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

Optimal. Leaf size=46 \[ -\frac{8}{3} b x^{3/2} \tanh ^{-1}(\tanh (a+b x))+2 \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^2+\frac{16}{15} b^2 x^{5/2} \]

[Out]

(16*b^2*x^(5/2))/15 - (8*b*x^(3/2)*ArcTanh[Tanh[a + b*x]])/3 + 2*Sqrt[x]*ArcTanh[Tanh[a + b*x]]^2

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Rubi [A]  time = 0.0208972, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2168, 30} \[ -\frac{8}{3} b x^{3/2} \tanh ^{-1}(\tanh (a+b x))+2 \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^2+\frac{16}{15} b^2 x^{5/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*b^2*x^(5/2))/15 - (8*b*x^(3/2)*ArcTanh[Tanh[a + b*x]])/3 + 2*Sqrt[x]*ArcTanh[Tanh[a + b*x]]^2

Rule 2168

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(u^(m + 1)*v^
n)/(a*(m + 1)), x] - Dist[(b*n)/(a*(m + 1)), Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m
, n}, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0] &&  !(ILtQ[m + n, -2] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\tanh ^{-1}(\tanh (a+b x))^2}{\sqrt{x}} \, dx &=2 \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^2-(4 b) \int \sqrt{x} \tanh ^{-1}(\tanh (a+b x)) \, dx\\ &=-\frac{8}{3} b x^{3/2} \tanh ^{-1}(\tanh (a+b x))+2 \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^2+\frac{1}{3} \left (8 b^2\right ) \int x^{3/2} \, dx\\ &=\frac{16}{15} b^2 x^{5/2}-\frac{8}{3} b x^{3/2} \tanh ^{-1}(\tanh (a+b x))+2 \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^2\\ \end{align*}

Mathematica [A]  time = 0.0311231, size = 40, normalized size = 0.87 \[ \frac{2}{15} \sqrt{x} \left (-20 b x \tanh ^{-1}(\tanh (a+b x))+15 \tanh ^{-1}(\tanh (a+b x))^2+8 b^2 x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*(8*b^2*x^2 - 20*b*x*ArcTanh[Tanh[a + b*x]] + 15*ArcTanh[Tanh[a + b*x]]^2))/15

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Maple [A]  time = 0.04, size = 47, normalized size = 1. \begin{align*}{\frac{2\,{b}^{2}}{5}{x}^{{\frac{5}{2}}}}+{\frac{ \left ( 4\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -4\,bx \right ) b}{3}{x}^{{\frac{3}{2}}}}+2\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}\sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*b^2*x^(5/2)+4/3*(arctanh(tanh(b*x+a))-b*x)*b*x^(3/2)+2*(arctanh(tanh(b*x+a))-b*x)^2*x^(1/2)

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Maxima [A]  time = 1.02751, size = 49, normalized size = 1.07 \begin{align*} \frac{16}{15} \, b^{2} x^{\frac{5}{2}} - \frac{8}{3} \, b x^{\frac{3}{2}} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right ) + 2 \, \sqrt{x} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/15*b^2*x^(5/2) - 8/3*b*x^(3/2)*arctanh(tanh(b*x + a)) + 2*sqrt(x)*arctanh(tanh(b*x + a))^2

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Fricas [A]  time = 1.99867, size = 62, normalized size = 1.35 \begin{align*} \frac{2}{15} \,{\left (3 \, b^{2} x^{2} + 10 \, a b x + 15 \, a^{2}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*b^2*x^2 + 10*a*b*x + 15*a^2)*sqrt(x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1664, size = 32, normalized size = 0.7 \begin{align*} \frac{2}{5} \, b^{2} x^{\frac{5}{2}} + \frac{4}{3} \, a b x^{\frac{3}{2}} + 2 \, a^{2} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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