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

Optimal result

Integrand size = 11, antiderivative size = 21 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x} \, dx=b x-\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2189, 29} \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x} \, dx=b x-\log (x) \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \]

[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*} \text {integral}& = 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*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x} \, dx=b x+\left (-b x+\coth ^{-1}(\tanh (a+b x))\right ) \log (x) \]

[In]

Integrate[ArcCoth[Tanh[a + b*x]]/x,x]

[Out]

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

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00

method	result	size
default	\(\ln \left (x \right ) \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )+b \left (-x \ln \left (x \right )+x \right )\)	\(21\)
parts	\(\ln \left (x \right ) \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )+b \left (-x \ln \left (x \right )+x \right )\)	\(21\)
risch	\(\ln \left (x \right ) \ln \left ({\mathrm e}^{b x +a}\right )-b \ln \left (x \right ) x +b x -\frac {i \pi \left (\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )-\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+2 \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}-2 \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )-2 \,\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+2\right ) \ln \left (x \right )}{4}\)	\(314\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x} \, dx=b x + \frac {1}{2} \, {\left (i \, \pi + 2 \, a\right )} \log \left (x\right ) \]

[In]

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

[Out]

b*x + 1/2*(I*pi + 2*a)*log(x)

Sympy [F]

\[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x} \, dx=\int \frac {\operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{x}\, dx \]

[In]

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

[Out]

Integral(acoth(tanh(a + b*x))/x, x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x} \, dx=-b {\left (x + \frac {a}{b}\right )} \log \left (x\right ) + b {\left (x + \frac {a \log \left (x\right )}{b}\right )} + \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right ) \log \left (x\right ) \]

[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)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.31 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x} \, dx=b x + \frac {1}{2} \, {\left (i \, \pi + 2 \, a\right )} \log \left (x\right ) \]

[In]

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

[Out]

b*x + 1/2*(I*pi + 2*a)*log(x)

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.81 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x} \, dx=b\,x-\ln \left (x\right )\,\left (\frac {\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{2}-\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{2}+b\,x\right ) \]

[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)