3.143 \(\int \frac{\coth ^{-1}(\tanh (a+b x))^2}{x^3} \, dx\)
Optimal. Leaf size=36 \[ -\frac{\coth ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac{b \coth ^{-1}(\tanh (a+b x))}{x}+b^2 \log (x) \]
[Out]
-((b*ArcCoth[Tanh[a + b*x]])/x) - ArcCoth[Tanh[a + b*x]]^2/(2*x^2) + b^2*Log[x]
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Rubi [A] time = 0.0228216, antiderivative size = 36, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used =
{2168, 29} \[ -\frac{\coth ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac{b \coth ^{-1}(\tanh (a+b x))}{x}+b^2 \log (x) \]
Antiderivative was successfully verified.
[In]
Int[ArcCoth[Tanh[a + b*x]]^2/x^3,x]
[Out]
-((b*ArcCoth[Tanh[a + b*x]])/x) - ArcCoth[Tanh[a + b*x]]^2/(2*x^2) + b^2*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 29
Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]
Rubi steps
\begin{align*} \int \frac{\coth ^{-1}(\tanh (a+b x))^2}{x^3} \, dx &=-\frac{\coth ^{-1}(\tanh (a+b x))^2}{2 x^2}+b \int \frac{\coth ^{-1}(\tanh (a+b x))}{x^2} \, dx\\ &=-\frac{b \coth ^{-1}(\tanh (a+b x))}{x}-\frac{\coth ^{-1}(\tanh (a+b x))^2}{2 x^2}+b^2 \int \frac{1}{x} \, dx\\ &=-\frac{b \coth ^{-1}(\tanh (a+b x))}{x}-\frac{\coth ^{-1}(\tanh (a+b x))^2}{2 x^2}+b^2 \log (x)\\ \end{align*}
Mathematica [A] time = 0.0337682, size = 42, normalized size = 1.17 \[ -\frac{2 b x \coth ^{-1}(\tanh (a+b x))+\coth ^{-1}(\tanh (a+b x))^2-b^2 x^2 (2 \log (x)+3)}{2 x^2} \]
Antiderivative was successfully verified.
[In]
Integrate[ArcCoth[Tanh[a + b*x]]^2/x^3,x]
[Out]
-(2*b*x*ArcCoth[Tanh[a + b*x]] + ArcCoth[Tanh[a + b*x]]^2 - b^2*x^2*(3 + 2*Log[x]))/(2*x^2)
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Maple [C] time = 0.375, size = 3213, normalized size = 89.3 \begin{align*} \text{output 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^3,x)
[Out]
-1/2/x^2*ln(exp(b*x+a))^2-1/4*(4*b*x-2*I*Pi*csgn(I/(exp(2*b*x+2*a)+1))^3+I*Pi*csgn(I*exp(2*b*x+2*a))*csgn(I*ex
p(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-I*Pi*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))+2*I*Pi*csgn(I/(exp(2*b*x+2
*a)+1))^2-2*I*Pi-I*Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3-I*Pi*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*I*Pi*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2-I*Pi*c
sgn(I*exp(2*b*x+2*a))^3+I*Pi*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)/x^2*ln(ex
p(b*x+a))+1/32*(6*Pi^2*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))^4+8*I*Pi*b*x*csgn(I/(exp(2*b*x+2*a)+1))*csg
n(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))+4*Pi^2+4*Pi^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2
*a)+1))^3+Pi^2*csgn(I*exp(2*b*x+2*a))^6+4*Pi^2*csgn(I*exp(2*b*x+2*a))^3-4*Pi^2*csgn(I*exp(2*b*x+2*a))*csgn(I*e
xp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+8*I*Pi*b*x*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+8*I*Pi*b*x*csgn(I*e
xp(2*b*x+2*a))^3-4*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))^4*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+4*Pi^2*csgn(I
/(exp(2*b*x+2*a)+1))^3*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+4*Pi^2*csgn(I*exp(2*b*x+2*a))^3*csgn(I/(exp
(2*b*x+2*a)+1))^3+Pi^2*csgn(I/(exp(2*b*x+2*a)+1))^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^4+32*b^2*x^2*ln(
x)+4*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))^6-2*Pi^2*csgn(I*exp(2*b*x+2*a))^4*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1
))^2+2*Pi^2*csgn(I*exp(2*b*x+2*a))^3*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3-4*Pi^2*csgn(I*exp(2*b*x+2*a))
^3*csgn(I/(exp(2*b*x+2*a)+1))^2+4*Pi^2*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)+1))^2-8*Pi^2*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2-8*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))^
2+8*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))^3-4*Pi^2*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^5-2*Pi^2*csgn(I*exp(2*b
*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^5+Pi^2*csgn(I*exp(2*b*x+2*a))^2*csgn(I*exp(2*b*x+2*a)/(exp(
2*b*x+2*a)+1))^4-2*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^5+16*I*Pi*x*b+4*P
i^2*csgn(I/(exp(2*b*x+2*a)+1))^3*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-4*Pi^2*csgn(I*exp(2*b*x+2*a)/(exp
(2*b*x+2*a)+1))^3*csgn(I/(exp(2*b*x+2*a)+1))^2+4*Pi^2*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-16*I*Pi*x*b*
csgn(I/(exp(2*b*x+2*a)+1))^2+16*I*Pi*x*b*csgn(I/(exp(2*b*x+2*a)+1))^3-4*Pi^2*csgn(I*exp(b*x+a))^3*csgn(I*exp(2
*b*x+2*a))^3-4*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-4*Pi^2*csgn(I/(exp(
2*b*x+2*a)+1))^3*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))+Pi^2*csgn(I*exp(b*x+a))^4*cs
gn(I*exp(2*b*x+2*a))^2-8*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))^5+4*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))^4-2*Pi^2*csgn(I/(
exp(2*b*x+2*a)+1))^2*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+2*Pi^2*csgn(I/(exp(2*b
*x+2*a)+1))*csgn(I*exp(2*b*x+2*a))^4*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))-2*Pi^2*csgn(I/(exp(2*b*x+2*a)+1
))*csgn(I*exp(2*b*x+2*a))^3*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-2*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn
(I*exp(2*b*x+2*a))^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+4*Pi^2*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*
a))^3*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+8*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))^2*csgn(I*exp(b*x+a))*csgn(
I*exp(2*b*x+2*a))^2+4*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))^4*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*
x+2*a)+1))+4*Pi^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))-8*Pi^2*csgn(I/(exp(
2*b*x+2*a)+1))^3*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2-4*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))^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+Pi^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^6+4*Pi^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))-2*Pi^2*csgn(I*exp(
b*x+a))^2*csgn(I*exp(2*b*x+2*a))^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-16*I*Pi*b*x*csgn(I*exp(b*x+a))*
csgn(I*exp(2*b*x+2*a))^2-8*I*Pi*b*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+8*I*Pi*
x*b*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-4*Pi^2*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2*csgn(I*exp(
2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+2*Pi^2*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp
(2*b*x+2*a)+1))^3-8*I*Pi*b*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))^2-4*Pi^2*csg
n(I/(exp(2*b*x+2*a)+1))^2*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))+4*Pi^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))^4+Pi^2*csgn(I/(exp(2*b*x+2*a)+1))^2*csgn(I*exp(2*b*
x+2*a))^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-2*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(b*x+a))^2*c
sgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+4*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp
(b*x+a))*csgn(I*exp(2*b*x+2*a))^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-4*Pi^2*csgn(I/(exp(2*b*x+2*a)+1)
)*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^3*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))+2*Pi^2*csgn(I/(exp(2*b
*x+2*a)+1))*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1)))/x^2
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Maxima [A] time = 1.40233, size = 46, normalized size = 1.28 \begin{align*} b^{2} \log \left (x\right ) - \frac{b \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )}{x} - \frac{\operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}{2 \, x^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(arccoth(tanh(b*x+a))^2/x^3,x, algorithm="maxima")
[Out]
b^2*log(x) - b*arccoth(tanh(b*x + a))/x - 1/2*arccoth(tanh(b*x + a))^2/x^2
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Fricas [A] time = 1.6086, size = 73, normalized size = 2.03 \begin{align*} \frac{8 \, b^{2} x^{2} \log \left (x\right ) - 16 \, a b x + \pi ^{2} - 4 \, a^{2}}{8 \, x^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(arccoth(tanh(b*x+a))^2/x^3,x, algorithm="fricas")
[Out]
1/8*(8*b^2*x^2*log(x) - 16*a*b*x + pi^2 - 4*a^2)/x^2
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Sympy [A] time = 0.752501, size = 32, normalized size = 0.89 \begin{align*} b^{2} \log{\left (x \right )} - \frac{b \operatorname{acoth}{\left (\tanh{\left (a + b x \right )} \right )}}{x} - \frac{\operatorname{acoth}^{2}{\left (\tanh{\left (a + b x \right )} \right )}}{2 x^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(acoth(tanh(b*x+a))**2/x**3,x)
[Out]
b**2*log(x) - b*acoth(tanh(a + b*x))/x - acoth(tanh(a + b*x))**2/(2*x**2)
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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^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(arccoth(tanh(b*x+a))^2/x^3,x, algorithm="giac")
[Out]
integrate(arccoth(tanh(b*x + a))^2/x^3, x)