\(\int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^3} \, dx\) [154]

Optimal result

Integrand size = 13, antiderivative size = 60 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^3} \, dx=3 b^3 x-\frac {3 b \coth ^{-1}(\tanh (a+b x))^2}{2 x}-\frac {\coth ^{-1}(\tanh (a+b x))^3}{2 x^2}-3 b^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2199, 2189, 29} \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^3} \, dx=-3 b^2 \log (x) \left (b x-\coth ^{-1}(\tanh (a+b x))\right )-\frac {\coth ^{-1}(\tanh (a+b x))^3}{2 x^2}-\frac {3 b \coth ^{-1}(\tanh (a+b x))^2}{2 x}+3 b^3 x \]

[In]

Int[ArcCoth[Tanh[a + b*x]]^3/x^3,x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2189

Int[(v_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[b*(x/a), x] - Dist[(b*u
- a*v)/a, Int[1/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x]

Rule 2199

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[u^(m + 1)*(v^
n/(a*(m + 1))), x] - Dist[b*(n/(a*(m + 1))), Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m
, n}, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0] &&  !(ILtQ[m + n, -2] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

Rubi steps \begin{align*} \text {integral}& = -\frac {\coth ^{-1}(\tanh (a+b x))^3}{2 x^2}+\frac {1}{2} (3 b) \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^2} \, dx \\ & = -\frac {3 b \coth ^{-1}(\tanh (a+b x))^2}{2 x}-\frac {\coth ^{-1}(\tanh (a+b x))^3}{2 x^2}+\left (3 b^2\right ) \int \frac {\coth ^{-1}(\tanh (a+b x))}{x} \, dx \\ & = 3 b^3 x-\frac {3 b \coth ^{-1}(\tanh (a+b x))^2}{2 x}-\frac {\coth ^{-1}(\tanh (a+b x))^3}{2 x^2}-\left (3 b^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )\right ) \int \frac {1}{x} \, dx \\ & = 3 b^3 x-\frac {3 b \coth ^{-1}(\tanh (a+b x))^2}{2 x}-\frac {\coth ^{-1}(\tanh (a+b x))^3}{2 x^2}-3 b^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.10 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^3} \, dx=b^3 x-\frac {3 b \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2}{x}-\frac {\left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^3}{2 x^2}+3 b^2 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right ) \log (x) \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.28 (sec) , antiderivative size = 3227, normalized size of antiderivative = 53.78

method	result	size
risch	\(\text {Expression too large to display}\)	\(3227\)

[In]

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

[Out]

-1/2/x^2*ln(exp(b*x+a))^3-3/2/x*ln(exp(b*x+a))^2*b-3*ln(x)*x*b^3+3*ln(exp(b*x+a))*ln(x)*b^2+3*b^3*x+(-3/4*Pi^2
*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+
1))^2+3/4*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))^2*csgn(I*exp(2*b*x+2*a))^3-3/2*Pi^2*csgn(I*exp(b*x+a))*csgn(I*exp(2*
b*x+2*a))^2*csgn(I/(exp(2*b*x+2*a)+1))^2-3/4*Pi^2-3/4*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))^4*csgn(I*exp(2*b*x+2*a))
*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))-3/4*Pi^2*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*
a)+1))^2*csgn(I/(exp(2*b*x+2*a)+1))^2+3/4*Pi^2*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))*csgn(I/(exp(2*b*x+2
*a)+1))^2+3/8*Pi^2*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^5-3/4*csgn(I*exp(2*b*x+2*a
))*csgn(I*exp(b*x+a))^2*Pi^2+3/2*csgn(I*exp(2*b*x+2*a))^2*csgn(I*exp(b*x+a))*Pi^2-3/4*Pi^2*csgn(I/(exp(2*b*x+2
*a)+1))*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^4-3/8*Pi^2*csgn(I*exp(b*x+a))^2*csgn(
I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+3/4*Pi^2*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a
))^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3-3/16*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))^2*csgn(I*exp(2*b*x+2*a)/
(exp(2*b*x+2*a)+1))^4+3/8*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^5-3/16*Pi^
2*csgn(I*exp(b*x+a))^4*csgn(I*exp(2*b*x+2*a))^2+3/4*Pi^2*csgn(I*exp(b*x+a))^3*csgn(I*exp(2*b*x+2*a))^3-3/4*Pi^
2*csgn(I/(exp(2*b*x+2*a)+1))^3*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+3/8*Pi^2*csgn(I*exp(2*b*x+2*a))^4*c
sgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-3/16*Pi^2*csgn(I*exp(2*b*x+2*a))^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x
+2*a)+1))^4+3/4*Pi^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3*csgn(I/(exp(2*b*x+2*a)+1))^2-3/4*Pi^2*csgn(I*
exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2*csgn(I/(exp(2*b*x+2*a)+1))^3+3/4*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))^4*csgn(I
*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+3/4*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a
)+1))^2-3/4*Pi^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+3/2*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))^5-9/8*Pi^2*cs
gn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))^4+3/4*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))^3*csgn(I*exp(2*b*x+2*a))*csgn(
I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))-3/4*Pi^2*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))*csgn(I/(exp(2*b*x+2*
a)+1))^3+3/4*Pi^2*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-3/4*Pi^2*csgn(I*exp(2*b*x
+2*a))^3*csgn(I/(exp(2*b*x+2*a)+1))^3-3/8*Pi^2*csgn(I*exp(2*b*x+2*a))^3*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+
1))^3+3/8*Pi^2*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-3/4*P
i^2*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^3*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-3/4*Pi^2*csgn(I/(e
xp(2*b*x+2*a)+1))^6-3/16*Pi^2*csgn(I*exp(2*b*x+2*a))^6-3/4*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))^4-3/16*Pi^2*csgn(I*
exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^6-3/2*csgn(I/(exp(2*b*x+2*a)+1))^3*Pi^2-3/8*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))
*csgn(I*exp(2*b*x+2*a))^4*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))+3/8*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I
*exp(2*b*x+2*a))^3*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+3/8*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(
2*b*x+2*a))^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3-3/4*csgn(I*exp(2*b*x+2*a))^3*Pi^2+3/4*Pi^2*csgn(I/(e
xp(2*b*x+2*a)+1))^3*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+3/4*Pi^2*csgn(I*exp(b*x
+a))*csgn(I*exp(2*b*x+2*a))^5+3/2*csgn(I/(exp(2*b*x+2*a)+1))^2*Pi^2+3/2*Pi^2*csgn(I*exp(b*x+a))*csgn(I*exp(2*b
*x+2*a))^2*csgn(I/(exp(2*b*x+2*a)+1))^3-3/4*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(
2*b*x+2*a)/(exp(2*b*x+2*a)+1))-3/16*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))^2*csgn(I*exp(2*b*x+2*a))^2*csgn(I*exp(2*b*
x+2*a)/(exp(2*b*x+2*a)+1))^2+3/8*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))^2*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a
)/(exp(2*b*x+2*a)+1))^3-3/8*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))^2*csgn
(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))+3/8*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*
x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+3/4*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(b*x+a))*csg
n(I*exp(2*b*x+2*a))^3*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1)))*(-1/2/x^2*ln(exp(b*x+a))-1/2*b/x)-1/128*I*Pi^
3*(csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))-csgn(I/(exp(2*b
*x+2*a)+1))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+2*csgn(I/(exp(2*b*x+2*a)+1))^3-2*csgn(I/(exp(2*b*x+2*a
)+1))^2+csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-2*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2+csgn(I*exp(2
*b*x+2*a))^3-csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+csgn(I*exp(2*b*x+2*a)/(exp(2*b
*x+2*a)+1))^3+2)^3/x^2-3/4*I*Pi*(csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(
2*b*x+2*a)+1))-csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+2*csgn(I/(exp(2*b*x+2*a)
+1))^3-2*csgn(I/(exp(2*b*x+2*a)+1))^2+csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-2*csgn(I*exp(b*x+a))*csgn(I*
exp(2*b*x+2*a))^2+csgn(I*exp(2*b*x+2*a))^3-csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+
csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+2)*(-1/2/x^2*ln(exp(b*x+a))^2+b*(-1/x*ln(exp(b*x+a))+b*ln(x)))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.28 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^3} \, dx=\frac {16 \, b^{3} x^{3} - 48 \, a^{2} b x + i \, \pi ^{3} + 6 \, \pi ^{2} {\left (2 \, b x + a\right )} - 8 \, a^{3} - 12 i \, \pi {\left (4 \, a b x + a^{2}\right )} - 24 \, {\left (-i \, \pi b^{2} x^{2} - 2 \, a b^{2} x^{2}\right )} \log \left (x\right )}{16 \, x^{2}} \]

[In]

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

[Out]

1/16*(16*b^3*x^3 - 48*a^2*b*x + I*pi^3 + 6*pi^2*(2*b*x + a) - 8*a^3 - 12*I*pi*(4*a*b*x + a^2) - 24*(-I*pi*b^2*
x^2 - 2*a*b^2*x^2)*log(x))/x^2

Sympy [F]

\[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^3} \, dx=\int \frac {\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{3}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.20 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^3} \, dx=3 \, {\left (b \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right ) \log \left (x\right ) - {\left (b {\left (x + \frac {a}{b}\right )} \log \left (x\right ) - b {\left (x + \frac {a \log \left (x\right )}{b}\right )}\right )} b\right )} b - \frac {3 \, b \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}{2 \, x} - \frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{3}}{2 \, x^{2}} \]

[In]

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

[Out]

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.18 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^3} \, dx=b^{3} x + \frac {3}{2} \, {\left (i \, \pi b^{2} + 2 \, a b^{2}\right )} \log \left (x\right ) + \frac {12 \, \pi ^{2} b x - 48 i \, \pi a b x - 48 \, a^{2} b x + i \, \pi ^{3} + 6 \, \pi ^{2} a - 12 i \, \pi a^{2} - 8 \, a^{3}}{16 \, x^{2}} \]

[In]

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

[Out]

b^3*x + 3/2*(I*pi*b^2 + 2*a*b^2)*log(x) + 1/16*(12*pi^2*b*x - 48*I*pi*a*b*x - 48*a^2*b*x + I*pi^3 + 6*pi^2*a -
12*I*pi*a^2 - 8*a^3)/x^2

Mupad [B] (verification not implemented)

Time = 3.99 (sec) , antiderivative size = 383, normalized size of antiderivative = 6.38 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^3} \, dx=\frac {{\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^3}{16\,x^2}-\frac {{\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^3}{16\,x^2}+\frac {9\,b^2\,\ln \left (\frac {{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{4}-\frac {9\,b^2\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{4}-\frac {3\,b^3\,x}{2}-\frac {3\,b\,{\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2}{8\,x}+\frac {3\,b^2\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,\ln \left (x\right )}{2}-3\,b^3\,x\,\ln \left (x\right )-\frac {3\,b\,{\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2}{8\,x}-\frac {3\,b^2\,\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,\ln \left (x\right )}{2}-\frac {3\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,{\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2}{16\,x^2}+\frac {3\,{\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2\,\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{16\,x^2}+\frac {3\,b\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{4\,x} \]

[In]

int(acoth(tanh(a + b*x))^3/x^3,x)

[Out]

log(-2/(exp(2*a)*exp(2*b*x) - 1))^3/(16*x^2) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))^3/(16*x^
2) + (9*b^2*log(exp(2*b*x)/(exp(2*a)*exp(2*b*x) - 1)))/4 - (9*b^2*log(1/(exp(2*a)*exp(2*b*x) - 1)))/4 - (3*b^3
*x)/2 - (3*b*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))^2)/(8*x) + (3*b^2*log((2*exp(2*a)*exp(2*b*
x))/(exp(2*a)*exp(2*b*x) - 1))*log(x))/2 - 3*b^3*x*log(x) - (3*b*log(-2/(exp(2*a)*exp(2*b*x) - 1))^2)/(8*x) -
(3*b^2*log(-2/(exp(2*a)*exp(2*b*x) - 1))*log(x))/2 - (3*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))
*log(-2/(exp(2*a)*exp(2*b*x) - 1))^2)/(16*x^2) + (3*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))^2*l
og(-2/(exp(2*a)*exp(2*b*x) - 1)))/(16*x^2) + (3*b*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))*log(-
2/(exp(2*a)*exp(2*b*x) - 1)))/(4*x)