3.285 \(\int \sinh ^{\frac{5}{2}}(a+\frac{2 \log (c x^n)}{n}) \, dx\)

Optimal. Leaf size=209 \[ -\frac{5 e^{-2 a} x \left (c x^n\right )^{-4/n} \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )}{4 \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )^2}-\frac{1}{4} x \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )+\frac{5 x \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )}{12 \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )}-\frac{5 e^{-3 a} x \left (c x^n\right )^{-6/n} \csc ^{-1}\left (e^a \left (c x^n\right )^{2/n}\right ) \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )}{4 \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )^{5/2}} \]

[Out]

-(x*Sinh[a + (2*Log[c*x^n])/n]^(5/2))/4 - (5*x*Sinh[a + (2*Log[c*x^n])/n]^(5/2))/(4*E^(2*a)*(c*x^n)^(4/n)*(1 -
1/(E^(2*a)*(c*x^n)^(4/n)))^2) + (5*x*Sinh[a + (2*Log[c*x^n])/n]^(5/2))/(12*(1 - 1/(E^(2*a)*(c*x^n)^(4/n)))) -
(5*x*ArcCsc[E^a*(c*x^n)^(2/n)]*Sinh[a + (2*Log[c*x^n])/n]^(5/2))/(4*E^(3*a)*(c*x^n)^(6/n)*(1 - 1/(E^(2*a)*(c*
x^n)^(4/n)))^(5/2))

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Rubi [A]  time = 0.160211, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {5525, 5533, 353, 349, 345, 242, 277, 216} \[ -\frac{5 e^{-2 a} x \left (c x^n\right )^{-4/n} \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )}{4 \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )^2}-\frac{1}{4} x \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )+\frac{5 x \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )}{12 \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )}-\frac{5 e^{-3 a} x \left (c x^n\right )^{-6/n} \csc ^{-1}\left (e^a \left (c x^n\right )^{2/n}\right ) \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )}{4 \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + (2*Log[c*x^n])/n]^(5/2),x]

[Out]

-(x*Sinh[a + (2*Log[c*x^n])/n]^(5/2))/4 - (5*x*Sinh[a + (2*Log[c*x^n])/n]^(5/2))/(4*E^(2*a)*(c*x^n)^(4/n)*(1 -
1/(E^(2*a)*(c*x^n)^(4/n)))^2) + (5*x*Sinh[a + (2*Log[c*x^n])/n]^(5/2))/(12*(1 - 1/(E^(2*a)*(c*x^n)^(4/n)))) -
(5*x*ArcCsc[E^a*(c*x^n)^(2/n)]*Sinh[a + (2*Log[c*x^n])/n]^(5/2))/(4*E^(3*a)*(c*x^n)^(6/n)*(1 - 1/(E^(2*a)*(c*
x^n)^(4/n)))^(5/2))

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]

Rule 353

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

Rule 349

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

Rule 345

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

Rule 242

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^2, x], x, 1/x] /; FreeQ[{a, b, p},
x] && ILtQ[n, 0]

Rule 277

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

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int \sinh ^{\frac{5}{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 x^{-1+\frac{1}{n}} \sinh ^{\frac{5}{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} \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )\right ) \operatorname{Subst}\left (\int x^{-1+\frac{6}{n}} \left (1-e^{-2 a} x^{-4/n}\right )^{5/2} \, dx,x,c x^n\right )}{n \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )^{5/2}}\\ &=-\frac{1}{4} x \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )+\frac{\left (5 x \left (c x^n\right )^{-6/n} \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )\right ) \operatorname{Subst}\left (\int x^{-1+\frac{6}{n}} \left (1-e^{-2 a} x^{-4/n}\right )^{3/2} \, dx,x,c x^n\right )}{2 n \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )^{5/2}}\\ &=-\frac{1}{4} x \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )+\frac{5 x \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )}{12 \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )}-\frac{\left (5 e^{-2 a} x \left (c x^n\right )^{-6/n} \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )\right ) \operatorname{Subst}\left (\int x^{-1+\frac{2}{n}} \sqrt{1-e^{-2 a} x^{-4/n}} \, dx,x,c x^n\right )}{2 n \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )^{5/2}}\\ &=-\frac{1}{4} x \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )+\frac{5 x \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )}{12 \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )}-\frac{\left (5 e^{-2 a} x \left (c x^n\right )^{-6/n} \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )\right ) \operatorname{Subst}\left (\int \sqrt{1-\frac{e^{-2 a}}{x^2}} \, dx,x,\left (c x^n\right )^{2/n}\right )}{4 \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )^{5/2}}\\ &=-\frac{1}{4} x \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )+\frac{5 x \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )}{12 \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )}+\frac{\left (5 e^{-2 a} x \left (c x^n\right )^{-6/n} \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1-e^{-2 a} x^2}}{x^2} \, dx,x,\left (c x^n\right )^{-2/n}\right )}{4 \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )^{5/2}}\\ &=-\frac{1}{4} x \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )-\frac{5 e^{-2 a} x \left (c x^n\right )^{-4/n} \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )}{4 \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )^2}+\frac{5 x \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )}{12 \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )}-\frac{\left (5 e^{-4 a} x \left (c x^n\right )^{-6/n} \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-e^{-2 a} x^2}} \, dx,x,\left (c x^n\right )^{-2/n}\right )}{4 \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )^{5/2}}\\ &=-\frac{1}{4} x \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )-\frac{5 e^{-2 a} x \left (c x^n\right )^{-4/n} \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )}{4 \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )^2}+\frac{5 x \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )}{12 \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )}-\frac{5 e^{-3 a} x \left (c x^n\right )^{-6/n} \sin ^{-1}\left (e^{-a} \left (c x^n\right )^{-2/n}\right ) \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right )}{4 \left (1-e^{-2 a} \left (c x^n\right )^{-4/n}\right )^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.4182, size = 86, normalized size = 0.41 \[ \frac{1}{14} e^{2 a} x \left (c x^n\right )^{4/n} \left (e^{2 a} \left (c x^n\right )^{4/n}-1\right ) \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};1-e^{2 a} \left (c x^n\right )^{4/n}\right ) \sinh ^{\frac{5}{2}}\left (a+\frac{2 \log \left (c x^n\right )}{n}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sinh[a + (2*Log[c*x^n])/n]^(5/2),x]

[Out]

(E^(2*a)*x*(c*x^n)^(4/n)*(-1 + E^(2*a)*(c*x^n)^(4/n))*Hypergeometric2F1[2, 7/2, 9/2, 1 - E^(2*a)*(c*x^n)^(4/n)
]*Sinh[a + (2*Log[c*x^n])/n]^(5/2))/14

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Maple [F]  time = 0.362, size = 0, normalized size = 0. \begin{align*} \int \left ( \sinh \left ( a+2\,{\frac{\ln \left ( c{x}^{n} \right ) }{n}} \right ) \right ) ^{{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a+2*ln(c*x^n)/n)^(5/2),x)

[Out]

int(sinh(a+2*ln(c*x^n)/n)^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sinh \left (a + \frac{2 \, \log \left (c x^{n}\right )}{n}\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+2*log(c*x^n)/n)^(5/2),x, algorithm="maxima")

[Out]

integrate(sinh(a + 2*log(c*x^n)/n)^(5/2), x)

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Fricas [A]  time = 2.18024, size = 420, normalized size = 2.01 \begin{align*} \frac{{\left (15 \, \sqrt{2} x^{3} \arctan \left (\sqrt{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}}}\right ) e^{\left (\frac{3 \,{\left (a n + 2 \, \log \left (c\right )\right )}}{2 \, n}\right )} + 2 \, \sqrt{\frac{1}{2}}{\left (2 \, x^{8} e^{\left (\frac{4 \,{\left (a n + 2 \, \log \left (c\right )\right )}}{n}\right )} - 14 \, x^{4} e^{\left (\frac{2 \,{\left (a n + 2 \, \log \left (c\right )\right )}}{n}\right )} - 3\right )} \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 )}\right )} e^{\left (-\frac{2 \,{\left (a n + 2 \, \log \left (c\right )\right )}}{n}\right )}}{96 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+2*log(c*x^n)/n)^(5/2),x, algorithm="fricas")

[Out]

1/96*(15*sqrt(2)*x^3*arctan(sqrt(2)*sqrt(1/2)*x*sqrt((x^4*e^(2*(a*n + 2*log(c))/n) - 1)/x^2))*e^(3/2*(a*n + 2*
log(c))/n) + 2*sqrt(1/2)*(2*x^8*e^(4*(a*n + 2*log(c))/n) - 14*x^4*e^(2*(a*n + 2*log(c))/n) - 3)*sqrt((x^4*e^(2
*(a*n + 2*log(c))/n) - 1)/x^2)*e^(-1/2*(a*n + 2*log(c))/n))*e^(-2*(a*n + 2*log(c))/n)/x^3

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+2*ln(c*x**n)/n)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sinh \left (a + \frac{2 \, \log \left (c x^{n}\right )}{n}\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+2*log(c*x^n)/n)^(5/2),x, algorithm="giac")

[Out]

integrate(sinh(a + 2*log(c*x^n)/n)^(5/2), x)