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

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Rubi [A]  time = 0.03, antiderivative size = 39, normalized size of antiderivative = 1.00, 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 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))^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*}

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Mathematica [A]  time = 0.05, 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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fricas [A]  time = 0.58, size = 29, normalized size = 0.74 \[ \frac {4 \, b^{2} x^{2} + 8 \, a b x \log \relax (x) + \pi ^{2} - 4 \, a^{2}}{4 \, x} \]

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

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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maple [C]  time = 0.27, size = 1095, normalized size = 28.08 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^2*x-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
(b*x+a))^2*csgn(I*exp(2*b*x+2*a))+I*Pi*ln(x)*b*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2+I*Pi*ln(exp(b*x+a))
/x-1/x*ln(exp(b*x+a))^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))^
2-1/2*I*Pi*ln(x)*b*csgn(I*exp(b*x+a))^2*csgn(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)
)*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-2*ln(x)*x*b^2+2*ln(x)*ln(exp(b*x+a))*b+1/2*I*Pi*ln(x)*b*csgn(I*e
xp(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(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)+1))*csgn(I*exp(2*b*x+2*a
))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))-I*Pi*ln(x)*b+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))+1/16*Pi^2*(-2*csgn(I/(exp(2*b*x+2*a)+1))^2+c
sgn(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*exp(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(ex
p(b*x+a))/x*csgn(I*exp(2*b*x+2*a))^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*P
i*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))^3-1/2*I*Pi*ln(x)*b*csg
n(I*exp(2*b*x+2*a))^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(x)*b*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3

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maxima [A]  time = 0.38, size = 54, normalized size = 1.38 \[ 2 \, b \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right ) \log \relax (x) - 2 \, {\left (b {\left (x + \frac {a}{b}\right )} \log \relax (x) - b {\left (x + \frac {a \log \relax (x)}{b}\right )}\right )} b - \frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}{x} \]

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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mupad [B]  time = 1.25, size = 207, normalized size = 5.31 \[ b\,\ln \left (\frac {{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\frac {{\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2}{4\,x}-b\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\frac {{\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2}{4\,x}+b\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,\ln \relax (x)-b\,\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,\ln \relax (x)+\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{2\,x}-2\,b^2\,x\,\ln \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(tanh(a + b*x))^2/x^2,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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