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

Optimal. Leaf size=70 \[ -\frac {1}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {\log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}-\frac {\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 70, 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} \[ -\frac {1}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {\log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}-\frac {\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*ArcCoth[Tanh[a + b*x]]^2),x]

[Out]

-(1/((b*x - ArcCoth[Tanh[a + b*x]])*ArcCoth[Tanh[a + b*x]])) + Log[x]/(b*x - ArcCoth[Tanh[a + b*x]])^2 - Log[A
rcCoth[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 \coth ^{-1}(\tanh (a+b x))^2} \, dx &=-\frac {1}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {\int \frac {1}{x \coth ^{-1}(\tanh (a+b x))} \, dx}{-b x+\coth ^{-1}(\tanh (a+b x))}\\ &=-\frac {1}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}-\frac {\int \frac {1}{x} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )}+\frac {b \int \frac {1}{\coth ^{-1}(\tanh (a+b x))} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac {1}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {\log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac {\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 ) \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac {1}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {\log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}-\frac {\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 = 0.07, size = 53, normalized size = 0.76 \[ \frac {\coth ^{-1}(\tanh (a+b x)) \left (-\log \left (\coth ^{-1}(\tanh (a+b x))\right )+\log (b x)+1\right )-b x}{\coth ^{-1}(\tanh (a+b x)) \left (\coth ^{-1}(\tanh (a+b x))-b x\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.46, size = 306, normalized size = 4.37 \[ -\frac {2 \, {\left (2 \, \pi ^{4} - 32 \, a^{4} - 8 \, {\left (\pi ^{2} a b + 4 \, a^{3} b\right )} x + 16 \, {\left (4 \, \pi a b^{2} x^{2} + 8 \, \pi a^{2} b x + \pi ^{3} a + 4 \, \pi a^{3}\right )} \arctan \left (-\frac {2 \, b x + 2 \, a - \sqrt {4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}}}{\pi }\right ) - {\left (\pi ^{4} - 16 \, a^{4} + 4 \, {\left (\pi ^{2} b^{2} - 4 \, a^{2} b^{2}\right )} x^{2} + 8 \, {\left (\pi ^{2} a b - 4 \, a^{3} b\right )} x\right )} \log \left (4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}\right ) + 2 \, {\left (\pi ^{4} - 16 \, a^{4} + 4 \, {\left (\pi ^{2} b^{2} - 4 \, a^{2} b^{2}\right )} x^{2} + 8 \, {\left (\pi ^{2} a b - 4 \, a^{3} b\right )} x\right )} \log \relax (x)\right )}}{\pi ^{6} + 12 \, \pi ^{4} a^{2} + 48 \, \pi ^{2} a^{4} + 64 \, a^{6} + 4 \, {\left (\pi ^{4} b^{2} + 8 \, \pi ^{2} a^{2} b^{2} + 16 \, a^{4} b^{2}\right )} x^{2} + 8 \, {\left (\pi ^{4} a b + 8 \, \pi ^{2} a^{3} b + 16 \, a^{5} b\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(2*pi^4 - 32*a^4 - 8*(pi^2*a*b + 4*a^3*b)*x + 16*(4*pi*a*b^2*x^2 + 8*pi*a^2*b*x + pi^3*a + 4*pi*a^3)*arctan
(-(2*b*x + 2*a - sqrt(4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a^2))/pi) - (pi^4 - 16*a^4 + 4*(pi^2*b^2 - 4*a^2*b^2)*x^2
+ 8*(pi^2*a*b - 4*a^3*b)*x)*log(4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a^2) + 2*(pi^4 - 16*a^4 + 4*(pi^2*b^2 - 4*a^2*
b^2)*x^2 + 8*(pi^2*a*b - 4*a^3*b)*x)*log(x))/(pi^6 + 12*pi^4*a^2 + 48*pi^2*a^4 + 64*a^6 + 4*(pi^4*b^2 + 8*pi^2
*a^2*b^2 + 16*a^4*b^2)*x^2 + 8*(pi^4*a*b + 8*pi^2*a^3*b + 16*a^5*b)*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.77, size = 77, normalized size = 1.10 \[ \frac {4 \, \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{\pi ^{2} + 4 i \, \pi a - 4 \, a^{2}} - \frac {4 \, \log \relax (x)}{\pi ^{2} + 4 i \, \pi a - 4 \, a^{2}} - \frac {4}{\pi ^{2} + 4 i \, \pi a - 4 \, a^{2} + {\left (2 i \, \pi b - 4 \, a b\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.03, size = 421, normalized size = 6.01 \[ -\frac {4\,\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-4\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+8\,b\,x+\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,\mathrm {atan}\left (\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,1{}\mathrm {i}-\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}{\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x}\right )\,8{}\mathrm {i}-\mathrm {atan}\left (\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,1{}\mathrm {i}-\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}{\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x}\right )\,\left (\ln \relax (2)+\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\right )\,8{}\mathrm {i}}{\left (\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\right )\,{\left (\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*acoth(tanh(a + b*x))^2),x)

[Out]

-(4*log(-1/(exp(2*a)*exp(2*b*x) - 1)) - 4*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + 8*b*x + log((
2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))*atan((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 - atan((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))*(log(2) + log(-1/(exp(2*a)*exp(2*b*x) - 1)))*8i)/((log((exp(2*a)*
exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) - log(-1/(exp(2*a)*exp(2*b*x) - 1)))*(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 \[ \int \frac {1}{x \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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