Integrand size = 15, antiderivative size = 137 \[ \int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^3} \, dx=-\frac {\left (c^4+\frac {1}{x^4}\right ) \sqrt {\text {sech}(2 \log (c x))}}{c^2+\frac {1}{x^2}}+c \sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) x E\left (2 \cot ^{-1}(c x)|\frac {1}{2}\right ) \sqrt {\text {sech}(2 \log (c x))}-\frac {1}{2} c \sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) x \operatorname {EllipticF}\left (2 \cot ^{-1}(c x),\frac {1}{2}\right ) \sqrt {\text {sech}(2 \log (c x))} \]
-(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*arcc ot(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*arc cot(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)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.15 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.43 \[ \int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^3} \, dx=-\frac {c^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-c^4 x^4\right )}{\sqrt {1+c^4 x^4} \sqrt {\frac {c^2 x^2}{2+2 c^4 x^4}}} \]
-((c^2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(c^4*x^4)])/(Sqrt[1 + c^4*x^4]*S qrt[(c^2*x^2)/(2 + 2*c^4*x^4)]))
Time = 0.35 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.39, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6085, 6083, 858, 834, 761, 1510}
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.
\(\displaystyle \int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^3} \, dx\) |
\(\Big \downarrow \) 6085 |
\(\displaystyle c^2 \int \frac {\sqrt {\text {sech}(2 \log (c x))}}{c^3 x^3}d(c x)\) |
\(\Big \downarrow \) 6083 |
\(\displaystyle c^3 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))} \int \frac {1}{c^4 \sqrt {1+\frac {1}{c^4 x^4}} x^4}d(c x)\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -c^3 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))} \int \frac {c^2 x^2}{\sqrt {c^4 x^4+1}}d\frac {1}{c x}\) |
\(\Big \downarrow \) 834 |
\(\displaystyle -c^3 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))} \left (\int \frac {1}{\sqrt {c^4 x^4+1}}d\frac {1}{c x}-\int \frac {1-c^2 x^2}{\sqrt {c^4 x^4+1}}d\frac {1}{c x}\right )\) |
\(\Big \downarrow \) 761 |
\(\displaystyle -c^3 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))} \left (\frac {\left (c^2 x^2+1\right ) \sqrt {\frac {c^4 x^4+1}{\left (c^2 x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {1}{c x}\right ),\frac {1}{2}\right )}{2 \sqrt {c^4 x^4+1}}-\int \frac {1-c^2 x^2}{\sqrt {c^4 x^4+1}}d\frac {1}{c x}\right )\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle -c^3 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))} \left (\frac {\left (c^2 x^2+1\right ) \sqrt {\frac {c^4 x^4+1}{\left (c^2 x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {1}{c x}\right ),\frac {1}{2}\right )}{2 \sqrt {c^4 x^4+1}}-\frac {\left (c^2 x^2+1\right ) \sqrt {\frac {c^4 x^4+1}{\left (c^2 x^2+1\right )^2}} E\left (2 \arctan \left (\frac {1}{c x}\right )|\frac {1}{2}\right )}{\sqrt {c^4 x^4+1}}+\frac {\sqrt {c^4 x^4+1}}{c x \left (c^2 x^2+1\right )}\right )\) |
-(c^3*Sqrt[1 + 1/(c^4*x^4)]*x*(Sqrt[1 + c^4*x^4]/(c*x*(1 + c^2*x^2)) - ((1 + c^2*x^2)*Sqrt[(1 + c^4*x^4)/(1 + c^2*x^2)^2]*EllipticE[2*ArcTan[1/(c*x) ], 1/2])/Sqrt[1 + c^4*x^4] + ((1 + c^2*x^2)*Sqrt[(1 + c^4*x^4)/(1 + c^2*x^ 2)^2]*EllipticF[2*ArcTan[1/(c*x)], 1/2])/(2*Sqrt[1 + c^4*x^4]))*Sqrt[Sech[ 2*Log[c*x]]])
3.2.66.3.1 Defintions of rubi rules used
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)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S imp[1/q Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
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[((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)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Int[((e_.)*(x_))^(m_.)*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[(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]
Int[((e_.)*(x_))^(m_.)*Sech[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p _.), x_Symbol] :> Simp[(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])
Result contains complex when optimal does not.
Time = 0.66 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.98
method | result | size |
risch | \(-\frac {\left (c^{4} x^{4}+1\right ) \sqrt {2}\, \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}{x^{2}}+\frac {i c^{2} \sqrt {-i c^{2} x^{2}+1}\, \sqrt {i c^{2} x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {i c^{2}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {i c^{2}}, i\right )\right ) \sqrt {2}\, \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}{\sqrt {i c^{2}}\, x}\) | \(134\) |
-(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)-Ellip ticE(x*(I*c^2)^(1/2),I))*2^(1/2)*(c^2*x^2/(c^4*x^4+1))^(1/2)/x
Time = 0.08 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.69 \[ \int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^3} \, dx=-\frac {\sqrt {2} \left (-c^{4}\right )^{\frac {3}{4}} c x^{2} E(\arcsin \left (\left (-c^{4}\right )^{\frac {1}{4}} x\right )\,|\,-1) - \sqrt {2} \left (-c^{4}\right )^{\frac {3}{4}} c x^{2} F(\arcsin \left (\left (-c^{4}\right )^{\frac {1}{4}} x\right )\,|\,-1) + \sqrt {2} {\left (c^{4} x^{4} + 1\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}}}{x^{2}} \]
-(sqrt(2)*(-c^4)^(3/4)*c*x^2*elliptic_e(arcsin((-c^4)^(1/4)*x), -1) - sqrt (2)*(-c^4)^(3/4)*c*x^2*elliptic_f(arcsin((-c^4)^(1/4)*x), -1) + sqrt(2)*(c ^4*x^4 + 1)*sqrt(c^2*x^2/(c^4*x^4 + 1)))/x^2
\[ \int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^3} \, dx=\int \frac {\sqrt {\operatorname {sech}{\left (2 \log {\left (c x \right )} \right )}}}{x^{3}}\, dx \]
\[ \int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^3} \, dx=\int { \frac {\sqrt {\operatorname {sech}\left (2 \, \log \left (c x\right )\right )}}{x^{3}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^3} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sqrt {\text {sech}(2 \log (c x))}}{x^3} \, dx=\int \frac {\sqrt {\frac {1}{\mathrm {cosh}\left (2\,\ln \left (c\,x\right )\right )}}}{x^3} \,d x \]