∫ x coth−1(tanh(a + bx))2 dx [139]

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

Optimal. Leaf size=34 \[ \frac {x \coth ^{-1}(\tanh (a+b x))^3}{3 b}-\frac {\coth ^{-1}(\tanh (a+b x))^4}{12 b^2} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2199, 2188, 30} \begin {gather*} \frac {x \coth ^{-1}(\tanh (a+b x))^3}{3 b}-\frac {\coth ^{-1}(\tanh (a+b x))^4}{12 b^2} \end {gather*}

Antiderivative was successfully veriﬁed.

[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*} \int x \coth ^{-1}(\tanh (a+b x))^2 \, dx &=\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*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(74\) vs. \(2(34)=68\).
time = 0.04, size = 74, normalized size = 2.18 \begin {gather*} \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} \end {gather*}

Antiderivative was successfully veriﬁed.

[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)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 33.90, size = 3418, normalized size = 100.53


method	result	size

risch	\(\text {Expression too large to display}\)	\(3418\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*I*Pi*b*x^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))+1/

2*x^2*ln(exp(b*x+a))^2+1/12*b^2*x^4-1/8*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(b*x+a))^2*Pi^2*x^2+1/4*csgn(I*exp(2*

b*x+2*a))^2*csgn(I*exp(b*x+a))*Pi^2*x^2+1/4*csgn(I/(exp(2*b*x+2*a)+1))^2*Pi^2*x^2-1/16*Pi^2*x^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))^3-1/8*Pi^2*x^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))^2-1/8*Pi^2*x^2*csgn(I*exp(2*b*x+2*a))*csgn(I*ex

p(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2*csgn(I/(exp(2*b*x+2*a)+1))^2+1/8*Pi^2*x^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+(-1/3*b*x^3+1/4*I*Pi*x^2*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/

(exp(2*b*x+2*a)+1))^2-1/2*I*Pi*x^2*csgn(I/(exp(2*b*x+2*a)+1))^3-1/2*I*Pi*x^2+1/2*I*Pi*x^2*csgn(I/(exp(2*b*x+2*

a)+1))^2-1/4*I*Pi*x^2*csgn(I*exp(2*b*x+2*a))^3-1/4*I*Pi*x^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))+1/2*I*Pi*x^2*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2-1/4*I*Pi*x^

2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3-1/4*I*Pi*x^2*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))+1/4*I*P

i*x^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)*ln(exp(b*x+a))-1/32*Pi^2*x^2*csg

n(I*exp(2*b*x+2*a))^6+1/16*Pi^2*x^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+1/12*I*Pi*b*x^3*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3-1/4*csgn(I/(exp(2*b*x+2*a)+1

))^3*Pi^2*x^2+1/4*Pi^2*x^2*csgn(I/(exp(2*b*x+2*a)+1))^5-3/16*Pi^2*x^2*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*

a))^4+1/8*Pi^2*x^2*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^5+1/16*Pi^2*x^2*csgn(I*exp(2*b*x+2*a))^4*csgn(I*e

xp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-1/16*Pi^2*x^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))^3-1/32*Pi^2*x^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+1/16*Pi^2*x^2*csgn

(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^5-1/8*Pi^2*x^2*csgn(I/(exp(2*b*x+2*a)+1))^6-1/8*P

i^2*x^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))^2+1/8*Pi^2*x^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))^3+1/8*Pi^2*x^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+1/8*Pi^2*x^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))-1/32*Pi^2*x^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))^2-1/8*csgn(I/(exp(2*b*x+2*a)+1))^4*Pi^2*x^2+1/6*I*Pi*b*x^3+1/8*Pi^2*x^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+1/8*Pi^2*x^2*csgn(I*exp(2*b*x+2*a))^3*csgn(I/(e

xp(2*b*x+2*a)+1))^2-1/8*csgn(I*exp(2*b*x+2*a))^3*Pi^2*x^2-1/8*Pi^2*x^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))-1/8*Pi^2*x^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^

3-1/8*Pi^2*x^2-1/8*Pi^2*x^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+1/16*Pi^2*x^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+1/8*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2*csgn(I*exp(2*b*x+2*a))*Pi^2*x^2-1/8*Pi^2*x^2*csgn(I/(exp

(2*b*x+2*a)+1))^3*csgn(I*exp(2*b*x+2*a))^3-1/8*Pi^2*x^2*csgn(I/(exp(2*b*x+2*a)+1))^3*csgn(I*exp(2*b*x+2*a)/(ex

p(2*b*x+2*a)+1))^3+1/16*Pi^2*x^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)/(ex

p(2*b*x+2*a)+1))^2+1/16*Pi^2*x^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)/(ex

p(2*b*x+2*a)+1))^3+1/12*I*Pi*b*x^3*csgn(I*exp(2*b*x+2*a))^3-1/16*Pi^2*x^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*ex

p(2*b*x+2*a))^4*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))+1/8*Pi^2*x^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))^3-1/8*Pi^2*x^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+1/4*Pi^2*x^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

+1/8*Pi^2*x^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))^3-

1/32*Pi^2*x^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+1/16*Pi^2*x^2*csgn(I/(e

xp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^5-1/32*Pi^2*x^2*csgn(I*exp(b*x+a))^4*csgn(I*exp(2*

b*x+2*a))^2+1/8*Pi^2*x^2*csgn(I*exp(b*x+a))^3*csgn(I*exp(2*b*x+2*a))^3-1/8*Pi^2*x^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))-1/16*Pi^2*x^2*csgn(I/(exp(2*b*x+2*a)+1))*csg

n(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))+1/16*Pi^2*x^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+1/8*Pi

^2*x^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))-1/8*Pi^2*x^2*csgn(I/(exp(2*b*x+2*a)+1...

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Maxima [A]
time = 0.34, size = 36, normalized size = 1.06 \begin {gather*} \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} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]
time = 0.33, size = 30, normalized size = 0.88 \begin {gather*} \frac {1}{4} \, b^{2} x^{4} + \frac {2}{3} \, a b x^{3} - \frac {1}{8} \, {\left (\pi ^{2} - 4 \, a^{2}\right )} x^{2} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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 - 4*a^2)*x^2

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Sympy [A]
time = 0.14, size = 41, normalized size = 1.21 \begin {gather*} \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} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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))

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Giac [C] Result contains complex when optimal does not.
time = 0.40, size = 41, normalized size = 1.21 \begin {gather*} \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} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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

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Mupad [B]
time = 1.15, size = 36, normalized size = 1.06 \begin {gather*} \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} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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

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