∫ x3 coth−1(tanh(a + bx))2 dx

3.137 \(\int x^3 \coth ^{-1}(\tanh (a+b x))^2 \, dx\)

Optimal. Leaf size=42 \[ -\frac {1}{10} b x^5 \coth ^{-1}(\tanh (a+b x))+\frac {1}{4} x^4 \coth ^{-1}(\tanh (a+b x))^2+\frac {b^2 x^6}{60} \]

[Out]

1/60*b^2*x^6-1/10*b*x^5*arccoth(tanh(b*x+a))+1/4*x^4*arccoth(tanh(b*x+a))^2

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Rubi [A] time = 0.03, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2168, 30} \[ -\frac {1}{10} b x^5 \coth ^{-1}(\tanh (a+b x))+\frac {1}{4} x^4 \coth ^{-1}(\tanh (a+b x))^2+\frac {b^2 x^6}{60} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(b^2*x^6)/60 - (b*x^5*ArcCoth[Tanh[a + b*x]])/10 + (x^4*ArcCoth[Tanh[a + b*x]]^2)/4

Rule 30

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

Rule 2168

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^3 \coth ^{-1}(\tanh (a+b x))^2 \, dx &=\frac {1}{4} x^4 \coth ^{-1}(\tanh (a+b x))^2-\frac {1}{2} b \int x^4 \coth ^{-1}(\tanh (a+b x)) \, dx\\ &=-\frac {1}{10} b x^5 \coth ^{-1}(\tanh (a+b x))+\frac {1}{4} x^4 \coth ^{-1}(\tanh (a+b x))^2+\frac {1}{10} b^2 \int x^5 \, dx\\ &=\frac {b^2 x^6}{60}-\frac {1}{10} b x^5 \coth ^{-1}(\tanh (a+b x))+\frac {1}{4} x^4 \coth ^{-1}(\tanh (a+b x))^2\\ \end {align*}

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Mathematica [A] time = 0.03, size = 37, normalized size = 0.88 \[ \frac {1}{60} x^4 \left (-6 b x \coth ^{-1}(\tanh (a+b x))+15 \coth ^{-1}(\tanh (a+b x))^2+b^2 x^2\right ) \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(x^4*(b^2*x^2 - 6*b*x*ArcCoth[Tanh[a + b*x]] + 15*ArcCoth[Tanh[a + b*x]]^2))/60

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fricas [A] time = 0.58, size = 30, normalized size = 0.71 \[ \frac {1}{6} \, b^{2} x^{6} + \frac {2}{5} \, a b x^{5} - \frac {1}{16} \, {\left (\pi ^{2} - 4 \, a^{2}\right )} x^{4} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b^2*x^6 + 2/5*a*b*x^5 - 1/16*(pi^2 - 4*a^2)*x^4

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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2}\,{d x} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*arccoth(tanh(b*x + a))^2, x)

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maple [C] time = 0.39, size = 3418, normalized size = 81.38 \[ \text {output too large to display} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arccoth(tanh(b*x+a))^2,x)

[Out]

1/40*I*Pi*b*x^5*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^4*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+1/

16*Pi^2*x^4*csgn(I*exp(2*b*x+2*a))^3*csgn(I/(exp(2*b*x+2*a)+1))^2-1/32*Pi^2*x^4*csgn(I*exp(2*b*x+2*a))^3*csgn(

I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3-1/64*Pi^2*x^4*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+1/32*Pi^2*x^4*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^4*csgn

(I/(exp(2*b*x+2*a)+1))^5+1/8*csgn(I/(exp(2*b*x+2*a)+1))^2*Pi^2*x^4+(-1/10*b*x^5+1/8*I*Pi*x^4*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*I*Pi*x^4*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a

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

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

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

xp(b*x+a))-1/16*csgn(I*exp(2*b*x+2*a))^3*Pi^2*x^4-1/16*Pi^2*x^4*csgn(I/(exp(2*b*x+2*a)+1))^6-1/8*csgn(I/(exp(2

*b*x+2*a)+1))^3*Pi^2*x^4-1/32*Pi^2*x^4*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))+1/32*Pi^2*x^4*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+1/32*Pi^2*x^4*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)

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

1/16*Pi^2*x^4*csgn(I/(exp(2*b*x+2*a)+1))^3*csgn(I*exp(2*b*x+2*a))^3-1/16*Pi^2*x^4*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+1/32*Pi^2*x^4*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a)/(e

xp(2*b*x+2*a)+1))^5-1/64*Pi^2*x^4*csgn(I*exp(b*x+a))^4*csgn(I*exp(2*b*x+2*a))^2+1/16*Pi^2*x^4*csgn(I*exp(b*x+a

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

a))^2-1/32*Pi^2*x^4*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))+1/32*Pi^2*x^4*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+1/16*Pi^2*x^4*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))-1/16*Pi^2*x^4*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*e

xp(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))^2-1/16*Pi^2*x^4*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))^4-1/32*Pi^2*x^4*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))^3-3/32*Pi^2*x^4*csgn(I*exp(b*x+a))^2*csgn(

I*exp(2*b*x+2*a))^4+1/16*Pi^2*x^4*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^5+1/32*Pi^2*x^4*csgn(I*exp(2*b*x+2

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

/16*Pi^2*x^4*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))+1/1

6*Pi^2*x^4*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))-1/16*Pi^2*x^4*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))^2-1/20*I*Pi*b*x^5*csgn(I/(exp(2*b*x+

2*a)+1))^2+1/20*I*Pi*b*x^5*csgn(I/(exp(2*b*x+2*a)+1))^3-1/16*Pi^2*x^4-1/64*Pi^2*x^4*csgn(I*exp(2*b*x+2*a))^6-1

/64*Pi^2*x^4*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^6+1/4*x^4*ln(exp(b*x+a))^2-1/16*Pi^2*x^4*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))+1/20*I*Pi*b*x^5-1/16*Pi^2*x^4*cs

gn(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))+1/32*Pi^2*x^4*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))^2-1/40*I*Pi*b*x^5*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+1/40*I*Pi*b*x^5*csgn(I*exp(b*x+a))^2*csgn(I*exp

(2*b*x+2*a))-1/20*I*Pi*b*x^5*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2-1/40*I*Pi*b*x^5*csgn(I*exp(2*b*x+2*a)

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

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

+2*a)+1))^2-1/64*Pi^2*x^4*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+1/60*b^2*x^

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

16*Pi^2*x^4*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))^2-1/

64*Pi^2*x^4*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+

1/32*Pi^2*x^4*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-

1/16*Pi^2*x^4*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+1/16*Pi^2*x^4*csgn(I/(e

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

n(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))+1/40*I*Pi*b*x^5*csgn(I*exp(2*b*x+2*a))^3+1/40*I*Pi*b*x^5*csgn(I*exp(2

*b*x+2*a)/(exp(2*b*x+2*a)+1))^3

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maxima [A] time = 0.45, size = 36, normalized size = 0.86 \[ \frac {1}{60} \, b^{2} x^{6} - \frac {1}{10} \, b x^{5} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right ) + \frac {1}{4} \, x^{4} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*b^2*x^6 - 1/10*b*x^5*arccoth(tanh(b*x + a)) + 1/4*x^4*arccoth(tanh(b*x + a))^2

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mupad [B] time = 1.18, size = 36, normalized size = 0.86 \[ \frac {b^2\,x^6}{60}-\frac {b\,x^5\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{10}+\frac {x^4\,{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{4} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*acoth(tanh(a + b*x))^2,x)

[Out]

(x^4*acoth(tanh(a + b*x))^2)/4 + (b^2*x^6)/60 - (b*x^5*acoth(tanh(a + b*x)))/10

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sympy [A] time = 2.62, size = 78, normalized size = 1.86 \[ \begin {cases} \frac {x^{3} \operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{3 b} - \frac {x^{2} \operatorname {acoth}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{4 b^{2}} + \frac {x \operatorname {acoth}^{5}{\left (\tanh {\left (a + b x \right )} \right )}}{10 b^{3}} - \frac {\operatorname {acoth}^{6}{\left (\tanh {\left (a + b x \right )} \right )}}{60 b^{4}} & \text {for}\: b \neq 0 \\\frac {x^{4} \operatorname {acoth}^{2}{\left (\tanh {\relax (a )} \right )}}{4} & \text {otherwise} \end {cases} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

**5/(10*b**3) - acoth(tanh(a + b*x))**6/(60*b**4), Ne(b, 0)), (x**4*acoth(tanh(a))**2/4, True))

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