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

Optimal. Leaf size=31 \[ \frac{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^2}+\frac{x}{b} \]

[Out]

x/b + ((b*x - ArcTanh[Tanh[a + b*x]])*Log[ArcTanh[Tanh[a + b*x]]])/b^2

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Rubi [A]  time = 0.0150204, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {2158, 2157, 29} \[ \frac{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^2}+\frac{x}{b} \]

Antiderivative was successfully verified.

[In]

Int[x/ArcTanh[Tanh[a + b*x]],x]

[Out]

x/b + ((b*x - ArcTanh[Tanh[a + b*x]])*Log[ArcTanh[Tanh[a + b*x]]])/b^2

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 2157

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

Rule 29

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

Rubi steps

\begin{align*} \int \frac{x}{\tanh ^{-1}(\tanh (a+b x))} \, dx &=\frac{x}{b}-\frac{\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right ) \int \frac{1}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b}\\ &=\frac{x}{b}-\frac{\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b^2}\\ &=\frac{x}{b}+\frac{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^2}\\ \end{align*}

Mathematica [A]  time = 0.0232616, size = 31, normalized size = 1. \[ \frac{x}{b}-\frac{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right ) \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x/ArcTanh[Tanh[a + b*x]],x]

[Out]

x/b - ((-(b*x) + ArcTanh[Tanh[a + b*x]])*Log[ArcTanh[Tanh[a + b*x]]])/b^2

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Maple [A]  time = 0.033, size = 49, normalized size = 1.6 \begin{align*}{\frac{x}{b}}-{\frac{\ln \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) a}{{b}^{2}}}-{\frac{\ln \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/b-1/b^2*ln(arctanh(tanh(b*x+a)))*a-1/b^2*ln(arctanh(tanh(b*x+a)))*(arctanh(tanh(b*x+a))-b*x-a)

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Maxima [A]  time = 1.76601, size = 24, normalized size = 0.77 \begin{align*} \frac{x}{b} - \frac{a \log \left (b x + a\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/b - a*log(b*x + a)/b^2

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Fricas [A]  time = 1.53216, size = 38, normalized size = 1.23 \begin{align*} \frac{b x - a \log \left (b x + a\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x - a*log(b*x + a))/b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/atanh(tanh(a + b*x)), x)

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Giac [A]  time = 1.13733, size = 26, normalized size = 0.84 \begin{align*} \frac{x}{b} - \frac{a \log \left ({\left | b x + a \right |}\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/b - a*log(abs(b*x + a))/b^2