\(\int \frac {\sqrt {\text {csch}(2 \log (c x))}}{x^5} \, dx\) [142]

Optimal result

Integrand size = 15, antiderivative size = 64 \[ \int \frac {\sqrt {\text {csch}(2 \log (c x))}}{x^5} \, dx=\frac {1}{3} \left (c^4-\frac {1}{x^4}\right ) \sqrt {\text {csch}(2 \log (c x))}-\frac {1}{3} c^5 \sqrt {1-\frac {1}{c^4 x^4}} x \sqrt {\text {csch}(2 \log (c x))} \operatorname {EllipticF}\left (\csc ^{-1}(c x),-1\right ) \]

[Out]

1/3*(c^4-1/x^4)*csch(2*ln(c*x))^(1/2)-1/3*c^5*x*EllipticF(1/c/x,I)*(1-1/c^4/x^4)^(1/2)*csch(2*ln(c*x))^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 64, 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 = {5671, 5669, 342, 327, 227} \[ \int \frac {\sqrt {\text {csch}(2 \log (c x))}}{x^5} \, dx=\frac {1}{3} \left (c^4-\frac {1}{x^4}\right ) \sqrt {\text {csch}(2 \log (c x))}-\frac {1}{3} c^5 x \sqrt {1-\frac {1}{c^4 x^4}} \operatorname {EllipticF}\left (\csc ^{-1}(c x),-1\right ) \sqrt {\text {csch}(2 \log (c x))} \]

[In]

Int[Sqrt[Csch[2*Log[c*x]]]/x^5,x]

[Out]

((c^4 - x^(-4))*Sqrt[Csch[2*Log[c*x]]])/3 - (c^5*Sqrt[1 - 1/(c^4*x^4)]*x*Sqrt[Csch[2*Log[c*x]]]*EllipticF[ArcC
sc[c*x], -1])/3

Rule 227

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[Rt[-b, 4]*(x/Rt[a, 4])], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 327

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*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 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 5669

Int[Csch[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[Csch[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 5671

Int[Csch[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[(e*x)^(m + 1
)/(e*n*(c*x^n)^((m + 1)/n)), Subst[Int[x^((m + 1)/n - 1)*Csch[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^4 \text {Subst}\left (\int \frac {\sqrt {\text {csch}(2 \log (x))}}{x^5} \, dx,x,c x\right ) \\ & = \left (c^5 \sqrt {1-\frac {1}{c^4 x^4}} x \sqrt {\text {csch}(2 \log (c x))}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {1}{x^4}} x^6} \, dx,x,c x\right ) \\ & = -\left (\left (c^5 \sqrt {1-\frac {1}{c^4 x^4}} x \sqrt {\text {csch}(2 \log (c x))}\right ) \text {Subst}\left (\int \frac {x^4}{\sqrt {1-x^4}} \, dx,x,\frac {1}{c x}\right )\right ) \\ & = \frac {1}{3} \left (c^4-\frac {1}{x^4}\right ) \sqrt {\text {csch}(2 \log (c x))}-\frac {1}{3} \left (c^5 \sqrt {1-\frac {1}{c^4 x^4}} x \sqrt {\text {csch}(2 \log (c x))}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^4}} \, dx,x,\frac {1}{c x}\right ) \\ & = \frac {1}{3} \left (c^4-\frac {1}{x^4}\right ) \sqrt {\text {csch}(2 \log (c x))}-\frac {1}{3} c^5 \sqrt {1-\frac {1}{c^4 x^4}} x \sqrt {\text {csch}(2 \log (c x))} \operatorname {EllipticF}\left (\csc ^{-1}(c x),-1\right ) \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {\text {csch}(2 \log (c x))}}{x^5} \, dx=-\frac {\sqrt {2-2 c^4 x^4} \sqrt {\frac {c^2 x^2}{-1+c^4 x^4}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {1}{4},c^4 x^4\right )}{3 x^4} \]

[In]

Integrate[Sqrt[Csch[2*Log[c*x]]]/x^5,x]

[Out]

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

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.75

method	result	size
risch	\(\frac {\left (c^{4} x^{4}-1\right ) \sqrt {2}\, \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}-1}}}{3 x^{4}}+\frac {c^{4} \sqrt {c^{2} x^{2}+1}\, \sqrt {-c^{2} x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-c^{2}}, i\right ) \sqrt {2}\, \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}-1}}}{3 \sqrt {-c^{2}}\, x}\)	\(112\)

[In]

int(csch(2*ln(c*x))^(1/2)/x^5,x,method=_RETURNVERBOSE)

[Out]

1/3*(c^4*x^4-1)/x^4*2^(1/2)*(c^2*x^2/(c^4*x^4-1))^(1/2)+1/3*c^4/(-c^2)^(1/2)*(c^2*x^2+1)^(1/2)*(-c^2*x^2+1)^(1
/2)*EllipticF(x*(-c^2)^(1/2),I)*2^(1/2)*(c^2*x^2/(c^4*x^4-1))^(1/2)/x

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt {\text {csch}(2 \log (c x))}}{x^5} \, dx=\frac {-i \, \sqrt {2} c^{4} x^{4} F(\arcsin \left (c x\right )\,|\,-1) + \sqrt {2} {\left (c^{4} x^{4} - 1\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} - 1}}}{3 \, x^{4}} \]

[In]

integrate(csch(2*log(c*x))^(1/2)/x^5,x, algorithm="fricas")

[Out]

1/3*(-I*sqrt(2)*c^4*x^4*elliptic_f(arcsin(c*x), -1) + sqrt(2)*(c^4*x^4 - 1)*sqrt(c^2*x^2/(c^4*x^4 - 1)))/x^4

Sympy [F]

\[ \int \frac {\sqrt {\text {csch}(2 \log (c x))}}{x^5} \, dx=\int \frac {\sqrt {\operatorname {csch}{\left (2 \log {\left (c x \right )} \right )}}}{x^{5}}\, dx \]

[In]

integrate(csch(2*ln(c*x))**(1/2)/x**5,x)

[Out]

Integral(sqrt(csch(2*log(c*x)))/x**5, x)

Maxima [F]

\[ \int \frac {\sqrt {\text {csch}(2 \log (c x))}}{x^5} \, dx=\int { \frac {\sqrt {\operatorname {csch}\left (2 \, \log \left (c x\right )\right )}}{x^{5}} \,d x } \]

[In]

integrate(csch(2*log(c*x))^(1/2)/x^5,x, algorithm="maxima")

[Out]

integrate(sqrt(csch(2*log(c*x)))/x^5, x)

Giac [F(-1)]

Timed out. \[ \int \frac {\sqrt {\text {csch}(2 \log (c x))}}{x^5} \, dx=\text {Timed out} \]

[In]

integrate(csch(2*log(c*x))^(1/2)/x^5,x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {\text {csch}(2 \log (c x))}}{x^5} \, dx=\int \frac {\sqrt {\frac {1}{\mathrm {sinh}\left (2\,\ln \left (c\,x\right )\right )}}}{x^5} \,d x \]

[In]

int((1/sinh(2*log(c*x)))^(1/2)/x^5,x)

[Out]

int((1/sinh(2*log(c*x)))^(1/2)/x^5, x)