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

Optimal. Leaf size=65 \[ \frac{1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{b \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0443764, antiderivative size = 65, 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} \[ \frac{1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{b \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*ArcTanh[Tanh[a + b*x]]),x]

[Out]

1/(x*(b*x - ArcTanh[Tanh[a + b*x]])) - (b*Log[x])/(b*x - ArcTanh[Tanh[a + b*x]])^2 + (b*Log[ArcTanh[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^2 \tanh ^{-1}(\tanh (a+b x))} \, dx &=\frac{1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{b \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))} \, dx}{b x-\tanh ^{-1}(\tanh (a+b x))}\\ &=\frac{1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{b \int \frac{1}{x} \, dx}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{b^2 \int \frac{1}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac{1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{b \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 )^2}\\ &=\frac{1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{b \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 = 0.0257177, size = 45, normalized size = 0.69 \[ \frac{b x \left (\log \left (\tanh ^{-1}(\tanh (a+b x))\right )-\log (x)+1\right )-\tanh ^{-1}(\tanh (a+b x))}{x \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.079, size = 64, normalized size = 1. \begin{align*}{\frac{b\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 ) x}}-{\frac{b\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^2/arctanh(tanh(b*x+a)),x)

[Out]

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

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Maxima [A]  time = 1.79131, size = 38, normalized size = 0.58 \begin{align*} \frac{b \log \left (b x + a\right )}{a^{2}} - \frac{b \log \left (x\right )}{a^{2}} - \frac{1}{a x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.5147, size = 61, normalized size = 0.94 \begin{align*} \frac{b x \log \left (b x + a\right ) - b x \log \left (x\right ) - a}{a^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.14328, size = 41, normalized size = 0.63 \begin{align*} \frac{b \log \left ({\left | b x + a \right |}\right )}{a^{2}} - \frac{b \log \left ({\left | x \right |}\right )}{a^{2}} - \frac{1}{a x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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