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

Optimal. Leaf size=21 \[ b x-\log (x) \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \]

[Out]

b*x-(b*x-arccoth(tanh(b*x+a)))*ln(x)

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Rubi [A]  time = 0.04, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2158, 29} \[ b x-\log (x) \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Tanh[a + b*x]]/x,x]

[Out]

b*x - (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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 19, normalized size = 0.90 \[ \log (x) \left (\coth ^{-1}(\tanh (a+b x))-b x\right )+b x \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[Tanh[a + b*x]]/x,x]

[Out]

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

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fricas [A]  time = 0.48, size = 8, normalized size = 0.38 \[ b x + a \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*x + a*log(x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.42, size = 354, normalized size = 16.86 \[ \ln \relax (x ) \ln \left ({\mathrm e}^{b x +a}\right )-\ln \relax (x ) x b +b x -\frac {i \ln \relax (x ) \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}}{2}+\frac {i \ln \relax (x ) \pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}}{4}+\frac {i \ln \relax (x ) \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}}{4}-\frac {i \ln \relax (x ) \pi \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )}{4}-\frac {i \pi \ln \relax (x )}{2}-\frac {i \ln \relax (x ) \pi \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}}{4}+\frac {i \ln \relax (x ) \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}}{2}+\frac {i \ln \relax (x ) \pi \,\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}}{2}-\frac {i \ln \relax (x ) \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}}{4}-\frac {i \ln \relax (x ) \pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(x)*ln(exp(b*x+a))-ln(x)*x*b+b*x-1/2*I*ln(x)*Pi*csgn(I/(exp(2*b*x+2*a)+1))^3+1/4*I*ln(x)*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+1/4*I*ln(x)*Pi*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+
2*a)/(exp(2*b*x+2*a)+1))^2-1/4*I*ln(x)*Pi*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-1/2*I*Pi*ln(x)-1/4*I*ln(
x)*Pi*csgn(I*exp(2*b*x+2*a))^3+1/2*I*ln(x)*Pi*csgn(I/(exp(2*b*x+2*a)+1))^2+1/2*I*ln(x)*Pi*csgn(I*exp(b*x+a))*c
sgn(I*exp(2*b*x+2*a))^2-1/4*I*ln(x)*Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3-1/4*I*ln(x)*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))

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maxima [A]  time = 0.33, size = 34, normalized size = 1.62 \[ -b {\left (x + \frac {a}{b}\right )} \log \relax (x) + b {\left (x + \frac {a \log \relax (x)}{b}\right )} + \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right ) \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.18, size = 59, normalized size = 2.81 \[ b\,x-\ln \relax (x)\,\left (\frac {\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{2}-\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{2}+b\,x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(tanh(a + b*x))/x,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acoth(tanh(a + b*x))/x, x)

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