3.78 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^4}{x^6} \, dx\)

Optimal. Leaf size=31 \[ \frac {\tanh ^{-1}(\tanh (a+b x))^5}{5 x^5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

[Out]

1/5*arctanh(tanh(b*x+a))^5/x^5/(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} \[ \frac {\tanh ^{-1}(\tanh (a+b x))^5}{5 x^5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^4/x^6,x]

[Out]

ArcTanh[Tanh[a + b*x]]^5/(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]

Rubi steps

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

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Mathematica [B]  time = 0.06, size = 66, normalized size = 2.13 \[ -\frac {b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2+b x \tanh ^{-1}(\tanh (a+b x))^3+\tanh ^{-1}(\tanh (a+b x))^4+b^4 x^4}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^4/x^6,x]

[Out]

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

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fricas [A]  time = 0.49, size = 44, normalized size = 1.42 \[ -\frac {5 \, b^{4} x^{4} + 10 \, a b^{3} x^{3} + 10 \, a^{2} b^{2} x^{2} + 5 \, a^{3} b x + a^{4}}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^6,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 44, normalized size = 1.42 \[ -\frac {5 \, b^{4} x^{4} + 10 \, a b^{3} x^{3} + 10 \, a^{2} b^{2} x^{2} + 5 \, a^{3} b x + a^{4}}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^6,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.15, size = 74, normalized size = 2.39 \[ -\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{4}}{5 x^{5}}+\frac {4 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{4 x^{4}}+\frac {3 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{3 x^{3}}+\frac {2 b \left (-\frac {b}{2 x}-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )}{2 x^{2}}\right )}{3}\right )}{4}\right )}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^4/x^6,x)

[Out]

-1/5*arctanh(tanh(b*x+a))^4/x^5+4/5*b*(-1/4*arctanh(tanh(b*x+a))^3/x^4+3/4*b*(-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 [B]  time = 0.60, size = 70, normalized size = 2.26 \[ -\frac {1}{5} \, {\left (b {\left (\frac {b^{2}}{x} + \frac {b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}{x^{2}}\right )} + \frac {b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{x^{3}}\right )} b - \frac {b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{5 \, x^{4}} - \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{4}}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^6,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.20, size = 64, normalized size = 2.06 \[ -\frac {b^4\,x^4+b^3\,x^3\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )+b^2\,x^2\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2+b\,x\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3+{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^4}{5\,x^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^4/x^6,x)

[Out]

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

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sympy [B]  time = 2.10, size = 75, normalized size = 2.42 \[ - \frac {b^{4}}{5 x} - \frac {b^{3} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{5 x^{2}} - \frac {b^{2} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{5 x^{3}} - \frac {b \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{5 x^{4}} - \frac {\operatorname {atanh}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{5 x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**4/x**6,x)

[Out]

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

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