3.79 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^4}{x^7} \, dx\)

Optimal. Leaf size=64 \[ \frac {\tanh ^{-1}(\tanh (a+b x))^5}{6 x^6 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {b \tanh ^{-1}(\tanh (a+b x))^5}{30 x^5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

[Out]

1/30*b*arctanh(tanh(b*x+a))^5/x^5/(b*x-arctanh(tanh(b*x+a)))^2+1/6*arctanh(tanh(b*x+a))^5/x^6/(b*x-arctanh(tan
h(b*x+a)))

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Rubi [A]  time = 0.03, antiderivative size = 64, normalized size of antiderivative = 1.00, 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} \[ \frac {\tanh ^{-1}(\tanh (a+b x))^5}{6 x^6 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {b \tanh ^{-1}(\tanh (a+b x))^5}{30 x^5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^4/x^7,x]

[Out]

(b*ArcTanh[Tanh[a + b*x]]^5)/(30*x^5*(b*x - ArcTanh[Tanh[a + b*x]])^2) + ArcTanh[Tanh[a + b*x]]^5/(6*x^6*(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^7} \, dx &=\frac {\tanh ^{-1}(\tanh (a+b x))^5}{6 x^6 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {b \int \frac {\tanh ^{-1}(\tanh (a+b x))^4}{x^6} \, dx}{6 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {b \tanh ^{-1}(\tanh (a+b x))^5}{30 x^5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {\tanh ^{-1}(\tanh (a+b x))^5}{6 x^6 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 71, normalized size = 1.11 \[ -\frac {2 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+3 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2+4 b x \tanh ^{-1}(\tanh (a+b x))^3+5 \tanh ^{-1}(\tanh (a+b x))^4+b^4 x^4}{30 x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^4/x^7,x]

[Out]

-1/30*(b^4*x^4 + 2*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 3*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 + 4*b*x*ArcTanh[Tanh[a
+ b*x]]^3 + 5*ArcTanh[Tanh[a + b*x]]^4)/x^6

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fricas [A]  time = 0.40, size = 46, normalized size = 0.72 \[ -\frac {15 \, b^{4} x^{4} + 40 \, a b^{3} x^{3} + 45 \, a^{2} b^{2} x^{2} + 24 \, a^{3} b x + 5 \, a^{4}}{30 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^7,x, algorithm="fricas")

[Out]

-1/30*(15*b^4*x^4 + 40*a*b^3*x^3 + 45*a^2*b^2*x^2 + 24*a^3*b*x + 5*a^4)/x^6

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giac [A]  time = 0.16, size = 46, normalized size = 0.72 \[ -\frac {15 \, b^{4} x^{4} + 40 \, a b^{3} x^{3} + 45 \, a^{2} b^{2} x^{2} + 24 \, a^{3} b x + 5 \, a^{4}}{30 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^7,x, algorithm="giac")

[Out]

-1/30*(15*b^4*x^4 + 40*a*b^3*x^3 + 45*a^2*b^2*x^2 + 24*a^3*b*x + 5*a^4)/x^6

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maple [A]  time = 0.15, size = 74, normalized size = 1.16 \[ -\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{4}}{6 x^{6}}+\frac {2 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{5 x^{5}}+\frac {3 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{4 x^{4}}+\frac {b \left (-\frac {b}{6 x^{2}}-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )}{3 x^{3}}\right )}{2}\right )}{5}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^4/x^7,x)

[Out]

-1/6*arctanh(tanh(b*x+a))^4/x^6+2/3*b*(-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 = 0.60, size = 72, normalized size = 1.12 \[ -\frac {1}{30} \, {\left (b {\left (\frac {b^{2}}{x^{2}} + \frac {2 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}{x^{3}}\right )} + \frac {3 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{x^{4}}\right )} b - \frac {2 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{15 \, x^{5}} - \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{4}}{6 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^7,x, algorithm="maxima")

[Out]

-1/30*(b*(b^2/x^2 + 2*b*arctanh(tanh(b*x + a))/x^3) + 3*b*arctanh(tanh(b*x + a))^2/x^4)*b - 2/15*b*arctanh(tan
h(b*x + a))^3/x^5 - 1/6*arctanh(tanh(b*x + a))^4/x^6

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mupad [B]  time = 1.04, size = 70, normalized size = 1.09 \[ -\frac {{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^4}{6\,x^6}-\frac {b^4}{30\,x^2}-\frac {b^2\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{10\,x^4}-\frac {b^3\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{15\,x^3}-\frac {2\,b\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3}{15\,x^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^4/x^7,x)

[Out]

- atanh(tanh(a + b*x))^4/(6*x^6) - b^4/(30*x^2) - (b^2*atanh(tanh(a + b*x))^2)/(10*x^4) - (b^3*atanh(tanh(a +
b*x)))/(15*x^3) - (2*b*atanh(tanh(a + b*x))^3)/(15*x^5)

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sympy [A]  time = 3.29, size = 78, normalized size = 1.22 \[ - \frac {b^{4}}{30 x^{2}} - \frac {b^{3} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{15 x^{3}} - \frac {b^{2} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{10 x^{4}} - \frac {2 b \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{15 x^{5}} - \frac {\operatorname {atanh}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{6 x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**4/x**7,x)

[Out]

-b**4/(30*x**2) - b**3*atanh(tanh(a + b*x))/(15*x**3) - b**2*atanh(tanh(a + b*x))**2/(10*x**4) - 2*b*atanh(tan
h(a + b*x))**3/(15*x**5) - atanh(tanh(a + b*x))**4/(6*x**6)

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