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

Optimal. Leaf size=38 \[ \frac{2 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b}-\frac{4 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2} \]

[Out]

(2*x*ArcTanh[Tanh[a + b*x]]^(5/2))/(5*b) - (4*ArcTanh[Tanh[a + b*x]]^(7/2))/(35*b^2)

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Rubi [A]  time = 0.0141962, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2168, 2157, 30} \[ \frac{2 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b}-\frac{4 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcTanh[Tanh[a + b*x]]^(3/2),x]

[Out]

(2*x*ArcTanh[Tanh[a + b*x]]^(5/2))/(5*b) - (4*ArcTanh[Tanh[a + b*x]]^(7/2))/(35*b^2)

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

Mathematica [A]  time = 0.0568584, size = 32, normalized size = 0.84 \[ \frac{2 \left (7 b x-2 \tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}{35 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.036, size = 42, normalized size = 1.1 \begin{align*} 2\,{\frac{1/7\, \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 ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{5/2}}{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.76248, size = 42, normalized size = 1.11 \begin{align*} \frac{2 \,{\left (5 \, b^{2} x^{2} + 3 \, a b x - 2 \, a^{2}\right )}{\left (b x + a\right )}^{\frac{3}{2}}}{35 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.92149, size = 92, normalized size = 2.42 \begin{align*} \frac{2 \,{\left (5 \, b^{3} x^{3} + 8 \, a b^{2} x^{2} + a^{2} b x - 2 \, a^{3}\right )} \sqrt{b x + a}}{35 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(5*b^3*x^3 + 8*a*b^2*x^2 + a^2*b*x - 2*a^3)*sqrt(b*x + a)/b^2

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Sympy [A]  time = 49.9938, size = 49, normalized size = 1.29 \begin{align*} \begin{cases} \frac{2 x \operatorname{atanh}^{\frac{5}{2}}{\left (\tanh{\left (a + b x \right )} \right )}}{5 b} - \frac{4 \operatorname{atanh}^{\frac{7}{2}}{\left (\tanh{\left (a + b x \right )} \right )}}{35 b^{2}} & \text{for}\: b \neq 0 \\\frac{x^{2} \operatorname{atanh}^{\frac{3}{2}}{\left (\tanh{\left (a \right )} \right )}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*x*atanh(tanh(a + b*x))**(5/2)/(5*b) - 4*atanh(tanh(a + b*x))**(7/2)/(35*b**2), Ne(b, 0)), (x**2*a
tanh(tanh(a))**(3/2)/2, True))

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Giac [B]  time = 1.15601, size = 104, normalized size = 2.74 \begin{align*} \frac{\sqrt{2}{\left (\frac{7 \, \sqrt{2}{\left (3 \,{\left (b x + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x + a\right )}^{\frac{3}{2}} a\right )} a}{b} + \frac{\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 )}}{b}\right )}}{105 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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