\(\int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^4} \, dx\) [144]

Optimal result

Integrand size = 13, antiderivative size = 31 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^4} \, dx=\frac {\coth ^{-1}(\tanh (a+b x))^3}{3 x^3 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )} \]

[Out]

1/3*arccoth(tanh(b*x+a))^3/x^3/(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} \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^4} \, dx=\frac {\coth ^{-1}(\tanh (a+b x))^3}{3 x^3 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )} \]

[In]

Int[ArcCoth[Tanh[a + b*x]]^2/x^4,x]

[Out]

ArcCoth[Tanh[a + b*x]]^3/(3*x^3*(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))^3}{3 x^3 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^4} \, dx=-\frac {b^2 x^2+b x \coth ^{-1}(\tanh (a+b x))+\coth ^{-1}(\tanh (a+b x))^2}{3 x^3} \]

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^2/x^4,x]

[Out]

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

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06

method	result	size
parallelrisch	\(-\frac {b^{2} x^{2}+b x \,\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )+\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2}}{3 x^{3}}\)	\(33\)
risch	\(\text {Expression too large to display}\)	\(3217\)

[In]

int(arccoth(tanh(b*x+a))^2/x^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.29 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^4} \, dx=-\frac {12 \, b^{2} x^{2} + 12 \, a b x - \pi ^{2} + 2 i \, \pi {\left (3 \, b x + 2 \, a\right )} + 4 \, a^{2}}{12 \, x^{3}} \]

[In]

integrate(arccoth(tanh(b*x+a))^2/x^4,x, algorithm="fricas")

[Out]

-1/12*(12*b^2*x^2 + 12*a*b*x - pi^2 + 2*I*pi*(3*b*x + 2*a) + 4*a^2)/x^3

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^4} \, dx=- \frac {b^{2}}{3 x} - \frac {b \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{3 x^{2}} - \frac {\operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{3 x^{3}} \]

[In]

integrate(acoth(tanh(b*x+a))**2/x**4,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^4} \, dx=-\frac {b^{2}}{3 \, x} - \frac {b \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )}{3 \, x^{2}} - \frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}{3 \, x^{3}} \]

[In]

integrate(arccoth(tanh(b*x+a))^2/x^4,x, algorithm="maxima")

[Out]

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^4} \, dx=-\frac {12 \, b^{2} x^{2} + 6 i \, \pi b x + 12 \, a b x - \pi ^{2} + 4 i \, \pi a + 4 \, a^{2}}{12 \, x^{3}} \]

[In]

integrate(arccoth(tanh(b*x+a))^2/x^4,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 3.81 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^4} \, dx=-\frac {b^2\,x^2+b\,x\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )+{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{3\,x^3} \]

[In]

int(acoth(tanh(a + b*x))^2/x^4,x)

[Out]

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