\(\int \frac {1}{\sqrt {\text {sech}(2 \log (c x))}} \, dx\) [163]

Optimal result

Integrand size = 11, antiderivative size = 59 \[ \int \frac {1}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\frac {x}{2 \sqrt {\text {sech}(2 \log (c x))}}-\frac {\text {csch}^{-1}\left (c^2 x^2\right )}{2 c^2 \sqrt {1+\frac {1}{c^4 x^4}} x \sqrt {\text {sech}(2 \log (c x))}} \]

[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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {5664, 5662, 342, 281, 283, 221} \[ \int \frac {1}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\frac {x}{2 \sqrt {\text {sech}(2 \log (c x))}}-\frac {\text {csch}^{-1}\left (c^2 x^2\right )}{2 c^2 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}} \]

[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 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 283

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 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 5662

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 5664

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*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\sqrt {\text {sech}(2 \log (x))}} \, dx,x,c x\right )}{c} \\ & = \frac {\text {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 {\text {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 {\text {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 {\text {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*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.27 \[ \int \frac {1}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\frac {x \left (\sqrt {1+c^4 x^4}-\text {arctanh}\left (\sqrt {1+c^4 x^4}\right )\right )}{2 \sqrt {2} \sqrt {\frac {c^2 x^2}{1+c^4 x^4}} \sqrt {1+c^4 x^4}} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {1}{\sqrt {\operatorname {sech}\left (2 \ln \left (c x \right )\right )}}d x\]

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (49) = 98\).

Time = 0.27 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.69 \[ \int \frac {1}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\frac {\sqrt {2} c x \log \left (\frac {c^{5} x^{5} + 2 \, c x - 2 \, {\left (c^{4} x^{4} + 1\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}}}{c x^{5}}\right ) + 2 \, \sqrt {2} {\left (c^{4} x^{4} + 1\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}}}{8 \, c^{2} x} \]

[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)

Sympy [F]

\[ \int \frac {1}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\int \frac {1}{\sqrt {\operatorname {sech}{\left (2 \log {\left (c x \right )} \right )}}}\, dx \]

[In]

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

[Out]

Integral(1/sqrt(sech(2*log(c*x))), x)

Maxima [F]

\[ \int \frac {1}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\int { \frac {1}{\sqrt {\operatorname {sech}\left (2 \, \log \left (c x\right )\right )}} \,d x } \]

[In]

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

[Out]

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

Giac [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

int(1/(1/cosh(2*log(c*x)))^(1/2), x)