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

Optimal. Leaf size=71 \[ \frac{4 \sqrt{x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{2 \sqrt{x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

[Out]

(-2*Sqrt[x])/(3*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(3/2)) + (4*Sqrt[x])/(3*(b*x - ArcTanh[T
anh[a + b*x]])^2*Sqrt[ArcTanh[Tanh[a + b*x]]])

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Rubi [A]  time = 0.0315466, antiderivative size = 71, 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} \[ \frac{4 \sqrt{x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{2 \sqrt{x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0353378, size = 47, normalized size = 0.66 \[ -\frac{2 \sqrt{x} \left (b x-3 \tanh ^{-1}(\tanh (a+b x))\right )}{3 \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.043, size = 58, normalized size = 0.8 \begin{align*}{\frac{2}{3\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -3\,bx}\sqrt{x} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{-{\frac{3}{2}}}}+{\frac{4}{3\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}}\sqrt{x}{\frac{1}{\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^(1/2)/arctanh(tanh(b*x+a))^(5/2),x)

[Out]

2/3*x^(1/2)/(arctanh(tanh(b*x+a))-b*x)/arctanh(tanh(b*x+a))^(3/2)+4/3/(arctanh(tanh(b*x+a))-b*x)^2*x^(1/2)/arc
tanh(tanh(b*x+a))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{x} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(x)*arctanh(tanh(b*x + a))^(5/2)), x)

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Fricas [A]  time = 2.07893, size = 99, normalized size = 1.39 \begin{align*} \frac{2 \,{\left (2 \, b x + 3 \, a\right )} \sqrt{b x + a} \sqrt{x}}{3 \,{\left (a^{2} b^{2} x^{2} + 2 \, a^{3} b x + a^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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**(1/2)/atanh(tanh(b*x+a))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.21801, size = 34, normalized size = 0.48 \begin{align*} \frac{2 \, \sqrt{x}{\left (\frac{2 \, b x}{a^{2}} + \frac{3}{a}\right )}}{3 \,{\left (b x + a\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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