3.176 \(\int \frac{x^4}{\coth ^{-1}(\tanh (a+b x))^3} \, dx\)

Optimal. Leaf size=92 \[ -\frac{2 x^3}{b^2 \coth ^{-1}(\tanh (a+b x))}+\frac{6 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^4}+\frac{6 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^5}-\frac{x^4}{2 b \coth ^{-1}(\tanh (a+b x))^2}+\frac{3 x^2}{b^3} \]

[Out]

(3*x^2)/b^3 + (6*x*(b*x - ArcCoth[Tanh[a + b*x]]))/b^4 - x^4/(2*b*ArcCoth[Tanh[a + b*x]]^2) - (2*x^3)/(b^2*Arc
Coth[Tanh[a + b*x]]) + (6*(b*x - ArcCoth[Tanh[a + b*x]])^2*Log[ArcCoth[Tanh[a + b*x]]])/b^5

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Rubi [A]  time = 0.0712986, antiderivative size = 92, 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} \[ -\frac{2 x^3}{b^2 \coth ^{-1}(\tanh (a+b x))}+\frac{6 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^4}+\frac{6 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^5}-\frac{x^4}{2 b \coth ^{-1}(\tanh (a+b x))^2}+\frac{3 x^2}{b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^4/ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

(3*x^2)/b^3 + (6*x*(b*x - ArcCoth[Tanh[a + b*x]]))/b^4 - x^4/(2*b*ArcCoth[Tanh[a + b*x]]^2) - (2*x^3)/(b^2*Arc
Coth[Tanh[a + b*x]]) + (6*(b*x - ArcCoth[Tanh[a + b*x]])^2*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))^3} \, dx &=-\frac{x^4}{2 b \coth ^{-1}(\tanh (a+b x))^2}+\frac{2 \int \frac{x^3}{\coth ^{-1}(\tanh (a+b x))^2} \, dx}{b}\\ &=-\frac{x^4}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac{2 x^3}{b^2 \coth ^{-1}(\tanh (a+b x))}+\frac{6 \int \frac{x^2}{\coth ^{-1}(\tanh (a+b x))} \, dx}{b^2}\\ &=\frac{3 x^2}{b^3}-\frac{x^4}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac{2 x^3}{b^2 \coth ^{-1}(\tanh (a+b x))}-\frac{\left (6 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )\right ) \int \frac{x}{\coth ^{-1}(\tanh (a+b x))} \, dx}{b^3}\\ &=\frac{3 x^2}{b^3}+\frac{6 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac{x^4}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac{2 x^3}{b^2 \coth ^{-1}(\tanh (a+b x))}+\frac{\left (6 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac{1}{\coth ^{-1}(\tanh (a+b x))} \, dx}{b^4}\\ &=\frac{3 x^2}{b^3}+\frac{6 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac{x^4}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac{2 x^3}{b^2 \coth ^{-1}(\tanh (a+b x))}+\frac{\left (6 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{b^5}\\ &=\frac{3 x^2}{b^3}+\frac{6 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac{x^4}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac{2 x^3}{b^2 \coth ^{-1}(\tanh (a+b x))}+\frac{6 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^5}\\ \end{align*}

Mathematica [A]  time = 0.0418133, size = 114, normalized size = 1.24 \[ -\frac{\left (\coth ^{-1}(\tanh (a+b x))-b x\right )^4}{2 b^5 \coth ^{-1}(\tanh (a+b x))^2}+\frac{4 \left (\coth ^{-1}(\tanh (a+b x))-b x\right )^3}{b^5 \coth ^{-1}(\tanh (a+b x))}-\frac{3 x \left (\coth ^{-1}(\tanh (a+b x))-b x\right )}{b^4}+\frac{6 \left (\coth ^{-1}(\tanh (a+b x))-b x\right )^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^5}+\frac{x^2}{2 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

x^2/(2*b^3) - (3*x*(-(b*x) + ArcCoth[Tanh[a + b*x]]))/b^4 + (4*(-(b*x) + ArcCoth[Tanh[a + b*x]])^3)/(b^5*ArcCo
th[Tanh[a + b*x]]) - (-(b*x) + ArcCoth[Tanh[a + b*x]])^4/(2*b^5*ArcCoth[Tanh[a + b*x]]^2) + (6*(-(b*x) + ArcCo
th[Tanh[a + b*x]])^2*Log[ArcCoth[Tanh[a + b*x]]])/b^5

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Maple [C]  time = 1.396, size = 29456, normalized size = 320.2 \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))^3,x)

[Out]

result too large to display

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Maxima [C]  time = 3.54373, size = 266, normalized size = 2.89 \begin{align*} \frac{8 \,{\left (16 \, b^{4} x^{4} + 7 \, \pi ^{4} + 56 i \, \pi ^{3} a - 168 \, \pi ^{2} a^{2} - 224 i \, \pi a^{3} + 112 \, a^{4} +{\left (32 i \, \pi b^{3} - 64 \, a b^{3}\right )} x^{3} +{\left (44 \, \pi ^{2} b^{2} + 176 i \, \pi a b^{2} - 176 \, a^{2} b^{2}\right )} x^{2} +{\left (4 i \, \pi ^{3} b - 24 \, \pi ^{2} a b - 48 i \, \pi a^{2} b + 32 \, a^{3} b\right )} x\right )}}{256 \, b^{7} x^{2} - 64 \, \pi ^{2} b^{5} - 256 i \, \pi a b^{5} + 256 \, a^{2} b^{5} +{\left (-256 i \, \pi b^{6} + 512 \, a b^{6}\right )} x} - \frac{{\left (3 \, \pi ^{2} + 12 i \, \pi a - 12 \, a^{2}\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))^3,x, algorithm="maxima")

[Out]

8*(16*b^4*x^4 + 7*pi^4 + 56*I*pi^3*a - 168*pi^2*a^2 - 224*I*pi*a^3 + 112*a^4 + (32*I*pi*b^3 - 64*a*b^3)*x^3 +
(44*pi^2*b^2 + 176*I*pi*a*b^2 - 176*a^2*b^2)*x^2 + (4*I*pi^3*b - 24*pi^2*a*b - 48*I*pi*a^2*b + 32*a^3*b)*x)/(2
56*b^7*x^2 - 64*pi^2*b^5 - 256*I*pi*a*b^5 + 256*a^2*b^5 + (-256*I*pi*b^6 + 512*a*b^6)*x) - 1/2*(3*pi^2 + 12*I*
pi*a - 12*a^2)*log(-I*pi + 2*b*x + 2*a)/b^5

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Fricas [B]  time = 1.79415, size = 1100, normalized size = 11.96 \begin{align*} \frac{64 \, b^{6} x^{6} - 128 \, a b^{5} x^{5} - 7 \, \pi ^{6} - 28 \, \pi ^{4} a^{2} + 112 \, \pi ^{2} a^{4} + 448 \, a^{6} + 32 \,{\left (\pi ^{2} b^{4} - 36 \, a^{2} b^{4}\right )} x^{4} - 512 \,{\left (\pi ^{2} a b^{3} + 3 \, a^{3} b^{3}\right )} x^{3} - 32 \,{\left (\pi ^{4} b^{2} + 32 \, \pi ^{2} a^{2} b^{2}\right )} x^{2} - 32 \,{\left (5 \, \pi ^{4} a b + 12 \, \pi ^{2} a^{3} b - 32 \, a^{5} b\right )} x - 96 \,{\left (16 \, \pi a b^{4} x^{4} + 64 \, \pi a^{2} b^{3} x^{3} + \pi ^{5} a + 8 \, \pi ^{3} a^{3} + 16 \, \pi a^{5} + 8 \,{\left (\pi ^{3} a b^{2} + 12 \, \pi a^{3} b^{2}\right )} x^{2} + 16 \,{\left (\pi ^{3} a^{2} b + 4 \, \pi a^{4} 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 (\pi ^{6} + 4 \, \pi ^{4} a^{2} - 16 \, \pi ^{2} a^{4} - 64 \, a^{6} + 16 \,{\left (\pi ^{2} b^{4} - 4 \, a^{2} b^{4}\right )} x^{4} + 64 \,{\left (\pi ^{2} a b^{3} - 4 \, a^{3} b^{3}\right )} x^{3} + 8 \,{\left (\pi ^{4} b^{2} + 8 \, \pi ^{2} a^{2} b^{2} - 48 \, a^{4} b^{2}\right )} x^{2} + 16 \,{\left (\pi ^{4} a b - 16 \, a^{5} b\right )} x\right )} \log \left (4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}\right )}{8 \,{\left (16 \, b^{9} x^{4} + 64 \, a b^{8} x^{3} + \pi ^{4} b^{5} + 8 \, \pi ^{2} a^{2} b^{5} + 16 \, a^{4} b^{5} + 8 \,{\left (\pi ^{2} b^{7} + 12 \, a^{2} b^{7}\right )} x^{2} + 16 \,{\left (\pi ^{2} a b^{6} + 4 \, a^{3} b^{6}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arccoth(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

1/8*(64*b^6*x^6 - 128*a*b^5*x^5 - 7*pi^6 - 28*pi^4*a^2 + 112*pi^2*a^4 + 448*a^6 + 32*(pi^2*b^4 - 36*a^2*b^4)*x
^4 - 512*(pi^2*a*b^3 + 3*a^3*b^3)*x^3 - 32*(pi^4*b^2 + 32*pi^2*a^2*b^2)*x^2 - 32*(5*pi^4*a*b + 12*pi^2*a^3*b -
32*a^5*b)*x - 96*(16*pi*a*b^4*x^4 + 64*pi*a^2*b^3*x^3 + pi^5*a + 8*pi^3*a^3 + 16*pi*a^5 + 8*(pi^3*a*b^2 + 12*
pi*a^3*b^2)*x^2 + 16*(pi^3*a^2*b + 4*pi*a^4*b)*x)*arctan(-(2*b*x + 2*a - sqrt(4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a
^2))/pi) - 6*(pi^6 + 4*pi^4*a^2 - 16*pi^2*a^4 - 64*a^6 + 16*(pi^2*b^4 - 4*a^2*b^4)*x^4 + 64*(pi^2*a*b^3 - 4*a^
3*b^3)*x^3 + 8*(pi^4*b^2 + 8*pi^2*a^2*b^2 - 48*a^4*b^2)*x^2 + 16*(pi^4*a*b - 16*a^5*b)*x)*log(4*b^2*x^2 + 8*a*
b*x + pi^2 + 4*a^2))/(16*b^9*x^4 + 64*a*b^8*x^3 + pi^4*b^5 + 8*pi^2*a^2*b^5 + 16*a^4*b^5 + 8*(pi^2*b^7 + 12*a^
2*b^7)*x^2 + 16*(pi^2*a*b^6 + 4*a^3*b^6)*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\operatorname{acoth}^{3}{\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))**3,x)

[Out]

Integral(x**4/acoth(tanh(a + b*x))**3, 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 )^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arccoth(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

integrate(x^4/arccoth(tanh(b*x + a))^3, x)