\(\int x \coth ^{-1}(\tanh (a+b x))^2 \, dx\) [139]

Optimal result

Integrand size = 11, antiderivative size = 34 \[ \int x \coth ^{-1}(\tanh (a+b x))^2 \, dx=\frac {x \coth ^{-1}(\tanh (a+b x))^3}{3 b}-\frac {\coth ^{-1}(\tanh (a+b x))^4}{12 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (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} \[ \int x \coth ^{-1}(\tanh (a+b x))^2 \, dx=\frac {x \coth ^{-1}(\tanh (a+b x))^3}{3 b}-\frac {\coth ^{-1}(\tanh (a+b x))^4}{12 b^2} \]

[In]

Int[x*ArcCoth[Tanh[a + b*x]]^2,x]

[Out]

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

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(74\) vs. \(2(34)=68\).

Time = 0.16 (sec) , antiderivative size = 74, normalized size of antiderivative = 2.18 \[ \int x \coth ^{-1}(\tanh (a+b x))^2 \, dx=\frac {(a+b x) \left (-\left ((3 a-b x) (a+b x)^2\right )+4 \left (2 a^2+a b x-b^2 x^2\right ) \coth ^{-1}(\tanh (a+b x))-6 (a-b x) \coth ^{-1}(\tanh (a+b x))^2\right )}{12 b^2} \]

[In]

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

[Out]

((a + b*x)*(-((3*a - b*x)*(a + b*x)^2) + 4*(2*a^2 + a*b*x - b^2*x^2)*ArcCoth[Tanh[a + b*x]] - 6*(a - b*x)*ArcC
oth[Tanh[a + b*x]]^2))/(12*b^2)

Maple [A] (verified)

Time = 24.16 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09

method	result	size
parallelrisch	\(-\frac {b \,x^{3} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}{3}+\frac {x^{2} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2}}{2}+\frac {b^{2} x^{4}}{12}\)	\(37\)
risch	\(\text {Expression too large to display}\)	\(3418\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.41 \[ \int x \coth ^{-1}(\tanh (a+b x))^2 \, dx=\frac {1}{4} \, b^{2} x^{4} + \frac {2}{3} \, a b x^{3} - \frac {1}{8} \, \pi ^{2} x^{2} + \frac {1}{2} \, a^{2} x^{2} + \frac {1}{6} i \, \pi {\left (2 \, b x^{3} + 3 \, a x^{2}\right )} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int x \coth ^{-1}(\tanh (a+b x))^2 \, dx=\begin {cases} \frac {x \operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{3 b} - \frac {\operatorname {acoth}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{12 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \operatorname {acoth}^{2}{\left (\tanh {\left (a \right )} \right )}}{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x*acoth(tanh(a + b*x))**3/(3*b) - acoth(tanh(a + b*x))**4/(12*b**2), Ne(b, 0)), (x**2*acoth(tanh(a)
)**2/2, True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int x \coth ^{-1}(\tanh (a+b x))^2 \, dx=\frac {1}{12} \, b^{2} x^{4} - \frac {1}{3} \, b x^{3} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right ) + \frac {1}{2} \, x^{2} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2} \]

[In]

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

[Out]

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int x \coth ^{-1}(\tanh (a+b x))^2 \, dx=\frac {1}{4} \, b^{2} x^{4} - \frac {1}{3} \, {\left (-i \, \pi b - 2 \, a b\right )} x^{3} - \frac {1}{8} \, {\left (\pi ^{2} - 4 i \, \pi a - 4 \, a^{2}\right )} x^{2} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 3.80 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int x \coth ^{-1}(\tanh (a+b x))^2 \, dx=\frac {b^2\,x^4}{12}-\frac {b\,x^3\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{3}+\frac {x^2\,{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{2} \]

[In]

int(x*acoth(tanh(a + b*x))^2,x)

[Out]

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