\(\int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))^2} \, dx\) [173]

Optimal result

Integrand size = 13, antiderivative size = 102 \[ \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))^2} \, dx=-\frac {2 b}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))}+\frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {2 b \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3}-\frac {2 b \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3} \]

[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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2202, 2194, 2191, 2188, 29} \[ \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))^2} \, dx=-\frac {2 b}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))}+\frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {2 b \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3}-\frac {2 b \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3} \]

[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 2188

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 2191

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 2194

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 2202

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*} \text {integral}& = \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) \text {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*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.69 \[ \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))^2} \, dx=\frac {-b^2 x^2+\coth ^{-1}(\tanh (a+b x))^2+2 b x \coth ^{-1}(\tanh (a+b x)) \left (\log (x)-\log \left (\coth ^{-1}(\tanh (a+b x))\right )\right )}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))} \]

[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]])

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.35 (sec) , antiderivative size = 5357, normalized size of antiderivative = 52.52

\[\text {output too large to display}\]

[In]

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

[Out]

result too large to display

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.54 \[ \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))^2} \, dx=-\frac {4 \, {\left (8 \, a b x - \pi ^{2} + 4 i \, \pi {\left (b x + a\right )} + 4 \, a^{2} - 4 \, {\left (2 \, b^{2} x^{2} + i \, \pi b x + 2 \, a b x\right )} \log \left (i \, \pi + 2 \, b x + 2 \, a\right ) + 4 \, {\left (2 \, b^{2} x^{2} + i \, \pi b x + 2 \, a b x\right )} \log \left (x\right )\right )}}{16 \, a^{3} b x^{2} + \pi ^{4} x + 16 \, a^{4} x - 2 i \, \pi ^{3} {\left (b x^{2} + 4 \, a x\right )} - 12 \, \pi ^{2} {\left (a b x^{2} + 2 \, a^{2} x\right )} + 8 i \, \pi {\left (3 \, a^{2} b x^{2} + 4 \, a^{3} x\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))^2} \, dx=\int \frac {1}{x^{2} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

[In]

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

[Out]

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.51 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.32 \[ \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))^2} \, dx=-\frac {16 \, b \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{-i \, \pi ^{3} + 6 \, \pi ^{2} a + 12 i \, \pi a^{2} - 8 \, a^{3}} + \frac {16 \, b \log \left (x\right )}{-i \, \pi ^{3} + 6 \, \pi ^{2} a + 12 i \, \pi a^{2} - 8 \, a^{3}} - \frac {4 \, {\left (i \, \pi - 4 \, b x - 2 \, a\right )}}{2 \, {\left (\pi ^{2} b + 4 i \, \pi a b - 4 \, a^{2} b\right )} x^{2} - {\left (i \, \pi ^{3} - 6 \, \pi ^{2} a - 12 i \, \pi a^{2} + 8 \, a^{3}\right )} x} \]

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.33 \[ \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))^2} \, dx=\frac {16 i \, b \log \left (i \, \pi + 2 \, b x + 2 \, a\right )}{\pi ^{3} - 6 i \, \pi ^{2} a - 12 \, \pi a^{2} + 8 i \, a^{3}} - \frac {16 i \, b \log \left (x\right )}{\pi ^{3} - 6 i \, \pi ^{2} a - 12 \, \pi a^{2} + 8 i \, a^{3}} + \frac {8 \, b}{2 \, \pi ^{2} b x - 8 i \, \pi a b x - 8 \, a^{2} b x + i \, \pi ^{3} + 6 \, \pi ^{2} a - 12 i \, \pi a^{2} - 8 \, a^{3}} + \frac {4}{\pi ^{2} x - 4 i \, \pi a x - 4 \, a^{2} x} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.68 (sec) , antiderivative size = 453, normalized size of antiderivative = 4.44 \[ \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))^2} \, dx=\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} \]

[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)