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

Optimal. Leaf size=39 \[ -\frac{\coth ^{-1}(\tanh (a+b x))^2}{x}-2 b \log (x) \left (b x-\coth ^{-1}(\tanh (a+b x))\right )+2 b^2 x \]

[Out]

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

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Rubi [A]  time = 0.0258467, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2168, 2158, 29} \[ -\frac{\coth ^{-1}(\tanh (a+b x))^2}{x}-2 b \log (x) \left (b x-\coth ^{-1}(\tanh (a+b x))\right )+2 b^2 x \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Tanh[a + b*x]]^2/x^2,x]

[Out]

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

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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rubi steps

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

Mathematica [A]  time = 0.0489712, size = 37, normalized size = 0.95 \[ -\frac{\coth ^{-1}(\tanh (a+b x))^2}{x}+2 b (\log (x)+1) \coth ^{-1}(\tanh (a+b x))-2 b^2 x \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.21, size = 1095, normalized size = 28.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*Pi*ln(exp(b*x+a))/x*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2-1/2*I*Pi*ln(exp(b*x+a))/x*csgn(I*exp(2*b*x+
2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+1/2*I*Pi*ln(x)*b*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x
+2*a)/(exp(2*b*x+2*a)+1))^2+I*Pi*ln(x)*b*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2+2*b^2*x-1/2*I*Pi*ln(x)*b*
csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))+1/2*I*Pi*ln(exp(b*
x+a))/x*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))-2*ln(x)*x*
b^2+2*ln(x)*ln(exp(b*x+a))*b-I*Pi*ln(x)*b*csgn(I/(exp(2*b*x+2*a)+1))^3-I*Pi*ln(exp(b*x+a))/x*csgn(I/(exp(2*b*x
+2*a)+1))^2+I*Pi*ln(x)*b*csgn(I/(exp(2*b*x+2*a)+1))^2+1/2*I*Pi*ln(exp(b*x+a))/x*csgn(I*exp(2*b*x+2*a))^3+1/16*
Pi^2*(-2*csgn(I/(exp(2*b*x+2*a)+1))^2+csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/
(exp(2*b*x+2*a)+1))-csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+2*csgn(I/(exp(2*b*x
+2*a)+1))^3+csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-2*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2+csgn(I*e
xp(2*b*x+2*a))^3-csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+csgn(I*exp(2*b*x+2*a)/(exp
(2*b*x+2*a)+1))^3+2)^2/x+1/2*I*Pi*ln(exp(b*x+a))/x*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))+1/2*I*Pi*ln(x)*
b*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-1/2*I*Pi*ln(x)*b*csgn(I*exp(b*x+a))^2*csg
n(I*exp(2*b*x+2*a))-1/2*I*Pi*ln(exp(b*x+a))/x*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)
+1))^2-1/2*I*Pi*ln(x)*b*csgn(I*exp(2*b*x+2*a))^3-1/2*I*Pi*ln(x)*b*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+
1/2*I*Pi*ln(exp(b*x+a))/x*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+I*Pi*ln(exp(b*x+a))/x-I*Pi*ln(x)*b+I*Pi*
ln(exp(b*x+a))/x*csgn(I/(exp(2*b*x+2*a)+1))^3-1/x*ln(exp(b*x+a))^2

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Maxima [A]  time = 1.19632, size = 73, normalized size = 1.87 \begin{align*} 2 \, b \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right ) \log \left (x\right ) - 2 \,{\left (b{\left (x + \frac{a}{b}\right )} \log \left (x\right ) - b{\left (x + \frac{a \log \left (x\right )}{b}\right )}\right )} b - \frac{\operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.67534, size = 69, normalized size = 1.77 \begin{align*} \frac{4 \, b^{2} x^{2} + 8 \, a b x \log \left (x\right ) + \pi ^{2} - 4 \, a^{2}}{4 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{acoth}^{2}{\left (\tanh{\left (a + b x \right )} \right )}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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