3.153 \(\int \frac{\coth ^{-1}(\tanh (a+b x))^3}{x^2} \, dx\)

Optimal. Leaf size=68 \[ -3 b^2 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )-\frac{\coth ^{-1}(\tanh (a+b x))^3}{x}+\frac{3}{2} b \coth ^{-1}(\tanh (a+b x))^2+3 b \log (x) \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \]

[Out]

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

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Rubi [A]  time = 0.0435156, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {2168, 2159, 2158, 29} \[ -3 b^2 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )-\frac{\coth ^{-1}(\tanh (a+b x))^3}{x}+\frac{3}{2} b \coth ^{-1}(\tanh (a+b x))^2+3 b \log (x) \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Tanh[a + b*x]]^3/x^2,x]

[Out]

-3*b^2*x*(b*x - ArcCoth[Tanh[a + b*x]]) + (3*b*ArcCoth[Tanh[a + b*x]]^2)/2 - ArcCoth[Tanh[a + b*x]]^3/x + 3*b*
(b*x - ArcCoth[Tanh[a + b*x]])^2*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 2159

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^n/(a*n), x] - Dis
t[(b*u - a*v)/a, Int[v^(n - 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && GtQ[n, 0] && Ne
Q[n, 1]

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

Mathematica [A]  time = 0.039466, size = 62, normalized size = 0.91 \[ -6 b^2 x \log (x) \coth ^{-1}(\tanh (a+b x))-\frac{\coth ^{-1}(\tanh (a+b x))^3}{x}+3 b (\log (x)+1) \coth ^{-1}(\tanh (a+b x))^2+\frac{3}{2} b^3 x^2 (2 \log (x)-1) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^3/x^2,x]

[Out]

-(ArcCoth[Tanh[a + b*x]]^3/x) - 6*b^2*x*ArcCoth[Tanh[a + b*x]]*Log[x] + 3*b*ArcCoth[Tanh[a + b*x]]^2*(1 + Log[
x]) + (3*b^3*x^2*(-1 + 2*Log[x]))/2

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Maple [C]  time = 0.415, size = 7683, normalized size = 113. \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [C]  time = 2.08232, size = 166, normalized size = 2.44 \begin{align*} 3 \, b \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{2} \log \left (x\right ) - \frac{3}{2} \,{\left (2 \, \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{2} \log \left (x\right ) -{\left (b x^{2} +{\left (2 i \, \pi + 4 \, a\right )} x + 2 \,{\left (-\frac{i \, \pi{\left (b x + a\right )}}{b} - \frac{{\left (b x + a\right )}^{2}}{b}\right )} \log \left (x\right ) + \frac{2 \, \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{2} \log \left (x\right )}{b} + \frac{2 \,{\left (i \, \pi a + a^{2}\right )} \log \left (x\right )}{b}\right )} b\right )} b - \frac{\operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{3}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*b*arccoth(tanh(b*x + a))^2*log(x) - 3/2*(2*arccoth(tanh(b*x + a))^2*log(x) - (b*x^2 + (2*I*pi + 4*a)*x + 2*(
-I*pi*(b*x + a)/b - (b*x + a)^2/b)*log(x) + 2*arccoth(tanh(b*x + a))^2*log(x)/b + 2*(I*pi*a + a^2)*log(x)/b)*b
)*b - arccoth(tanh(b*x + a))^3/x

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Fricas [A]  time = 1.72542, size = 115, normalized size = 1.69 \begin{align*} \frac{2 \, b^{3} x^{3} + 12 \, a b^{2} x^{2} + 3 \, \pi ^{2} a - 4 \, a^{3} - 3 \,{\left (\pi ^{2} b - 4 \, a^{2} b\right )} x \log \left (x\right )}{4 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*b^3*x^3 + 12*a*b^2*x^2 + 3*pi^2*a - 4*a^3 - 3*(pi^2*b - 4*a^2*b)*x*log(x))/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{3}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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