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

Optimal. Leaf size=60 \[ -3 b^2 \log (x) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )-\frac{\tanh ^{-1}(\tanh (a+b x))^3}{2 x^2}-\frac{3 b \tanh ^{-1}(\tanh (a+b x))^2}{2 x}+3 b^3 x \]

[Out]

3*b^3*x - (3*b*ArcTanh[Tanh[a + b*x]]^2)/(2*x) - ArcTanh[Tanh[a + b*x]]^3/(2*x^2) - 3*b^2*(b*x - ArcTanh[Tanh[
a + b*x]])*Log[x]

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Rubi [A]  time = 0.0396589, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2168, 2158, 29} \[ -3 b^2 \log (x) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )-\frac{\tanh ^{-1}(\tanh (a+b x))^3}{2 x^2}-\frac{3 b \tanh ^{-1}(\tanh (a+b x))^2}{2 x}+3 b^3 x \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^3/x^3,x]

[Out]

3*b^3*x - (3*b*ArcTanh[Tanh[a + b*x]]^2)/(2*x) - ArcTanh[Tanh[a + b*x]]^3/(2*x^2) - 3*b^2*(b*x - ArcTanh[Tanh[
a + b*x]])*Log[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 2158

Int[(v_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(b*x)/a, x] - Dist[(b*u
- a*v)/a, Int[1/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rubi steps

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

Mathematica [A]  time = 0.0444846, size = 66, normalized size = 1.1 \[ 3 b^2 \log (x) \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )-\frac{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3}{2 x^2}-\frac{3 b \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}{x}+b^3 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

b^3*x - (3*b*(-(b*x) + ArcTanh[Tanh[a + b*x]])^2)/x - (-(b*x) + ArcTanh[Tanh[a + b*x]])^3/(2*x^2) + 3*b^2*(-(b
*x) + ArcTanh[Tanh[a + b*x]])*Log[x]

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Maple [A]  time = 0.042, size = 59, normalized size = 1. \begin{align*} -{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3}}{2\,{x}^{2}}}-{\frac{3\,b \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{2\,x}}-3\,\ln \left ( x \right ) x{b}^{3}+3\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \ln \left ( x \right ){b}^{2}+3\,{b}^{3}x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*arctanh(tanh(b*x+a))^3/x^2-3/2*b*arctanh(tanh(b*x+a))^2/x-3*ln(x)*x*b^3+3*arctanh(tanh(b*x+a))*ln(x)*b^2+
3*b^3*x

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Maxima [A]  time = 1.42748, size = 97, normalized size = 1.62 \begin{align*} 3 \,{\left (b \operatorname{artanh}\left (\tanh \left (b x + a\right )\right ) \log \left (x\right ) -{\left (b{\left (x + \frac{a}{b}\right )} \log \left (x\right ) - b{\left (x + \frac{a \log \left (x\right )}{b}\right )}\right )} b\right )} b - \frac{3 \, b \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{2 \, x} - \frac{\operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*(b*arctanh(tanh(b*x + a))*log(x) - (b*(x + a/b)*log(x) - b*(x + a*log(x)/b))*b)*b - 3/2*b*arctanh(tanh(b*x +
a))^2/x - 1/2*arctanh(tanh(b*x + a))^3/x^2

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Fricas [A]  time = 1.52351, size = 81, normalized size = 1.35 \begin{align*} \frac{2 \, b^{3} x^{3} + 6 \, a b^{2} x^{2} \log \left (x\right ) - 6 \, a^{2} b x - a^{3}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b^3*x^3 + 6*a*b^2*x^2*log(x) - 6*a^2*b*x - a^3)/x^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14616, size = 42, normalized size = 0.7 \begin{align*} b^{3} x + 3 \, a b^{2} \log \left ({\left | x \right |}\right ) - \frac{6 \, a^{2} b x + a^{3}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^3*x + 3*a*b^2*log(abs(x)) - 1/2*(6*a^2*b*x + a^3)/x^2