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

Optimal. Leaf size=68 \[ -3 b^2 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )-\frac {\coth ^{-1}(\tanh (a+b x))^3}{x}+\frac {3}{2} b \coth ^{-1}(\tanh (a+b x))^2+3 b \log (x) \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2168, 2159, 2158, 29} \[ -3 b^2 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )-\frac {\coth ^{-1}(\tanh (a+b x))^3}{x}+\frac {3}{2} b \coth ^{-1}(\tanh (a+b x))^2+3 b \log (x) \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Tanh[a + b*x]]^3/x^2,x]

[Out]

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

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Mathematica [A]  time = 0.04, size = 62, normalized size = 0.91 \[ -6 b^2 x \log (x) \coth ^{-1}(\tanh (a+b x))-\frac {\coth ^{-1}(\tanh (a+b x))^3}{x}+3 b (\log (x)+1) \coth ^{-1}(\tanh (a+b x))^2+\frac {3}{2} b^3 x^2 (2 \log (x)-1) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcCoth[Tanh[a + b*x]]^3/x) - 6*b^2*x*ArcCoth[Tanh[a + b*x]]*Log[x] + 3*b*ArcCoth[Tanh[a + b*x]]^2*(1 + Log[
x]) + (3*b^3*x^2*(-1 + 2*Log[x]))/2

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fricas [A]  time = 0.46, size = 51, normalized size = 0.75 \[ \frac {2 \, b^{3} x^{3} + 12 \, a b^{2} x^{2} + 3 \, \pi ^{2} a - 4 \, a^{3} - 3 \, {\left (\pi ^{2} b - 4 \, a^{2} b\right )} x \log \relax (x)}{4 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.46, size = 7683, normalized size = 112.99 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [C]  time = 0.61, size = 124, normalized size = 1.82 \[ 3 \, b \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2} \log \relax (x) - \frac {3}{2} \, {\left (2 \, \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2} \log \relax (x) - {\left (b x^{2} - 2 \, {\left (-i \, \pi - 2 \, a\right )} x + 2 \, {\left (-\frac {i \, \pi {\left (b x + a\right )}}{b} - \frac {{\left (b x + a\right )}^{2}}{b}\right )} \log \relax (x) + \frac {2 \, \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2} \log \relax (x)}{b} + \frac {2 \, {\left (i \, \pi a + a^{2}\right )} \log \relax (x)}{b}\right )} b\right )} b - \frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{3}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.20, size = 372, normalized size = 5.47 \[ \ln \relax (x)\,\left (3\,a^2\,b+\frac {3\,b\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2}{4}-3\,a\,b\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )\right )+\frac {{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^3-8\,a^3-6\,a\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2+12\,a^2\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}{8\,x}+\frac {b^3\,x^2}{2}-\frac {3\,b^2\,x\,\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(tanh(a + b*x))^3/x^2,x)

[Out]

log(x)*(3*a^2*b + (3*b*(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)/4 - 3*a*b*(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*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))/(8*x) + (b^3*x^2)/2 - (3*b^2*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))/2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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