3.80 \(\int \frac{\tanh ^{-1}(\tanh (a+b x))^4}{x^8} \, dx\)

Optimal. Leaf size=98 \[ \frac{b^2 \tanh ^{-1}(\tanh (a+b x))^5}{105 x^5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{\tanh ^{-1}(\tanh (a+b x))^5}{7 x^7 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{b \tanh ^{-1}(\tanh (a+b x))^5}{21 x^6 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

[Out]

(b^2*ArcTanh[Tanh[a + b*x]]^5)/(105*x^5*(b*x - ArcTanh[Tanh[a + b*x]])^3) + (b*ArcTanh[Tanh[a + b*x]]^5)/(21*x
^6*(b*x - ArcTanh[Tanh[a + b*x]])^2) + ArcTanh[Tanh[a + b*x]]^5/(7*x^7*(b*x - ArcTanh[Tanh[a + b*x]]))

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Rubi [A]  time = 0.0531644, antiderivative size = 98, 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 = {2171, 2167} \[ \frac{b^2 \tanh ^{-1}(\tanh (a+b x))^5}{105 x^5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{\tanh ^{-1}(\tanh (a+b x))^5}{7 x^7 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{b \tanh ^{-1}(\tanh (a+b x))^5}{21 x^6 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^4/x^8,x]

[Out]

(b^2*ArcTanh[Tanh[a + b*x]]^5)/(105*x^5*(b*x - ArcTanh[Tanh[a + b*x]])^3) + (b*ArcTanh[Tanh[a + b*x]]^5)/(21*x
^6*(b*x - ArcTanh[Tanh[a + b*x]])^2) + ArcTanh[Tanh[a + b*x]]^5/(7*x^7*(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))^4}{x^8} \, dx &=\frac{\tanh ^{-1}(\tanh (a+b x))^5}{7 x^7 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{(2 b) \int \frac{\tanh ^{-1}(\tanh (a+b x))^4}{x^7} \, dx}{7 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{b \tanh ^{-1}(\tanh (a+b x))^5}{21 x^6 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{\tanh ^{-1}(\tanh (a+b x))^5}{7 x^7 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{b^2 \int \frac{\tanh ^{-1}(\tanh (a+b x))^4}{x^6} \, dx}{21 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac{b^2 \tanh ^{-1}(\tanh (a+b x))^5}{105 x^5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{b \tanh ^{-1}(\tanh (a+b x))^5}{21 x^6 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{\tanh ^{-1}(\tanh (a+b x))^5}{7 x^7 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end{align*}

Mathematica [A]  time = 0.033165, size = 71, normalized size = 0.72 \[ -\frac{3 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+6 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2+10 b x \tanh ^{-1}(\tanh (a+b x))^3+15 \tanh ^{-1}(\tanh (a+b x))^4+b^4 x^4}{105 x^7} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^4/x^8,x]

[Out]

-(b^4*x^4 + 3*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 6*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 + 10*b*x*ArcTanh[Tanh[a + b*
x]]^3 + 15*ArcTanh[Tanh[a + b*x]]^4)/(105*x^7)

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Maple [A]  time = 0.04, size = 74, normalized size = 0.8 \begin{align*} -{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{4}}{7\,{x}^{7}}}+{\frac{4\,b}{7} \left ( -{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3}}{6\,{x}^{6}}}+{\frac{b}{2} \left ( -{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{5\,{x}^{5}}}+{\frac{2\,b}{5} \left ( -{\frac{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{4\,{x}^{4}}}-{\frac{b}{12\,{x}^{3}}} \right ) } \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^4/x^8,x)

[Out]

-1/7*arctanh(tanh(b*x+a))^4/x^7+4/7*b*(-1/6*arctanh(tanh(b*x+a))^3/x^6+1/2*b*(-1/5*arctanh(tanh(b*x+a))^2/x^5+
2/5*b*(-1/4*arctanh(tanh(b*x+a))/x^4-1/12*b/x^3)))

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Maxima [A]  time = 1.79776, size = 97, normalized size = 0.99 \begin{align*} -\frac{1}{105} \,{\left (b{\left (\frac{b^{2}}{x^{3}} + \frac{3 \, b \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )}{x^{4}}\right )} + \frac{6 \, b \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{x^{5}}\right )} b - \frac{2 \, b \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{21 \, x^{6}} - \frac{\operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{4}}{7 \, x^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^8,x, algorithm="maxima")

[Out]

-1/105*(b*(b^2/x^3 + 3*b*arctanh(tanh(b*x + a))/x^4) + 6*b*arctanh(tanh(b*x + a))^2/x^5)*b - 2/21*b*arctanh(ta
nh(b*x + a))^3/x^6 - 1/7*arctanh(tanh(b*x + a))^4/x^7

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Fricas [A]  time = 1.52279, size = 109, normalized size = 1.11 \begin{align*} -\frac{35 \, b^{4} x^{4} + 105 \, a b^{3} x^{3} + 126 \, a^{2} b^{2} x^{2} + 70 \, a^{3} b x + 15 \, a^{4}}{105 \, x^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^8,x, algorithm="fricas")

[Out]

-1/105*(35*b^4*x^4 + 105*a*b^3*x^3 + 126*a^2*b^2*x^2 + 70*a^3*b*x + 15*a^4)/x^7

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Sympy [A]  time = 11.6407, size = 80, normalized size = 0.82 \begin{align*} - \frac{b^{4}}{105 x^{3}} - \frac{b^{3} \operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}{35 x^{4}} - \frac{2 b^{2} \operatorname{atanh}^{2}{\left (\tanh{\left (a + b x \right )} \right )}}{35 x^{5}} - \frac{2 b \operatorname{atanh}^{3}{\left (\tanh{\left (a + b x \right )} \right )}}{21 x^{6}} - \frac{\operatorname{atanh}^{4}{\left (\tanh{\left (a + b x \right )} \right )}}{7 x^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**4/x**8,x)

[Out]

-b**4/(105*x**3) - b**3*atanh(tanh(a + b*x))/(35*x**4) - 2*b**2*atanh(tanh(a + b*x))**2/(35*x**5) - 2*b*atanh(
tanh(a + b*x))**3/(21*x**6) - atanh(tanh(a + b*x))**4/(7*x**7)

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Giac [A]  time = 1.15779, size = 62, normalized size = 0.63 \begin{align*} -\frac{35 \, b^{4} x^{4} + 105 \, a b^{3} x^{3} + 126 \, a^{2} b^{2} x^{2} + 70 \, a^{3} b x + 15 \, a^{4}}{105 \, x^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^8,x, algorithm="giac")

[Out]

-1/105*(35*b^4*x^4 + 105*a*b^3*x^3 + 126*a^2*b^2*x^2 + 70*a^3*b*x + 15*a^4)/x^7