\(\int \text {csch}^3(a+2 \log (c \sqrt {x})) \, dx\) [161]

Optimal result

Integrand size = 15, antiderivative size = 26 \[ \int \text {csch}^3\left (a+2 \log \left (c \sqrt {x}\right )\right ) \, dx=-\frac {2 c^6 e^{-a}}{\left (c^4-\frac {e^{-2 a}}{x^2}\right )^2} \]

[Out]

-2*c^6/exp(a)/(c^4-1/exp(2*a)/x^2)^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 26, 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 = {5665, 5667, 267} \[ \int \text {csch}^3\left (a+2 \log \left (c \sqrt {x}\right )\right ) \, dx=-\frac {2 e^{-a} c^6}{\left (c^4-\frac {e^{-2 a}}{x^2}\right )^2} \]

[In]

Int[Csch[a + 2*Log[c*Sqrt[x]]]^3,x]

[Out]

(-2*c^6)/(E^a*(c^4 - 1/(E^(2*a)*x^2))^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 5665

Int[Csch[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[
x^(1/n - 1)*Csch[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])

Rule 5667

Int[Csch[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[2^p/E^(a*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}, x] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \text {Subst}\left (\int x \text {csch}^3(a+2 \log (x)) \, dx,x,c \sqrt {x}\right )}{c^2} \\ & = \frac {\left (16 e^{-3 a}\right ) \text {Subst}\left (\int \frac {1}{\left (1-\frac {e^{-2 a}}{x^4}\right )^3 x^5} \, dx,x,c \sqrt {x}\right )}{c^2} \\ & = -\frac {2 c^6 e^{-a}}{\left (c^4-\frac {e^{-2 a}}{x^2}\right )^2} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(62\) vs. \(2(26)=52\).

Time = 0.14 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.38 \[ \int \text {csch}^3\left (a+2 \log \left (c \sqrt {x}\right )\right ) \, dx=\frac {2 (\cosh (a)-\sinh (a)) \left (-2 c^4 x^2+\cosh ^2(a)-2 \cosh (a) \sinh (a)+\sinh ^2(a)\right )}{c^2 \left (\left (-1+c^4 x^2\right ) \cosh (a)+\left (1+c^4 x^2\right ) \sinh (a)\right )^2} \]

[In]

Integrate[Csch[a + 2*Log[c*Sqrt[x]]]^3,x]

[Out]

(2*(Cosh[a] - Sinh[a])*(-2*c^4*x^2 + Cosh[a]^2 - 2*Cosh[a]*Sinh[a] + Sinh[a]^2))/(c^2*((-1 + c^4*x^2)*Cosh[a]
+ (1 + c^4*x^2)*Sinh[a])^2)

Maple [F]

\[\int \operatorname {csch}\left (a +2 \ln \left (c \sqrt {x}\right )\right )^{3}d x\]

[In]

int(csch(a+2*ln(c*x^(1/2)))^3,x)

[Out]

int(csch(a+2*ln(c*x^(1/2)))^3,x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.85 \[ \int \text {csch}^3\left (a+2 \log \left (c \sqrt {x}\right )\right ) \, dx=-\frac {2 \, {\left (2 \, c^{4} x^{2} e^{\left (2 \, a\right )} - 1\right )}}{c^{10} x^{4} e^{\left (5 \, a\right )} - 2 \, c^{6} x^{2} e^{\left (3 \, a\right )} + c^{2} e^{a}} \]

[In]

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

[Out]

-2*(2*c^4*x^2*e^(2*a) - 1)/(c^10*x^4*e^(5*a) - 2*c^6*x^2*e^(3*a) + c^2*e^a)

Sympy [F]

\[ \int \text {csch}^3\left (a+2 \log \left (c \sqrt {x}\right )\right ) \, dx=\int \operatorname {csch}^{3}{\left (a + 2 \log {\left (c \sqrt {x} \right )} \right )}\, dx \]

[In]

integrate(csch(a+2*ln(c*x**(1/2)))**3,x)

[Out]

Integral(csch(a + 2*log(c*sqrt(x)))**3, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (24) = 48\).

Time = 0.19 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.92 \[ \int \text {csch}^3\left (a+2 \log \left (c \sqrt {x}\right )\right ) \, dx=-\frac {2 \, {\left (\frac {2 \, c^{4} x^{2} e^{\left (2 \, a\right )}}{c^{8} x^{4} e^{\left (5 \, a\right )} - 2 \, c^{4} x^{2} e^{\left (3 \, a\right )} + e^{a}} - \frac {1}{c^{8} x^{4} e^{\left (5 \, a\right )} - 2 \, c^{4} x^{2} e^{\left (3 \, a\right )} + e^{a}}\right )}}{c^{2}} \]

[In]

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

[Out]

-2*(2*c^4*x^2*e^(2*a)/(c^8*x^4*e^(5*a) - 2*c^4*x^2*e^(3*a) + e^a) - 1/(c^8*x^4*e^(5*a) - 2*c^4*x^2*e^(3*a) + e
^a))/c^2

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \text {csch}^3\left (a+2 \log \left (c \sqrt {x}\right )\right ) \, dx=-\frac {2 \, {\left (2 \, c^{4} x^{2} e^{\left (2 \, a\right )} - 1\right )} e^{\left (-a\right )}}{{\left (c^{4} x^{2} e^{\left (2 \, a\right )} - 1\right )}^{2} c^{2}} \]

[In]

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

[Out]

-2*(2*c^4*x^2*e^(2*a) - 1)*e^(-a)/((c^4*x^2*e^(2*a) - 1)^2*c^2)

Mupad [B] (verification not implemented)

Time = 2.38 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.85 \[ \int \text {csch}^3\left (a+2 \log \left (c \sqrt {x}\right )\right ) \, dx=\frac {\frac {2\,{\mathrm {e}}^{-a}}{c^2}-4\,c^2\,x^2\,{\mathrm {e}}^a}{{\mathrm {e}}^{4\,a}\,c^8\,x^4-2\,{\mathrm {e}}^{2\,a}\,c^4\,x^2+1} \]

[In]

int(1/sinh(a + 2*log(c*x^(1/2)))^3,x)

[Out]

((2*exp(-a))/c^2 - 4*c^2*x^2*exp(a))/(c^8*x^4*exp(4*a) - 2*c^4*x^2*exp(2*a) + 1)