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

Optimal result

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

[Out]

-1/3*(c^4+1/x^4)*sech(2*ln(c*x))^(1/2)+1/6*c^3*(c^2+1/x^2)*x*(cos(2*arccot(c*x))^2)^(1/2)/cos(2*arccot(c*x))*E
llipticF(sin(2*arccot(c*x)),1/2*2^(1/2))*((c^4+1/x^4)/(c^2+1/x^2)^2)^(1/2)*sech(2*ln(c*x))^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5670, 5668, 342, 327, 226} \[ \int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^5} \, dx=\frac {1}{6} c^3 x \sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) \sqrt {\text {sech}(2 \log (c x))} \operatorname {EllipticF}\left (2 \cot ^{-1}(c x),\frac {1}{2}\right )-\frac {1}{3} \left (c^4+\frac {1}{x^4}\right ) \sqrt {\text {sech}(2 \log (c x))} \]

[In]

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

[Out]

-1/3*((c^4 + x^(-4))*Sqrt[Sech[2*Log[c*x]]]) + (c^3*Sqrt[(c^4 + x^(-4))/(c^2 + x^(-2))^2]*(c^2 + x^(-2))*x*Ell
ipticF[2*ArcCot[c*x], 1/2]*Sqrt[Sech[2*Log[c*x]]])/6

Rule 226

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 327

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] - Dist[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 342

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] && IntegerQ[m]

Rule 5668

Int[((e_.)*(x_))^(m_.)*Sech[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[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 5670

Int[((e_.)*(x_))^(m_.)*Sech[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[(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])

Rubi steps \begin{align*} \text {integral}& = c^4 \text {Subst}\left (\int \frac {\sqrt {\text {sech}(2 \log (x))}}{x^5} \, dx,x,c x\right ) \\ & = \left (c^5 \sqrt {1+\frac {1}{c^4 x^4}} x \sqrt {\text {sech}(2 \log (c x))}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {1}{x^4}} x^6} \, dx,x,c x\right ) \\ & = -\left (\left (c^5 \sqrt {1+\frac {1}{c^4 x^4}} x \sqrt {\text {sech}(2 \log (c x))}\right ) \text {Subst}\left (\int \frac {x^4}{\sqrt {1+x^4}} \, dx,x,\frac {1}{c x}\right )\right ) \\ & = -\frac {1}{3} \left (c^4+\frac {1}{x^4}\right ) \sqrt {\text {sech}(2 \log (c x))}+\frac {1}{3} \left (c^5 \sqrt {1+\frac {1}{c^4 x^4}} x \sqrt {\text {sech}(2 \log (c x))}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,\frac {1}{c x}\right ) \\ & = -\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))} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.13 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^5} \, dx=-\frac {\sqrt {2} \sqrt {\frac {c^2 x^2}{1+c^4 x^4}} \sqrt {1+c^4 x^4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {1}{4},-c^4 x^4\right )}{3 x^4} \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.60 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.46

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

[In]

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

[Out]

-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)

none

Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^5} \, dx=\frac {\sqrt {2} \left (-c^{4}\right )^{\frac {3}{4}} c x^{4} F(\arcsin \left (\left (-c^{4}\right )^{\frac {1}{4}} x\right )\,|\,-1) - \sqrt {2} {\left (c^{4} x^{4} + 1\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}}}{3 \, x^{4}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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