3.81 \(\int \frac{\tanh ^{-1}(\tanh (a+b x))^4}{x^9} \, dx\)

Optimal. Leaf size=80 \[ -\frac{b^3 \tanh ^{-1}(\tanh (a+b x))}{70 x^5}-\frac{b^2 \tanh ^{-1}(\tanh (a+b x))^2}{28 x^6}-\frac{b \tanh ^{-1}(\tanh (a+b x))^3}{14 x^7}-\frac{\tanh ^{-1}(\tanh (a+b x))^4}{8 x^8}-\frac{b^4}{280 x^4} \]

[Out]

-b^4/(280*x^4) - (b^3*ArcTanh[Tanh[a + b*x]])/(70*x^5) - (b^2*ArcTanh[Tanh[a + b*x]]^2)/(28*x^6) - (b*ArcTanh[
Tanh[a + b*x]]^3)/(14*x^7) - ArcTanh[Tanh[a + b*x]]^4/(8*x^8)

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Rubi [A]  time = 0.0754094, antiderivative size = 132, normalized size of antiderivative = 1.65, number of steps used = 4, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2171, 2167} \[ \frac{b^2 \tanh ^{-1}(\tanh (a+b x))^5}{56 x^6 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{b^3 \tanh ^{-1}(\tanh (a+b x))^5}{280 x^5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac{\tanh ^{-1}(\tanh (a+b x))^5}{8 x^8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{3 b \tanh ^{-1}(\tanh (a+b x))^5}{56 x^7 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^4/x^9,x]

[Out]

(b^3*ArcTanh[Tanh[a + b*x]]^5)/(280*x^5*(b*x - ArcTanh[Tanh[a + b*x]])^4) + (b^2*ArcTanh[Tanh[a + b*x]]^5)/(56
*x^6*(b*x - ArcTanh[Tanh[a + b*x]])^3) + (3*b*ArcTanh[Tanh[a + b*x]]^5)/(56*x^7*(b*x - ArcTanh[Tanh[a + b*x]])
^2) + ArcTanh[Tanh[a + b*x]]^5/(8*x^8*(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^9} \, dx &=\frac{\tanh ^{-1}(\tanh (a+b x))^5}{8 x^8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{(3 b) \int \frac{\tanh ^{-1}(\tanh (a+b x))^4}{x^8} \, dx}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{3 b \tanh ^{-1}(\tanh (a+b x))^5}{56 x^7 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{\tanh ^{-1}(\tanh (a+b x))^5}{8 x^8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{\left (3 b^2\right ) \int \frac{\tanh ^{-1}(\tanh (a+b x))^4}{x^7} \, dx}{28 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac{b^2 \tanh ^{-1}(\tanh (a+b x))^5}{56 x^6 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{3 b \tanh ^{-1}(\tanh (a+b x))^5}{56 x^7 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{\tanh ^{-1}(\tanh (a+b x))^5}{8 x^8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{b^3 \int \frac{\tanh ^{-1}(\tanh (a+b x))^4}{x^6} \, dx}{56 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}\\ &=\frac{b^3 \tanh ^{-1}(\tanh (a+b x))^5}{280 x^5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac{b^2 \tanh ^{-1}(\tanh (a+b x))^5}{56 x^6 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{3 b \tanh ^{-1}(\tanh (a+b x))^5}{56 x^7 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{\tanh ^{-1}(\tanh (a+b x))^5}{8 x^8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end{align*}

Mathematica [A]  time = 0.0335632, size = 71, normalized size = 0.89 \[ -\frac{4 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+10 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2+20 b x \tanh ^{-1}(\tanh (a+b x))^3+35 \tanh ^{-1}(\tanh (a+b x))^4+b^4 x^4}{280 x^8} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^4/x^9,x]

[Out]

-(b^4*x^4 + 4*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 10*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 + 20*b*x*ArcTanh[Tanh[a + b
*x]]^3 + 35*ArcTanh[Tanh[a + b*x]]^4)/(280*x^8)

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Maple [A]  time = 0.04, size = 74, normalized size = 0.9 \begin{align*} -{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{4}}{8\,{x}^{8}}}+{\frac{b}{2} \left ( -{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3}}{7\,{x}^{7}}}+{\frac{3\,b}{7} \left ( -{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{6\,{x}^{6}}}+{\frac{b}{3} \left ( -{\frac{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{5\,{x}^{5}}}-{\frac{b}{20\,{x}^{4}}} \right ) } \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^4/x^9,x)

[Out]

-1/8*arctanh(tanh(b*x+a))^4/x^8+1/2*b*(-1/7/x^7*arctanh(tanh(b*x+a))^3+3/7*b*(-1/6/x^6*arctanh(tanh(b*x+a))^2+
1/3*b*(-1/5/x^5*arctanh(tanh(b*x+a))-1/20/x^4*b)))

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Maxima [A]  time = 1.79259, size = 97, normalized size = 1.21 \begin{align*} -\frac{1}{280} \,{\left (b{\left (\frac{b^{2}}{x^{4}} + \frac{4 \, b \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )}{x^{5}}\right )} + \frac{10 \, b \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{x^{6}}\right )} b - \frac{b \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{14 \, x^{7}} - \frac{\operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{4}}{8 \, x^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^9,x, algorithm="maxima")

[Out]

-1/280*(b*(b^2/x^4 + 4*b*arctanh(tanh(b*x + a))/x^5) + 10*b*arctanh(tanh(b*x + a))^2/x^6)*b - 1/14*b*arctanh(t
anh(b*x + a))^3/x^7 - 1/8*arctanh(tanh(b*x + a))^4/x^8

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Fricas [A]  time = 1.40823, size = 111, normalized size = 1.39 \begin{align*} -\frac{70 \, b^{4} x^{4} + 224 \, a b^{3} x^{3} + 280 \, a^{2} b^{2} x^{2} + 160 \, a^{3} b x + 35 \, a^{4}}{280 \, x^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^9,x, algorithm="fricas")

[Out]

-1/280*(70*b^4*x^4 + 224*a*b^3*x^3 + 280*a^2*b^2*x^2 + 160*a^3*b*x + 35*a^4)/x^8

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Sympy [A]  time = 16.1373, size = 76, normalized size = 0.95 \begin{align*} - \frac{b^{4}}{280 x^{4}} - \frac{b^{3} \operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}{70 x^{5}} - \frac{b^{2} \operatorname{atanh}^{2}{\left (\tanh{\left (a + b x \right )} \right )}}{28 x^{6}} - \frac{b \operatorname{atanh}^{3}{\left (\tanh{\left (a + b x \right )} \right )}}{14 x^{7}} - \frac{\operatorname{atanh}^{4}{\left (\tanh{\left (a + b x \right )} \right )}}{8 x^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**4/x**9,x)

[Out]

-b**4/(280*x**4) - b**3*atanh(tanh(a + b*x))/(70*x**5) - b**2*atanh(tanh(a + b*x))**2/(28*x**6) - b*atanh(tanh
(a + b*x))**3/(14*x**7) - atanh(tanh(a + b*x))**4/(8*x**8)

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Giac [A]  time = 1.10287, size = 62, normalized size = 0.78 \begin{align*} -\frac{70 \, b^{4} x^{4} + 224 \, a b^{3} x^{3} + 280 \, a^{2} b^{2} x^{2} + 160 \, a^{3} b x + 35 \, a^{4}}{280 \, x^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^9,x, algorithm="giac")

[Out]

-1/280*(70*b^4*x^4 + 224*a*b^3*x^3 + 280*a^2*b^2*x^2 + 160*a^3*b*x + 35*a^4)/x^8