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

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

[Out]

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

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Rubi [A]  time = 0.0286383, antiderivative size = 47, normalized size of antiderivative = 1., 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 = {2168, 2157, 29} \[ -\frac{x}{b^2 \coth ^{-1}(\tanh (a+b x))}+\frac{\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac{x^2}{2 b \coth ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

Int[x^2/ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

-x^2/(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 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]))

Rule 2157

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 29

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

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{\operatorname{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*}

Mathematica [A]  time = 0.0312588, size = 49, normalized size = 1.04 \[ \frac{-\frac{b^2 x^2}{\coth ^{-1}(\tanh (a+b x))^2}-\frac{2 b x}{\coth ^{-1}(\tanh (a+b x))}+2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )+3}{2 b^3} \]

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]  time = 0.195, size = 952, normalized size = 20.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*I*(-2*Pi*x*csgn(I/(exp(2*b*x+2*a)+1))^2+Pi*x*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))-Pi*x*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
*x*csgn(I/(exp(2*b*x+2*a)+1))^3+Pi*x*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-2*Pi*x*csgn(I*exp(b*x+a))*csg
n(I*exp(2*b*x+2*a))^2+Pi*x*csgn(I*exp(2*b*x+2*a))^3-Pi*x*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*x*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+4*I*x*ln(exp(b*x+a))+2*Pi*x+2*I*x^2*b)/b^2/(-2*
Pi*csgn(I/(exp(2*b*x+2*a)+1))^2+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)/(ex
p(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))^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
+Pi*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*e
xp(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*(-2*csgn(I/(ex
p(2*b*x+2*a)+1))^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))
-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))^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-cs
gn(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]  time = 3.57015, size = 127, normalized size = 2.7 \begin{align*} -\frac{8 \,{\left (3 \, \pi ^{2} + 12 i \, \pi a - 12 \, a^{2} +{\left (8 i \, \pi b - 16 \, a b\right )} x\right )}}{64 \, b^{5} x^{2} - 16 \, \pi ^{2} b^{3} - 64 i \, \pi a b^{3} + 64 \, a^{2} b^{3} +{\left (-64 i \, \pi b^{4} + 128 \, a b^{4}\right )} x} + \frac{\log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*(3*pi^2 + 12*I*pi*a - 12*a^2 + (8*I*pi*b - 16*a*b)*x)/(64*b^5*x^2 - 16*pi^2*b^3 - 64*I*pi*a*b^3 + 64*a^2*b^
3 + (-64*I*pi*b^4 + 128*a*b^4)*x) + log(-I*pi + 2*b*x + 2*a)/b^3

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Fricas [B]  time = 1.78249, size = 548, normalized size = 11.66 \begin{align*} \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{align*}

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 = 24.5137, size = 54, normalized size = 1.15 \begin{align*} \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{align*}

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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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