\(\int \frac {\coth ^{-1}(\coth (a+b x))}{x} \, dx\) [197]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.04 (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} \[ \int \frac {\coth ^{-1}(\coth (a+b x))}{x} \, dx=b x-\log (x) \left (b x-\coth ^{-1}(\coth (a+b x))\right ) \]

[In]

Int[ArcCoth[Coth[a + b*x]]/x,x]

[Out]

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

Mathematica [A] (verified)

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

[In]

Integrate[ArcCoth[Coth[a + b*x]]/x,x]

[Out]

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

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00

method	result	size
default	\(\ln \left (x \right ) \operatorname {arccoth}\left (\coth \left (b x +a \right )\right )+b \left (-x \ln \left (x \right )+x \right )\)	\(21\)
parts	\(\ln \left (x \right ) \operatorname {arccoth}\left (\coth \left (b x +a \right )\right )+b \left (-x \ln \left (x \right )+x \right )\)	\(21\)
risch	\(\ln \left (x \right ) \ln \left ({\mathrm e}^{b x +a}\right )-b \ln \left (x \right ) x +b x -\frac {i \pi \left (\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )-2 \,\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}+\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}-1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}-1}\right )-\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}-1}\right )^{2}-\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}-1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}-1}\right )^{2}+\operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}-1}\right )^{3}\right ) \ln \left (x \right )}{4}\)	\(271\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.38 \[ \int \frac {\coth ^{-1}(\coth (a+b x))}{x} \, dx=b x + a \log \left (x\right ) \]

[In]

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

[Out]

b*x + a*log(x)

Sympy [F]

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

[In]

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

[Out]

Integral(acoth(coth(a + b*x))/x, x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.38 \[ \int \frac {\coth ^{-1}(\coth (a+b x))}{x} \, dx=b x + a \log \left (x\right ) \]

[In]

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

[Out]

b*x + a*log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.43 \[ \int \frac {\coth ^{-1}(\coth (a+b x))}{x} \, dx=b x + a \log \left ({\left | x \right |}\right ) \]

[In]

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

[Out]

b*x + a*log(abs(x))

Mupad [B] (verification not implemented)

Time = 4.26 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.76 \[ \int \frac {\coth ^{-1}(\coth (a+b x))}{x} \, dx=b\,x-\frac {\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (x\right )}{2}+\frac {\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (x\right )}{2}-b\,x\,\ln \left (x\right ) \]

[In]

int(acoth(coth(a + b*x))/x,x)

[Out]

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