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[a + b*x]])*Log[x]

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Rubi [A]  time = 0.0439774, antiderivative size = 21, normalized size of antiderivative = 1., 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 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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], 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*}

Mathematica [A]  time = 0.0161214, size = 19, normalized size = 0.9 \[ \log (x) \left (\coth ^{-1}(\tanh (a+b x))-b x\right )+b x \]

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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Maple [C]  time = 0.189, size = 354, normalized size = 16.9 \begin{align*} \ln \left ( x \right ) \ln \left ({{\rm e}^{bx+a}} \right ) -\ln \left ( x \right ) xb+bx+{\frac{i}{2}}\pi \,\ln \left ( x \right ){\it csgn} \left ( i{{\rm e}^{bx+a}} \right ) \left ({\it csgn} \left ( i{{\rm e}^{2\,bx+2\,a}} \right ) \right ) ^{2}-{\frac{i}{2}}\pi \,\ln \left ( x \right ) +{\frac{i}{4}}\pi \,\ln \left ( x \right ){\it csgn} \left ({\frac{i}{{{\rm e}^{2\,bx+2\,a}}+1}} \right ) \left ({\it csgn} \left ({\frac{i{{\rm e}^{2\,bx+2\,a}}}{{{\rm e}^{2\,bx+2\,a}}+1}} \right ) \right ) ^{2}+{\frac{i}{4}}\pi \,\ln \left ( x \right ){\it csgn} \left ( i{{\rm e}^{2\,bx+2\,a}} \right ) \left ({\it csgn} \left ({\frac{i{{\rm e}^{2\,bx+2\,a}}}{{{\rm e}^{2\,bx+2\,a}}+1}} \right ) \right ) ^{2}-{\frac{i}{4}}\pi \,\ln \left ( x \right ) \left ({\it csgn} \left ( i{{\rm e}^{2\,bx+2\,a}} \right ) \right ) ^{3}-{\frac{i}{4}}\pi \,\ln \left ( x \right ) \left ({\it csgn} \left ({\frac{i{{\rm e}^{2\,bx+2\,a}}}{{{\rm e}^{2\,bx+2\,a}}+1}} \right ) \right ) ^{3}+{\frac{i}{2}}\pi \,\ln \left ( x \right ) \left ({\it csgn} \left ({\frac{i}{{{\rm e}^{2\,bx+2\,a}}+1}} \right ) \right ) ^{2}-{\frac{i}{2}}\pi \,\ln \left ( x \right ) \left ({\it csgn} \left ({\frac{i}{{{\rm e}^{2\,bx+2\,a}}+1}} \right ) \right ) ^{3}-{\frac{i}{4}}\pi \,\ln \left ( x \right ){\it csgn} \left ({\frac{i}{{{\rm e}^{2\,bx+2\,a}}+1}} \right ){\it csgn} \left ( i{{\rm e}^{2\,bx+2\,a}} \right ){\it csgn} \left ({\frac{i{{\rm e}^{2\,bx+2\,a}}}{{{\rm e}^{2\,bx+2\,a}}+1}} \right ) -{\frac{i}{4}}\pi \,\ln \left ( x \right ) \left ({\it csgn} \left ( i{{\rm e}^{bx+a}} \right ) \right ) ^{2}{\it csgn} \left ( i{{\rm e}^{2\,bx+2\,a}} \right ) \end{align*}

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*Pi*ln(x)*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2-1/2*I*Pi*ln(x)+1
/4*I*Pi*ln(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+1/4*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))^3-1/4*I*Pi*ln(x
)*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+1/2*I*Pi*ln(x)*csgn(I/(exp(2*b*x+2*a)+1))^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(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))-1/4*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 = 0.977923, size = 46, normalized size = 2.19 \begin{align*} -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{align*}

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 = 1.62091, size = 22, normalized size = 1.05 \begin{align*} b x + a \log \left (x\right ) \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{acoth}{\left (\tanh{\left (a + b x \right )} \right )}}{x}\, dx \end{align*}

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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )}{x}\,{d x} \end{align*}

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)