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(b*x+a)))-b*ln(x)/(b*x-arccoth(tanh(b*x+a)))^2+b*ln(arccoth(tanh(b*x+a)))/(b*x-arccoth(ta
nh(b*x+a)))^2

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Rubi [A]  time = 0.04, antiderivative size = 65, normalized size of antiderivative = 1.00, 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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], 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 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 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]

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*}

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Mathematica [A]  time = 0.03, 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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fricas [B]  time = 0.59, size = 137, normalized size = 2.11 \[ \frac {2 \, {\left (16 \, \pi a b x \arctan \left (-\frac {2 \, b x + 2 \, a - \sqrt {4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}}}{\pi }\right ) - 2 \, \pi ^{2} a - 8 \, a^{3} - {\left (\pi ^{2} b - 4 \, a^{2} b\right )} x \log \left (4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}\right ) + 2 \, {\left (\pi ^{2} b - 4 \, a^{2} b\right )} x \log \relax (x)\right )}}{{\left (\pi ^{4} + 8 \, \pi ^{2} a^{2} + 16 \, a^{4}\right )} x} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{2} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )}\,{d x} \]

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)

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{2} \mathrm {arccoth}\left (\tanh \left (b x +a \right )\right )}\, dx \]

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 = 0.52, size = 65, normalized size = 1.00 \[ -\frac {4 \, b \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{\pi ^{2} + 4 i \, \pi a - 4 \, a^{2}} + \frac {4 \, b \log \relax (x)}{\pi ^{2} + 4 i \, \pi a - 4 \, a^{2}} + \frac {2}{{\left (i \, \pi - 2 \, a\right )} x} \]

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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mupad [B]  time = 3.12, size = 220, normalized size = 3.38 \[ \frac {2\,\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-2\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+4\,b\,x+b\,x\,\mathrm {atan}\left (\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,1{}\mathrm {i}-\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}{\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x}\right )\,8{}\mathrm {i}}{x\,{\left (\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*acoth(tanh(a + b*x))),x)

[Out]

(2*log(-1/(exp(2*a)*exp(2*b*x) - 1)) - 2*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + 4*b*x + b*x*at
an((log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))*1i - log(-2/(exp(2*a)*exp(2*b*x) - 1))*1i + b*x*2i)
/(log(-2/(exp(2*a)*exp(2*b*x) - 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x))*8i)/(x*
(log(-1/(exp(2*a)*exp(2*b*x) - 1)) - log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{2} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

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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