\(\int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^6} \, dx\) [157]

Optimal result

Integrand size = 13, antiderivative size = 64 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^6} \, dx=\frac {b \coth ^{-1}(\tanh (a+b x))^4}{20 x^4 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac {\coth ^{-1}(\tanh (a+b x))^4}{5 x^5 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )} \]

[Out]

1/20*b*arccoth(tanh(b*x+a))^4/x^4/(b*x-arccoth(tanh(b*x+a)))^2+1/5*arccoth(tanh(b*x+a))^4/x^5/(b*x-arccoth(tan
h(b*x+a)))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2202, 2198} \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^6} \, dx=\frac {\coth ^{-1}(\tanh (a+b x))^4}{5 x^5 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}+\frac {b \coth ^{-1}(\tanh (a+b x))^4}{20 x^4 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2} \]

[In]

Int[ArcCoth[Tanh[a + b*x]]^3/x^6,x]

[Out]

(b*ArcCoth[Tanh[a + b*x]]^4)/(20*x^4*(b*x - ArcCoth[Tanh[a + b*x]])^2) + ArcCoth[Tanh[a + b*x]]^4/(5*x^5*(b*x
- ArcCoth[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*} \text {integral}& = \frac {\coth ^{-1}(\tanh (a+b x))^4}{5 x^5 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}+\frac {b \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^5} \, dx}{5 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )} \\ & = \frac {b \coth ^{-1}(\tanh (a+b x))^4}{20 x^4 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac {\coth ^{-1}(\tanh (a+b x))^4}{5 x^5 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.84 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^6} \, dx=-\frac {b^3 x^3+2 b^2 x^2 \coth ^{-1}(\tanh (a+b x))+3 b x \coth ^{-1}(\tanh (a+b x))^2+4 \coth ^{-1}(\tanh (a+b x))^3}{20 x^5} \]

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^3/x^6,x]

[Out]

-1/20*(b^3*x^3 + 2*b^2*x^2*ArcCoth[Tanh[a + b*x]] + 3*b*x*ArcCoth[Tanh[a + b*x]]^2 + 4*ArcCoth[Tanh[a + b*x]]^
3)/x^5

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.83

method	result	size
parallelrisch	\(-\frac {b^{3} x^{3}+2 b^{2} x^{2} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )+3 b x \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2}+4 \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{3}}{20 x^{5}}\)	\(53\)
risch	\(\text {Expression too large to display}\)	\(17234\)

[In]

int(arccoth(tanh(b*x+a))^3/x^6,x,method=_RETURNVERBOSE)

[Out]

-1/20*(b^3*x^3+2*b^2*x^2*arccoth(tanh(b*x+a))+3*b*x*arccoth(tanh(b*x+a))^2+4*arccoth(tanh(b*x+a))^3)/x^5

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.17 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^6} \, dx=-\frac {40 \, b^{3} x^{3} + 80 \, a b^{2} x^{2} + 60 \, a^{2} b x - 2 i \, \pi ^{3} - 3 \, \pi ^{2} {\left (5 \, b x + 4 \, a\right )} + 16 \, a^{3} + 4 i \, \pi {\left (10 \, b^{2} x^{2} + 15 \, a b x + 6 \, a^{2}\right )}}{80 \, x^{5}} \]

[In]

integrate(arccoth(tanh(b*x+a))^3/x^6,x, algorithm="fricas")

[Out]

-1/80*(40*b^3*x^3 + 80*a*b^2*x^2 + 60*a^2*b*x - 2*I*pi^3 - 3*pi^2*(5*b*x + 4*a) + 16*a^3 + 4*I*pi*(10*b^2*x^2
+ 15*a*b*x + 6*a^2))/x^5

Sympy [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.94 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^6} \, dx=- \frac {b^{3}}{20 x^{2}} - \frac {b^{2} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{10 x^{3}} - \frac {3 b \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{20 x^{4}} - \frac {\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{5 x^{5}} \]

[In]

integrate(acoth(tanh(b*x+a))**3/x**6,x)

[Out]

-b**3/(20*x**2) - b**2*acoth(tanh(a + b*x))/(10*x**3) - 3*b*acoth(tanh(a + b*x))**2/(20*x**4) - acoth(tanh(a +
b*x))**3/(5*x**5)

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.84 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^6} \, dx=-\frac {1}{20} \, b {\left (\frac {b^{2}}{x^{2}} + \frac {2 \, b \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )}{x^{3}}\right )} - \frac {3 \, b \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}{20 \, x^{4}} - \frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{3}}{5 \, x^{5}} \]

[In]

integrate(arccoth(tanh(b*x+a))^3/x^6,x, algorithm="maxima")

[Out]

-1/20*b*(b^2/x^2 + 2*b*arccoth(tanh(b*x + a))/x^3) - 3/20*b*arccoth(tanh(b*x + a))^2/x^4 - 1/5*arccoth(tanh(b*
x + a))^3/x^5

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.33 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.16 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^6} \, dx=-\frac {40 \, b^{3} x^{3} + 40 i \, \pi b^{2} x^{2} + 80 \, a b^{2} x^{2} - 15 \, \pi ^{2} b x + 60 i \, \pi a b x + 60 \, a^{2} b x - 2 i \, \pi ^{3} - 12 \, \pi ^{2} a + 24 i \, \pi a^{2} + 16 \, a^{3}}{80 \, x^{5}} \]

[In]

integrate(arccoth(tanh(b*x+a))^3/x^6,x, algorithm="giac")

[Out]

-1/80*(40*b^3*x^3 + 40*I*pi*b^2*x^2 + 80*a*b^2*x^2 - 15*pi^2*b*x + 60*I*pi*a*b*x + 60*a^2*b*x - 2*I*pi^3 - 12*
pi^2*a + 24*I*pi*a^2 + 16*a^3)/x^5

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.83 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^6} \, dx=-\frac {{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3}{5\,x^5}-\frac {b^3}{20\,x^2}-\frac {b^2\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{10\,x^3}-\frac {3\,b\,{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{20\,x^4} \]

[In]

int(acoth(tanh(a + b*x))^3/x^6,x)

[Out]

- acoth(tanh(a + b*x))^3/(5*x^5) - b^3/(20*x^2) - (b^2*acoth(tanh(a + b*x)))/(10*x^3) - (3*b*acoth(tanh(a + b*
x))^2)/(20*x^4)