\(\int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^3} \, dx\) [180]

Optimal result

Integrand size = 15, antiderivative size = 92 \[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^3} \, dx=\frac {1}{2} \left (c^4+\frac {1}{x^4}\right ) x^2 \text {sech}^{\frac {3}{2}}(2 \log (c x))-\frac {\left (c^4+\frac {1}{x^4}\right ) \sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) x^3 \operatorname {EllipticF}\left (2 \cot ^{-1}(c x),\frac {1}{2}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x))}{4 c} \]

[Out]

1/2*(c^4+1/x^4)*x^2*sech(2*ln(c*x))^(3/2)-1/4*(c^4+1/x^4)*(c^2+1/x^2)*x^3*(cos(2*arccot(c*x))^2)^(1/2)/cos(2*a
rccot(c*x))*EllipticF(sin(2*arccot(c*x)),1/2*2^(1/2))*sech(2*ln(c*x))^(3/2)*((c^4+1/x^4)/(c^2+1/x^2)^2)^(1/2)/
c

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5670, 5668, 342, 294, 226} \[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^3} \, dx=\frac {1}{2} x^2 \left (c^4+\frac {1}{x^4}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x))-\frac {x^3 \left (c^4+\frac {1}{x^4}\right ) \sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x)) \operatorname {EllipticF}\left (2 \cot ^{-1}(c x),\frac {1}{2}\right )}{4 c} \]

[In]

Int[Sech[2*Log[c*x]]^(3/2)/x^3,x]

[Out]

((c^4 + x^(-4))*x^2*Sech[2*Log[c*x]]^(3/2))/2 - ((c^4 + x^(-4))*Sqrt[(c^4 + x^(-4))/(c^2 + x^(-2))^2]*(c^2 + x
^(-2))*x^3*EllipticF[2*ArcCot[c*x], 1/2]*Sech[2*Log[c*x]]^(3/2))/(4*c)

Rule 226

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]

Rule 294

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] - Dist[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]

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 5668

Int[((e_.)*(x_))^(m_.)*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[(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]

Rule 5670

Int[((e_.)*(x_))^(m_.)*Sech[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[(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])

Rubi steps \begin{align*} \text {integral}& = c^2 \text {Subst}\left (\int \frac {\text {sech}^{\frac {3}{2}}(2 \log (x))}{x^3} \, dx,x,c x\right ) \\ & = \left (c^5 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {1}{x^4}\right )^{3/2} x^6} \, dx,x,c x\right ) \\ & = -\left (\left (c^5 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))\right ) \text {Subst}\left (\int \frac {x^4}{\left (1+x^4\right )^{3/2}} \, dx,x,\frac {1}{c x}\right )\right ) \\ & = \frac {1}{2} \left (c^4+\frac {1}{x^4}\right ) x^2 \text {sech}^{\frac {3}{2}}(2 \log (c x))-\frac {1}{2} \left (c^5 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,\frac {1}{c x}\right ) \\ & = \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} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.14 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.71 \[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^3} \, dx=\sqrt {2} c^2 \sqrt {\frac {c^2 x^2}{1+c^4 x^4}} \left (1+\sqrt {1+c^4 x^4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-c^4 x^4\right )\right ) \]

[In]

Integrate[Sech[2*Log[c*x]]^(3/2)/x^3,x]

[Out]

Sqrt[2]*c^2*Sqrt[(c^2*x^2)/(1 + c^4*x^4)]*(1 + Sqrt[1 + c^4*x^4]*Hypergeometric2F1[1/4, 1/2, 5/4, -(c^4*x^4)])

Maple [F]

\[\int \frac {\operatorname {sech}\left (2 \ln \left (c x \right )\right )^{\frac {3}{2}}}{x^{3}}d x\]

[In]

int(sech(2*ln(c*x))^(3/2)/x^3,x)

[Out]

int(sech(2*ln(c*x))^(3/2)/x^3,x)

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.61 \[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^3} \, dx=\frac {\sqrt {2} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}} c^{3} - \sqrt {2} \left (-c^{4}\right )^{\frac {3}{4}} F(\arcsin \left (\left (-c^{4}\right )^{\frac {1}{4}} x\right )\,|\,-1)}{c} \]

[In]

integrate(sech(2*log(c*x))^(3/2)/x^3,x, algorithm="fricas")

[Out]

(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

Sympy [F]

\[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^3} \, dx=\int \frac {\operatorname {sech}^{\frac {3}{2}}{\left (2 \log {\left (c x \right )} \right )}}{x^{3}}\, dx \]

[In]

integrate(sech(2*ln(c*x))**(3/2)/x**3,x)

[Out]

Integral(sech(2*log(c*x))**(3/2)/x**3, x)

Maxima [F]

\[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^3} \, dx=\int { \frac {\operatorname {sech}\left (2 \, \log \left (c x\right )\right )^{\frac {3}{2}}}{x^{3}} \,d x } \]

[In]

integrate(sech(2*log(c*x))^(3/2)/x^3,x, algorithm="maxima")

[Out]

integrate(sech(2*log(c*x))^(3/2)/x^3, x)

Giac [F(-1)]

Timed out. \[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^3} \, dx=\text {Timed out} \]

[In]

integrate(sech(2*log(c*x))^(3/2)/x^3,x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^3} \, dx=\int \frac {{\left (\frac {1}{\mathrm {cosh}\left (2\,\ln \left (c\,x\right )\right )}\right )}^{3/2}}{x^3} \,d x \]

[In]

int((1/cosh(2*log(c*x)))^(3/2)/x^3,x)

[Out]

int((1/cosh(2*log(c*x)))^(3/2)/x^3, x)