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

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

[Out]

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

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Rubi [A]  time = 0.0378546, 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} \[ \frac{1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac{b \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac{b \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*ArcCoth[Tanh[a + b*x]]),x]

[Out]

1/(x*(b*x - ArcCoth[Tanh[a + b*x]])) - (b*Log[x])/(b*x - ArcCoth[Tanh[a + b*x]])^2 + (b*Log[ArcCoth[Tanh[a + b
*x]]])/(b*x - ArcCoth[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 \coth ^{-1}(\tanh (a+b x))} \, dx &=\frac{1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}+\frac{b \int \frac{1}{x \coth ^{-1}(\tanh (a+b x))} \, dx}{b x-\coth ^{-1}(\tanh (a+b x))}\\ &=\frac{1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac{b \int \frac{1}{x} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac{b^2 \int \frac{1}{\coth ^{-1}(\tanh (a+b x))} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac{1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac{b \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac{b \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac{1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac{b \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac{b \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}\\ \end{align*}

Mathematica [A]  time = 0.024122, size = 45, normalized size = 0.69 \[ \frac{b x \left (\log \left (\coth ^{-1}(\tanh (a+b x))\right )-\log (x)+1\right )-\coth ^{-1}(\tanh (a+b x))}{x \left (\coth ^{-1}(\tanh (a+b x))-b x\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}{\rm arccoth} \left (\tanh \left ( bx+a \right ) \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C]  time = 1.7929, size = 88, normalized size = 1.35 \begin{align*} -\frac{4 \, b \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{\pi ^{2} + 4 i \, \pi a - 4 \, a^{2}} + \frac{4 \, b \log \left (x\right )}{\pi ^{2} + 4 i \, \pi a - 4 \, a^{2}} + \frac{2}{{\left (i \, \pi - 2 \, a\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.72745, size = 315, normalized size = 4.85 \begin{align*} \frac{2 \,{\left (16 \, \pi a b x \arctan \left (-\frac{2 \, b x + 2 \, a - \sqrt{4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}}}{\pi }\right ) - 2 \, \pi ^{2} a - 8 \, a^{3} -{\left (\pi ^{2} b - 4 \, a^{2} b\right )} x \log \left (4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}\right ) + 2 \,{\left (\pi ^{2} b - 4 \, a^{2} b\right )} x \log \left (x\right )\right )}}{{\left (\pi ^{4} + 8 \, \pi ^{2} a^{2} + 16 \, a^{4}\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(16*pi*a*b*x*arctan(-(2*b*x + 2*a - sqrt(4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a^2))/pi) - 2*pi^2*a - 8*a^3 - (pi^2
*b - 4*a^2*b)*x*log(4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a^2) + 2*(pi^2*b - 4*a^2*b)*x*log(x))/((pi^4 + 8*pi^2*a^2 +
16*a^4)*x)

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(x^2*arccoth(tanh(b*x + a))), x)