\(\int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^5} \, dx\) [156]

Optimal result

Integrand size = 13, antiderivative size = 31 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^5} \, dx=\frac {\coth ^{-1}(\tanh (a+b x))^4}{4 x^4 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )} \]

[Out]

1/4*arccoth(tanh(b*x+a))^4/x^4/(b*x-arccoth(tanh(b*x+a)))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2198} \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^5} \, dx=\frac {\coth ^{-1}(\tanh (a+b x))^4}{4 x^4 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )} \]

[In]

Int[ArcCoth[Tanh[a + b*x]]^3/x^5,x]

[Out]

ArcCoth[Tanh[a + b*x]]^4/(4*x^4*(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]

Rubi steps \begin{align*} \text {integral}& = \frac {\coth ^{-1}(\tanh (a+b x))^4}{4 x^4 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.61 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^5} \, dx=-\frac {b^3 x^3+b^2 x^2 \coth ^{-1}(\tanh (a+b x))+b x \coth ^{-1}(\tanh (a+b x))^2+\coth ^{-1}(\tanh (a+b x))^3}{4 x^4} \]

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^3/x^5,x]

[Out]

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

Maple [A] (verified)

Time = 1.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58

method	result	size
parallelrisch	\(-\frac {b^{3} x^{3}+b^{2} x^{2} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )+b x \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2}+\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{3}}{4 x^{4}}\)	\(49\)
risch	\(\text {Expression too large to display}\)	\(17235\)

[In]

int(arccoth(tanh(b*x+a))^3/x^5,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.42 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^5} \, dx=-\frac {32 \, b^{3} x^{3} + 48 \, a b^{2} x^{2} + 32 \, a^{2} b x - i \, \pi ^{3} - 2 \, \pi ^{2} {\left (4 \, b x + 3 \, a\right )} + 8 \, a^{3} + 4 i \, \pi {\left (6 \, b^{2} x^{2} + 8 \, a b x + 3 \, a^{2}\right )}}{32 \, x^{4}} \]

[In]

integrate(arccoth(tanh(b*x+a))^3/x^5,x, algorithm="fricas")

[Out]

-1/32*(32*b^3*x^3 + 48*a*b^2*x^2 + 32*a^2*b*x - I*pi^3 - 2*pi^2*(4*b*x + 3*a) + 8*a^3 + 4*I*pi*(6*b^2*x^2 + 8*
a*b*x + 3*a^2))/x^4

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (26) = 52\).

Time = 0.29 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.81 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^5} \, dx=- \frac {b^{3}}{4 x} - \frac {b^{2} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{4 x^{2}} - \frac {b \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{4 x^{3}} - \frac {\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{4 x^{4}} \]

[In]

integrate(acoth(tanh(b*x+a))**3/x**5,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.71 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^5} \, dx=-\frac {1}{4} \, b {\left (\frac {b^{2}}{x} + \frac {b \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )}{x^{2}}\right )} - \frac {b \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}{4 \, x^{3}} - \frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{3}}{4 \, x^{4}} \]

[In]

integrate(arccoth(tanh(b*x+a))^3/x^5,x, algorithm="maxima")

[Out]

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 74, normalized size of antiderivative = 2.39 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^5} \, dx=-\frac {32 \, b^{3} x^{3} + 24 i \, \pi b^{2} x^{2} + 48 \, a b^{2} x^{2} - 8 \, \pi ^{2} b x + 32 i \, \pi a b x + 32 \, a^{2} b x - i \, \pi ^{3} - 6 \, \pi ^{2} a + 12 i \, \pi a^{2} + 8 \, a^{3}}{32 \, x^{4}} \]

[In]

integrate(arccoth(tanh(b*x+a))^3/x^5,x, algorithm="giac")

[Out]

-1/32*(32*b^3*x^3 + 24*I*pi*b^2*x^2 + 48*a*b^2*x^2 - 8*pi^2*b*x + 32*I*pi*a*b*x + 32*a^2*b*x - I*pi^3 - 6*pi^2
*a + 12*I*pi*a^2 + 8*a^3)/x^4

Mupad [B] (verification not implemented)

Time = 4.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.55 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^5} \, dx=-\frac {b^3\,x^3+b^2\,x^2\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )+b\,x\,{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2+{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3}{4\,x^4} \]

[In]

int(acoth(tanh(a + b*x))^3/x^5,x)

[Out]

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