Integrand size = 11, antiderivative size = 59 \[ \int \frac {1}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\frac {x}{2 \sqrt {\text {sech}(2 \log (c x))}}-\frac {\text {csch}^{-1}\left (c^2 x^2\right )}{2 c^2 \sqrt {1+\frac {1}{c^4 x^4}} x \sqrt {\text {sech}(2 \log (c x))}} \] Output:
1/2*x/sech(2*ln(c*x))^(1/2)-1/2*arccsch(c^2*x^2)/c^2/(1+1/c^4/x^4)^(1/2)/x /sech(2*ln(c*x))^(1/2)
Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.27 \[ \int \frac {1}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\frac {x \left (\sqrt {1+c^4 x^4}-\text {arctanh}\left (\sqrt {1+c^4 x^4}\right )\right )}{2 \sqrt {2} \sqrt {\frac {c^2 x^2}{1+c^4 x^4}} \sqrt {1+c^4 x^4}} \] Input:
Integrate[1/Sqrt[Sech[2*Log[c*x]]],x]
Output:
(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])
Time = 0.26 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {6079, 6077, 858, 807, 247, 222}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\sqrt {\text {sech}(2 \log (c x))}} \, dx\) |
\(\Big \downarrow \) 6079 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {\text {sech}(2 \log (c x))}}d(c x)}{c}\) |
\(\Big \downarrow \) 6077 |
\(\displaystyle \frac {\int c \sqrt {1+\frac {1}{c^4 x^4}} xd(c x)}{c^2 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\frac {\int \frac {\sqrt {c^4 x^4+1}}{c^3 x^3}d\frac {1}{c x}}{c^2 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle -\frac {\int \frac {\sqrt {c^2 x^2+1}}{c^2 x^2}d\left (c^2 x^2\right )}{2 c^2 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}}\) |
\(\Big \downarrow \) 247 |
\(\displaystyle -\frac {\int \frac {1}{\sqrt {c^2 x^2+1}}d\left (c^2 x^2\right )-c^2 x^2 \sqrt {c^2 x^2+1}}{2 c^2 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle -\frac {\text {arcsinh}\left (c^2 x^2\right )-c^2 x^2 \sqrt {c^2 x^2+1}}{2 c^2 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}}\) |
Input:
Int[1/Sqrt[Sech[2*Log[c*x]]],x]
Output:
-1/2*(-(c^2*x^2*Sqrt[1 + c^2*x^2]) + ArcSinh[c^2*x^2])/(c^2*Sqrt[1 + 1/(c^ 4*x^4)]*x*Sqrt[Sech[2*Log[c*x]]])
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]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[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]
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] && Int egerQ[m]
Int[Sech[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Simp[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]
Int[Sech[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> S imp[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] )
\[\int \frac {1}{\sqrt {\operatorname {sech}\left (2 \ln \left (x c \right )\right )}}d x\]
Input:
int(1/sech(2*ln(x*c))^(1/2),x)
Output:
int(1/sech(2*ln(x*c))^(1/2),x)
Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (49) = 98\).
Time = 0.20 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.69 \[ \int \frac {1}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\frac {\sqrt {2} c x \log \left (\frac {c^{5} x^{5} + 2 \, c x - 2 \, {\left (c^{4} x^{4} + 1\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}}}{c x^{5}}\right ) + 2 \, \sqrt {2} {\left (c^{4} x^{4} + 1\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}}}{8 \, c^{2} x} \] Input:
integrate(1/sech(2*log(c*x))^(1/2),x, algorithm="fricas")
Output:
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)
\[ \int \frac {1}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\int \frac {1}{\sqrt {\operatorname {sech}{\left (2 \log {\left (c x \right )} \right )}}}\, dx \] Input:
integrate(1/sech(2*ln(c*x))**(1/2),x)
Output:
Integral(1/sqrt(sech(2*log(c*x))), x)
\[ \int \frac {1}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\int { \frac {1}{\sqrt {\operatorname {sech}\left (2 \, \log \left (c x\right )\right )}} \,d x } \] Input:
integrate(1/sech(2*log(c*x))^(1/2),x, algorithm="maxima")
Output:
integrate(1/sqrt(sech(2*log(c*x))), x)
Timed out. \[ \int \frac {1}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\text {Timed out} \] Input:
integrate(1/sech(2*log(c*x))^(1/2),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {1}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\int \frac {1}{\sqrt {\frac {1}{\mathrm {cosh}\left (2\,\ln \left (c\,x\right )\right )}}} \,d x \] Input:
int(1/(1/cosh(2*log(c*x)))^(1/2),x)
Output:
int(1/(1/cosh(2*log(c*x)))^(1/2), x)
\[ \int \frac {1}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\int \frac {\sqrt {\mathrm {sech}\left (2 \,\mathrm {log}\left (c x \right )\right )}}{\mathrm {sech}\left (2 \,\mathrm {log}\left (c x \right )\right )}d x \] Input:
int(1/sech(2*log(c*x))^(1/2),x)
Output:
int(sqrt(sech(2*log(c*x)))/sech(2*log(c*x)),x)