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

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

[Out]

-1/280*b^4/x^4-1/70*b^3*arctanh(tanh(b*x+a))/x^5-1/28*b^2*arctanh(tanh(b*x+a))^2/x^6-1/14*b*arctanh(tanh(b*x+a
))^3/x^7-1/8*arctanh(tanh(b*x+a))^4/x^8

________________________________________________________________________________________

Rubi [A]  time = 0.08, 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 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))^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 = 0.04, 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]

-1/280*(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)/x^8

________________________________________________________________________________________

fricas [A]  time = 0.47, size = 46, normalized size = 0.58 \[ -\frac {70 \, b^{4} x^{4} + 224 \, a b^{3} x^{3} + 280 \, a^{2} b^{2} x^{2} + 160 \, a^{3} b x + 35 \, a^{4}}{280 \, x^{8}} \]

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

________________________________________________________________________________________

giac [A]  time = 0.15, size = 46, normalized size = 0.58 \[ -\frac {70 \, b^{4} x^{4} + 224 \, a b^{3} x^{3} + 280 \, a^{2} b^{2} x^{2} + 160 \, a^{3} b x + 35 \, a^{4}}{280 \, x^{8}} \]

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

________________________________________________________________________________________

maple [A]  time = 0.15, size = 74, normalized size = 0.92 \[ -\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{4}}{8 x^{8}}+\frac {b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{7 x^{7}}+\frac {3 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{6 x^{6}}+\frac {b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )}{5 x^{5}}-\frac {b}{20 x^{4}}\right )}{3}\right )}{7}\right )}{2} \]

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)))

________________________________________________________________________________________

maxima [A]  time = 0.60, size = 72, normalized size = 0.90 \[ -\frac {1}{280} \, {\left (b {\left (\frac {b^{2}}{x^{4}} + \frac {4 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}{x^{5}}\right )} + \frac {10 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{x^{6}}\right )} b - \frac {b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{14 \, x^{7}} - \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{4}}{8 \, x^{8}} \]

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

________________________________________________________________________________________

mupad [B]  time = 1.07, size = 70, normalized size = 0.88 \[ -\frac {{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^4}{8\,x^8}-\frac {b^4}{280\,x^4}-\frac {b^2\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{28\,x^6}-\frac {b^3\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{70\,x^5}-\frac {b\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3}{14\,x^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^4/x^9,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 8.25, size = 76, normalized size = 0.95 \[ - \frac {b^{4}}{280 x^{4}} - \frac {b^{3} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{70 x^{5}} - \frac {b^{2} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{28 x^{6}} - \frac {b \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{14 x^{7}} - \frac {\operatorname {atanh}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{8 x^{8}} \]

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)

________________________________________________________________________________________