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

Optimal. Leaf size=59 \[ \frac{16 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{315 b^3}-\frac{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2}+\frac{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b} \]

[Out]

(2*x^2*ArcTanh[Tanh[a + b*x]]^(5/2))/(5*b) - (8*x*ArcTanh[Tanh[a + b*x]]^(7/2))/(35*b^2) + (16*ArcTanh[Tanh[a
+ b*x]]^(9/2))/(315*b^3)

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Rubi [A]  time = 0.0292661, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2157, 30} \[ \frac{16 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{315 b^3}-\frac{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2}+\frac{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^2*ArcTanh[Tanh[a + b*x]]^(5/2))/(5*b) - (8*x*ArcTanh[Tanh[a + b*x]]^(7/2))/(35*b^2) + (16*ArcTanh[Tanh[a
+ b*x]]^(9/2))/(315*b^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 2157

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

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

Mathematica [A]  time = 0.0303672, size = 49, normalized size = 0.83 \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2} \left (-36 b x \tanh ^{-1}(\tanh (a+b x))+8 \tanh ^{-1}(\tanh (a+b x))^2+63 b^2 x^2\right )}{315 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(5/2)*(63*b^2*x^2 - 36*b*x*ArcTanh[Tanh[a + b*x]] + 8*ArcTanh[Tanh[a + b*x]]^2))/(31
5*b^3)

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Maple [A]  time = 0.033, size = 69, normalized size = 1.2 \begin{align*} 2\,{\frac{1/9\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{9/2}+1/7\, \left ( -2\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +2\,bx \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{7/2}+1/5\, \left ( bx-{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{5/2}}{{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.76832, size = 57, normalized size = 0.97 \begin{align*} \frac{2 \,{\left (35 \, b^{3} x^{3} + 15 \, a b^{2} x^{2} - 12 \, a^{2} b x + 8 \, a^{3}\right )}{\left (b x + a\right )}^{\frac{3}{2}}}{315 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*b^3*x^3 + 15*a*b^2*x^2 - 12*a^2*b*x + 8*a^3)*(b*x + a)^(3/2)/b^3

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Fricas [A]  time = 1.78853, size = 120, normalized size = 2.03 \begin{align*} \frac{2 \,{\left (35 \, b^{4} x^{4} + 50 \, a b^{3} x^{3} + 3 \, a^{2} b^{2} x^{2} - 4 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt{b x + a}}{315 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*b^4*x^4 + 50*a*b^3*x^3 + 3*a^2*b^2*x^2 - 4*a^3*b*x + 8*a^4)*sqrt(b*x + a)/b^3

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

[Out]

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

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Giac [B]  time = 1.14326, size = 136, normalized size = 2.31 \begin{align*} \frac{\sqrt{2}{\left (\frac{3 \, \sqrt{2}{\left (15 \,{\left (b x + a\right )}^{\frac{7}{2}} - 42 \,{\left (b x + a\right )}^{\frac{5}{2}} a + 35 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{2}\right )} a}{b^{2}} + \frac{\sqrt{2}{\left (35 \,{\left (b x + a\right )}^{\frac{9}{2}} - 135 \,{\left (b x + a\right )}^{\frac{7}{2}} a + 189 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{2} - 105 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{3}\right )}}{b^{2}}\right )}}{315 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/315*sqrt(2)*(3*sqrt(2)*(15*(b*x + a)^(7/2) - 42*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2)*a/b^2 + sqrt(2)*
(35*(b*x + a)^(9/2) - 135*(b*x + a)^(7/2)*a + 189*(b*x + a)^(5/2)*a^2 - 105*(b*x + a)^(3/2)*a^3)/b^2)/b