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

Optimal. Leaf size=36 \[ -\frac{\tanh ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac{b \tanh ^{-1}(\tanh (a+b x))}{x}+b^2 \log (x) \]

[Out]

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

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Rubi [A]  time = 0.0211679, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2168, 29} \[ -\frac{\tanh ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac{b \tanh ^{-1}(\tanh (a+b x))}{x}+b^2 \log (x) \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^2/x^3,x]

[Out]

-((b*ArcTanh[Tanh[a + b*x]])/x) - ArcTanh[Tanh[a + b*x]]^2/(2*x^2) + b^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 29

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

Rubi steps

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

Mathematica [A]  time = 0.0326993, size = 42, normalized size = 1.17 \[ -\frac{2 b x \tanh ^{-1}(\tanh (a+b x))+\tanh ^{-1}(\tanh (a+b x))^2-b^2 x^2 (2 \log (x)+3)}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.036, size = 35, normalized size = 1. \begin{align*} -{\frac{b{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{x}}-{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{2\,{x}^{2}}}+{b}^{2}\ln \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.37652, size = 46, normalized size = 1.28 \begin{align*} b^{2} \log \left (x\right ) - \frac{b \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )}{x} - \frac{\operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.5044, size = 59, normalized size = 1.64 \begin{align*} \frac{2 \, b^{2} x^{2} \log \left (x\right ) - 4 \, a b x - a^{2}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.822608, size = 32, normalized size = 0.89 \begin{align*} b^{2} \log{\left (x \right )} - \frac{b \operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}{x} - \frac{\operatorname{atanh}^{2}{\left (\tanh{\left (a + b x \right )} \right )}}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b**2*log(x) - b*atanh(tanh(a + b*x))/x - atanh(tanh(a + b*x))**2/(2*x**2)

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Giac [A]  time = 1.18836, size = 30, normalized size = 0.83 \begin{align*} b^{2} \log \left ({\left | x \right |}\right ) - \frac{4 \, a b x + a^{2}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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