3.2.78 \(\int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))^3} \, dx\) [178]

Optimal. Leaf size=47 \[ -\frac {x^2}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac {x}{b^2 \coth ^{-1}(\tanh (a+b x))}+\frac {\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^3} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2199, 2188, 29} \begin {gather*} \frac {\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac {x}{b^2 \coth ^{-1}(\tanh (a+b x))}-\frac {x^2}{2 b \coth ^{-1}(\tanh (a+b x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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

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Mathematica [A]
time = 0.03, size = 49, normalized size = 1.04 \begin {gather*} \frac {3-\frac {b^2 x^2}{\coth ^{-1}(\tanh (a+b x))^2}-\frac {2 b x}{\coth ^{-1}(\tanh (a+b x))}+2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{2 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.17, size = 952, normalized size = 20.26

method	result	size
risch	\(-\frac {4 i \left (\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right ) x -\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2} x -2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2} x +2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3} x +\pi \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) x -2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2} x +\pi \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3} x -\pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2} x +\pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3} x +4 i x \ln \left ({\mathrm e}^{b x +a}\right )+2 \pi x +2 i b \,x^{2}\right )}{b^{2} \left (\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}-2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+\pi \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )-2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \mathrm {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 )\right )^{2}}+\frac {\ln \left (\ln \left ({\mathrm e}^{b x +a}\right )-\frac {i \pi \left (\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )-\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}-2 \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+2 \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )-2 \,\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+2\right )}{4}\right )}{b^{3}}\)	\(952\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/arccoth(tanh(b*x+a))^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.86, size = 96, normalized size = 2.04 \begin {gather*} -\frac {3 \, \pi ^{2} + 12 i \, \pi a - 12 \, a^{2} - 8 \, {\left (-i \, \pi b + 2 \, a b\right )} x}{2 \, {\left (4 \, b^{5} x^{2} - \pi ^{2} b^{3} - 4 i \, \pi a b^{3} + 4 \, a^{2} b^{3} - 4 \, {\left (i \, \pi b^{4} - 2 \, a b^{4}\right )} x\right )}} + \frac {\log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(3*pi^2 + 12*I*pi*a - 12*a^2 - 8*(-I*pi*b + 2*a*b)*x)/(4*b^5*x^2 - pi^2*b^3 - 4*I*pi*a*b^3 + 4*a^2*b^3 -
4*(I*pi*b^4 - 2*a*b^4)*x) + log(-I*pi + 2*b*x + 2*a)/b^3

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (45) = 90\).
time = 0.38, size = 250, normalized size = 5.32 \begin {gather*} \frac {64 \, a b^{3} x^{3} + 3 \, \pi ^{4} + 24 \, \pi ^{2} a^{2} + 48 \, a^{4} + 4 \, {\left (5 \, \pi ^{2} b^{2} + 44 \, a^{2} b^{2}\right )} x^{2} + 40 \, {\left (\pi ^{2} a b + 4 \, a^{3} b\right )} x + {\left (16 \, b^{4} x^{4} + 64 \, a b^{3} x^{3} + \pi ^{4} + 8 \, \pi ^{2} a^{2} + 16 \, a^{4} + 8 \, {\left (\pi ^{2} b^{2} + 12 \, a^{2} b^{2}\right )} x^{2} + 16 \, {\left (\pi ^{2} a b + 4 \, a^{3} b\right )} x\right )} \log \left (4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}\right )}{2 \, {\left (16 \, b^{7} x^{4} + 64 \, a b^{6} x^{3} + \pi ^{4} b^{3} + 8 \, \pi ^{2} a^{2} b^{3} + 16 \, a^{4} b^{3} + 8 \, {\left (\pi ^{2} b^{5} + 12 \, a^{2} b^{5}\right )} x^{2} + 16 \, {\left (\pi ^{2} a b^{4} + 4 \, a^{3} b^{4}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(64*a*b^3*x^3 + 3*pi^4 + 24*pi^2*a^2 + 48*a^4 + 4*(5*pi^2*b^2 + 44*a^2*b^2)*x^2 + 40*(pi^2*a*b + 4*a^3*b)*
x + (16*b^4*x^4 + 64*a*b^3*x^3 + pi^4 + 8*pi^2*a^2 + 16*a^4 + 8*(pi^2*b^2 + 12*a^2*b^2)*x^2 + 16*(pi^2*a*b + 4
*a^3*b)*x)*log(4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a^2))/(16*b^7*x^4 + 64*a*b^6*x^3 + pi^4*b^3 + 8*pi^2*a^2*b^3 + 1
6*a^4*b^3 + 8*(pi^2*b^5 + 12*a^2*b^5)*x^2 + 16*(pi^2*a*b^4 + 4*a^3*b^4)*x)

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Sympy [A]
time = 33.10, size = 54, normalized size = 1.15 \begin {gather*} \begin {cases} - \frac {x^{2}}{2 b \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}} - \frac {x}{b^{2} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}} + \frac {\log {\left (\operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )} \right )}}{b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3}}{3 \operatorname {acoth}^{3}{\left (\tanh {\left (a \right )} \right )}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C] Result contains complex when optimal does not.
time = 0.41, size = 91, normalized size = 1.94 \begin {gather*} \frac {8 \, \pi b x - 16 i \, a b x + 3 i \, \pi ^{2} + 12 \, \pi a - 12 i \, a^{2}}{-8 i \, b^{5} x^{2} + 8 \, \pi b^{4} x - 16 i \, a b^{4} x + 2 i \, \pi ^{2} b^{3} + 8 \, \pi a b^{3} - 8 i \, a^{2} b^{3}} + \frac {\log \left (i \, \pi + 2 \, b x + 2 \, a\right )}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8*pi*b*x - 16*I*a*b*x + 3*I*pi^2 + 12*pi*a - 12*I*a^2)/(-8*I*b^5*x^2 + 8*pi*b^4*x - 16*I*a*b^4*x + 2*I*pi^2*b
^3 + 8*pi*a*b^3 - 8*I*a^2*b^3) + log(I*pi + 2*b*x + 2*a)/b^3

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Mupad [B]
time = 1.22, size = 46, normalized size = 0.98 \begin {gather*} \frac {\ln \left (\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )\right )}{b^3}-\frac {\frac {b^2\,x^2}{2}+b\,x\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{b^3\,{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/acoth(tanh(a + b*x))^3,x)

[Out]

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

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