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

Optimal. Leaf size=75 \[ \frac {3 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^4}+\frac {3 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac {x^3}{b \coth ^{-1}(\tanh (a+b x))}+\frac {3 x^2}{2 b^2} \]

[Out]

3/2*x^2/b^2+3*x*(b*x-arccoth(tanh(b*x+a)))/b^3-x^3/b/arccoth(tanh(b*x+a))+3*(b*x-arccoth(tanh(b*x+a)))^2*ln(ar
ccoth(tanh(b*x+a)))/b^4

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Rubi [A]  time = 0.05, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \[ \frac {3 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac {3 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac {x^3}{b \coth ^{-1}(\tanh (a+b x))}+\frac {3 x^2}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3/ArcCoth[Tanh[a + b*x]]^2,x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], 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 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 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 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 \frac {x^3}{\coth ^{-1}(\tanh (a+b x))^2} \, dx &=-\frac {x^3}{b \coth ^{-1}(\tanh (a+b x))}+\frac {3 \int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))} \, dx}{b}\\ &=\frac {3 x^2}{2 b^2}-\frac {x^3}{b \coth ^{-1}(\tanh (a+b x))}-\frac {\left (3 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )\right ) \int \frac {x}{\coth ^{-1}(\tanh (a+b x))} \, dx}{b^2}\\ &=\frac {3 x^2}{2 b^2}+\frac {3 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac {x^3}{b \coth ^{-1}(\tanh (a+b x))}+\frac {\left (3 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac {1}{\coth ^{-1}(\tanh (a+b x))} \, dx}{b^3}\\ &=\frac {3 x^2}{2 b^2}+\frac {3 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac {x^3}{b \coth ^{-1}(\tanh (a+b x))}+\frac {\left (3 \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^4}\\ &=\frac {3 x^2}{2 b^2}+\frac {3 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac {x^3}{b \coth ^{-1}(\tanh (a+b x))}+\frac {3 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^4}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 83, normalized size = 1.11 \[ \frac {\left (\coth ^{-1}(\tanh (a+b x))-b x\right )^3}{b^4 \coth ^{-1}(\tanh (a+b x))}+\frac {3 \left (\coth ^{-1}(\tanh (a+b x))-b x\right )^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac {2 x \left (\coth ^{-1}(\tanh (a+b x))-b x\right )}{b^3}+\frac {x^2}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.50, size = 244, normalized size = 3.25 \[ \frac {16 \, b^{4} x^{4} - 32 \, a b^{3} x^{3} - 2 \, \pi ^{4} + 32 \, a^{4} + 4 \, {\left (\pi ^{2} b^{2} - 28 \, a^{2} b^{2}\right )} x^{2} - 8 \, {\left (5 \, \pi ^{2} a b + 4 \, a^{3} b\right )} x - 48 \, {\left (4 \, \pi a b^{2} x^{2} + 8 \, \pi a^{2} b x + \pi ^{3} a + 4 \, \pi a^{3}\right )} \arctan \left (-\frac {2 \, b x + 2 \, a - \sqrt {4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}}}{\pi }\right ) - 3 \, {\left (\pi ^{4} - 16 \, a^{4} + 4 \, {\left (\pi ^{2} b^{2} - 4 \, a^{2} b^{2}\right )} x^{2} + 8 \, {\left (\pi ^{2} a b - 4 \, a^{3} b\right )} x\right )} \log \left (4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}\right )}{8 \, {\left (4 \, b^{6} x^{2} + 8 \, a b^{5} x + \pi ^{2} b^{4} + 4 \, a^{2} b^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(16*b^4*x^4 - 32*a*b^3*x^3 - 2*pi^4 + 32*a^4 + 4*(pi^2*b^2 - 28*a^2*b^2)*x^2 - 8*(5*pi^2*a*b + 4*a^3*b)*x
- 48*(4*pi*a*b^2*x^2 + 8*pi*a^2*b*x + pi^3*a + 4*pi*a^3)*arctan(-(2*b*x + 2*a - sqrt(4*b^2*x^2 + 8*a*b*x + pi^
2 + 4*a^2))/pi) - 3*(pi^4 - 16*a^4 + 4*(pi^2*b^2 - 4*a^2*b^2)*x^2 + 8*(pi^2*a*b - 4*a^3*b)*x)*log(4*b^2*x^2 +
8*a*b*x + pi^2 + 4*a^2))/(4*b^6*x^2 + 8*a*b^5*x + pi^2*b^4 + 4*a^2*b^4)

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

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 = 1.48, size = 29109, normalized size = 388.12 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [C]  time = 0.76, size = 124, normalized size = 1.65 \[ \frac {4 \, {\left (4 \, b^{3} x^{3} + i \, \pi ^{3} - 6 \, \pi ^{2} a - 12 i \, \pi a^{2} + 8 \, a^{3} + {\left (6 i \, \pi b^{2} - 12 \, a b^{2}\right )} x^{2} + {\left (4 \, \pi ^{2} b + 16 i \, \pi a b - 16 \, a^{2} b\right )} x\right )}}{32 \, b^{5} x - 16 i \, \pi b^{4} + 32 \, a b^{4}} - \frac {{\left (3 \, \pi ^{2} + 12 i \, \pi a - 12 \, a^{2}\right )} \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{4 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*(4*b^3*x^3 + I*pi^3 - 6*pi^2*a - 12*I*pi*a^2 + 8*a^3 + (6*I*pi*b^2 - 12*a*b^2)*x^2 + (4*pi^2*b + 16*I*pi*a*b
- 16*a^2*b)*x)/(32*b^5*x - 16*I*pi*b^4 + 32*a*b^4) - 1/4*(3*pi^2 + 12*I*pi*a - 12*a^2)*log(-I*pi + 2*b*x + 2*
a)/b^4

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mupad [B]  time = 0.17, size = 490, normalized size = 6.53 \[ \frac {x^2}{2\,b^2}+\frac {\ln \left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\right )\,\left (3\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2-12\,a\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )+12\,a^2\right )}{4\,b^4}-\frac {{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^3-8\,a^3-6\,a\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2+12\,a^2\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}{4\,b\,\left (2\,a\,b^3+2\,b^4\,x-b^3\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )\right )}+\frac {x\,\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}{b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/acoth(tanh(a + b*x))^2,x)

[Out]

x^2/(2*b^2) + (log(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) - log(-2/(exp(2*a)*exp(2*b*x) - 1)))
*(3*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)
^2 - 12*a*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x) - 1)) +
2*b*x) + 12*a^2))/(4*b^4) - ((2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*
exp(2*b*x) - 1)) + 2*b*x)^3 - 8*a^3 - 6*a*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(
-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)^2 + 12*a^2*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)
) + log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x))/(4*b*(2*a*b^3 + 2*b^4*x - b^3*(2*a - log((2*exp(2*a)*exp(2*b*x
))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x))) + (x*(log(-2/(exp(2*a)*exp(2*b*x)
- 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x))/b^3

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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