Integrand size = 15, antiderivative size = 92 \[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^3} \, dx=\frac {1}{2} \left (c^4+\frac {1}{x^4}\right ) x^2 \text {sech}^{\frac {3}{2}}(2 \log (c x))-\frac {\left (c^4+\frac {1}{x^4}\right ) \sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) x^3 \operatorname {EllipticF}\left (2 \cot ^{-1}(c x),\frac {1}{2}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x))}{4 c} \] Output:
1/2*(c^4+1/x^4)*x^2*sech(2*ln(c*x))^(3/2)-1/4*(c^4+1/x^4)*((c^4+1/x^4)/(c^ 2+1/x^2)^2)^(1/2)*(c^2+1/x^2)*x^3*InverseJacobiAM(2*arccot(c*x),1/2*2^(1/2 ))*sech(2*ln(c*x))^(3/2)/c
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.09 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.71 \[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^3} \, dx=\sqrt {2} c^2 \sqrt {\frac {c^2 x^2}{1+c^4 x^4}} \left (1+\sqrt {1+c^4 x^4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-c^4 x^4\right )\right ) \] Input:
Integrate[Sech[2*Log[c*x]]^(3/2)/x^3,x]
Output:
Sqrt[2]*c^2*Sqrt[(c^2*x^2)/(1 + c^4*x^4)]*(1 + Sqrt[1 + c^4*x^4]*Hypergeom etric2F1[1/4, 1/2, 5/4, -(c^4*x^4)])
Time = 0.32 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.32, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6085, 6083, 858, 817, 761}
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.
\(\displaystyle \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^3} \, dx\) |
\(\Big \downarrow \) 6085 |
\(\displaystyle c^2 \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{c^3 x^3}d(c x)\) |
\(\Big \downarrow \) 6083 |
\(\displaystyle c^5 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x)) \int \frac {1}{c^6 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^6}d(c x)\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -c^5 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x)) \int \frac {c^4 x^4}{\left (c^4 x^4+1\right )^{3/2}}d\frac {1}{c x}\) |
\(\Big \downarrow \) 817 |
\(\displaystyle -c^5 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x)) \left (\frac {1}{2} \int \frac {1}{\sqrt {c^4 x^4+1}}d\frac {1}{c x}-\frac {1}{2 c x \sqrt {c^4 x^4+1}}\right )\) |
\(\Big \downarrow \) 761 |
\(\displaystyle -c^5 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \left (\frac {\left (c^2 x^2+1\right ) \sqrt {\frac {c^4 x^4+1}{\left (c^2 x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {1}{c x}\right ),\frac {1}{2}\right )}{4 \sqrt {c^4 x^4+1}}-\frac {1}{2 c x \sqrt {c^4 x^4+1}}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x))\) |
Input:
Int[Sech[2*Log[c*x]]^(3/2)/x^3,x]
Output:
-(c^5*(1 + 1/(c^4*x^4))^(3/2)*x^3*(-1/2*1/(c*x*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])/(4*Sqrt[1 + c^4*x^4]))*Sech[2*Log[c*x]]^(3/2))
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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^( n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] - Simp[c^n *((m - n + 1)/(b*n*(p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x ] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] && ! ILtQ[(m + n*(p + 1) + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
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[((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])
\[\int \frac {\operatorname {sech}\left (2 \ln \left (x c \right )\right )^{\frac {3}{2}}}{x^{3}}d x\]
Input:
int(sech(2*ln(x*c))^(3/2)/x^3,x)
Output:
int(sech(2*ln(x*c))^(3/2)/x^3,x)
Time = 0.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.61 \[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^3} \, dx=\frac {\sqrt {2} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}} c^{3} - \sqrt {2} \left (-c^{4}\right )^{\frac {3}{4}} F(\arcsin \left (\left (-c^{4}\right )^{\frac {1}{4}} x\right )\,|\,-1)}{c} \] Input:
integrate(sech(2*log(c*x))^(3/2)/x^3,x, algorithm="fricas")
Output:
(sqrt(2)*sqrt(c^2*x^2/(c^4*x^4 + 1))*c^3 - sqrt(2)*(-c^4)^(3/4)*elliptic_f (arcsin((-c^4)^(1/4)*x), -1))/c
\[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^3} \, dx=\int \frac {\operatorname {sech}^{\frac {3}{2}}{\left (2 \log {\left (c x \right )} \right )}}{x^{3}}\, dx \] Input:
integrate(sech(2*ln(c*x))**(3/2)/x**3,x)
Output:
Integral(sech(2*log(c*x))**(3/2)/x**3, x)
\[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^3} \, dx=\int { \frac {\operatorname {sech}\left (2 \, \log \left (c x\right )\right )^{\frac {3}{2}}}{x^{3}} \,d x } \] Input:
integrate(sech(2*log(c*x))^(3/2)/x^3,x, algorithm="maxima")
Output:
integrate(sech(2*log(c*x))^(3/2)/x^3, x)
Timed out. \[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^3} \, dx=\text {Timed out} \] Input:
integrate(sech(2*log(c*x))^(3/2)/x^3,x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^3} \, dx=\int \frac {{\left (\frac {1}{\mathrm {cosh}\left (2\,\ln \left (c\,x\right )\right )}\right )}^{3/2}}{x^3} \,d x \] Input:
int((1/cosh(2*log(c*x)))^(3/2)/x^3,x)
Output:
int((1/cosh(2*log(c*x)))^(3/2)/x^3, x)
\[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^3} \, dx=\int \frac {\sqrt {\mathrm {sech}\left (2 \,\mathrm {log}\left (c x \right )\right )}\, \mathrm {sech}\left (2 \,\mathrm {log}\left (c x \right )\right )}{x^{3}}d x \] Input:
int(sech(2*log(c*x))^(3/2)/x^3,x)
Output:
int((sqrt(sech(2*log(c*x)))*sech(2*log(c*x)))/x**3,x)