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

Optimal. Leaf size=102 \[ -\frac{2 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))}+\frac{1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))}+\frac{2 b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac{2 b \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]

[Out]

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

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Rubi [A]  time = 0.0716322, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{2 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))}+\frac{1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))}+\frac{2 b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac{2 b \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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))^2} \, dx &=\frac{1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))}-\frac{(2 b) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^2} \, dx}{-b x+\tanh ^{-1}(\tanh (a+b x))}\\ &=-\frac{2 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))}+\frac{1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))}-\frac{(2 b) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))} \, dx}{\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac{2 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))}+\frac{1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))}+\frac{(2 b) \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 )^2}-\frac{\left (2 b^2\right ) \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 )^2}\\ &=-\frac{2 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))}+\frac{1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))}+\frac{2 b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac{(2 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 ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac{2 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))}+\frac{1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))}+\frac{2 b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac{2 b \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 = 0.0604331, size = 70, normalized size = 0.69 \[ \frac{\tanh ^{-1}(\tanh (a+b x))^2+2 b x \tanh ^{-1}(\tanh (a+b x)) \left (\log (x)-\log \left (\tanh ^{-1}(\tanh (a+b x))\right )\right )-b^2 x^2}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.09, size = 91, normalized size = 0.9 \begin{align*} -{\frac{b}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}+2\,{\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 ) ^{3}}}-{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}x}}-2\,{\frac{b\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^2/arctanh(tanh(b*x+a))^2,x)

[Out]

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

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Maxima [A]  time = 2.42605, size = 61, normalized size = 0.6 \begin{align*} -\frac{2 \, b x + a}{a^{2} b x^{2} + a^{3} x} + \frac{2 \, b \log \left (b x + a\right )}{a^{3}} - \frac{2 \, b \log \left (x\right )}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.50922, size = 138, normalized size = 1.35 \begin{align*} -\frac{2 \, a b x + a^{2} - 2 \,{\left (b^{2} x^{2} + a b x\right )} \log \left (b x + a\right ) + 2 \,{\left (b^{2} x^{2} + a b x\right )} \log \left (x\right )}{a^{3} b x^{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))^2,x, algorithm="fricas")

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.14788, size = 61, normalized size = 0.6 \begin{align*} \frac{2 \, b \log \left ({\left | b x + a \right |}\right )}{a^{3}} - \frac{2 \, b \log \left ({\left | x \right |}\right )}{a^{3}} - \frac{2 \, b x + a}{{\left (b x^{2} + a x\right )} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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