\(\int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^2} \, dx\) [153]

Optimal result

Integrand size = 13, antiderivative size = 68 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^2} \, dx=-3 b^2 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )+\frac {3}{2} b \coth ^{-1}(\tanh (a+b x))^2-\frac {\coth ^{-1}(\tanh (a+b x))^3}{x}+3 b \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2199, 2190, 2189, 29} \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^2} \, dx=-3 b^2 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )-\frac {\coth ^{-1}(\tanh (a+b x))^3}{x}+\frac {3}{2} b \coth ^{-1}(\tanh (a+b x))^2+3 b \log (x) \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \]

[In]

Int[ArcCoth[Tanh[a + b*x]]^3/x^2,x]

[Out]

-3*b^2*x*(b*x - ArcCoth[Tanh[a + b*x]]) + (3*b*ArcCoth[Tanh[a + b*x]]^2)/2 - ArcCoth[Tanh[a + b*x]]^3/x + 3*b*
(b*x - ArcCoth[Tanh[a + b*x]])^2*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 2190

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^n/(a*n), x] - Dis
t[(b*u - a*v)/a, Int[v^(n - 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && GtQ[n, 0] && Ne
Q[n, 1]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.91 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^2} \, dx=-\frac {\coth ^{-1}(\tanh (a+b x))^3}{x}-6 b^2 x \coth ^{-1}(\tanh (a+b x)) \log (x)+3 b \coth ^{-1}(\tanh (a+b x))^2 (1+\log (x))+\frac {3}{2} b^3 x^2 (-1+2 \log (x)) \]

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^3/x^2,x]

[Out]

-(ArcCoth[Tanh[a + b*x]]^3/x) - 6*b^2*x*ArcCoth[Tanh[a + b*x]]*Log[x] + 3*b*ArcCoth[Tanh[a + b*x]]^2*(1 + Log[
x]) + (3*b^3*x^2*(-1 + 2*Log[x]))/2

Maple [C] (warning: unable to verify)

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

Time = 0.26 (sec) , antiderivative size = 3248, normalized size of antiderivative = 47.76

method	result	size
risch	\(\text {Expression too large to display}\)	\(3248\)

[In]

int(arccoth(tanh(b*x+a))^3/x^2,x,method=_RETURNVERBOSE)

[Out]

-1/x*ln(exp(b*x+a))^3+3*ln(x)*ln(exp(b*x+a))^2*b+3*b^3*x^2*ln(x)-9/2*b^3*x^2-6*b^2*ln(x)*ln(exp(b*x+a))*x+6*b^
2*ln(exp(b*x+a))*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*ex
p(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*csg
n(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*csg
n(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*csgn(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*csg
n(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*csgn(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*ex
p(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*Pi^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/(exp(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/(e
xp(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/(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))^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))*csgn(I*exp(2*b*x+2*a))^3*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1)))*(-1/x*ln(exp(b*x
+a))+b*ln(x))-1/64*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-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/x*ln(exp(b*x+a))^2+2*b*(ln(x)*ln(exp(
b*x+a))-b*(x*ln(x)-x)))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.16 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^2} \, dx=\frac {4 \, b^{3} x^{3} + 24 \, a b^{2} x^{2} + i \, \pi ^{3} + 6 \, \pi ^{2} a - 8 \, a^{3} + 12 i \, \pi {\left (b^{2} x^{2} - a^{2}\right )} - 6 \, {\left (\pi ^{2} b x - 4 i \, \pi a b x - 4 \, a^{2} b x\right )} \log \left (x\right )}{8 \, x} \]

[In]

integrate(arccoth(tanh(b*x+a))^3/x^2,x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.43 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.82 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^2} \, dx=3 \, b \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2} \log \left (x\right ) - \frac {3}{2} \, {\left (2 \, \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2} \log \left (x\right ) - {\left (b x^{2} - 2 \, {\left (-i \, \pi - 2 \, a\right )} x + 2 \, {\left (-\frac {i \, \pi {\left (b x + a\right )}}{b} - \frac {{\left (b x + a\right )}^{2}}{b}\right )} \log \left (x\right ) + \frac {2 \, \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2} \log \left (x\right )}{b} + \frac {2 \, {\left (i \, \pi a + a^{2}\right )} \log \left (x\right )}{b}\right )} b\right )} b - \frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{3}}{x} \]

[In]

integrate(arccoth(tanh(b*x+a))^3/x^2,x, algorithm="maxima")

[Out]

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.09 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^2} \, dx=\frac {1}{2} \, b^{3} x^{2} - \frac {3}{2} \, {\left (-i \, \pi b^{2} - 2 \, a b^{2}\right )} x - \frac {3}{4} \, {\left (\pi ^{2} b - 4 i \, \pi a b - 4 \, a^{2} b\right )} \log \left (x\right ) - \frac {-i \, \pi ^{3} - 6 \, \pi ^{2} a + 12 i \, \pi a^{2} + 8 \, a^{3}}{8 \, x} \]

[In]

integrate(arccoth(tanh(b*x+a))^3/x^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 3.88 (sec) , antiderivative size = 372, normalized size of antiderivative = 5.47 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^2} \, dx=\ln \left (x\right )\,\left (3\,a^2\,b+\frac {3\,b\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2}{4}-3\,a\,b\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )\right )+\frac {{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^3-8\,a^3-6\,a\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2+12\,a^2\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}{8\,x}+\frac {b^3\,x^2}{2}-\frac {3\,b^2\,x\,\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}{2} \]

[In]

int(acoth(tanh(a + b*x))^3/x^2,x)

[Out]

log(x)*(3*a^2*b + (3*b*(2*a - 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*b*x)^2)/4 - 3*a*b*(2*a - 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*b*x)) + ((2*a - 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*b*x)^3 - 8*a^3 - 6*a*(2*a - 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*b*x)^2 + 12*a^2*(2*a - 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*b*x))/(8*x) + (b^3*x^2)/2 - (3*b^2*x*(log(-2/(exp(2*a)*exp(2*b*
x) - 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x))/2