3.167 \(\int \frac{x^4}{\coth ^{-1}(\tanh (a+b x))^2} \, dx\)

Optimal. Leaf size=98 \[ \frac{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac{4 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}{b^4}+\frac{4 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^5}-\frac{x^4}{b \coth ^{-1}(\tanh (a+b x))}+\frac{4 x^3}{3 b^2} \]

[Out]

(4*x^3)/(3*b^2) + (2*x^2*(b*x - ArcCoth[Tanh[a + b*x]]))/b^3 + (4*x*(b*x - ArcCoth[Tanh[a + b*x]])^2)/b^4 - x^
4/(b*ArcCoth[Tanh[a + b*x]]) + (4*(b*x - ArcCoth[Tanh[a + b*x]])^3*Log[ArcCoth[Tanh[a + b*x]]])/b^5

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Rubi [A]  time = 0.0808205, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {2168, 2159, 2158, 2157, 29} \[ \frac{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac{4 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}{b^4}+\frac{4 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^5}-\frac{x^4}{b \coth ^{-1}(\tanh (a+b x))}+\frac{4 x^3}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^4/ArcCoth[Tanh[a + b*x]]^2,x]

[Out]

(4*x^3)/(3*b^2) + (2*x^2*(b*x - ArcCoth[Tanh[a + b*x]]))/b^3 + (4*x*(b*x - ArcCoth[Tanh[a + b*x]])^2)/b^4 - x^
4/(b*ArcCoth[Tanh[a + b*x]]) + (4*(b*x - ArcCoth[Tanh[a + b*x]])^3*Log[ArcCoth[Tanh[a + b*x]]])/b^5

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 2159

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^n/(a*n), x] - Dis
t[(b*u - a*v)/a, Int[v^(n - 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && GtQ[n, 0] && Ne
Q[n, 1]

Rule 2158

Int[(v_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(b*x)/a, x] - Dist[(b*u
- a*v)/a, Int[1/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x]

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^4}{\coth ^{-1}(\tanh (a+b x))^2} \, dx &=-\frac{x^4}{b \coth ^{-1}(\tanh (a+b x))}+\frac{4 \int \frac{x^3}{\coth ^{-1}(\tanh (a+b x))} \, dx}{b}\\ &=\frac{4 x^3}{3 b^2}-\frac{x^4}{b \coth ^{-1}(\tanh (a+b x))}-\frac{\left (4 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )\right ) \int \frac{x^2}{\coth ^{-1}(\tanh (a+b x))} \, dx}{b^2}\\ &=\frac{4 x^3}{3 b^2}+\frac{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac{x^4}{b \coth ^{-1}(\tanh (a+b x))}+\frac{\left (4 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac{x}{\coth ^{-1}(\tanh (a+b x))} \, dx}{b^3}\\ &=\frac{4 x^3}{3 b^2}+\frac{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac{4 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}{b^4}-\frac{x^4}{b \coth ^{-1}(\tanh (a+b x))}-\frac{\left (4 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^3\right ) \int \frac{1}{\coth ^{-1}(\tanh (a+b x))} \, dx}{b^4}\\ &=\frac{4 x^3}{3 b^2}+\frac{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac{4 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}{b^4}-\frac{x^4}{b \coth ^{-1}(\tanh (a+b x))}-\frac{\left (4 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^3\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{b^5}\\ &=\frac{4 x^3}{3 b^2}+\frac{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac{4 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}{b^4}-\frac{x^4}{b \coth ^{-1}(\tanh (a+b x))}+\frac{4 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^5}\\ \end{align*}

Mathematica [A]  time = 0.0974712, size = 106, normalized size = 1.08 \[ -\frac{x^2 \left (\coth ^{-1}(\tanh (a+b x))-b x\right )}{b^3}-\frac{\left (\coth ^{-1}(\tanh (a+b x))-b x\right )^4}{b^5 \coth ^{-1}(\tanh (a+b x))}+\frac{3 x \left (\coth ^{-1}(\tanh (a+b x))-b x\right )^2}{b^4}-\frac{4 \left (\coth ^{-1}(\tanh (a+b x))-b x\right )^3 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^5}+\frac{x^3}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/ArcCoth[Tanh[a + b*x]]^2,x]

[Out]

x^3/(3*b^2) - (x^2*(-(b*x) + ArcCoth[Tanh[a + b*x]]))/b^3 + (3*x*(-(b*x) + ArcCoth[Tanh[a + b*x]])^2)/b^4 - (-
(b*x) + ArcCoth[Tanh[a + b*x]])^4/(b^5*ArcCoth[Tanh[a + b*x]]) - (4*(-(b*x) + ArcCoth[Tanh[a + b*x]])^3*Log[Ar
cCoth[Tanh[a + b*x]]])/b^5

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Maple [C]  time = 6.12, size = 131085, normalized size = 1337.6 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/arccoth(tanh(b*x+a))^2,x)

[Out]

result too large to display

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Maxima [C]  time = 2.48721, size = 240, normalized size = 2.45 \begin{align*} \frac{4 \,{\left (16 \, b^{4} x^{4} - 3 \, \pi ^{4} - 24 i \, \pi ^{3} a + 72 \, \pi ^{2} a^{2} + 96 i \, \pi a^{3} - 48 \, a^{4} +{\left (16 i \, \pi b^{3} - 32 \, a b^{3}\right )} x^{3} -{\left (24 \, \pi ^{2} b^{2} + 96 i \, \pi a b^{2} - 96 \, a^{2} b^{2}\right )} x^{2} +{\left (18 i \, \pi ^{3} b - 108 \, \pi ^{2} a b - 216 i \, \pi a^{2} b + 144 \, a^{3} b\right )} x\right )}}{192 \, b^{6} x - 96 i \, \pi b^{5} + 192 \, a b^{5}} - \frac{{\left (i \, \pi ^{3} - 6 \, \pi ^{2} a - 12 i \, \pi a^{2} + 8 \, a^{3}\right )} \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{2 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arccoth(tanh(b*x+a))^2,x, algorithm="maxima")

[Out]

4*(16*b^4*x^4 - 3*pi^4 - 24*I*pi^3*a + 72*pi^2*a^2 + 96*I*pi*a^3 - 48*a^4 + (16*I*pi*b^3 - 32*a*b^3)*x^3 - (24
*pi^2*b^2 + 96*I*pi*a*b^2 - 96*a^2*b^2)*x^2 + (18*I*pi^3*b - 108*pi^2*a*b - 216*I*pi*a^2*b + 144*a^3*b)*x)/(19
2*b^6*x - 96*I*pi*b^5 + 192*a*b^5) - 1/2*(I*pi^3 - 6*pi^2*a - 12*I*pi*a^2 + 8*a^3)*log(-I*pi + 2*b*x + 2*a)/b^
5

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Fricas [B]  time = 1.66643, size = 729, normalized size = 7.44 \begin{align*} \frac{16 \, b^{5} x^{5} - 16 \, a b^{4} x^{4} + 9 \, \pi ^{4} a + 24 \, \pi ^{2} a^{3} - 48 \, a^{5} - 32 \,{\left (\pi ^{2} b^{3} - 2 \, a^{2} b^{3}\right )} x^{3} - 12 \,{\left (7 \, \pi ^{2} a b^{2} - 20 \, a^{3} b^{2}\right )} x^{2} - 12 \,{\left (\pi ^{4} b - 6 \, \pi ^{2} a^{2} b - 8 \, a^{4} b\right )} x - 12 \,{\left (\pi ^{5} - 8 \, \pi ^{3} a^{2} - 48 \, \pi a^{4} + 4 \,{\left (\pi ^{3} b^{2} - 12 \, \pi a^{2} b^{2}\right )} x^{2} + 8 \,{\left (\pi ^{3} a b - 12 \, \pi a^{3} b\right )} x\right )} \arctan \left (-\frac{2 \, b x + 2 \, a - \sqrt{4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}}}{\pi }\right ) + 6 \,{\left (3 \, \pi ^{4} a + 8 \, \pi ^{2} a^{3} - 16 \, a^{5} + 4 \,{\left (3 \, \pi ^{2} a b^{2} - 4 \, a^{3} b^{2}\right )} x^{2} + 8 \,{\left (3 \, \pi ^{2} a^{2} b - 4 \, a^{4} b\right )} x\right )} \log \left (4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}\right )}{12 \,{\left (4 \, b^{7} x^{2} + 8 \, a b^{6} x + \pi ^{2} b^{5} + 4 \, a^{2} b^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arccoth(tanh(b*x+a))^2,x, algorithm="fricas")

[Out]

1/12*(16*b^5*x^5 - 16*a*b^4*x^4 + 9*pi^4*a + 24*pi^2*a^3 - 48*a^5 - 32*(pi^2*b^3 - 2*a^2*b^3)*x^3 - 12*(7*pi^2
*a*b^2 - 20*a^3*b^2)*x^2 - 12*(pi^4*b - 6*pi^2*a^2*b - 8*a^4*b)*x - 12*(pi^5 - 8*pi^3*a^2 - 48*pi*a^4 + 4*(pi^
3*b^2 - 12*pi*a^2*b^2)*x^2 + 8*(pi^3*a*b - 12*pi*a^3*b)*x)*arctan(-(2*b*x + 2*a - sqrt(4*b^2*x^2 + 8*a*b*x + p
i^2 + 4*a^2))/pi) + 6*(3*pi^4*a + 8*pi^2*a^3 - 16*a^5 + 4*(3*pi^2*a*b^2 - 4*a^3*b^2)*x^2 + 8*(3*pi^2*a^2*b - 4
*a^4*b)*x)*log(4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a^2))/(4*b^7*x^2 + 8*a*b^6*x + pi^2*b^5 + 4*a^2*b^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\operatorname{acoth}^{2}{\left (\tanh{\left (a + b x \right )} \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/acoth(tanh(b*x+a))**2,x)

[Out]

Integral(x**4/acoth(tanh(a + b*x))**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arccoth(tanh(b*x+a))^2,x, algorithm="giac")

[Out]

integrate(x^4/arccoth(tanh(b*x + a))^2, x)