3.92 \(\int \frac{1}{x^3 \tanh ^{-1}(\tanh (a+b x))} \, dx\)

Optimal. Leaf size=92 \[ -\frac{b^2 \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{b^2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{1}{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{b}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

[Out]

b/(x*(b*x - ArcTanh[Tanh[a + b*x]])^2) + 1/(2*x^2*(b*x - ArcTanh[Tanh[a + b*x]])) - (b^2*Log[x])/(b*x - ArcTan
h[Tanh[a + b*x]])^3 + (b^2*Log[ArcTanh[Tanh[a + b*x]]])/(b*x - ArcTanh[Tanh[a + b*x]])^3

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Rubi [A]  time = 0.0681549, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {2163, 2160, 2157, 29} \[ -\frac{b^2 \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{b^2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{1}{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{b}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*ArcTanh[Tanh[a + b*x]]),x]

[Out]

b/(x*(b*x - ArcTanh[Tanh[a + b*x]])^2) + 1/(2*x^2*(b*x - ArcTanh[Tanh[a + b*x]])) - (b^2*Log[x])/(b*x - ArcTan
h[Tanh[a + b*x]])^3 + (b^2*Log[ArcTanh[Tanh[a + b*x]]])/(b*x - ArcTanh[Tanh[a + b*x]])^3

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^3 \tanh ^{-1}(\tanh (a+b x))} \, dx &=\frac{1}{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{b \int \frac{1}{x^2 \tanh ^{-1}(\tanh (a+b x))} \, dx}{-b x+\tanh ^{-1}(\tanh (a+b x))}\\ &=\frac{b}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{1}{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{b^2 \int \frac{1}{x \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{b}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{1}{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{b^2 \int \frac{1}{x} \, dx}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{b^3 \int \frac{1}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{b}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{1}{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{b^2 \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac{b^2 \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 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{b}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{1}{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{b^2 \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{b^2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}\\ \end{align*}

Mathematica [A]  time = 0.0341613, size = 66, normalized size = 0.72 \[ \frac{b^2 x^2 \left (2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )-2 \log (x)+3\right )-4 b x \tanh ^{-1}(\tanh (a+b x))+\tanh ^{-1}(\tanh (a+b x))^2}{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*ArcTanh[Tanh[a + b*x]]),x]

[Out]

(-4*b*x*ArcTanh[Tanh[a + b*x]] + ArcTanh[Tanh[a + b*x]]^2 + b^2*x^2*(3 - 2*Log[x] + 2*Log[ArcTanh[Tanh[a + b*x
]]]))/(2*x^2*(b*x - ArcTanh[Tanh[a + b*x]])^3)

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Maple [A]  time = 0.081, size = 87, normalized size = 1. \begin{align*} -{\frac{{b}^{2}\ln \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) }{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3}}}-{\frac{1}{ \left ( 2\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -2\,bx \right ){x}^{2}}}+{\frac{{b}^{2}\ln \left ( x \right ) }{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3}}}+{\frac{b}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/arctanh(tanh(b*x+a)),x)

[Out]

-1/(arctanh(tanh(b*x+a))-b*x)^3*b^2*ln(arctanh(tanh(b*x+a)))-1/2/(arctanh(tanh(b*x+a))-b*x)/x^2+1/(arctanh(tan
h(b*x+a))-b*x)^3*b^2*ln(x)+1/(arctanh(tanh(b*x+a))-b*x)^2*b/x

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Maxima [A]  time = 1.79578, size = 54, normalized size = 0.59 \begin{align*} -\frac{b^{2} \log \left (b x + a\right )}{a^{3}} + \frac{b^{2} \log \left (x\right )}{a^{3}} + \frac{2 \, b x - a}{2 \, a^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/arctanh(tanh(b*x+a)),x, algorithm="maxima")

[Out]

-b^2*log(b*x + a)/a^3 + b^2*log(x)/a^3 + 1/2*(2*b*x - a)/(a^2*x^2)

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Fricas [A]  time = 1.51525, size = 103, normalized size = 1.12 \begin{align*} -\frac{2 \, b^{2} x^{2} \log \left (b x + a\right ) - 2 \, b^{2} x^{2} \log \left (x\right ) - 2 \, a b x + a^{2}}{2 \, a^{3} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/arctanh(tanh(b*x+a)),x, algorithm="fricas")

[Out]

-1/2*(2*b^2*x^2*log(b*x + a) - 2*b^2*x^2*log(x) - 2*a*b*x + a^2)/(a^3*x^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \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**3/atanh(tanh(b*x+a)),x)

[Out]

Integral(1/(x**3*atanh(tanh(a + b*x))), x)

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Giac [A]  time = 1.13528, size = 61, normalized size = 0.66 \begin{align*} -\frac{b^{2} \log \left ({\left | b x + a \right |}\right )}{a^{3}} + \frac{b^{2} \log \left ({\left | x \right |}\right )}{a^{3}} + \frac{2 \, a b x - a^{2}}{2 \, a^{3} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/arctanh(tanh(b*x+a)),x, algorithm="giac")

[Out]

-b^2*log(abs(b*x + a))/a^3 + b^2*log(abs(x))/a^3 + 1/2*(2*a*b*x - a^2)/(a^3*x^2)