\(\int \frac {1}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx\) [177]

Optimal result

Integrand size = 11, antiderivative size = 92 \[ \int \frac {1}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=-\frac {3}{4 \left (c^4+\frac {1}{x^4}\right ) x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {x}{4 \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {3 \text {arctanh}\left (\sqrt {1+\frac {1}{c^4 x^4}}\right )}{4 c^4 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))} \]

[Out]

-3/4/(c^4+1/x^4)/x^3/sech(2*ln(c*x))^(3/2)+1/4*x/sech(2*ln(c*x))^(3/2)+3/4*arctanh((1+1/c^4/x^4)^(1/2))/c^4/(1
+1/c^4/x^4)^(3/2)/x^3/sech(2*ln(c*x))^(3/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {5664, 5662, 272, 43, 52, 65, 213} \[ \int \frac {1}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {3 \text {arctanh}\left (\sqrt {\frac {1}{c^4 x^4}+1}\right )}{4 c^4 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}-\frac {3}{4 x^3 \left (c^4+\frac {1}{x^4}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {x}{4 \text {sech}^{\frac {3}{2}}(2 \log (c x))} \]

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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}{\text {sech}^{\frac {3}{2}}(2 \log (x))} \, dx,x,c x\right )}{c} \\ & = \frac {\text {Subst}\left (\int \left (1+\frac {1}{x^4}\right )^{3/2} x^3 \, dx,x,c x\right )}{c^4 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))} \\ & = -\frac {\text {Subst}\left (\int \frac {(1+x)^{3/2}}{x^2} \, dx,x,\frac {1}{c^4 x^4}\right )}{4 c^4 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))} \\ & = \frac {x}{4 \text {sech}^{\frac {3}{2}}(2 \log (c x))}-\frac {3 \text {Subst}\left (\int \frac {\sqrt {1+x}}{x} \, dx,x,\frac {1}{c^4 x^4}\right )}{8 c^4 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))} \\ & = -\frac {3}{4 \left (c^4+\frac {1}{x^4}\right ) x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {x}{4 \text {sech}^{\frac {3}{2}}(2 \log (c x))}-\frac {3 \text {Subst}\left (\int \frac {1}{x \sqrt {1+x}} \, dx,x,\frac {1}{c^4 x^4}\right )}{8 c^4 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))} \\ & = -\frac {3}{4 \left (c^4+\frac {1}{x^4}\right ) x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {x}{4 \text {sech}^{\frac {3}{2}}(2 \log (c x))}-\frac {3 \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {1+\frac {1}{c^4 x^4}}\right )}{4 c^4 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))} \\ & = -\frac {3}{4 \left (c^4+\frac {1}{x^4}\right ) x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {x}{4 \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {3 \text {arctanh}\left (\sqrt {1+\frac {1}{c^4 x^4}}\right )}{4 c^4 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.13 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=-\frac {\sqrt {1+c^4 x^4} \sqrt {\frac {c^2 x^2}{2+2 c^4 x^4}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{2},\frac {1}{2},-c^4 x^4\right )}{4 c^4 x^3} \]

[In]

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

[Out]

-1/4*(Sqrt[1 + c^4*x^4]*Sqrt[(c^2*x^2)/(2 + 2*c^4*x^4)]*Hypergeometric2F1[-3/2, -1/2, 1/2, -(c^4*x^4)])/(c^4*x
^3)

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.42

method	result	size
risch	\(\frac {\left (c^{8} x^{8}-c^{4} x^{4}-2\right ) \sqrt {2}}{16 x \left (c^{4} x^{4}+1\right ) c^{2} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}+\frac {3 c^{2} \ln \left (\frac {c^{4} x^{2}}{\sqrt {c^{4}}}+\sqrt {c^{4} x^{4}+1}\right ) \sqrt {2}\, x}{16 \sqrt {c^{4}}\, \sqrt {c^{4} x^{4}+1}\, \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}\)	\(131\)

[In]

int(1/sech(2*ln(c*x))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/16*(c^8*x^8-c^4*x^4-2)/x/(c^4*x^4+1)*2^(1/2)/c^2/(c^2*x^2/(c^4*x^4+1))^(1/2)+3/16*c^2*ln(c^4*x^2/(c^4)^(1/2)
+(c^4*x^4+1)^(1/2))/(c^4)^(1/2)*2^(1/2)*x/(c^4*x^4+1)^(1/2)/(c^2*x^2/(c^4*x^4+1))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.15 \[ \int \frac {1}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {3 \, \sqrt {2} c^{3} x^{3} \log \left (-2 \, c^{4} x^{4} - 2 \, {\left (c^{5} x^{5} + c x\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}} - 1\right ) + 2 \, \sqrt {2} {\left (c^{8} x^{8} - c^{4} x^{4} - 2\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}}}{32 \, c^{4} x^{3}} \]

[In]

integrate(1/sech(2*log(c*x))^(3/2),x, algorithm="fricas")

[Out]

1/32*(3*sqrt(2)*c^3*x^3*log(-2*c^4*x^4 - 2*(c^5*x^5 + c*x)*sqrt(c^2*x^2/(c^4*x^4 + 1)) - 1) + 2*sqrt(2)*(c^8*x
^8 - c^4*x^4 - 2)*sqrt(c^2*x^2/(c^4*x^4 + 1)))/(c^4*x^3)

Sympy [F]

\[ \int \frac {1}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {1}{\operatorname {sech}^{\frac {3}{2}}{\left (2 \log {\left (c x \right )} \right )}}\, dx \]

[In]

integrate(1/sech(2*ln(c*x))**(3/2),x)

[Out]

Integral(sech(2*log(c*x))**(-3/2), x)

Maxima [F]

\[ \int \frac {1}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int { \frac {1}{\operatorname {sech}\left (2 \, \log \left (c x\right )\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/sech(2*log(c*x))^(3/2),x, algorithm="maxima")

[Out]

integrate(sech(2*log(c*x))^(-3/2), x)

Giac [F(-1)]

Timed out. \[ \int \frac {1}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\text {Timed out} \]

[In]

integrate(1/sech(2*log(c*x))^(3/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

int(1/(1/cosh(2*log(c*x)))^(3/2), x)