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

Optimal. Leaf size=35 \[ -\frac{2 x^{3/2}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0121499, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {2167} \[ -\frac{2 x^{3/2}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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{\sqrt{x}}{\tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx &=-\frac{2 x^{3/2}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0415808, size = 34, normalized size = 0.97 \[ \frac{2 x^{3/2}}{3 \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.118, size = 92, normalized size = 2.6 \begin{align*} -{\frac{1}{b}\sqrt{x} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{-{\frac{3}{2}}}}+{\frac{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx}{b} \left ({\frac{1}{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{2}{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 ) }}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^(1/2)/b/arctanh(tanh(b*x+a))^(3/2)+(arctanh(tanh(b*x+a))-b*x)/b*(1/3*x^(1/2)/(arctanh(tanh(b*x+a))-b*x)/arc
tanh(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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\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(x^(1/2)/arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.09297, size = 77, normalized size = 2.2 \begin{align*} \frac{2 \, \sqrt{b x + a} x^{\frac{3}{2}}}{3 \,{\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.16159, size = 20, normalized size = 0.57 \begin{align*} \frac{2 \, x^{\frac{3}{2}}}{3 \,{\left (b x + a\right )}^{\frac{3}{2}} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*x^(3/2)/((b*x + a)^(3/2)*a)