3.2.37 \(\int \frac {1}{\sqrt {\text {csch}(2 \log (c x))}} \, dx\) [137]

3.2.37.1 Optimal result
3.2.37.2 Mathematica [A] (verified)
3.2.37.3 Rubi [A] (warning: unable to verify)
3.2.37.4 Maple [F]
3.2.37.5 Fricas [A] (verification not implemented)
3.2.37.6 Sympy [F]
3.2.37.7 Maxima [F]
3.2.37.8 Giac [F(-1)]
3.2.37.9 Mupad [F(-1)]

3.2.37.1 Optimal result

Integrand size = 11, antiderivative size = 60 \[ \int \frac {1}{\sqrt {\text {csch}(2 \log (c x))}} \, dx=\frac {x}{2 \sqrt {\text {csch}(2 \log (c x))}}+\frac {\csc ^{-1}\left (c^2 x^2\right )}{2 c^2 \sqrt {1-\frac {1}{c^4 x^4}} x \sqrt {\text {csch}(2 \log (c x))}} \]

output
1/2*x/csch(2*ln(c*x))^(1/2)+1/2*arccsc(c^2*x^2)/c^2/x/(1-1/c^4/x^4)^(1/2)/ 
csch(2*ln(c*x))^(1/2)
 
3.2.37.2 Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.25 \[ \int \frac {1}{\sqrt {\text {csch}(2 \log (c x))}} \, dx=\frac {x \left (\sqrt {-1+c^4 x^4}-\arctan \left (\sqrt {-1+c^4 x^4}\right )\right )}{2 \sqrt {2} \sqrt {\frac {c^2 x^2}{-1+c^4 x^4}} \sqrt {-1+c^4 x^4}} \]

input
Integrate[1/Sqrt[Csch[2*Log[c*x]]],x]
 
output
(x*(Sqrt[-1 + c^4*x^4] - ArcTan[Sqrt[-1 + c^4*x^4]]))/(2*Sqrt[2]*Sqrt[(c^2 
*x^2)/(-1 + c^4*x^4)]*Sqrt[-1 + c^4*x^4])
 
3.2.37.3 Rubi [A] (warning: unable to verify)

Time = 0.26 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {6080, 6078, 858, 807, 247, 223}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{\sqrt {\text {csch}(2 \log (c x))}} \, dx\)

\(\Big \downarrow \) 6080

\(\displaystyle \frac {\int \frac {1}{\sqrt {\text {csch}(2 \log (c x))}}d(c x)}{c}\)

\(\Big \downarrow \) 6078

\(\displaystyle \frac {\int c \sqrt {1-\frac {1}{c^4 x^4}} xd(c x)}{c^2 x \sqrt {1-\frac {1}{c^4 x^4}} \sqrt {\text {csch}(2 \log (c x))}}\)

\(\Big \downarrow \) 858

\(\displaystyle -\frac {\int \frac {\sqrt {1-c^4 x^4}}{c^3 x^3}d\frac {1}{c x}}{c^2 x \sqrt {1-\frac {1}{c^4 x^4}} \sqrt {\text {csch}(2 \log (c x))}}\)

\(\Big \downarrow \) 807

\(\displaystyle -\frac {\int \frac {\sqrt {1-c^2 x^2}}{c^2 x^2}d\left (c^2 x^2\right )}{2 c^2 x \sqrt {1-\frac {1}{c^4 x^4}} \sqrt {\text {csch}(2 \log (c x))}}\)

\(\Big \downarrow \) 247

\(\displaystyle -\frac {-\int \frac {1}{\sqrt {1-c^2 x^2}}d\left (c^2 x^2\right )-c^2 x^2 \sqrt {1-c^2 x^2}}{2 c^2 x \sqrt {1-\frac {1}{c^4 x^4}} \sqrt {\text {csch}(2 \log (c x))}}\)

\(\Big \downarrow \) 223

\(\displaystyle -\frac {-\arcsin \left (c^2 x^2\right )-c^2 x^2 \sqrt {1-c^2 x^2}}{2 c^2 x \sqrt {1-\frac {1}{c^4 x^4}} \sqrt {\text {csch}(2 \log (c x))}}\)

input
Int[1/Sqrt[Csch[2*Log[c*x]]],x]
 
output
-1/2*(-(c^2*x^2*Sqrt[1 - c^2*x^2]) - ArcSin[c^2*x^2])/(c^2*Sqrt[1 - 1/(c^4 
*x^4)]*x*Sqrt[Csch[2*Log[c*x]]])
 

3.2.37.3.1 Defintions of rubi rules used

rule 223
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt 
[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
 

rule 247
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1)))   Int[ 
(c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 
0] && LtQ[m, -1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, 
m, p, x]
 

rule 807
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 858
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

rule 6078
Int[Csch[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Simp[Csch[d*(a 
+ b*Log[x])]^p*((1 - 1/(E^(2*a*d)*x^(2*b*d)))^p/x^((-b)*d*p))   Int[1/(x^(b 
*d*p)*(1 - 1/(E^(2*a*d)*x^(2*b*d)))^p), x], x] /; FreeQ[{a, b, d, p}, x] && 
  !IntegerQ[p]
 

rule 6080
Int[Csch[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> S 
imp[x/(n*(c*x^n)^(1/n))   Subst[Int[x^(1/n - 1)*Csch[d*(a + b*Log[x])]^p, x 
], x, c*x^n], x] /; FreeQ[{a, b, c, d, n, p}, x] && (NeQ[c, 1] || NeQ[n, 1] 
)
 
3.2.37.4 Maple [F]

\[\int \frac {1}{\sqrt {\operatorname {csch}\left (2 \ln \left (c x \right )\right )}}d x\]

input
int(1/csch(2*ln(c*x))^(1/2),x)
 
output
int(1/csch(2*ln(c*x))^(1/2),x)
 
3.2.37.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.43 \[ \int \frac {1}{\sqrt {\text {csch}(2 \log (c x))}} \, dx=-\frac {\sqrt {2} c x \arctan \left (\frac {{\left (c^{4} x^{4} - 1\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} - 1}}}{c x}\right ) - \sqrt {2} {\left (c^{4} x^{4} - 1\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} - 1}}}{4 \, c^{2} x} \]

input
integrate(1/csch(2*log(c*x))^(1/2),x, algorithm="fricas")
 
output
-1/4*(sqrt(2)*c*x*arctan((c^4*x^4 - 1)*sqrt(c^2*x^2/(c^4*x^4 - 1))/(c*x)) 
- sqrt(2)*(c^4*x^4 - 1)*sqrt(c^2*x^2/(c^4*x^4 - 1)))/(c^2*x)
 
3.2.37.6 Sympy [F]

\[ \int \frac {1}{\sqrt {\text {csch}(2 \log (c x))}} \, dx=\int \frac {1}{\sqrt {\operatorname {csch}{\left (2 \log {\left (c x \right )} \right )}}}\, dx \]

input
integrate(1/csch(2*ln(c*x))**(1/2),x)
 
output
Integral(1/sqrt(csch(2*log(c*x))), x)
 
3.2.37.7 Maxima [F]

\[ \int \frac {1}{\sqrt {\text {csch}(2 \log (c x))}} \, dx=\int { \frac {1}{\sqrt {\operatorname {csch}\left (2 \, \log \left (c x\right )\right )}} \,d x } \]

input
integrate(1/csch(2*log(c*x))^(1/2),x, algorithm="maxima")
 
output
integrate(1/sqrt(csch(2*log(c*x))), x)
 
3.2.37.8 Giac [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {\text {csch}(2 \log (c x))}} \, dx=\text {Timed out} \]

input
integrate(1/csch(2*log(c*x))^(1/2),x, algorithm="giac")
 
output
Timed out
 
3.2.37.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {\text {csch}(2 \log (c x))}} \, dx=\int \frac {1}{\sqrt {\frac {1}{\mathrm {sinh}\left (2\,\ln \left (c\,x\right )\right )}}} \,d x \]

input
int(1/(1/sinh(2*log(c*x)))^(1/2),x)
 
output
int(1/(1/sinh(2*log(c*x)))^(1/2), x)