3.288 \(\int \frac {1}{\sinh ^{\frac {7}{2}}(a+\frac {2 \log (c x^n)}{n})} \, dx\)

Optimal. Leaf size=103 \[ \frac {e^{-2 a} x \left (c x^n\right )^{-4/n} \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )}{15 \sinh ^{\frac {7}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )}-\frac {x \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )}{6 \sinh ^{\frac {7}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \]

[Out]

-1/6*x*(1-1/exp(2*a)/((c*x^n)^(4/n)))/sinh(a+2*ln(c*x^n)/n)^(7/2)+1/15*x*(1-1/exp(2*a)/((c*x^n)^(4/n)))/exp(2*
a)/((c*x^n)^(4/n))/sinh(a+2*ln(c*x^n)/n)^(7/2)

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Rubi [A]  time = 0.08, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5525, 5533, 271, 264} \[ \frac {e^{-2 a} x \left (c x^n\right )^{-4/n} \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )}{15 \sinh ^{\frac {7}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )}-\frac {x \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )}{6 \sinh ^{\frac {7}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + (2*Log[c*x^n])/n]^(-7/2),x]

[Out]

-(x*(1 - 1/(E^(2*a)*(c*x^n)^(4/n))))/(6*Sinh[a + (2*Log[c*x^n])/n]^(7/2)) + (x*(1 - 1/(E^(2*a)*(c*x^n)^(4/n)))
)/(15*E^(2*a)*(c*x^n)^(4/n)*Sinh[a + (2*Log[c*x^n])/n]^(7/2))

Rule 264

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 271

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

Rule 5525

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 5533

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*} \int \frac {1}{\sinh ^{\frac {7}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx &=\frac {\left (x \left (c x^n\right )^{-1/n}\right ) \operatorname {Subst}\left (\int \frac {x^{-1+\frac {1}{n}}}{\sinh ^{\frac {7}{2}}\left (a+\frac {2 \log (x)}{n}\right )} \, dx,x,c x^n\right )}{n}\\ &=\frac {\left (x \left (c x^n\right )^{6/n} \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )^{7/2}\right ) \operatorname {Subst}\left (\int \frac {x^{-1-\frac {6}{n}}}{\left (1-e^{-2 a} x^{-4/n}\right )^{7/2}} \, dx,x,c x^n\right )}{n \sinh ^{\frac {7}{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 )}{6 \sinh ^{\frac {7}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )}-\frac {\left (2 e^{-2 a} x \left (c x^n\right )^{6/n} \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )^{7/2}\right ) \operatorname {Subst}\left (\int \frac {x^{-1-\frac {10}{n}}}{\left (1-e^{-2 a} x^{-4/n}\right )^{7/2}} \, dx,x,c x^n\right )}{3 n \sinh ^{\frac {7}{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 )}{6 \sinh ^{\frac {7}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )}+\frac {e^{-2 a} x \left (c x^n\right )^{-4/n} \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )}{15 \sinh ^{\frac {7}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 121, normalized size = 1.17 \[ \frac {\left (\left (5 x^4+2\right ) \sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}-2 \log (x)\right )+\left (5 x^4-2\right ) \cosh \left (a+\frac {2 \log \left (c x^n\right )}{n}-2 \log (x)\right )\right ) \left (\sinh \left (2 a+\frac {4 \log \left (c x^n\right )}{n}-4 \log (x)\right )-\cosh \left (2 a+\frac {4 \log \left (c x^n\right )}{n}-4 \log (x)\right )\right )}{15 x^5 \sinh ^{\frac {5}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + (2*Log[c*x^n])/n]^(-7/2),x]

[Out]

(((-2 + 5*x^4)*Cosh[a - 2*Log[x] + (2*Log[c*x^n])/n] + (2 + 5*x^4)*Sinh[a - 2*Log[x] + (2*Log[c*x^n])/n])*(-Co
sh[2*a - 4*Log[x] + (4*Log[c*x^n])/n] + Sinh[2*a - 4*Log[x] + (4*Log[c*x^n])/n]))/(15*x^5*Sinh[a + (2*Log[c*x^
n])/n]^(5/2))

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fricas [A]  time = 0.55, size = 128, normalized size = 1.24 \[ -\frac {8 \, \sqrt {\frac {1}{2}} {\left (5 \, x^{5} e^{\left (\frac {2 \, {\left (a n + 2 \, \log \relax (c)\right )}}{n}\right )} - 2 \, x\right )} \sqrt {\frac {x^{4} e^{\left (\frac {2 \, {\left (a n + 2 \, \log \relax (c)\right )}}{n}\right )} - 1}{x^{2}}} e^{\left (-\frac {a n + 2 \, \log \relax (c)}{2 \, n}\right )}}{15 \, {\left (x^{12} e^{\left (\frac {6 \, {\left (a n + 2 \, \log \relax (c)\right )}}{n}\right )} - 3 \, x^{8} e^{\left (\frac {4 \, {\left (a n + 2 \, \log \relax (c)\right )}}{n}\right )} + 3 \, x^{4} e^{\left (\frac {2 \, {\left (a n + 2 \, \log \relax (c)\right )}}{n}\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sinh(a+2*log(c*x^n)/n)^(7/2),x, algorithm="fricas")

[Out]

-8/15*sqrt(1/2)*(5*x^5*e^(2*(a*n + 2*log(c))/n) - 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^12*e^(6*(a*n + 2*log(c))/n) - 3*x^8*e^(4*(a*n + 2*log(c))/n) + 3*x^4*e^(2*(a*n + 2*log(c)
)/n) - 1)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sinh(a+2*log(c*x^n)/n)^(7/2),x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 0.23, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sinh \left (a +\frac {2 \ln \left (c \,x^{n}\right )}{n}\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sinh(a+2*ln(c*x^n)/n)^(7/2),x)

[Out]

int(1/sinh(a+2*ln(c*x^n)/n)^(7/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sinh \left (a + \frac {2 \, \log \left (c x^{n}\right )}{n}\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sinh(a+2*log(c*x^n)/n)^(7/2),x, algorithm="maxima")

[Out]

integrate(sinh(a + 2*log(c*x^n)/n)^(-7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\mathrm {sinh}\left (a+\frac {2\,\ln \left (c\,x^n\right )}{n}\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sinh(a + (2*log(c*x^n))/n)^(7/2),x)

[Out]

int(1/sinh(a + (2*log(c*x^n))/n)^(7/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sinh(a+2*ln(c*x**n)/n)**(7/2),x)

[Out]

Timed out

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