\(\int x \coth ^{-1}(\tanh (a+b x)) \, dx\) [130]

Optimal result

Integrand size = 9, antiderivative size = 23 \[ \int x \coth ^{-1}(\tanh (a+b x)) \, dx=-\frac {b x^3}{6}+\frac {1}{2} x^2 \coth ^{-1}(\tanh (a+b x)) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6375, 30} \[ \int x \coth ^{-1}(\tanh (a+b x)) \, dx=\frac {1}{2} x^2 \coth ^{-1}(\tanh (a+b x))-\frac {b x^3}{6} \]

[In]

Int[x*ArcCoth[Tanh[a + b*x]],x]

[Out]

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

Rule 30

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

Rule 6375

Int[ArcCoth[(c_.) + (d_.)*Tanh[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^(m
+ 1)*(ArcCoth[c + d*Tanh[a + b*x]]/(f*(m + 1))), x] + Dist[b/(f*(m + 1)), Int[(e + f*x)^(m + 1)/(c - d + c*E^
(2*a + 2*b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && EqQ[(c - d)^2, 1]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} x^2 \coth ^{-1}(\tanh (a+b x))-\frac {1}{2} b \int x^2 \, dx \\ & = -\frac {b x^3}{6}+\frac {1}{2} x^2 \coth ^{-1}(\tanh (a+b x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int x \coth ^{-1}(\tanh (a+b x)) \, dx=-\frac {1}{6} x^2 \left (b x-3 \coth ^{-1}(\tanh (a+b x))\right ) \]

[In]

Integrate[x*ArcCoth[Tanh[a + b*x]],x]

[Out]

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

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
default	\(-\frac {b \,x^{3}}{6}+\frac {x^{2} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}{2}\)	\(20\)
parallelrisch	\(-\frac {b \,x^{3}}{6}+\frac {x^{2} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}{2}\)	\(20\)
parts	\(-\frac {b \,x^{3}}{6}+\frac {x^{2} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}{2}\)	\(20\)
risch	\(\frac {x^{2} \ln \left ({\mathrm e}^{b x +a}\right )}{2}-\frac {b \,x^{3}}{6}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )}{8}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}}{4}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}}{8}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )}{8}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}}{4}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}}{4}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}}{8}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}}{8}-\frac {i \pi \,x^{2}}{4}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}}{8}\)	\(363\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int x \coth ^{-1}(\tanh (a+b x)) \, dx=\frac {1}{3} \, b x^{3} + \frac {1}{4} i \, \pi x^{2} + \frac {1}{2} \, a x^{2} \]

[In]

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

[Out]

1/3*b*x^3 + 1/4*I*pi*x^2 + 1/2*a*x^2

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (19) = 38\).

Time = 0.16 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70 \[ \int x \coth ^{-1}(\tanh (a+b x)) \, dx=\begin {cases} \frac {x \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{2 b} - \frac {\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{6 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \operatorname {acoth}{\left (\tanh {\left (a \right )} \right )}}{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x*acoth(tanh(a + b*x))**2/(2*b) - acoth(tanh(a + b*x))**3/(6*b**2), Ne(b, 0)), (x**2*acoth(tanh(a))
/2, True))

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int x \coth ^{-1}(\tanh (a+b x)) \, dx=-\frac {1}{6} \, b x^{3} + \frac {1}{2} \, x^{2} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right ) \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (19) = 38\).

Time = 0.28 (sec) , antiderivative size = 71, normalized size of antiderivative = 3.09 \[ \int x \coth ^{-1}(\tanh (a+b x)) \, dx=-\frac {1}{6} \, b x^{3} + \frac {1}{4} \, x^{2} \log \left (-\frac {\frac {e^{\left (2 \, b x + 2 \, a\right )} + 1}{e^{\left (2 \, b x + 2 \, a\right )} - 1} + 1}{\frac {e^{\left (2 \, b x + 2 \, a\right )} + 1}{e^{\left (2 \, b x + 2 \, a\right )} - 1} - 1}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int x \coth ^{-1}(\tanh (a+b x)) \, dx=\frac {x^2\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{2}-\frac {b\,x^3}{6} \]

[In]

int(x*acoth(tanh(a + b*x)),x)

[Out]

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