3.2.32 \(\int \frac {\coth ^{-1}(\tanh (a+b x))}{x} \, dx\) [132]

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

[Out]

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

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Rubi [A]
time = 0.03, 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 = {2189, 29} \begin {gather*} b x-\log (x) \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \end {gather*}

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 2189

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 \begin {gather*} b x+\left (-b x+\coth ^{-1}(\tanh (a+b x))\right ) \log (x) \end {gather*}

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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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.25, size = 354, normalized size = 16.86

method	result	size
risch	\(\ln \left (x \right ) \ln \left ({\mathrm e}^{b x +a}\right )-\ln \left (x \right ) x b +b x -\frac {i \pi \ln \left (x \right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}}{4}+\frac {i \pi \ln \left (x \right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}}{4}-\frac {i \pi \ln \left (x \right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )}{4}+\frac {i \pi \ln \left (x \right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}}{2}+\frac {i \pi \ln \left (x \right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}}{4}-\frac {i \pi \ln \left (x \right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}}{4}-\frac {i \pi \ln \left (x \right )}{2}+\frac {i \pi \ln \left (x \right ) \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}}{2}-\frac {i \pi \ln \left (x \right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}}{2}-\frac {i \pi \ln \left (x \right ) \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )}{4}\)	\(354\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 34, normalized size = 1.62 \begin {gather*} -b {\left (x + \frac {a}{b}\right )} \log \left (x\right ) + b {\left (x + \frac {a \log \left (x\right )}{b}\right )} + \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right ) \log \left (x\right ) \end {gather*}

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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Fricas [A]
time = 0.38, size = 8, normalized size = 0.38 \begin {gather*} b x + a \log \left (x\right ) \end {gather*}

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

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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Giac [C] Result contains complex when optimal does not.
time = 0.40, size = 15, normalized size = 0.71 \begin {gather*} b x + \frac {1}{2} \, {\left (i \, \pi + 2 \, a\right )} \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*x + 1/2*(I*pi + 2*a)*log(x)

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Mupad [B]
time = 0.18, size = 59, normalized size = 2.81 \begin {gather*} b\,x-\ln \left (x\right )\,\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 ) \end {gather*}

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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