3.177 \(\int x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2 \, dx\)

Optimal. Leaf size=48 \[ -\frac{8}{35} b x^{7/2} \tanh ^{-1}(\tanh (a+b x))+\frac{2}{5} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac{16}{315} b^2 x^{9/2} \]

[Out]

(16*b^2*x^(9/2))/315 - (8*b*x^(7/2)*ArcTanh[Tanh[a + b*x]])/35 + (2*x^(5/2)*ArcTanh[Tanh[a + b*x]]^2)/5

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Rubi [A]  time = 0.0228219, 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} \[ -\frac{8}{35} b x^{7/2} \tanh ^{-1}(\tanh (a+b x))+\frac{2}{5} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac{16}{315} b^2 x^{9/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*b^2*x^(9/2))/315 - (8*b*x^(7/2)*ArcTanh[Tanh[a + b*x]])/35 + (2*x^(5/2)*ArcTanh[Tanh[a + b*x]]^2)/5

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 x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2 \, dx &=\frac{2}{5} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2-\frac{1}{5} (4 b) \int x^{5/2} \tanh ^{-1}(\tanh (a+b x)) \, dx\\ &=-\frac{8}{35} b x^{7/2} \tanh ^{-1}(\tanh (a+b x))+\frac{2}{5} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac{1}{35} \left (8 b^2\right ) \int x^{7/2} \, dx\\ &=\frac{16}{315} b^2 x^{9/2}-\frac{8}{35} b x^{7/2} \tanh ^{-1}(\tanh (a+b x))+\frac{2}{5} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2\\ \end{align*}

Mathematica [A]  time = 0.0401329, size = 40, normalized size = 0.83 \[ \frac{2}{315} x^{5/2} \left (-36 b x \tanh ^{-1}(\tanh (a+b x))+63 \tanh ^{-1}(\tanh (a+b x))^2+8 b^2 x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(5/2)*(8*b^2*x^2 - 36*b*x*ArcTanh[Tanh[a + b*x]] + 63*ArcTanh[Tanh[a + b*x]]^2))/315

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Maple [A]  time = 0.039, size = 38, normalized size = 0.8 \begin{align*}{\frac{2\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{5}{x}^{{\frac{5}{2}}}}-{\frac{8\,b}{5} \left ({\frac{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{7}{x}^{{\frac{7}{2}}}}-{\frac{2\,b}{63}{x}^{{\frac{9}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*x^(5/2)*arctanh(tanh(b*x+a))^2-8/5*b*(1/7*x^(7/2)*arctanh(tanh(b*x+a))-2/63*b*x^(9/2))

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Maxima [A]  time = 1.01492, size = 49, normalized size = 1.02 \begin{align*} \frac{16}{315} \, b^{2} x^{\frac{9}{2}} - \frac{8}{35} \, b x^{\frac{7}{2}} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right ) + \frac{2}{5} \, x^{\frac{5}{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(x^(3/2)*arctanh(tanh(b*x+a))^2,x, algorithm="maxima")

[Out]

16/315*b^2*x^(9/2) - 8/35*b*x^(7/2)*arctanh(tanh(b*x + a)) + 2/5*x^(5/2)*arctanh(tanh(b*x + a))^2

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Fricas [A]  time = 2.0062, size = 73, normalized size = 1.52 \begin{align*} \frac{2}{315} \,{\left (35 \, b^{2} x^{4} + 90 \, a b x^{3} + 63 \, a^{2} x^{2}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*b^2*x^4 + 90*a*b*x^3 + 63*a^2*x^2)*sqrt(x)

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

[Out]

Integral(x**(3/2)*atanh(tanh(a + b*x))**2, x)

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Giac [A]  time = 1.12977, size = 32, normalized size = 0.67 \begin{align*} \frac{2}{9} \, b^{2} x^{\frac{9}{2}} + \frac{4}{7} \, a b x^{\frac{7}{2}} + \frac{2}{5} \, a^{2} x^{\frac{5}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*b^2*x^(9/2) + 4/7*a*b*x^(7/2) + 2/5*a^2*x^(5/2)