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]

(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]]))

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Rubi [A]  time = 0.0319722, antiderivative size = 64, normalized size of antiderivative = 1., 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 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]

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))^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*}

Mathematica [A]  time = 0.0398583, 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]

-(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)/(2
0*x^5)

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Maple [A]  time = 0.038, size = 56, normalized size = 0.9 \begin{align*} -{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3}}{5\,{x}^{5}}}+{\frac{3\,b}{5} \left ( -{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{4\,{x}^{4}}}+{\frac{b}{2} \left ( -{\frac{b}{6\,{x}^{2}}}-{\frac{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{3\,{x}^{3}}} \right ) } \right ) } \end{align*}

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 = 1.58905, size = 73, normalized size = 1.14 \begin{align*} -\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}} \end{align*}

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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Fricas [A]  time = 1.44183, size = 81, normalized size = 1.27 \begin{align*} -\frac{10 \, b^{3} x^{3} + 20 \, a b^{2} x^{2} + 15 \, a^{2} b x + 4 \, a^{3}}{20 \, x^{5}} \end{align*}

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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Sympy [A]  time = 3.85814, size = 60, normalized size = 0.94 \begin{align*} - \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}} \end{align*}

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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Giac [A]  time = 1.14204, size = 47, normalized size = 0.73 \begin{align*} -\frac{10 \, b^{3} x^{3} + 20 \, a b^{2} x^{2} + 15 \, a^{2} b x + 4 \, a^{3}}{20 \, x^{5}} \end{align*}

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