\(\int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^5} \, dx\) [168]

Optimal result

Integrand size = 15, antiderivative size = 80 \[ \int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^5} \, dx=-\frac {1}{3} \left (c^4+\frac {1}{x^4}\right ) \sqrt {\text {sech}(2 \log (c x))}+\frac {1}{6} c^3 \sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) x \operatorname {EllipticF}\left (2 \cot ^{-1}(c x),\frac {1}{2}\right ) \sqrt {\text {sech}(2 \log (c x))} \] Output:

-1/3*(c^4+1/x^4)*sech(2*ln(c*x))^(1/2)+1/6*c^3*((c^4+1/x^4)/(c^2+1/x^2)^2) 
^(1/2)*(c^2+1/x^2)*x*InverseJacobiAM(2*arccot(c*x),1/2*2^(1/2))*sech(2*ln( 
c*x))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^5} \, dx=-\frac {\sqrt {2} \sqrt {\frac {c^2 x^2}{1+c^4 x^4}} \sqrt {1+c^4 x^4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {1}{4},-c^4 x^4\right )}{3 x^4} \] Input:

Integrate[Sqrt[Sech[2*Log[c*x]]]/x^5,x]
 

Output:

-1/3*(Sqrt[2]*Sqrt[(c^2*x^2)/(1 + c^4*x^4)]*Sqrt[1 + c^4*x^4]*Hypergeometr 
ic2F1[-3/4, 1/2, 1/4, -(c^4*x^4)])/x^4
 

Rubi [A] (warning: unable to verify)

Time = 0.30 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.49, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6085, 6083, 858, 843, 761}

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.

\(\displaystyle \int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^5} \, dx\)

\(\Big \downarrow \) 6085

\(\displaystyle c^4 \int \frac {\sqrt {\text {sech}(2 \log (c x))}}{c^5 x^5}d(c x)\)

\(\Big \downarrow \) 6083

\(\displaystyle c^5 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))} \int \frac {1}{c^6 \sqrt {1+\frac {1}{c^4 x^4}} x^6}d(c x)\)

\(\Big \downarrow \) 858

\(\displaystyle -c^5 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))} \int \frac {c^4 x^4}{\sqrt {c^4 x^4+1}}d\frac {1}{c x}\)

\(\Big \downarrow \) 843

\(\displaystyle -c^5 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))} \left (\frac {\sqrt {c^4 x^4+1}}{3 c x}-\frac {1}{3} \int \frac {1}{\sqrt {c^4 x^4+1}}d\frac {1}{c x}\right )\)

\(\Big \downarrow \) 761

\(\displaystyle -c^5 x \sqrt {\frac {1}{c^4 x^4}+1} \left (\frac {\sqrt {c^4 x^4+1}}{3 c x}-\frac {\left (c^2 x^2+1\right ) \sqrt {\frac {c^4 x^4+1}{\left (c^2 x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {1}{c x}\right ),\frac {1}{2}\right )}{6 \sqrt {c^4 x^4+1}}\right ) \sqrt {\text {sech}(2 \log (c x))}\)

Input:

Int[Sqrt[Sech[2*Log[c*x]]]/x^5,x]
 

Output:

-(c^5*Sqrt[1 + 1/(c^4*x^4)]*x*(Sqrt[1 + c^4*x^4]/(3*c*x) - ((1 + c^2*x^2)* 
Sqrt[(1 + c^4*x^4)/(1 + c^2*x^2)^2]*EllipticF[2*ArcTan[1/(c*x)], 1/2])/(6* 
Sqrt[1 + c^4*x^4]))*Sqrt[Sech[2*Log[c*x]]])
 

Defintions of rubi rules used

rule 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 843

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

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 6083

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

rule 6085

Int[((e_.)*(x_))^(m_.)*Sech[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p 
_.), x_Symbol] :> Simp[(e*x)^(m + 1)/(e*n*(c*x^n)^((m + 1)/n))   Subst[Int[ 
x^((m + 1)/n - 1)*Sech[d*(a + b*Log[x])]^p, x], x, c*x^n], x] /; FreeQ[{a, 
b, c, d, e, m, n, p}, x] && (NeQ[c, 1] || NeQ[n, 1])
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.46

method	result	size
risch	\(-\frac {\left (c^{4} x^{4}+1\right ) \sqrt {2}\, \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}{3 x^{4}}-\frac {c^{4} \sqrt {-i c^{2} x^{2}+1}\, \sqrt {i c^{2} x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {i c^{2}}, i\right ) \sqrt {2}\, \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}{3 \sqrt {i c^{2}}\, x}\)	\(117\)

Input:

int(sech(2*ln(x*c))^(1/2)/x^5,x,method=_RETURNVERBOSE)
 

Output:

-1/3*(c^4*x^4+1)/x^4*2^(1/2)*(c^2*x^2/(c^4*x^4+1))^(1/2)-1/3*c^4/(I*c^2)^( 
1/2)*(1-I*c^2*x^2)^(1/2)*(1+I*c^2*x^2)^(1/2)*EllipticF(x*(I*c^2)^(1/2),I)* 
2^(1/2)*(c^2*x^2/(c^4*x^4+1))^(1/2)/x
 

Fricas [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^5} \, dx=\frac {\sqrt {2} \left (-c^{4}\right )^{\frac {3}{4}} c x^{4} F(\arcsin \left (\left (-c^{4}\right )^{\frac {1}{4}} x\right )\,|\,-1) - \sqrt {2} {\left (c^{4} x^{4} + 1\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}}}{3 \, x^{4}} \] Input:

integrate(sech(2*log(c*x))^(1/2)/x^5,x, algorithm="fricas")
 

Output:

1/3*(sqrt(2)*(-c^4)^(3/4)*c*x^4*elliptic_f(arcsin((-c^4)^(1/4)*x), -1) - s 
qrt(2)*(c^4*x^4 + 1)*sqrt(c^2*x^2/(c^4*x^4 + 1)))/x^4
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^5} \, dx=\int \frac {\sqrt {\operatorname {sech}{\left (2 \log {\left (c x \right )} \right )}}}{x^{5}}\, dx \] Input:

integrate(sech(2*ln(c*x))**(1/2)/x**5,x)
 

Output:

Integral(sqrt(sech(2*log(c*x)))/x**5, x)
 

Maxima [F]

\[ \int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^5} \, dx=\int { \frac {\sqrt {\operatorname {sech}\left (2 \, \log \left (c x\right )\right )}}{x^{5}} \,d x } \] Input:

integrate(sech(2*log(c*x))^(1/2)/x^5,x, algorithm="maxima")
 

Output:

integrate(sqrt(sech(2*log(c*x)))/x^5, x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^5} \, dx=\text {Timed out} \] Input:

integrate(sech(2*log(c*x))^(1/2)/x^5,x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^5} \, dx=\int \frac {\sqrt {\frac {1}{\mathrm {cosh}\left (2\,\ln \left (c\,x\right )\right )}}}{x^5} \,d x \] Input:

int((1/cosh(2*log(c*x)))^(1/2)/x^5,x)
 

Output:

int((1/cosh(2*log(c*x)))^(1/2)/x^5, x)
 

Reduce [F]

\[ \int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^5} \, dx=\int \frac {\sqrt {\mathrm {sech}\left (2 \,\mathrm {log}\left (c x \right )\right )}}{x^{5}}d x \] Input:

int(sech(2*log(c*x))^(1/2)/x^5,x)
 

Output:

int(sqrt(sech(2*log(c*x)))/x**5,x)