3.109 \(\int \frac{1}{x^2 \tanh ^{-1}(\tanh (a+b x))^3} \, dx\)

Optimal. Leaf size=131 \[ \frac{3 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))}-\frac{3 b}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^2}+\frac{1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2}-\frac{3 b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac{3 b \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4} \]

[Out]

(-3*b)/(2*(b*x - ArcTanh[Tanh[a + b*x]])^2*ArcTanh[Tanh[a + b*x]]^2) + 1/(x*(b*x - ArcTanh[Tanh[a + b*x]])*Arc
Tanh[Tanh[a + b*x]]^2) + (3*b)/((b*x - ArcTanh[Tanh[a + b*x]])^3*ArcTanh[Tanh[a + b*x]]) - (3*b*Log[x])/(b*x -
ArcTanh[Tanh[a + b*x]])^4 + (3*b*Log[ArcTanh[Tanh[a + b*x]]])/(b*x - ArcTanh[Tanh[a + b*x]])^4

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Rubi [A]  time = 0.0932824, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {2171, 2163, 2160, 2157, 29} \[ \frac{3 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))}-\frac{3 b}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^2}+\frac{1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2}-\frac{3 b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac{3 b \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*ArcTanh[Tanh[a + b*x]]^3),x]

[Out]

(-3*b)/(2*(b*x - ArcTanh[Tanh[a + b*x]])^2*ArcTanh[Tanh[a + b*x]]^2) + 1/(x*(b*x - ArcTanh[Tanh[a + b*x]])*Arc
Tanh[Tanh[a + b*x]]^2) + (3*b)/((b*x - ArcTanh[Tanh[a + b*x]])^3*ArcTanh[Tanh[a + b*x]]) - (3*b*Log[x])/(b*x -
ArcTanh[Tanh[a + b*x]])^4 + (3*b*Log[ArcTanh[Tanh[a + b*x]]])/(b*x - ArcTanh[Tanh[a + b*x]])^4

Rule 2171

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, -Simp[(u^(m + 1)*v^
(n + 1))/((m + 1)*(b*u - a*v)), x] + Dist[(b*(m + n + 2))/((m + 1)*(b*u - a*v)), Int[u^(m + 1)*v^n, x], x] /;
NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && LtQ[m, -1]

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))^3} \, dx &=\frac{1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2}-\frac{(3 b) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^3} \, dx}{-b x+\tanh ^{-1}(\tanh (a+b x))}\\ &=-\frac{3 b}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^2}+\frac{1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2}+\frac{(3 b) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^2} \, dx}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac{3 b}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^2}+\frac{1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2}+\frac{3 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))}+\frac{(3 b) \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 )^2}\\ &=-\frac{3 b}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^2}+\frac{1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2}+\frac{3 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))}-\frac{(3 b) \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 )^2}+\frac{\left (3 b^2\right ) \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 )^2}\\ &=-\frac{3 b}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^2}+\frac{1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2}+\frac{3 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))}-\frac{3 b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac{(3 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 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac{3 b}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^2}+\frac{1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2}+\frac{3 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))}-\frac{3 b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac{3 b \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}\\ \end{align*}

Mathematica [A]  time = 0.0428006, size = 93, normalized size = 0.71 \[ -\frac{-6 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+2 \tanh ^{-1}(\tanh (a+b x))^3+3 b x \tanh ^{-1}(\tanh (a+b x))^2 \left (-2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )+2 \log (x)+1\right )+b^3 x^3}{2 x \tanh ^{-1}(\tanh (a+b x))^2 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*ArcTanh[Tanh[a + b*x]]^3),x]

[Out]

-(b^3*x^3 - 6*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 2*ArcTanh[Tanh[a + b*x]]^3 + 3*b*x*ArcTanh[Tanh[a + b*x]]^2*(1
+ 2*Log[x] - 2*Log[ArcTanh[Tanh[a + b*x]]]))/(2*x*ArcTanh[Tanh[a + b*x]]^2*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4
)

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Maple [A]  time = 0.085, size = 117, normalized size = 0.9 \begin{align*} -{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3}x}}-3\,{\frac{b\ln \left ( x \right ) }{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{4}}}-{\frac{b}{2\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}}+3\,{\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 ) ^{4}}}-2\,{\frac{b}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/arctanh(tanh(b*x+a))^3,x)

[Out]

-1/(arctanh(tanh(b*x+a))-b*x)^3/x-3/(arctanh(tanh(b*x+a))-b*x)^4*b*ln(x)-1/2/(arctanh(tanh(b*x+a))-b*x)^2*b/ar
ctanh(tanh(b*x+a))^2+3/(arctanh(tanh(b*x+a))-b*x)^4*b*ln(arctanh(tanh(b*x+a)))-2/(arctanh(tanh(b*x+a))-b*x)^3*
b/arctanh(tanh(b*x+a))

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Maxima [A]  time = 3.52496, size = 93, normalized size = 0.71 \begin{align*} -\frac{6 \, b^{2} x^{2} + 9 \, a b x + 2 \, a^{2}}{2 \,{\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}} + \frac{3 \, b \log \left (b x + a\right )}{a^{4}} - \frac{3 \, b \log \left (x\right )}{a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/arctanh(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

-1/2*(6*b^2*x^2 + 9*a*b*x + 2*a^2)/(a^3*b^2*x^3 + 2*a^4*b*x^2 + a^5*x) + 3*b*log(b*x + a)/a^4 - 3*b*log(x)/a^4

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Fricas [A]  time = 1.52248, size = 232, normalized size = 1.77 \begin{align*} -\frac{6 \, a b^{2} x^{2} + 9 \, a^{2} b x + 2 \, a^{3} - 6 \,{\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \log \left (b x + a\right ) + 6 \,{\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \log \left (x\right )}{2 \,{\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/arctanh(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

-1/2*(6*a*b^2*x^2 + 9*a^2*b*x + 2*a^3 - 6*(b^3*x^3 + 2*a*b^2*x^2 + a^2*b*x)*log(b*x + a) + 6*(b^3*x^3 + 2*a*b^
2*x^2 + a^2*b*x)*log(x))/(a^4*b^2*x^3 + 2*a^5*b*x^2 + a^6*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \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**2/atanh(tanh(b*x+a))**3,x)

[Out]

Integral(1/(x**2*atanh(tanh(a + b*x))**3), x)

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Giac [A]  time = 1.144, size = 81, normalized size = 0.62 \begin{align*} \frac{3 \, b \log \left ({\left | b x + a \right |}\right )}{a^{4}} - \frac{3 \, b \log \left ({\left | x \right |}\right )}{a^{4}} - \frac{6 \, a b^{2} x^{2} + 9 \, a^{2} b x + 2 \, a^{3}}{2 \,{\left (b x + a\right )}^{2} a^{4} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/arctanh(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

3*b*log(abs(b*x + a))/a^4 - 3*b*log(abs(x))/a^4 - 1/2*(6*a*b^2*x^2 + 9*a^2*b*x + 2*a^3)/((b*x + a)^2*a^4*x)