\(\int \frac {1}{\sinh ^{\frac {3}{2}}(a+\frac {2 \log (c x^n)}{n})} \, dx\) [287]

Optimal result

Integrand size = 18, antiderivative size = 43 \[ \int \frac {1}{\sinh ^{\frac {3}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx=-\frac {x \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )}{2 \sinh ^{\frac {3}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \]

[Out]

-1/2*x*(1-1/exp(2*a)/((c*x^n)^(4/n)))/sinh(a+2*ln(c*x^n)/n)^(3/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5636, 5644, 270} \[ \int \frac {1}{\sinh ^{\frac {3}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx=-\frac {x \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )}{2 \sinh ^{\frac {3}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \]

[In]

Int[Sinh[a + (2*Log[c*x^n])/n]^(-3/2),x]

[Out]

-1/2*(x*(1 - 1/(E^(2*a)*(c*x^n)^(4/n))))/Sinh[a + (2*Log[c*x^n])/n]^(3/2)

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 5636

Int[Sinh[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[
x^(1/n - 1)*Sinh[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 5644

Int[((e_.)*(x_))^(m_.)*Sinh[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_), x_Symbol] :> Dist[Sinh[d*(a + b*Log[x])]^p/(x
^(b*d*p)*(1 - 1/(E^(2*a*d)*x^(2*b*d)))^p), Int[(e*x)^m*x^(b*d*p)*(1 - 1/(E^(2*a*d)*x^(2*b*d)))^p, x], x] /; Fr
eeQ[{a, b, d, e, m, p}, x] &&  !IntegerQ[p]

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.42 \[ \int \frac {1}{\sinh ^{\frac {3}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx=\frac {-\cosh \left (a-2 \log (x)+\frac {2 \log \left (c x^n\right )}{n}\right )+\sinh \left (a-2 \log (x)+\frac {2 \log \left (c x^n\right )}{n}\right )}{x \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )}} \]

[In]

Integrate[Sinh[a + (2*Log[c*x^n])/n]^(-3/2),x]

[Out]

(-Cosh[a - 2*Log[x] + (2*Log[c*x^n])/n] + Sinh[a - 2*Log[x] + (2*Log[c*x^n])/n])/(x*Sqrt[Sinh[a + (2*Log[c*x^n
])/n]])

Maple [F]

\[\int \frac {1}{{\sinh \left (a +\frac {2 \ln \left (c \,x^{n}\right )}{n}\right )}^{\frac {3}{2}}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.58 \[ \int \frac {1}{\sinh ^{\frac {3}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx=-\frac {2 \, \sqrt {\frac {1}{2}} x \sqrt {\frac {x^{4} e^{\left (\frac {2 \, {\left (a n + 2 \, \log \left (c\right )\right )}}{n}\right )} - 1}{x^{2}}} e^{\left (-\frac {a n + 2 \, \log \left (c\right )}{2 \, n}\right )}}{x^{4} e^{\left (\frac {2 \, {\left (a n + 2 \, \log \left (c\right )\right )}}{n}\right )} - 1} \]

[In]

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

[Out]

-2*sqrt(1/2)*x*sqrt((x^4*e^(2*(a*n + 2*log(c))/n) - 1)/x^2)*e^(-1/2*(a*n + 2*log(c))/n)/(x^4*e^(2*(a*n + 2*log
(c))/n) - 1)

Sympy [F]

\[ \int \frac {1}{\sinh ^{\frac {3}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx=\int \frac {1}{\sinh ^{\frac {3}{2}}{\left (a + \frac {2 \log {\left (c x^{n} \right )}}{n} \right )}}\, dx \]

[In]

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

[Out]

Integral(sinh(a + 2*log(c*x**n)/n)**(-3/2), x)

Maxima [F]

\[ \int \frac {1}{\sinh ^{\frac {3}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx=\int { \frac {1}{\sinh \left (a + \frac {2 \, \log \left (c x^{n}\right )}{n}\right )^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

integrate(sinh(a + 2*log(c*x^n)/n)^(-3/2), x)

Giac [A] (verification not implemented)

none

Time = 0.54 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95 \[ \int \frac {1}{\sinh ^{\frac {3}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx=-\frac {\sqrt {2}}{\sqrt {c^{\frac {4}{n}} e^{\left (3 \, a\right )} - \frac {e^{a}}{x^{4}}} c^{\left (\frac {1}{n}\right )} x^{2} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

-sqrt(2)/(sqrt(c^(4/n)*e^(3*a) - e^a/x^4)*c^(1/n)*x^2*sgn(x))

Mupad [F(-1)]

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

[In]

int(1/sinh(a + (2*log(c*x^n))/n)^(3/2),x)

[Out]

int(1/sinh(a + (2*log(c*x^n))/n)^(3/2), x)