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

Optimal. Leaf size=60 \[ -3 b^2 \log (x) \left (b x-\coth ^{-1}(\tanh (a+b x))\right )-\frac {\coth ^{-1}(\tanh (a+b x))^3}{2 x^2}-\frac {3 b \coth ^{-1}(\tanh (a+b x))^2}{2 x}+3 b^3 x \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 60, 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 = {2168, 2158, 29} \[ -3 b^2 \log (x) \left (b x-\coth ^{-1}(\tanh (a+b x))\right )-\frac {\coth ^{-1}(\tanh (a+b x))^3}{2 x^2}-\frac {3 b \coth ^{-1}(\tanh (a+b x))^2}{2 x}+3 b^3 x \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Tanh[a + b*x]]^3/x^3,x]

[Out]

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

Rule 29

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

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Mathematica [A]  time = 0.04, size = 66, normalized size = 1.10 \[ 3 b^2 \log (x) \left (\coth ^{-1}(\tanh (a+b x))-b x\right )-\frac {\left (\coth ^{-1}(\tanh (a+b x))-b x\right )^3}{2 x^2}-\frac {3 b \left (\coth ^{-1}(\tanh (a+b x))-b x\right )^2}{x}+b^3 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.51, size = 51, normalized size = 0.85 \[ \frac {8 \, b^{3} x^{3} + 24 \, a b^{2} x^{2} \log \relax (x) + 3 \, \pi ^{2} a - 4 \, a^{3} + 6 \, {\left (\pi ^{2} b - 4 \, a^{2} b\right )} x}{8 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(8*b^3*x^3 + 24*a*b^2*x^2*log(x) + 3*pi^2*a - 4*a^3 + 6*(pi^2*b - 4*a^2*b)*x)/x^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.55, size = 7366, normalized size = 122.77 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [A]  time = 0.46, size = 72, normalized size = 1.20 \[ 3 \, {\left (b \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right ) \log \relax (x) - {\left (b {\left (x + \frac {a}{b}\right )} \log \relax (x) - b {\left (x + \frac {a \log \relax (x)}{b}\right )}\right )} b\right )} b - \frac {3 \, b \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}{2 \, x} - \frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{3}}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.33, size = 383, normalized size = 6.38 \[ \frac {{\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^3}{16\,x^2}-\frac {{\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^3}{16\,x^2}+\frac {9\,b^2\,\ln \left (\frac {{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{4}-\frac {9\,b^2\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{4}-\frac {3\,b^3\,x}{2}-\frac {3\,b\,{\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2}{8\,x}+\frac {3\,b^2\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,\ln \relax (x)}{2}-3\,b^3\,x\,\ln \relax (x)-\frac {3\,b\,{\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2}{8\,x}-\frac {3\,b^2\,\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,\ln \relax (x)}{2}-\frac {3\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,{\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2}{16\,x^2}+\frac {3\,{\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2\,\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{16\,x^2}+\frac {3\,b\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{4\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(tanh(a + b*x))^3/x^3,x)

[Out]

log(-2/(exp(2*a)*exp(2*b*x) - 1))^3/(16*x^2) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))^3/(16*x^
2) + (9*b^2*log(exp(2*b*x)/(exp(2*a)*exp(2*b*x) - 1)))/4 - (9*b^2*log(1/(exp(2*a)*exp(2*b*x) - 1)))/4 - (3*b^3
*x)/2 - (3*b*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))^2)/(8*x) + (3*b^2*log((2*exp(2*a)*exp(2*b*
x))/(exp(2*a)*exp(2*b*x) - 1))*log(x))/2 - 3*b^3*x*log(x) - (3*b*log(-2/(exp(2*a)*exp(2*b*x) - 1))^2)/(8*x) -
(3*b^2*log(-2/(exp(2*a)*exp(2*b*x) - 1))*log(x))/2 - (3*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)/(16*x^2) + (3*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))^2*l
og(-2/(exp(2*a)*exp(2*b*x) - 1)))/(16*x^2) + (3*b*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)))/(4*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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