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

Optimal. Leaf size=80 \[ \frac {1}{6} c^3 x \sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) \sqrt {\text {sech}(2 \log (c x))} F\left (2 \cot ^{-1}(c x)|\frac {1}{2}\right )-\frac {1}{3} \left (c^4+\frac {1}{x^4}\right ) \sqrt {\text {sech}(2 \log (c x))} \]

[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)

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Rubi [A]  time = 0.07, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5551, 5549, 335, 321, 220} \[ \frac {1}{6} c^3 x \sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) \sqrt {\text {sech}(2 \log (c x))} F\left (2 \cot ^{-1}(c x)|\frac {1}{2}\right )-\frac {1}{3} \left (c^4+\frac {1}{x^4}\right ) \sqrt {\text {sech}(2 \log (c x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 321

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 335

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 5549

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/(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 5551

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*} \int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^5} \, dx &=c^4 \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 F\left (2 \cot ^{-1}(c x)|\frac {1}{2}\right ) \sqrt {\text {sech}(2 \log (c x))}\\ \end {align*}

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Mathematica [C]  time = 0.10, size = 65, normalized size = 0.81 \[ -\frac {\sqrt {2} \sqrt {\frac {c^2 x^2}{c^4 x^4+1}} \sqrt {c^4 x^4+1} \, _2F_1\left (-\frac {3}{4},\frac {1}{2};\frac {1}{4};-c^4 x^4\right )}{3 x^4} \]

Antiderivative was successfully verified.

[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

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\operatorname {sech}\left (2 \, \log \left (c x\right )\right )}}{x^{5}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [C]  time = 0.20, size = 117, normalized size = 1.46 \[ -\frac {\left (c^{4} x^{4}+1\right ) \sqrt {2}\, \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}{3 x^{4}}-\frac {c^{4} \sqrt {-i c^{2} x^{2}+1}\, \sqrt {i c^{2} x^{2}+1}\, \EllipticF \left (x \sqrt {i c^{2}}, i\right ) \sqrt {2}\, \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}{3 \sqrt {i c^{2}}\, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\operatorname {sech}\left (2 \, \log \left (c x\right )\right )}}{x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {\frac {1}{\mathrm {cosh}\left (2\,\ln \left (c\,x\right )\right )}}}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\operatorname {sech}{\left (2 \log {\left (c x \right )} \right )}}}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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