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

Optimal. Leaf size=69 \[ \frac{16}{105} b^2 x^{9/2} \tanh ^{-1}(\tanh (a+b x))-\frac{12}{35} b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac{2}{5} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^3-\frac{32 b^3 x^{11/2}}{1155} \]

[Out]

(-32*b^3*x^(11/2))/1155 + (16*b^2*x^(9/2)*ArcTanh[Tanh[a + b*x]])/105 - (12*b*x^(7/2)*ArcTanh[Tanh[a + b*x]]^2
)/35 + (2*x^(5/2)*ArcTanh[Tanh[a + b*x]]^3)/5

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Rubi [A]  time = 0.0374426, antiderivative size = 69, 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} \[ \frac{16}{105} b^2 x^{9/2} \tanh ^{-1}(\tanh (a+b x))-\frac{12}{35} b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac{2}{5} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^3-\frac{32 b^3 x^{11/2}}{1155} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-32*b^3*x^(11/2))/1155 + (16*b^2*x^(9/2)*ArcTanh[Tanh[a + b*x]])/105 - (12*b*x^(7/2)*ArcTanh[Tanh[a + b*x]]^2
)/35 + (2*x^(5/2)*ArcTanh[Tanh[a + b*x]]^3)/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))^3 \, dx &=\frac{2}{5} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^3-\frac{1}{5} (6 b) \int x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2 \, dx\\ &=-\frac{12}{35} b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac{2}{5} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^3+\frac{1}{35} \left (24 b^2\right ) \int x^{7/2} \tanh ^{-1}(\tanh (a+b x)) \, dx\\ &=\frac{16}{105} b^2 x^{9/2} \tanh ^{-1}(\tanh (a+b x))-\frac{12}{35} b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac{2}{5} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^3-\frac{1}{105} \left (16 b^3\right ) \int x^{9/2} \, dx\\ &=-\frac{32 b^3 x^{11/2}}{1155}+\frac{16}{105} b^2 x^{9/2} \tanh ^{-1}(\tanh (a+b x))-\frac{12}{35} b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac{2}{5} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^3\\ \end{align*}

Mathematica [A]  time = 0.0316131, size = 57, normalized size = 0.83 \[ -\frac{2 x^{5/2} \left (-88 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+198 b x \tanh ^{-1}(\tanh (a+b x))^2-231 \tanh ^{-1}(\tanh (a+b x))^3+16 b^3 x^3\right )}{1155} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x^(5/2)*(16*b^3*x^3 - 88*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 198*b*x*ArcTanh[Tanh[a + b*x]]^2 - 231*ArcTanh[T
anh[a + b*x]]^3))/1155

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Maple [A]  time = 0.046, size = 56, normalized size = 0.8 \begin{align*}{\frac{2\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3}}{5}{x}^{{\frac{5}{2}}}}-{\frac{12\,b}{5} \left ({\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{7}{x}^{{\frac{7}{2}}}}-{\frac{4\,b}{7} \left ({\frac{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{9}{x}^{{\frac{9}{2}}}}-{\frac{2\,b}{99}{x}^{{\frac{11}{2}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*x^(5/2)*arctanh(tanh(b*x+a))^3-12/5*b*(1/7*x^(7/2)*arctanh(tanh(b*x+a))^2-4/7*b*(1/9*x^(9/2)*arctanh(tanh(
b*x+a))-2/99*b*x^(11/2)))

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Maxima [A]  time = 1.06569, size = 74, normalized size = 1.07 \begin{align*} -\frac{12}{35} \, b x^{\frac{7}{2}} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2} + \frac{2}{5} \, x^{\frac{5}{2}} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{3} - \frac{16}{1155} \,{\left (2 \, b^{2} x^{\frac{11}{2}} - 11 \, b x^{\frac{9}{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(x^(3/2)*arctanh(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

-12/35*b*x^(7/2)*arctanh(tanh(b*x + a))^2 + 2/5*x^(5/2)*arctanh(tanh(b*x + a))^3 - 16/1155*(2*b^2*x^(11/2) - 1
1*b*x^(9/2)*arctanh(tanh(b*x + a)))*b

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Fricas [A]  time = 2.02683, size = 103, normalized size = 1.49 \begin{align*} \frac{2}{1155} \,{\left (105 \, b^{3} x^{5} + 385 \, a b^{2} x^{4} + 495 \, a^{2} b x^{3} + 231 \, a^{3} 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))^3,x, algorithm="fricas")

[Out]

2/1155*(105*b^3*x^5 + 385*a*b^2*x^4 + 495*a^2*b*x^3 + 231*a^3*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}^{3}{\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))**3,x)

[Out]

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

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Giac [A]  time = 1.17472, size = 47, normalized size = 0.68 \begin{align*} \frac{2}{11} \, b^{3} x^{\frac{11}{2}} + \frac{2}{3} \, a b^{2} x^{\frac{9}{2}} + \frac{6}{7} \, a^{2} b x^{\frac{7}{2}} + \frac{2}{5} \, a^{3} 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))^3,x, algorithm="giac")

[Out]

2/11*b^3*x^(11/2) + 2/3*a*b^2*x^(9/2) + 6/7*a^2*b*x^(7/2) + 2/5*a^3*x^(5/2)