3.163 \(\int \frac {1}{\sqrt {\text {sech}(2 \log (c x))}} \, dx\)

Optimal. Leaf size=59 \[ \frac {x}{2 \sqrt {\text {sech}(2 \log (c x))}}-\frac {\text {csch}^{-1}\left (c^2 x^2\right )}{2 c^2 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}} \]

[Out]

1/2*x/sech(2*ln(c*x))^(1/2)-1/2*arccsch(c^2*x^2)/c^2/x/(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 = 59, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {5545, 5543, 335, 275, 277, 215} \[ \frac {x}{2 \sqrt {\text {sech}(2 \log (c x))}}-\frac {\text {csch}^{-1}\left (c^2 x^2\right )}{2 c^2 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[Sech[2*Log[c*x]]],x]

[Out]

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

Rule 215

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 275

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 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 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 5543

Int[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[1/(x^(b*d*p)*(1 + 1/(E^(2*a*d)*x^(2*b*d)))^p), x], x] /; FreeQ[{a, b, d, p}, x
] &&  !IntegerQ[p]

Rule 5545

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

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 77, normalized size = 1.31 \[ \frac {x \left (2 \sqrt {c^4 x^4+1}-2 \tanh ^{-1}\left (\sqrt {c^4 x^4+1}\right )\right )}{4 \sqrt {2} \sqrt {\frac {c^2 x^2}{c^4 x^4+1}} \sqrt {c^4 x^4+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[Sech[2*Log[c*x]]],x]

[Out]

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

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fricas [B]  time = 0.41, size = 100, normalized size = 1.69 \[ \frac {\sqrt {2} c x \log \left (\frac {c^{5} x^{5} + 2 \, c x - 2 \, {\left (c^{4} x^{4} + 1\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}}}{c x^{5}}\right ) + 2 \, \sqrt {2} {\left (c^{4} x^{4} + 1\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}}}{8 \, c^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(sqrt(2)*c*x*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)) + 2*sqrt(2)*(c^4
*x^4 + 1)*sqrt(c^2*x^2/(c^4*x^4 + 1)))/(c^2*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(1/sech(2*log(c*x))^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 0.20, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {\mathrm {sech}\left (2 \ln \left (c x \right )\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {\operatorname {sech}\left (2 \, \log \left (c x\right )\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(sech(2*log(c*x))), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(1/cosh(2*log(c*x)))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {\operatorname {sech}{\left (2 \log {\left (c x \right )} \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(sech(2*log(c*x))), x)

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