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

Optimal. Leaf size=48 \[ -\frac{8}{15} b x^{5/2} \tanh ^{-1}(\tanh (a+b x))+\frac{2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac{16}{105} b^2 x^{7/2} \]

[Out]

(16*b^2*x^(7/2))/105 - (8*b*x^(5/2)*ArcTanh[Tanh[a + b*x]])/15 + (2*x^(3/2)*ArcTanh[Tanh[a + b*x]]^2)/3

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Rubi [A]  time = 0.0217704, antiderivative size = 48, 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} \[ -\frac{8}{15} b x^{5/2} \tanh ^{-1}(\tanh (a+b x))+\frac{2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac{16}{105} b^2 x^{7/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*b^2*x^(7/2))/105 - (8*b*x^(5/2)*ArcTanh[Tanh[a + b*x]])/15 + (2*x^(3/2)*ArcTanh[Tanh[a + b*x]]^2)/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 \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^2 \, dx &=\frac{2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2-\frac{1}{3} (4 b) \int x^{3/2} \tanh ^{-1}(\tanh (a+b x)) \, dx\\ &=-\frac{8}{15} b x^{5/2} \tanh ^{-1}(\tanh (a+b x))+\frac{2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac{1}{15} \left (8 b^2\right ) \int x^{5/2} \, dx\\ &=\frac{16}{105} b^2 x^{7/2}-\frac{8}{15} b x^{5/2} \tanh ^{-1}(\tanh (a+b x))+\frac{2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2\\ \end{align*}

Mathematica [A]  time = 0.0482864, size = 40, normalized size = 0.83 \[ \frac{2}{105} x^{3/2} \left (-28 b x \tanh ^{-1}(\tanh (a+b x))+35 \tanh ^{-1}(\tanh (a+b x))^2+8 b^2 x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(3/2)*(8*b^2*x^2 - 28*b*x*ArcTanh[Tanh[a + b*x]] + 35*ArcTanh[Tanh[a + b*x]]^2))/105

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Maple [A]  time = 0.04, size = 38, normalized size = 0.8 \begin{align*}{\frac{2\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{3}{x}^{{\frac{3}{2}}}}-{\frac{8\,b}{3} \left ({\frac{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{5}{x}^{{\frac{5}{2}}}}-{\frac{2\,b}{35}{x}^{{\frac{7}{2}}}} \right ) } \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]

2/3*x^(3/2)*arctanh(tanh(b*x+a))^2-8/3*b*(1/5*x^(5/2)*arctanh(tanh(b*x+a))-2/35*b*x^(7/2))

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Maxima [A]  time = 1.03271, size = 49, normalized size = 1.02 \begin{align*} \frac{16}{105} \, b^{2} x^{\frac{7}{2}} - \frac{8}{15} \, b x^{\frac{5}{2}} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right ) + \frac{2}{3} \, x^{\frac{3}{2}} \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]

16/105*b^2*x^(7/2) - 8/15*b*x^(5/2)*arctanh(tanh(b*x + a)) + 2/3*x^(3/2)*arctanh(tanh(b*x + a))^2

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Fricas [A]  time = 2.00495, size = 70, normalized size = 1.46 \begin{align*} \frac{2}{105} \,{\left (15 \, b^{2} x^{3} + 42 \, a b x^{2} + 35 \, a^{2} x\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]

2/105*(15*b^2*x^3 + 42*a*b*x^2 + 35*a^2*x)*sqrt(x)

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

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Giac [A]  time = 1.12525, size = 32, normalized size = 0.67 \begin{align*} \frac{2}{7} \, b^{2} x^{\frac{7}{2}} + \frac{4}{5} \, a b x^{\frac{5}{2}} + \frac{2}{3} \, a^{2} x^{\frac{3}{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="giac")

[Out]

2/7*b^2*x^(7/2) + 4/5*a*b*x^(5/2) + 2/3*a^2*x^(3/2)