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

Optimal. Leaf size=77 \[ b x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2-\frac {1}{2} \coth ^{-1}(\tanh (a+b x))^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )+\frac {1}{3} \coth ^{-1}(\tanh (a+b x))^3-\log (x) \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 77, 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 = {2159, 2158, 29} \[ b x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2-\frac {1}{2} \coth ^{-1}(\tanh (a+b x))^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )+\frac {1}{3} \coth ^{-1}(\tanh (a+b x))^3-\log (x) \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Tanh[a + b*x]]^3/x,x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 104, normalized size = 1.35 \[ (a+b x) \left (a^2-3 a \left (-\coth ^{-1}(\tanh (a+b x))+a+b x\right )+3 \left (-\coth ^{-1}(\tanh (a+b x))+a+b x\right )^2\right )+\frac {1}{3} (a+b x)^3-\frac {1}{2} (a+b x)^2 \left (-3 \coth ^{-1}(\tanh (a+b x))+2 a+3 b x\right )+\log (b x) \left (\coth ^{-1}(\tanh (a+b x))-b x\right )^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.67, size = 49, normalized size = 0.64 \[ \frac {1}{3} \, b^{3} x^{3} + \frac {3}{2} \, a b^{2} x^{2} - \frac {3}{4} \, {\left (\pi ^{2} b - 4 \, a^{2} b\right )} x - \frac {1}{4} \, {\left (3 \, \pi ^{2} a - 4 \, a^{3}\right )} \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.90, size = 21848, normalized size = 283.74 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [C]  time = 0.76, size = 75, normalized size = 0.97 \[ \frac {1}{3} \, b^{3} x^{3} + \frac {1}{24} \, {\left (-18 i \, \pi b^{2} + 36 \, a b^{2}\right )} x^{2} - \frac {1}{24} \, {\left (18 \, \pi ^{2} b + 72 i \, \pi a b - 72 \, a^{2} b\right )} x + \frac {1}{8} \, {\left (i \, \pi ^{3} - 6 \, \pi ^{2} a - 12 i \, \pi a^{2} + 8 \, a^{3}\right )} \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b^3*x^3 + 1/24*(-18*I*pi*b^2 + 36*a*b^2)*x^2 - 1/24*(18*pi^2*b + 72*I*pi*a*b - 72*a^2*b)*x + 1/8*(I*pi^3 -
6*pi^2*a - 12*I*pi*a^2 + 8*a^3)*log(x)

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mupad [B]  time = 0.14, size = 306, normalized size = 3.97 \[ \frac {b^3\,x^3}{3}-\ln \relax (x)\,\left (\frac {{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^3}{8}-a^3-\frac {3\,a\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2}{4}+\frac {3\,a^2\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}{2}\right )-\frac {3\,b^2\,x^2\,\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}{4}+\frac {3\,b\,x\,{\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(tanh(a + b*x))^3/x,x)

[Out]

(b^3*x^3)/3 - log(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 - (3*a*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(ex
p(2*a)*exp(2*b*x) - 1)) + 2*b*x)^2)/4 + (3*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))/2) - (3*b^2*x^2*(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))/4 + (3*b*x*(log(-2/(exp(2*a)*exp(2*b*x) - 1)) - log((2*ex
p(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)^2)/4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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