3.64 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^3}{x^6} \, dx\)

Optimal. Leaf size=64 \[ \frac {\tanh ^{-1}(\tanh (a+b x))^4}{5 x^5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {b \tanh ^{-1}(\tanh (a+b x))^4}{20 x^4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

[Out]

1/20*b*arctanh(tanh(b*x+a))^4/x^4/(b*x-arctanh(tanh(b*x+a)))^2+1/5*arctanh(tanh(b*x+a))^4/x^5/(b*x-arctanh(tan
h(b*x+a)))

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Rubi [A]  time = 0.03, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2171, 2167} \[ \frac {\tanh ^{-1}(\tanh (a+b x))^4}{5 x^5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {b \tanh ^{-1}(\tanh (a+b x))^4}{20 x^4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^3/x^6,x]

[Out]

(b*ArcTanh[Tanh[a + b*x]]^4)/(20*x^4*(b*x - ArcTanh[Tanh[a + b*x]])^2) + ArcTanh[Tanh[a + b*x]]^4/(5*x^5*(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]

Rule 2171

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] + Dist[(b*(m + n + 2))/((m + 1)*(b*u - a*v)), Int[u^(m + 1)*v^n, x], x] /;
NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}(\tanh (a+b x))^3}{x^6} \, dx &=\frac {\tanh ^{-1}(\tanh (a+b x))^4}{5 x^5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {b \int \frac {\tanh ^{-1}(\tanh (a+b x))^3}{x^5} \, dx}{5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {b \tanh ^{-1}(\tanh (a+b x))^4}{20 x^4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {\tanh ^{-1}(\tanh (a+b x))^4}{5 x^5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 54, normalized size = 0.84 \[ -\frac {2 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+3 b x \tanh ^{-1}(\tanh (a+b x))^2+4 \tanh ^{-1}(\tanh (a+b x))^3+b^3 x^3}{20 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^3/x^6,x]

[Out]

-1/20*(b^3*x^3 + 2*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 3*b*x*ArcTanh[Tanh[a + b*x]]^2 + 4*ArcTanh[Tanh[a + b*x]]^
3)/x^5

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fricas [A]  time = 0.37, size = 35, normalized size = 0.55 \[ -\frac {10 \, b^{3} x^{3} + 20 \, a b^{2} x^{2} + 15 \, a^{2} b x + 4 \, a^{3}}{20 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3/x^6,x, algorithm="fricas")

[Out]

-1/20*(10*b^3*x^3 + 20*a*b^2*x^2 + 15*a^2*b*x + 4*a^3)/x^5

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giac [A]  time = 0.16, size = 35, normalized size = 0.55 \[ -\frac {10 \, b^{3} x^{3} + 20 \, a b^{2} x^{2} + 15 \, a^{2} b x + 4 \, a^{3}}{20 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3/x^6,x, algorithm="giac")

[Out]

-1/20*(10*b^3*x^3 + 20*a*b^2*x^2 + 15*a^2*b*x + 4*a^3)/x^5

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maple [A]  time = 0.15, size = 56, normalized size = 0.88 \[ -\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{5 x^{5}}+\frac {3 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{4 x^{4}}+\frac {b \left (-\frac {b}{6 x^{2}}-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )}{3 x^{3}}\right )}{2}\right )}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^3/x^6,x)

[Out]

-1/5*arctanh(tanh(b*x+a))^3/x^5+3/5*b*(-1/4*arctanh(tanh(b*x+a))^2/x^4+1/2*b*(-1/6*b/x^2-1/3*arctanh(tanh(b*x+
a))/x^3))

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maxima [A]  time = 0.53, size = 54, normalized size = 0.84 \[ -\frac {1}{20} \, b {\left (\frac {b^{2}}{x^{2}} + \frac {2 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}{x^{3}}\right )} - \frac {3 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{20 \, x^{4}} - \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3/x^6,x, algorithm="maxima")

[Out]

-1/20*b*(b^2/x^2 + 2*b*arctanh(tanh(b*x + a))/x^3) - 3/20*b*arctanh(tanh(b*x + a))^2/x^4 - 1/5*arctanh(tanh(b*
x + a))^3/x^5

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mupad [B]  time = 0.99, size = 53, normalized size = 0.83 \[ -\frac {{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3}{5\,x^5}-\frac {b^3}{20\,x^2}-\frac {b^2\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{10\,x^3}-\frac {3\,b\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{20\,x^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^3/x^6,x)

[Out]

- atanh(tanh(a + b*x))^3/(5*x^5) - b^3/(20*x^2) - (b^2*atanh(tanh(a + b*x)))/(10*x^3) - (3*b*atanh(tanh(a + b*
x))^2)/(20*x^4)

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sympy [A]  time = 2.07, size = 60, normalized size = 0.94 \[ - \frac {b^{3}}{20 x^{2}} - \frac {b^{2} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{10 x^{3}} - \frac {3 b \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{20 x^{4}} - \frac {\operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{5 x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**3/x**6,x)

[Out]

-b**3/(20*x**2) - b**2*atanh(tanh(a + b*x))/(10*x**3) - 3*b*atanh(tanh(a + b*x))**2/(20*x**4) - atanh(tanh(a +
b*x))**3/(5*x**5)

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