3.187 \(\int \frac{\tanh ^{-1}(\tanh (a+b x))^3}{\sqrt{x}} \, dx\)

Optimal. Leaf size=65 \[ \frac{16}{5} b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))-4 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2+2 \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^3-\frac{32}{35} b^3 x^{7/2} \]

[Out]

(-32*b^3*x^(7/2))/35 + (16*b^2*x^(5/2)*ArcTanh[Tanh[a + b*x]])/5 - 4*b*x^(3/2)*ArcTanh[Tanh[a + b*x]]^2 + 2*Sq
rt[x]*ArcTanh[Tanh[a + b*x]]^3

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Rubi [A]  time = 0.0357662, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2168, 30} \[ \frac{16}{5} b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))-4 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2+2 \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^3-\frac{32}{35} b^3 x^{7/2} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^3/Sqrt[x],x]

[Out]

(-32*b^3*x^(7/2))/35 + (16*b^2*x^(5/2)*ArcTanh[Tanh[a + b*x]])/5 - 4*b*x^(3/2)*ArcTanh[Tanh[a + b*x]]^2 + 2*Sq
rt[x]*ArcTanh[Tanh[a + b*x]]^3

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))^3}{\sqrt{x}} \, dx &=2 \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^3-(6 b) \int \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^2 \, dx\\ &=-4 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2+2 \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^3+\left (8 b^2\right ) \int x^{3/2} \tanh ^{-1}(\tanh (a+b x)) \, dx\\ &=\frac{16}{5} b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))-4 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2+2 \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^3-\frac{1}{5} \left (16 b^3\right ) \int x^{5/2} \, dx\\ &=-\frac{32}{35} b^3 x^{7/2}+\frac{16}{5} b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))-4 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2+2 \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^3\\ \end{align*}

Mathematica [A]  time = 0.028104, size = 57, normalized size = 0.88 \[ \frac{2}{35} \sqrt{x} \left (56 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-70 b x \tanh ^{-1}(\tanh (a+b x))^2+35 \tanh ^{-1}(\tanh (a+b x))^3-16 b^3 x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^3/Sqrt[x],x]

[Out]

(2*Sqrt[x]*(-16*b^3*x^3 + 56*b^2*x^2*ArcTanh[Tanh[a + b*x]] - 70*b*x*ArcTanh[Tanh[a + b*x]]^2 + 35*ArcTanh[Tan
h[a + b*x]]^3))/35

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Maple [A]  time = 0.039, size = 69, normalized size = 1.1 \begin{align*}{\frac{2\,{b}^{3}}{7}{x}^{{\frac{7}{2}}}}+{\frac{ \left ( 6\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -6\,bx \right ){b}^{2}}{5}{x}^{{\frac{5}{2}}}}+2\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}b{x}^{3/2}+2\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3}\sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*b^3*x^(7/2)+6/5*(arctanh(tanh(b*x+a))-b*x)*b^2*x^(5/2)+2*(arctanh(tanh(b*x+a))-b*x)^2*b*x^(3/2)+2*(arctanh
(tanh(b*x+a))-b*x)^3*x^(1/2)

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Maxima [A]  time = 1.05474, size = 74, normalized size = 1.14 \begin{align*} -4 \, b x^{\frac{3}{2}} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2} + 2 \, \sqrt{x} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{3} - \frac{16}{35} \,{\left (2 \, b^{2} x^{\frac{7}{2}} - 7 \, b x^{\frac{5}{2}} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*b*x^(3/2)*arctanh(tanh(b*x + a))^2 + 2*sqrt(x)*arctanh(tanh(b*x + a))^3 - 16/35*(2*b^2*x^(7/2) - 7*b*x^(5/2
)*arctanh(tanh(b*x + a)))*b

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Fricas [A]  time = 1.9303, size = 85, normalized size = 1.31 \begin{align*} \frac{2}{35} \,{\left (5 \, b^{3} x^{3} + 21 \, a b^{2} x^{2} + 35 \, a^{2} b x + 35 \, a^{3}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(5*b^3*x^3 + 21*a*b^2*x^2 + 35*a^2*b*x + 35*a^3)*sqrt(x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{atanh}^{3}{\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))**3/x**(1/2),x)

[Out]

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

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Giac [A]  time = 1.12829, size = 47, normalized size = 0.72 \begin{align*} \frac{2}{7} \, b^{3} x^{\frac{7}{2}} + \frac{6}{5} \, a b^{2} x^{\frac{5}{2}} + 2 \, a^{2} b x^{\frac{3}{2}} + 2 \, a^{3} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*b^3*x^(7/2) + 6/5*a*b^2*x^(5/2) + 2*a^2*b*x^(3/2) + 2*a^3*sqrt(x)