3.106 \(\int \frac{x}{\tanh ^{-1}(\tanh (a+b x))^3} \, dx\)

Optimal. Leaf size=34 \[ -\frac{1}{2 b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac{x}{2 b \tanh ^{-1}(\tanh (a+b x))^2} \]

[Out]

-x/(2*b*ArcTanh[Tanh[a + b*x]]^2) - 1/(2*b^2*ArcTanh[Tanh[a + b*x]])

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Rubi [A]  time = 0.0144834, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {2168, 2157, 30} \[ -\frac{1}{2 b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac{x}{2 b \tanh ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

Int[x/ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

-x/(2*b*ArcTanh[Tanh[a + b*x]]^2) - 1/(2*b^2*ArcTanh[Tanh[a + b*x]])

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

Mathematica [A]  time = 0.0486065, size = 27, normalized size = 0.79 \[ -\frac{\tanh ^{-1}(\tanh (a+b x))+b x}{2 b^2 \tanh ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.037, size = 43, normalized size = 1.3 \begin{align*} -{\frac{bx-{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{2\,{b}^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}}-{\frac{1}{{b}^{2}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 3.53057, size = 43, normalized size = 1.26 \begin{align*} -\frac{2 \, b x + a}{2 \,{\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b*x + a)/(b^4*x^2 + 2*a*b^3*x + a^2*b^2)

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Fricas [A]  time = 1.55933, size = 68, normalized size = 2. \begin{align*} -\frac{2 \, b x + a}{2 \,{\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b*x + a)/(b^4*x^2 + 2*a*b^3*x + a^2*b^2)

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Sympy [A]  time = 22.1402, size = 42, normalized size = 1.24 \begin{align*} \begin{cases} - \frac{x}{2 b \operatorname{atanh}^{2}{\left (\tanh{\left (a + b x \right )} \right )}} - \frac{1}{2 b^{2} \operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}} & \text{for}\: b \neq 0 \\\frac{x^{2}}{2 \operatorname{atanh}^{3}{\left (\tanh{\left (a \right )} \right )}} & \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,x)

[Out]

Piecewise((-x/(2*b*atanh(tanh(a + b*x))**2) - 1/(2*b**2*atanh(tanh(a + b*x))), Ne(b, 0)), (x**2/(2*atanh(tanh(
a))**3), True))

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Giac [A]  time = 1.13751, size = 24, normalized size = 0.71 \begin{align*} -\frac{2 \, b x + a}{2 \,{\left (b x + a\right )}^{2} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b*x + a)/((b*x + a)^2*b^2)