3.2.65 \(\int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^2} \, dx\) [165]

Optimal. Leaf size=40 \[ -\frac {1}{2} c^2 \sqrt {1+\frac {1}{c^4 x^4}} x \text {csch}^{-1}\left (c^2 x^2\right ) \sqrt {\text {sech}(2 \log (c x))} \]

[Out]

-1/2*c^2*x*arccsch(c^2*x^2)*(1+1/c^4/x^4)^(1/2)*sech(2*ln(c*x))^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 40, 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 = {5670, 5668, 342, 281, 221} \begin {gather*} -\frac {1}{2} c^2 x \sqrt {\frac {1}{c^4 x^4}+1} \text {csch}^{-1}\left (c^2 x^2\right ) \sqrt {\text {sech}(2 \log (c x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(c^2*Sqrt[1 + 1/(c^4*x^4)]*x*ArcCsch[c^2*x^2]*Sqrt[Sech[2*Log[c*x]]])

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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

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Mathematica [A]
time = 0.09, size = 55, normalized size = 1.38 \begin {gather*} -\frac {\sqrt {1+c^4 x^4} \sqrt {\frac {c^2 x^2}{2+2 c^4 x^4}} \tanh ^{-1}\left (\sqrt {1+c^4 x^4}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[1 + c^4*x^4]*Sqrt[(c^2*x^2)/(2 + 2*c^4*x^4)]*ArcTanh[Sqrt[1 + c^4*x^4]])/x)

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Maple [F]
time = 0.88, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {\mathrm {sech}\left (2 \ln \left (c x \right )\right )}}{x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 57, normalized size = 1.42 \begin {gather*} \frac {1}{4} \, \sqrt {2} c \log \left (\frac {c^{5} x^{5} + 2 \, c x - 2 \, {\left (c^{4} x^{4} + 1\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}}}{c x^{5}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\sqrt {\frac {1}{\mathrm {cosh}\left (2\,\ln \left (c\,x\right )\right )}}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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