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

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

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 102, normalized size of antiderivative = 1.00, 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} \[ -\frac {2 b}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))}+\frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {2 b \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3}-\frac {2 b \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 70, normalized size = 0.69 \[ \frac {\coth ^{-1}(\tanh (a+b x))^2+2 b x \coth ^{-1}(\tanh (a+b x)) \left (\log (x)-\log \left (\coth ^{-1}(\tanh (a+b x))\right )\right )-b^2 x^2}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.68, size = 480, normalized size = 4.71 \[ \frac {4 \, {\left (\pi ^{6} + 4 \, \pi ^{4} a^{2} - 16 \, \pi ^{2} a^{4} - 64 \, a^{6} + 8 \, {\left (\pi ^{4} b^{2} - 16 \, a^{4} b^{2}\right )} x^{2} + 4 \, {\left (5 \, \pi ^{4} a b + 8 \, \pi ^{2} a^{3} b - 48 \, a^{5} b\right )} x - 8 \, {\left (4 \, {\left (\pi ^{3} b^{3} - 12 \, \pi a^{2} b^{3}\right )} x^{3} + 8 \, {\left (\pi ^{3} a b^{2} - 12 \, \pi a^{3} b^{2}\right )} x^{2} + {\left (\pi ^{5} b - 8 \, \pi ^{3} a^{2} b - 48 \, \pi a^{4} b\right )} x\right )} \arctan \left (-\frac {2 \, b x + 2 \, a - \sqrt {4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}}}{\pi }\right ) - 4 \, {\left (4 \, {\left (3 \, \pi ^{2} a b^{3} - 4 \, a^{3} b^{3}\right )} x^{3} + 8 \, {\left (3 \, \pi ^{2} a^{2} b^{2} - 4 \, a^{4} b^{2}\right )} x^{2} + {\left (3 \, \pi ^{4} a b + 8 \, \pi ^{2} a^{3} b - 16 \, a^{5} b\right )} x\right )} \log \left (4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}\right ) + 8 \, {\left (4 \, {\left (3 \, \pi ^{2} a b^{3} - 4 \, a^{3} b^{3}\right )} x^{3} + 8 \, {\left (3 \, \pi ^{2} a^{2} b^{2} - 4 \, a^{4} b^{2}\right )} x^{2} + {\left (3 \, \pi ^{4} a b + 8 \, \pi ^{2} a^{3} b - 16 \, a^{5} b\right )} x\right )} \log \relax (x)\right )}}{4 \, {\left (\pi ^{6} b^{2} + 12 \, \pi ^{4} a^{2} b^{2} + 48 \, \pi ^{2} a^{4} b^{2} + 64 \, a^{6} b^{2}\right )} x^{3} + 8 \, {\left (\pi ^{6} a b + 12 \, \pi ^{4} a^{3} b + 48 \, \pi ^{2} a^{5} b + 64 \, a^{7} b\right )} x^{2} + {\left (\pi ^{8} + 16 \, \pi ^{6} a^{2} + 96 \, \pi ^{4} a^{4} + 256 \, \pi ^{2} a^{6} + 256 \, a^{8}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*(pi^6 + 4*pi^4*a^2 - 16*pi^2*a^4 - 64*a^6 + 8*(pi^4*b^2 - 16*a^4*b^2)*x^2 + 4*(5*pi^4*a*b + 8*pi^2*a^3*b - 4
8*a^5*b)*x - 8*(4*(pi^3*b^3 - 12*pi*a^2*b^3)*x^3 + 8*(pi^3*a*b^2 - 12*pi*a^3*b^2)*x^2 + (pi^5*b - 8*pi^3*a^2*b
- 48*pi*a^4*b)*x)*arctan(-(2*b*x + 2*a - sqrt(4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a^2))/pi) - 4*(4*(3*pi^2*a*b^3 -
4*a^3*b^3)*x^3 + 8*(3*pi^2*a^2*b^2 - 4*a^4*b^2)*x^2 + (3*pi^4*a*b + 8*pi^2*a^3*b - 16*a^5*b)*x)*log(4*b^2*x^2
+ 8*a*b*x + pi^2 + 4*a^2) + 8*(4*(3*pi^2*a*b^3 - 4*a^3*b^3)*x^3 + 8*(3*pi^2*a^2*b^2 - 4*a^4*b^2)*x^2 + (3*pi^
4*a*b + 8*pi^2*a^3*b - 16*a^5*b)*x)*log(x))/(4*(pi^6*b^2 + 12*pi^4*a^2*b^2 + 48*pi^2*a^4*b^2 + 64*a^6*b^2)*x^3
+ 8*(pi^6*a*b + 12*pi^4*a^3*b + 48*pi^2*a^5*b + 64*a^7*b)*x^2 + (pi^8 + 16*pi^6*a^2 + 96*pi^4*a^4 + 256*pi^2*
a^6 + 256*a^8)*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.78, size = 135, normalized size = 1.32 \[ -\frac {16 \, b \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{-i \, \pi ^{3} + 6 \, \pi ^{2} a + 12 i \, \pi a^{2} - 8 \, a^{3}} + \frac {16 \, b \log \relax (x)}{-i \, \pi ^{3} + 6 \, \pi ^{2} a + 12 i \, \pi a^{2} - 8 \, a^{3}} - \frac {4 \, {\left (i \, \pi - 4 \, b x - 2 \, a\right )}}{{\left (2 \, \pi ^{2} b + 8 i \, \pi a b - 8 \, a^{2} b\right )} x^{2} - {\left (i \, \pi ^{3} - 6 \, \pi ^{2} a - 12 i \, \pi a^{2} + 8 \, a^{3}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16*b*log(-I*pi + 2*b*x + 2*a)/(-I*pi^3 + 6*pi^2*a + 12*I*pi*a^2 - 8*a^3) + 16*b*log(x)/(-I*pi^3 + 6*pi^2*a +
12*I*pi*a^2 - 8*a^3) - 4*(I*pi - 4*b*x - 2*a)/((2*pi^2*b + 8*I*pi*a*b - 8*a^2*b)*x^2 - (I*pi^3 - 6*pi^2*a - 12
*I*pi*a^2 + 8*a^3)*x)

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mupad [B]  time = 3.80, size = 453, normalized size = 4.44 \[ \frac {4\,{\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 )\,\left (8\,\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+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 )\,32{}\mathrm {i}\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 )}^2-16\,b^2\,x^2+b\,x\,\ln \left (-\frac {1}{{\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 )\,32{}\mathrm {i}}{x\,\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 )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*acoth(tanh(a + b*x))^2),x)

[Out]

(4*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))*(8*log(-1/(exp(2
*a)*exp(2*b*x) - 1)) + b*x*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))*32i) + 4*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))^2 - 16*b^2*x^2 + b*x*log(-
1/(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))*32i)/(x*(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
)^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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