3.99 \(\int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^2} \, dx\)

Optimal. Leaf size=70 \[ -\frac{1}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))}+\frac{\log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac{\log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0459563, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {2163, 2160, 2157, 29} \[ -\frac{1}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))}+\frac{\log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac{\log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*ArcTanh[Tanh[a + b*x]]^2),x]

[Out]

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

Rule 2163

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^(n + 1)/((n + 1)*
(b*u - a*v)), x] - Dist[(a*(n + 1))/((n + 1)*(b*u - a*v)), Int[v^(n + 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; Pi
ecewiseLinearQ[u, v, x] && LtQ[n, -1]

Rule 2160

Int[1/((u_)*(v_)), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Dist[b/(b*u - a*v), Int[1
/v, x], x] - Dist[a/(b*u - a*v), 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{1}{x \tanh ^{-1}(\tanh (a+b x))^2} \, dx &=-\frac{1}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))}+\frac{\int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))} \, dx}{-b x+\tanh ^{-1}(\tanh (a+b x))}\\ &=-\frac{1}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))}-\frac{\int \frac{1}{x} \, dx}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{b \int \frac{1}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac{1}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))}+\frac{\log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac{1}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))}+\frac{\log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac{\log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ \end{align*}

Mathematica [A]  time = 0.062484, size = 53, normalized size = 0.76 \[ \frac{\tanh ^{-1}(\tanh (a+b x)) \left (-\log \left (\tanh ^{-1}(\tanh (a+b x))\right )+\log (b x)+1\right )-b x}{\tanh ^{-1}(\tanh (a+b x)) \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*ArcTanh[Tanh[a + b*x]]^2),x]

[Out]

(-(b*x) + ArcTanh[Tanh[a + b*x]]*(1 + Log[b*x] - Log[ArcTanh[Tanh[a + b*x]]]))/(ArcTanh[Tanh[a + b*x]]*(-(b*x)
+ ArcTanh[Tanh[a + b*x]])^2)

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Maple [A]  time = 0.066, size = 67, normalized size = 1. \begin{align*} -{\frac{\ln \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) }{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}}}+{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ){\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}+{\frac{\ln \left ( x \right ) }{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/arctanh(tanh(b*x+a))^2,x)

[Out]

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

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Maxima [A]  time = 2.44898, size = 38, normalized size = 0.54 \begin{align*} \frac{1}{a b x + a^{2}} - \frac{\log \left (b x + a\right )}{a^{2}} + \frac{\log \left (x\right )}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/arctanh(tanh(b*x+a))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.59652, size = 89, normalized size = 1.27 \begin{align*} -\frac{{\left (b x + a\right )} \log \left (b x + a\right ) -{\left (b x + a\right )} \log \left (x\right ) - a}{a^{2} b x + a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/arctanh(tanh(b*x+a))^2,x, algorithm="fricas")

[Out]

-((b*x + a)*log(b*x + a) - (b*x + a)*log(x) - a)/(a^2*b*x + a^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \operatorname{atanh}^{2}{\left (\tanh{\left (a + b x \right )} \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/atanh(tanh(b*x+a))**2,x)

[Out]

Integral(1/(x*atanh(tanh(a + b*x))**2), x)

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Giac [A]  time = 1.15898, size = 42, normalized size = 0.6 \begin{align*} -\frac{\log \left ({\left | b x + a \right |}\right )}{a^{2}} + \frac{\log \left ({\left | x \right |}\right )}{a^{2}} + \frac{1}{{\left (b x + a\right )} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/arctanh(tanh(b*x+a))^2,x, algorithm="giac")

[Out]

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