\(\int \frac {1}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx\) [151]

Optimal result

Integrand size = 11, antiderivative size = 96 \[ \int \frac {1}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {3}{4 \left (c^4-\frac {1}{x^4}\right ) x^3 \text {csch}^{\frac {3}{2}}(2 \log (c x))}+\frac {x}{4 \text {csch}^{\frac {3}{2}}(2 \log (c x))}-\frac {3 \text {arctanh}\left (\sqrt {1-\frac {1}{c^4 x^4}}\right )}{4 c^4 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {csch}^{\frac {3}{2}}(2 \log (c x))} \] Output:

3/4/(c^4-1/x^4)/x^3/csch(2*ln(c*x))^(3/2)+1/4*x/csch(2*ln(c*x))^(3/2)-3/4* 
arctanh((1-1/c^4/x^4)^(1/2))/c^4/(1-1/c^4/x^4)^(3/2)/x^3/csch(2*ln(c*x))^( 
3/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.66 \[ \int \frac {1}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{2},\frac {1}{2},c^4 x^4\right )}{4 c^2 x \sqrt {2-2 c^4 x^4} \sqrt {\frac {c^2 x^2}{-1+c^4 x^4}}} \] Input:

Integrate[Csch[2*Log[c*x]]^(-3/2),x]
 

Output:

Hypergeometric2F1[-3/2, -1/2, 1/2, c^4*x^4]/(4*c^2*x*Sqrt[2 - 2*c^4*x^4]*S 
qrt[(c^2*x^2)/(-1 + c^4*x^4)])
 

Rubi [A] (warning: unable to verify)

Time = 0.28 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {6080, 6078, 798, 51, 60, 73, 219}

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.

\(\displaystyle \int \frac {1}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx\)

\(\Big \downarrow \) 6080

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

\(\Big \downarrow \) 6078

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

\(\Big \downarrow \) 798

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 219

\(\displaystyle -\frac {-\frac {3}{2} \left (2 \sqrt {1-\frac {1}{c^4 x^4}}-2 \text {arctanh}\left (\sqrt {1-\frac {1}{c^4 x^4}}\right )\right )-\frac {\left (1-\frac {1}{c^4 x^4}\right )^{3/2}}{c x}}{4 c^4 x^3 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} \text {csch}^{\frac {3}{2}}(2 \log (c x))}\)

Input:

Int[Csch[2*Log[c*x]]^(-3/2),x]
 

Output:

-1/4*(-((1 - 1/(c^4*x^4))^(3/2)/(c*x)) - (3*(2*Sqrt[1 - 1/(c^4*x^4)] - 2*A 
rcTanh[Sqrt[1 - 1/(c^4*x^4)]]))/2)/(c^4*(1 - 1/(c^4*x^4))^(3/2)*x^3*Csch[2 
*Log[c*x]]^(3/2))
 

Defintions of rubi rules used

rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 798

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n   Subst 
[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, 
b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
 

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] 
)
 

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.35

method	result	size
risch	\(\frac {\left (c^{8} x^{8}+c^{4} x^{4}-2\right ) \sqrt {2}}{16 x \left (c^{4} x^{4}-1\right ) c^{2} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}-1}}}-\frac {3 c^{2} \ln \left (\frac {c^{4} x^{2}}{\sqrt {c^{4}}}+\sqrt {c^{4} x^{4}-1}\right ) \sqrt {2}\, x}{16 \sqrt {c^{4}}\, \sqrt {c^{4} x^{4}-1}\, \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}-1}}}\)	\(130\)

Input:

int(1/csch(2*ln(x*c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/16*(c^8*x^8+c^4*x^4-2)/x/(c^4*x^4-1)*2^(1/2)/c^2/(c^2*x^2/(c^4*x^4-1))^( 
1/2)-3/16*c^2*ln(c^4*x^2/(c^4)^(1/2)+(c^4*x^4-1)^(1/2))/(c^4)^(1/2)*2^(1/2 
)*x/(c^4*x^4-1)^(1/2)/(c^2*x^2/(c^4*x^4-1))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.10 \[ \int \frac {1}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {3 \, \sqrt {2} c^{3} x^{3} \log \left (2 \, c^{4} x^{4} - 2 \, {\left (c^{5} x^{5} - c x\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} - 1}} - 1\right ) + 2 \, \sqrt {2} {\left (c^{8} x^{8} + c^{4} x^{4} - 2\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} - 1}}}{32 \, c^{4} x^{3}} \] Input:

integrate(1/csch(2*log(c*x))^(3/2),x, algorithm="fricas")
 

Output:

1/32*(3*sqrt(2)*c^3*x^3*log(2*c^4*x^4 - 2*(c^5*x^5 - c*x)*sqrt(c^2*x^2/(c^ 
4*x^4 - 1)) - 1) + 2*sqrt(2)*(c^8*x^8 + c^4*x^4 - 2)*sqrt(c^2*x^2/(c^4*x^4 
 - 1)))/(c^4*x^3)
 

Sympy [F]

\[ \int \frac {1}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {1}{\operatorname {csch}^{\frac {3}{2}}{\left (2 \log {\left (c x \right )} \right )}}\, dx \] Input:

integrate(1/csch(2*ln(c*x))**(3/2),x)
 

Output:

Integral(csch(2*log(c*x))**(-3/2), x)
 

Maxima [F]

\[ \int \frac {1}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int { \frac {1}{\operatorname {csch}\left (2 \, \log \left (c x\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/csch(2*log(c*x))^(3/2),x, algorithm="maxima")
 

Output:

integrate(csch(2*log(c*x))^(-3/2), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {1}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\text {Timed out} \] Input:

integrate(1/csch(2*log(c*x))^(3/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {1}{{\left (\frac {1}{\mathrm {sinh}\left (2\,\ln \left (c\,x\right )\right )}\right )}^{3/2}} \,d x \] Input:

int(1/(1/sinh(2*log(c*x)))^(3/2),x)
 

Output:

int(1/(1/sinh(2*log(c*x)))^(3/2), x)
 

Reduce [F]

\[ \int \frac {1}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {\sqrt {\mathrm {csch}\left (2 \,\mathrm {log}\left (c x \right )\right )}}{\mathrm {csch}\left (2 \,\mathrm {log}\left (c x \right )\right )^{2}}d x \] Input:

int(1/csch(2*log(c*x))^(3/2),x)
 

Output:

int(sqrt(csch(2*log(c*x)))/csch(2*log(c*x))**2,x)