3.52 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^2}{x^4} \, dx\)

Optimal. Leaf size=31 \[ \frac {\tanh ^{-1}(\tanh (a+b x))^3}{3 x^3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2167} \[ \frac {\tanh ^{-1}(\tanh (a+b x))^3}{3 x^3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^2/x^4,x]

[Out]

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

Rule 2167

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, -Simp[(u^(m + 1)*v^
(n + 1))/((m + 1)*(b*u - a*v)), x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n}, x] && PiecewiseLinearQ[u, v, x] && E
qQ[m + n + 2, 0] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}(\tanh (a+b x))^2}{x^4} \, dx &=\frac {\tanh ^{-1}(\tanh (a+b x))^3}{3 x^3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 34, normalized size = 1.10 \[ -\frac {b x \tanh ^{-1}(\tanh (a+b x))+\tanh ^{-1}(\tanh (a+b x))^2+b^2 x^2}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^2/x^4,x]

[Out]

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

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fricas [A]  time = 0.46, size = 22, normalized size = 0.71 \[ -\frac {3 \, b^{2} x^{2} + 3 \, a b x + a^{2}}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^2/x^4,x, algorithm="fricas")

[Out]

-1/3*(3*b^2*x^2 + 3*a*b*x + a^2)/x^3

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giac [A]  time = 0.81, size = 22, normalized size = 0.71 \[ -\frac {3 \, b^{2} x^{2} + 3 \, a b x + a^{2}}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^2/x^4,x, algorithm="giac")

[Out]

-1/3*(3*b^2*x^2 + 3*a*b*x + a^2)/x^3

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maple [A]  time = 0.14, size = 38, normalized size = 1.23 \[ -\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{3 x^{3}}+\frac {2 b \left (-\frac {b}{2 x}-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )}{2 x^{2}}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^2/x^4,x)

[Out]

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

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maxima [A]  time = 0.47, size = 36, normalized size = 1.16 \[ -\frac {b^{2}}{3 \, x} - \frac {b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}{3 \, x^{2}} - \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^2/x^4,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.94, size = 32, normalized size = 1.03 \[ -\frac {b^2\,x^2+b\,x\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )+{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{3\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^2/x^4,x)

[Out]

-(atanh(tanh(a + b*x))^2 + b^2*x^2 + b*x*atanh(tanh(a + b*x)))/(3*x^3)

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sympy [A]  time = 0.80, size = 37, normalized size = 1.19 \[ - \frac {b^{2}}{3 x} - \frac {b \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{3 x^{2}} - \frac {\operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**2/x**4,x)

[Out]

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

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