\(\int \frac {x}{\coth ^{-1}(\tanh (a+b x))^3} \, dx\) [179]

Optimal result

Integrand size = 11, antiderivative size = 34 \[ \int \frac {x}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=-\frac {x}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac {1}{2 b^2 \coth ^{-1}(\tanh (a+b x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2199, 2188, 30} \[ \int \frac {x}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=-\frac {1}{2 b^2 \coth ^{-1}(\tanh (a+b x))}-\frac {x}{2 b \coth ^{-1}(\tanh (a+b x))^2} \]

[In]

Int[x/ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2188

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, 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 {x}{2 b \coth ^{-1}(\tanh (a+b x))^2}+\frac {\int \frac {1}{\coth ^{-1}(\tanh (a+b x))^2} \, dx}{2 b} \\ & = -\frac {x}{2 b \coth ^{-1}(\tanh (a+b x))^2}+\frac {\text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{2 b^2} \\ & = -\frac {x}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac {1}{2 b^2 \coth ^{-1}(\tanh (a+b x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int \frac {x}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=-\frac {b x+\coth ^{-1}(\tanh (a+b x))}{2 b^2 \coth ^{-1}(\tanh (a+b x))^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76

method	result	size
parallelrisch	\(-\frac {b x +\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}{2 b^{2} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2}}\)	\(26\)
risch	\(-\frac {2 i \left (\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+2 \pi \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}-2 \pi \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )-2 \pi \,\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+\pi \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+2 \pi +4 i \ln \left ({\mathrm e}^{b x +a}\right )+4 i b x \right )}{b^{2} {\left (\pi \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-2 \pi \,\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )+\pi \operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )+\pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}-\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+2 \pi \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}-2 \pi \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+4 i \ln \left ({\mathrm e}^{b x +a}\right )+2 \pi \right )}^{2}}\)	\(634\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.76 \[ \int \frac {x}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=\frac {-i \, \pi - 4 \, b x - 2 \, a}{4 \, b^{4} x^{2} + 8 \, a b^{3} x - \pi ^{2} b^{2} + 4 \, a^{2} b^{2} + 4 i \, \pi {\left (b^{3} x + a b^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 24.81 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.24 \[ \int \frac {x}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=\begin {cases} - \frac {x}{2 b \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}} - \frac {1}{2 b^{2} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}} & \text {for}\: b \neq 0 \\\frac {x^{2}}{2 \operatorname {acoth}^{3}{\left (\tanh {\left (a \right )} \right )}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-x/(2*b*acoth(tanh(a + b*x))**2) - 1/(2*b**2*acoth(tanh(a + b*x))), Ne(b, 0)), (x**2/(2*acoth(tanh(
a))**3), True))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.77 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.85 \[ \int \frac {x}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=-\frac {-i \, \pi + 4 \, b x + 2 \, a}{4 \, b^{4} x^{2} - \pi ^{2} b^{2} - 4 i \, \pi a b^{2} + 4 \, a^{2} b^{2} - 4 \, {\left (i \, \pi b^{3} - 2 \, a b^{3}\right )} x} \]

[In]

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

[Out]

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.79 \[ \int \frac {x}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=-\frac {i \, \pi + 4 \, b x + 2 \, a}{4 \, b^{4} x^{2} + 4 i \, \pi b^{3} x + 8 \, a b^{3} x - \pi ^{2} b^{2} + 4 i \, \pi a b^{2} + 4 \, a^{2} b^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74 \[ \int \frac {x}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=-\frac {\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )+b\,x}{2\,b^2\,{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2} \]

[In]

int(x/acoth(tanh(a + b*x))^3,x)

[Out]

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