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

Optimal. Leaf size=97 \[ \frac{1}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))}-\frac{1}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2}-\frac{\log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{\log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]

[Out]

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

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Rubi [A]  time = 0.0663209, antiderivative size = 97, 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} \[ \frac{1}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))}-\frac{1}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2}-\frac{\log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{\log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*ArcTanh[Tanh[a + b*x]]^3),x]

[Out]

-1/(2*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^2) + 1/((b*x - ArcTanh[Tanh[a + b*x]])^2*ArcTanh[T
anh[a + b*x]]) - Log[x]/(b*x - ArcTanh[Tanh[a + b*x]])^3 + 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 \tanh ^{-1}(\tanh (a+b x))^3} \, dx &=-\frac{1}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2}-\frac{\int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^2} \, dx}{b x-\tanh ^{-1}(\tanh (a+b x))}\\ &=-\frac{1}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2}+\frac{1}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))}-\frac{\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{1}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2}+\frac{1}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))}+\frac{\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 \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{1}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2}+\frac{1}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))}-\frac{\log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}-\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 )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac{1}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2}+\frac{1}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))}-\frac{\log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{\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 = 0.0966306, size = 74, normalized size = 0.76 \[ \frac{-4 b x \tanh ^{-1}(\tanh (a+b x))+\tanh ^{-1}(\tanh (a+b x))^2 \left (-2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )+2 \log (b x)+3\right )+b^2 x^2}{2 \tanh ^{-1}(\tanh (a+b x))^2 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.068, size = 92, 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 ) ^{3}}}+{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}+{\frac{1}{ \left ( 2\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -2\,bx \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}}+{\frac{\ln \left ( x \right ) }{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 3.54231, size = 69, normalized size = 0.71 \begin{align*} \frac{2 \, b x + 3 \, a}{2 \,{\left (a^{2} b^{2} x^{2} + 2 \, a^{3} b x + a^{4}\right )}} - \frac{\log \left (b x + a\right )}{a^{3}} + \frac{\log \left (x\right )}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b*x + 3*a)/(a^2*b^2*x^2 + 2*a^3*b*x + a^4) - log(b*x + a)/a^3 + log(x)/a^3

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Fricas [A]  time = 1.52844, size = 182, normalized size = 1.88 \begin{align*} \frac{2 \, a b x + 3 \, a^{2} - 2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \log \left (b x + a\right ) + 2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \log \left (x\right )}{2 \,{\left (a^{3} b^{2} x^{2} + 2 \, a^{4} b x + a^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a*b*x + 3*a^2 - 2*(b^2*x^2 + 2*a*b*x + a^2)*log(b*x + a) + 2*(b^2*x^2 + 2*a*b*x + a^2)*log(x))/(a^3*b^2
*x^2 + 2*a^4*b*x + a^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \operatorname{atanh}^{3}{\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))**3,x)

[Out]

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

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Giac [A]  time = 1.15298, size = 58, normalized size = 0.6 \begin{align*} -\frac{\log \left ({\left | b x + a \right |}\right )}{a^{3}} + \frac{\log \left ({\left | x \right |}\right )}{a^{3}} + \frac{2 \, a b x + 3 \, a^{2}}{2 \,{\left (b x + a\right )}^{2} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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