3.83 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^4}{x^{11}} \, dx\)

Optimal. Leaf size=80 \[ -\frac {b^3 \tanh ^{-1}(\tanh (a+b x))}{210 x^7}-\frac {b^2 \tanh ^{-1}(\tanh (a+b x))^2}{60 x^8}-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{10 x^{10}}-\frac {2 b \tanh ^{-1}(\tanh (a+b x))^3}{45 x^9}-\frac {b^4}{1260 x^6} \]

[Out]

-1/1260*b^4/x^6-1/210*b^3*arctanh(tanh(b*x+a))/x^7-1/60*b^2*arctanh(tanh(b*x+a))^2/x^8-2/45*b*arctanh(tanh(b*x
+a))^3/x^9-1/10*arctanh(tanh(b*x+a))^4/x^10

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Rubi [A]  time = 0.06, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2168, 30} \[ -\frac {b^3 \tanh ^{-1}(\tanh (a+b x))}{210 x^7}-\frac {b^2 \tanh ^{-1}(\tanh (a+b x))^2}{60 x^8}-\frac {2 b \tanh ^{-1}(\tanh (a+b x))^3}{45 x^9}-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{10 x^{10}}-\frac {b^4}{1260 x^6} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^4/x^11,x]

[Out]

-b^4/(1260*x^6) - (b^3*ArcTanh[Tanh[a + b*x]])/(210*x^7) - (b^2*ArcTanh[Tanh[a + b*x]]^2)/(60*x^8) - (2*b*ArcT
anh[Tanh[a + b*x]]^3)/(45*x^9) - ArcTanh[Tanh[a + b*x]]^4/(10*x^10)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2168

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(u^(m + 1)*v^
n)/(a*(m + 1)), x] - Dist[(b*n)/(a*(m + 1)), Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m
, n}, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0] &&  !(ILtQ[m + n, -2] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}(\tanh (a+b x))^4}{x^{11}} \, dx &=-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{10 x^{10}}+\frac {1}{5} (2 b) \int \frac {\tanh ^{-1}(\tanh (a+b x))^3}{x^{10}} \, dx\\ &=-\frac {2 b \tanh ^{-1}(\tanh (a+b x))^3}{45 x^9}-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{10 x^{10}}+\frac {1}{15} \left (2 b^2\right ) \int \frac {\tanh ^{-1}(\tanh (a+b x))^2}{x^9} \, dx\\ &=-\frac {b^2 \tanh ^{-1}(\tanh (a+b x))^2}{60 x^8}-\frac {2 b \tanh ^{-1}(\tanh (a+b x))^3}{45 x^9}-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{10 x^{10}}+\frac {1}{30} b^3 \int \frac {\tanh ^{-1}(\tanh (a+b x))}{x^8} \, dx\\ &=-\frac {b^3 \tanh ^{-1}(\tanh (a+b x))}{210 x^7}-\frac {b^2 \tanh ^{-1}(\tanh (a+b x))^2}{60 x^8}-\frac {2 b \tanh ^{-1}(\tanh (a+b x))^3}{45 x^9}-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{10 x^{10}}+\frac {1}{210} b^4 \int \frac {1}{x^7} \, dx\\ &=-\frac {b^4}{1260 x^6}-\frac {b^3 \tanh ^{-1}(\tanh (a+b x))}{210 x^7}-\frac {b^2 \tanh ^{-1}(\tanh (a+b x))^2}{60 x^8}-\frac {2 b \tanh ^{-1}(\tanh (a+b x))^3}{45 x^9}-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{10 x^{10}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 71, normalized size = 0.89 \[ -\frac {6 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+21 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2+56 b x \tanh ^{-1}(\tanh (a+b x))^3+126 \tanh ^{-1}(\tanh (a+b x))^4+b^4 x^4}{1260 x^{10}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^4/x^11,x]

[Out]

-1/1260*(b^4*x^4 + 6*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 21*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 + 56*b*x*ArcTanh[Tan
h[a + b*x]]^3 + 126*ArcTanh[Tanh[a + b*x]]^4)/x^10

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fricas [A]  time = 0.50, size = 46, normalized size = 0.58 \[ -\frac {210 \, b^{4} x^{4} + 720 \, a b^{3} x^{3} + 945 \, a^{2} b^{2} x^{2} + 560 \, a^{3} b x + 126 \, a^{4}}{1260 \, x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^11,x, algorithm="fricas")

[Out]

-1/1260*(210*b^4*x^4 + 720*a*b^3*x^3 + 945*a^2*b^2*x^2 + 560*a^3*b*x + 126*a^4)/x^10

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giac [A]  time = 0.26, size = 46, normalized size = 0.58 \[ -\frac {210 \, b^{4} x^{4} + 720 \, a b^{3} x^{3} + 945 \, a^{2} b^{2} x^{2} + 560 \, a^{3} b x + 126 \, a^{4}}{1260 \, x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^11,x, algorithm="giac")

[Out]

-1/1260*(210*b^4*x^4 + 720*a*b^3*x^3 + 945*a^2*b^2*x^2 + 560*a^3*b*x + 126*a^4)/x^10

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maple [A]  time = 0.15, size = 74, normalized size = 0.92 \[ -\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{4}}{10 x^{10}}+\frac {2 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{9 x^{9}}+\frac {b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{8 x^{8}}+\frac {b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )}{7 x^{7}}-\frac {b}{42 x^{6}}\right )}{4}\right )}{3}\right )}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^4/x^11,x)

[Out]

-1/10*arctanh(tanh(b*x+a))^4/x^10+2/5*b*(-1/9/x^9*arctanh(tanh(b*x+a))^3+1/3*b*(-1/8/x^8*arctanh(tanh(b*x+a))^
2+1/4*b*(-1/7/x^7*arctanh(tanh(b*x+a))-1/42/x^6*b)))

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maxima [A]  time = 0.61, size = 72, normalized size = 0.90 \[ -\frac {1}{1260} \, {\left (b {\left (\frac {b^{2}}{x^{6}} + \frac {6 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}{x^{7}}\right )} + \frac {21 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{x^{8}}\right )} b - \frac {2 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{45 \, x^{9}} - \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{4}}{10 \, x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^11,x, algorithm="maxima")

[Out]

-1/1260*(b*(b^2/x^6 + 6*b*arctanh(tanh(b*x + a))/x^7) + 21*b*arctanh(tanh(b*x + a))^2/x^8)*b - 2/45*b*arctanh(
tanh(b*x + a))^3/x^9 - 1/10*arctanh(tanh(b*x + a))^4/x^10

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mupad [B]  time = 1.04, size = 70, normalized size = 0.88 \[ -\frac {{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^4}{10\,x^{10}}-\frac {b^4}{1260\,x^6}-\frac {b^2\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{60\,x^8}-\frac {b^3\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{210\,x^7}-\frac {2\,b\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3}{45\,x^9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^4/x^11,x)

[Out]

- atanh(tanh(a + b*x))^4/(10*x^10) - b^4/(1260*x^6) - (b^2*atanh(tanh(a + b*x))^2)/(60*x^8) - (b^3*atanh(tanh(
a + b*x)))/(210*x^7) - (2*b*atanh(tanh(a + b*x))^3)/(45*x^9)

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sympy [A]  time = 19.49, size = 78, normalized size = 0.98 \[ - \frac {b^{4}}{1260 x^{6}} - \frac {b^{3} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{210 x^{7}} - \frac {b^{2} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{60 x^{8}} - \frac {2 b \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{45 x^{9}} - \frac {\operatorname {atanh}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{10 x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**4/x**11,x)

[Out]

-b**4/(1260*x**6) - b**3*atanh(tanh(a + b*x))/(210*x**7) - b**2*atanh(tanh(a + b*x))**2/(60*x**8) - 2*b*atanh(
tanh(a + b*x))**3/(45*x**9) - atanh(tanh(a + b*x))**4/(10*x**10)

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