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

Optimal. Leaf size=106 \[ \frac{16 b \sqrt{x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{8 b \sqrt{x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{2}{\sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0548497, antiderivative size = 106, 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} \[ \frac{16 b \sqrt{x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{8 b \sqrt{x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{2}{\sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0522689, size = 66, normalized size = 0.62 \[ \frac{2 \left (6 b x \tanh ^{-1}(\tanh (a+b x))+3 \tanh ^{-1}(\tanh (a+b x))^2-b^2 x^2\right )}{3 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.119, size = 104, normalized size = 1. \begin{align*} -2\,{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) \sqrt{x} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}}}-8\,{\frac{b}{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx} \left ( 1/3\,{\frac{\sqrt{x}}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}}}+2/3\,{\frac{\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^(3/2)/arctanh(tanh(b*x+a))^(5/2),x)

[Out]

-2/(arctanh(tanh(b*x+a))-b*x)/x^(1/2)/arctanh(tanh(b*x+a))^(3/2)-8*b/(arctanh(tanh(b*x+a))-b*x)*(1/3*x^(1/2)/(
arctanh(tanh(b*x+a))-b*x)/arctanh(tanh(b*x+a))^(3/2)+2/3/(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 = 1.51431, size = 61, normalized size = 0.58 \begin{align*} -\frac{2 \,{\left (8 \, b^{3} x^{3} + 20 \, a b^{2} x^{2} + 15 \, a^{2} b x + 3 \, a^{3}\right )}}{3 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{3} \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))^(5/2),x, algorithm="maxima")

[Out]

-2/3*(8*b^3*x^3 + 20*a*b^2*x^2 + 15*a^2*b*x + 3*a^3)/((b*x + a)^(5/2)*a^3*sqrt(x))

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Fricas [A]  time = 2.16484, size = 128, normalized size = 1.21 \begin{align*} -\frac{2 \,{\left (8 \, b^{2} x^{2} + 12 \, a b x + 3 \, a^{2}\right )} \sqrt{b x + a} \sqrt{x}}{3 \,{\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(8*b^2*x^2 + 12*a*b*x + 3*a^2)*sqrt(b*x + a)*sqrt(x)/(a^3*b^2*x^3 + 2*a^4*b*x^2 + a^5*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))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.20973, size = 84, normalized size = 0.79 \begin{align*} -\frac{2 \, \sqrt{x}{\left (\frac{5 \, b^{2} x}{a^{3}} + \frac{6 \, b}{a^{2}}\right )}}{3 \,{\left (b x + a\right )}^{\frac{3}{2}}} + \frac{4 \, \sqrt{b}}{{\left ({\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{2} - a\right )} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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