3.166 \(\int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^3} \, dx\)

Optimal. Leaf size=137 \[ -\frac {\left (c^4+\frac {1}{x^4}\right ) \sqrt {\text {sech}(2 \log (c x))}}{c^2+\frac {1}{x^2}}-\frac {1}{2} c x \sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) \sqrt {\text {sech}(2 \log (c x))} F\left (2 \cot ^{-1}(c x)|\frac {1}{2}\right )+c x \sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) \sqrt {\text {sech}(2 \log (c x))} E\left (2 \cot ^{-1}(c x)|\frac {1}{2}\right ) \]

[Out]

-(c^4+1/x^4)*sech(2*ln(c*x))^(1/2)/(c^2+1/x^2)+c*(c^2+1/x^2)*x*(cos(2*arccot(c*x))^2)^(1/2)/cos(2*arccot(c*x))
*EllipticE(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)-1/2*c*(c^2+
1/x^2)*x*(cos(2*arccot(c*x))^2)^(1/2)/cos(2*arccot(c*x))*EllipticF(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 = 0.10, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5551, 5549, 335, 305, 220, 1196} \[ -\frac {\left (c^4+\frac {1}{x^4}\right ) \sqrt {\text {sech}(2 \log (c x))}}{c^2+\frac {1}{x^2}}-\frac {1}{2} c x \sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) \sqrt {\text {sech}(2 \log (c x))} F\left (2 \cot ^{-1}(c x)|\frac {1}{2}\right )+c x \sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) \sqrt {\text {sech}(2 \log (c x))} E\left (2 \cot ^{-1}(c x)|\frac {1}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Sech[2*Log[c*x]]]/x^3,x]

[Out]

-(((c^4 + x^(-4))*Sqrt[Sech[2*Log[c*x]]])/(c^2 + x^(-2))) + c*Sqrt[(c^4 + x^(-4))/(c^2 + x^(-2))^2]*(c^2 + x^(
-2))*x*EllipticE[2*ArcCot[c*x], 1/2]*Sqrt[Sech[2*Log[c*x]]] - (c*Sqrt[(c^4 + x^(-4))/(c^2 + x^(-2))^2]*(c^2 +
x^(-2))*x*EllipticF[2*ArcCot[c*x], 1/2]*Sqrt[Sech[2*Log[c*x]]])/2

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 305

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

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 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

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^3} \, dx &=c^2 \operatorname {Subst}\left (\int \frac {\sqrt {\text {sech}(2 \log (x))}}{x^3} \, dx,x,c x\right )\\ &=\left (c^3 \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^4} \, dx,x,c x\right )\\ &=-\left (\left (c^3 \sqrt {1+\frac {1}{c^4 x^4}} x \sqrt {\text {sech}(2 \log (c x))}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+x^4}} \, dx,x,\frac {1}{c x}\right )\right )\\ &=-\left (\left (c^3 \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 )\right )+\left (c^3 \sqrt {1+\frac {1}{c^4 x^4}} x \sqrt {\text {sech}(2 \log (c x))}\right ) \operatorname {Subst}\left (\int \frac {1-x^2}{\sqrt {1+x^4}} \, dx,x,\frac {1}{c x}\right )\\ &=-\frac {\left (c^4+\frac {1}{x^4}\right ) \sqrt {\text {sech}(2 \log (c x))}}{c^2+\frac {1}{x^2}}+c \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 E\left (2 \cot ^{-1}(c x)|\frac {1}{2}\right ) \sqrt {\text {sech}(2 \log (c x))}-\frac {1}{2} c \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 = 0.12, size = 59, normalized size = 0.43 \[ -\frac {c^2 \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-c^4 x^4\right )}{\sqrt {c^4 x^4+1} \sqrt {\frac {c^2 x^2}{2 c^4 x^4+2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Sech[2*Log[c*x]]]/x^3,x]

[Out]

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

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\operatorname {sech}\left (2 \, \log \left (c x\right )\right )}}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(sech(2*log(c*x)))/x^3, 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^3,x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 0.20, size = 134, normalized size = 0.98 \[ -\frac {\left (c^{4} x^{4}+1\right ) \sqrt {2}\, \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}{x^{2}}+\frac {i c^{2} \sqrt {-i c^{2} x^{2}+1}\, \sqrt {i c^{2} x^{2}+1}\, \left (\EllipticF \left (x \sqrt {i c^{2}}, i\right )-\EllipticE \left (x \sqrt {i c^{2}}, i\right )\right ) \sqrt {2}\, \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}{\sqrt {i c^{2}}\, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(2*ln(c*x))^(1/2)/x^3,x)

[Out]

-(c^4*x^4+1)/x^2*2^(1/2)*(c^2*x^2/(c^4*x^4+1))^(1/2)+I*c^2/(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)-EllipticE(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 \[ \int \frac {\sqrt {\operatorname {sech}\left (2 \, \log \left (c x\right )\right )}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(sech(2*log(c*x)))/x^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {\frac {1}{\mathrm {cosh}\left (2\,\ln \left (c\,x\right )\right )}}}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\operatorname {sech}{\left (2 \log {\left (c x \right )} \right )}}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(sech(2*log(c*x)))/x**3, x)

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