3.1.80 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^4}{x^8} \, dx\) [80]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 98, normalized size of antiderivative = 1.00, 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 = {2202, 2198} \begin {gather*} \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} \end {gather*}

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 2198

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] &&
EqQ[m + n + 2, 0] && NeQ[m, -1]

Rule 2202

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^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*}

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Mathematica [A]
time = 0.02, size = 71, normalized size = 0.72 \begin {gather*} -\frac {b^4 x^4+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}{105 x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/105*(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)/x^7

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Maple [A]
time = 3.01, size = 74, normalized size = 0.76

method	result	size
default	\(-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{4}}{7 x^{7}}+\frac {4 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{6 x^{6}}+\frac {b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{5 x^{5}}+\frac {2 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )}{4 x^{4}}-\frac {b}{12 x^{3}}\right )}{5}\right )}{2}\right )}{7}\)	\(74\)
risch	\(\text {Expression too large to display}\)	\(22625\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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/x^3*b)))

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Maxima [A]
time = 0.43, size = 72, normalized size = 0.73 \begin {gather*} -\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 {gather*}

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 = 0.31, size = 46, normalized size = 0.47 \begin {gather*} -\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 {gather*}

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 = 1.16, size = 80, normalized size = 0.82 \begin {gather*} - \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 {gather*}

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 = 0.39, size = 46, normalized size = 0.47 \begin {gather*} -\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 {gather*}

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

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Mupad [B]
time = 1.01, size = 70, normalized size = 0.71 \begin {gather*} -\frac {{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^4}{7\,x^7}-\frac {b^4}{105\,x^3}-\frac {2\,b^2\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{35\,x^5}-\frac {b^3\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{35\,x^4}-\frac {2\,b\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3}{21\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^4/x^8,x)

[Out]

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

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