\(\int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^5} \, dx\) [145]

Optimal result

Integrand size = 13, antiderivative size = 64 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^5} \, dx=\frac {b \coth ^{-1}(\tanh (a+b x))^3}{12 x^3 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac {\coth ^{-1}(\tanh (a+b x))^3}{4 x^4 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )} \]

[Out]

1/12*b*arccoth(tanh(b*x+a))^3/x^3/(b*x-arccoth(tanh(b*x+a)))^2+1/4*arccoth(tanh(b*x+a))^3/x^4/(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} \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^5} \, dx=\frac {\coth ^{-1}(\tanh (a+b x))^3}{4 x^4 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}+\frac {b \coth ^{-1}(\tanh (a+b x))^3}{12 x^3 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2} \]

[In]

Int[ArcCoth[Tanh[a + b*x]]^2/x^5,x]

[Out]

(b*ArcCoth[Tanh[a + b*x]]^3)/(12*x^3*(b*x - ArcCoth[Tanh[a + b*x]])^2) + ArcCoth[Tanh[a + b*x]]^3/(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]

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))^3}{4 x^4 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}+\frac {b \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^4} \, dx}{4 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )} \\ & = \frac {b \coth ^{-1}(\tanh (a+b x))^3}{12 x^3 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac {\coth ^{-1}(\tanh (a+b x))^3}{4 x^4 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.58 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^5} \, dx=-\frac {b^2 x^2+2 b x \coth ^{-1}(\tanh (a+b x))+3 \coth ^{-1}(\tanh (a+b x))^2}{12 x^4} \]

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^2/x^5,x]

[Out]

-1/12*(b^2*x^2 + 2*b*x*ArcCoth[Tanh[a + b*x]] + 3*ArcCoth[Tanh[a + b*x]]^2)/x^4

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.56

method	result	size
parallelrisch	\(-\frac {b^{2} x^{2}+2 b x \,\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )+3 \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2}}{12 x^{4}}\)	\(36\)
risch	\(\text {Expression too large to display}\)	\(3217\)

[In]

int(arccoth(tanh(b*x+a))^2/x^5,x,method=_RETURNVERBOSE)

[Out]

-1/12*(b^2*x^2+2*b*x*arccoth(tanh(b*x+a))+3*arccoth(tanh(b*x+a))^2)/x^4

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.62 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^5} \, dx=-\frac {24 \, b^{2} x^{2} + 32 \, a b x - 3 \, \pi ^{2} + 4 i \, \pi {\left (4 \, b x + 3 \, a\right )} + 12 \, a^{2}}{48 \, x^{4}} \]

[In]

integrate(arccoth(tanh(b*x+a))^2/x^5,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.61 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^5} \, dx=- \frac {b^{2}}{12 x^{2}} - \frac {b \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{6 x^{3}} - \frac {\operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{4 x^{4}} \]

[In]

integrate(acoth(tanh(b*x+a))**2/x**5,x)

[Out]

-b**2/(12*x**2) - b*acoth(tanh(a + b*x))/(6*x**3) - acoth(tanh(a + b*x))**2/(4*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.56 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^5} \, dx=-\frac {b^{2}}{12 \, x^{2}} - \frac {b \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )}{6 \, x^{3}} - \frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}{4 \, x^{4}} \]

[In]

integrate(arccoth(tanh(b*x+a))^2/x^5,x, algorithm="maxima")

[Out]

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.59 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^5} \, dx=-\frac {24 \, b^{2} x^{2} + 16 i \, \pi b x + 32 \, a b x - 3 \, \pi ^{2} + 12 i \, \pi a + 12 \, a^{2}}{48 \, x^{4}} \]

[In]

integrate(arccoth(tanh(b*x+a))^2/x^5,x, algorithm="giac")

[Out]

-1/48*(24*b^2*x^2 + 16*I*pi*b*x + 32*a*b*x - 3*pi^2 + 12*I*pi*a + 12*a^2)/x^4

Mupad [B] (verification not implemented)

Time = 3.81 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.56 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^5} \, dx=-\frac {{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{4\,x^4}-\frac {b^2}{12\,x^2}-\frac {b\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{6\,x^3} \]

[In]

int(acoth(tanh(a + b*x))^2/x^5,x)

[Out]

- acoth(tanh(a + b*x))^2/(4*x^4) - b^2/(12*x^2) - (b*acoth(tanh(a + b*x)))/(6*x^3)