-4\,{\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 ) }}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(arctanh(tanh(b*x+a))-b*x)/x^(1/2)/arctanh(tanh(b*x+a))^(1/2)-4*b/(arctanh(tanh(b*x+a))-b*x)^2*x^(1/2)/arct
anh(tanh(b*x+a))^(1/2)

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Maxima [A]  time = 1.49918, size = 43, normalized size = 0.63 \begin{align*} -\frac{2 \,{\left (2 \, b^{2} x^{2} + 3 \, a b x + a^{2}\right )}}{{\left (b x + a\right )}^{\frac{3}{2}} a^{2} \sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.03001, size = 78, normalized size = 1.15 \begin{align*} -\frac{2 \,{\left (2 \, b x + a\right )} \sqrt{b x + a} \sqrt{x}}{a^{2} b x^{2} + a^{3} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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**(3/2)/atanh(tanh(b*x+a))**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.17775, size = 68, normalized size = 1. \begin{align*} -\frac{2 \, b \sqrt{x}}{\sqrt{b x + a} a^{2}} + \frac{4 \, \sqrt{b}}{{\left ({\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{2} - a\right )} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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