\(\int \frac {\sqrt {\text {csch}(2 \log (c x))}}{x^4} \, dx\) [141]

Optimal result

Integrand size = 15, antiderivative size = 25 \[ \int \frac {\sqrt {\text {csch}(2 \log (c x))}}{x^4} \, dx=\frac {1}{2} \left (c^4-\frac {1}{x^4}\right ) x \sqrt {\text {csch}(2 \log (c x))} \]

[Out]

1/2*(c^4-1/x^4)*x*csch(2*ln(c*x))^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5671, 5669, 267} \[ \int \frac {\sqrt {\text {csch}(2 \log (c x))}}{x^4} \, dx=\frac {1}{2} x \left (c^4-\frac {1}{x^4}\right ) \sqrt {\text {csch}(2 \log (c x))} \]

[In]

Int[Sqrt[Csch[2*Log[c*x]]]/x^4,x]

[Out]

((c^4 - x^(-4))*x*Sqrt[Csch[2*Log[c*x]]])/2

Rule 267

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

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^3 \text {Subst}\left (\int \frac {\sqrt {\text {csch}(2 \log (x))}}{x^4} \, dx,x,c x\right ) \\ & = \left (c^4 \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^5} \, dx,x,c x\right ) \\ & = \frac {1}{2} \left (c^4-\frac {1}{x^4}\right ) x \sqrt {\text {csch}(2 \log (c x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {\sqrt {\text {csch}(2 \log (c x))}}{x^4} \, dx=\frac {c^2}{2 x \sqrt {\frac {c^2 x^2}{-2+2 c^4 x^4}}} \]

[In]

Integrate[Sqrt[Csch[2*Log[c*x]]]/x^4,x]

[Out]

c^2/(2*x*Sqrt[(c^2*x^2)/(-2 + 2*c^4*x^4)])

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52

method	result	size
risch	\(\frac {\left (c^{4} x^{4}-1\right ) \sqrt {2}\, \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}-1}}}{2 x^{3}}\)	\(38\)

[In]

int(csch(2*ln(c*x))^(1/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

1/2*(c^4*x^4-1)/x^3*2^(1/2)*(c^2*x^2/(c^4*x^4-1))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {\sqrt {\text {csch}(2 \log (c x))}}{x^4} \, dx=\frac {\sqrt {2} {\left (c^{4} x^{4} - 1\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} - 1}}}{2 \, x^{3}} \]

[In]

integrate(csch(2*log(c*x))^(1/2)/x^4,x, algorithm="fricas")

[Out]

1/2*sqrt(2)*(c^4*x^4 - 1)*sqrt(c^2*x^2/(c^4*x^4 - 1))/x^3

Sympy [F]

\[ \int \frac {\sqrt {\text {csch}(2 \log (c x))}}{x^4} \, dx=\int \frac {\sqrt {\operatorname {csch}{\left (2 \log {\left (c x \right )} \right )}}}{x^{4}}\, dx \]

[In]

integrate(csch(2*ln(c*x))**(1/2)/x**4,x)

[Out]

Integral(sqrt(csch(2*log(c*x)))/x**4, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 89 vs. \(2 (21) = 42\).

Time = 0.29 (sec) , antiderivative size = 89, normalized size of antiderivative = 3.56 \[ \int \frac {\sqrt {\text {csch}(2 \log (c x))}}{x^4} \, dx=\frac {1}{2} \, c^{3} {\left (\frac {\sqrt {2}}{\sqrt {\frac {1}{c x} + 1} \sqrt {-\frac {1}{c x} + 1} \sqrt {\frac {1}{c^{2} x^{2}} + 1}} - \frac {\sqrt {2}}{c^{4} x^{4} \sqrt {\frac {1}{c x} + 1} \sqrt {-\frac {1}{c x} + 1} \sqrt {\frac {1}{c^{2} x^{2}} + 1}}\right )} \]

[In]

integrate(csch(2*log(c*x))^(1/2)/x^4,x, algorithm="maxima")

[Out]

1/2*c^3*(sqrt(2)/(sqrt(1/(c*x) + 1)*sqrt(-1/(c*x) + 1)*sqrt(1/(c^2*x^2) + 1)) - sqrt(2)/(c^4*x^4*sqrt(1/(c*x)
+ 1)*sqrt(-1/(c*x) + 1)*sqrt(1/(c^2*x^2) + 1)))

Giac [F(-1)]

Timed out. \[ \int \frac {\sqrt {\text {csch}(2 \log (c x))}}{x^4} \, dx=\text {Timed out} \]

[In]

integrate(csch(2*log(c*x))^(1/2)/x^4,x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 2.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.32 \[ \int \frac {\sqrt {\text {csch}(2 \log (c x))}}{x^4} \, dx=\frac {c^4\,x\,\sqrt {\frac {2\,c^2\,x^2}{c^4\,x^4-1}}}{2}-\frac {\sqrt {\frac {2\,c^2\,x^2}{c^4\,x^4-1}}}{2\,x^3} \]

[In]

int((1/sinh(2*log(c*x)))^(1/2)/x^4,x)

[Out]

(c^4*x*((2*c^2*x^2)/(c^4*x^4 - 1))^(1/2))/2 - ((2*c^2*x^2)/(c^4*x^4 - 1))^(1/2)/(2*x^3)