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

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

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5525, 5533, 345, 242, 277, 216} \[ \frac {1}{2} x \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )}+\frac {e^{-a} x \left (c x^n\right )^{-2/n} \csc ^{-1}\left (e^a \left (c x^n\right )^{2/n}\right ) \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )}}{2 \sqrt {1-e^{-2 a} \left (c x^n\right )^{-4/n}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 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 \sqrt {\sinh \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}} \sqrt {\sinh \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} \sqrt {\sinh \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 )}{n \sqrt {1-e^{-2 a} \left (c x^n\right )^{-4/n}}}\\ &=\frac {\left (x \left (c x^n\right )^{-2/n} \sqrt {\sinh \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 )}{2 \sqrt {1-e^{-2 a} \left (c x^n\right )^{-4/n}}}\\ &=-\frac {\left (x \left (c x^n\right )^{-2/n} \sqrt {\sinh \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 )}{2 \sqrt {1-e^{-2 a} \left (c x^n\right )^{-4/n}}}\\ &=\frac {1}{2} x \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )}+\frac {\left (e^{-2 a} x \left (c x^n\right )^{-2/n} \sqrt {\sinh \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 )}{2 \sqrt {1-e^{-2 a} \left (c x^n\right )^{-4/n}}}\\ &=\frac {1}{2} x \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )}+\frac {e^{-a} x \left (c x^n\right )^{-2/n} \sin ^{-1}\left (e^{-a} \left (c x^n\right )^{-2/n}\right ) \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )}}{2 \sqrt {1-e^{-2 a} \left (c x^n\right )^{-4/n}}}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 74, normalized size = 0.72 \[ \frac {1}{2} x \left (1-\frac {\tan ^{-1}\left (\sqrt {e^{2 a} \left (c x^n\right )^{4/n}-1}\right )}{\sqrt {e^{2 a} \left (c x^n\right )^{4/n}-1}}\right ) \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 1.40, size = 117, normalized size = 1.14 \[ \frac {1}{4} \, {\left (2 \, \sqrt {\frac {1}{2}} x \sqrt {\frac {x^{4} e^{\left (\frac {2 \, {\left (a n + 2 \, \log \relax (c)\right )}}{n}\right )} - 1}{x^{2}}} e^{\left (\frac {a n + 2 \, \log \relax (c)}{2 \, n}\right )} - \sqrt {2} \arctan \left (\sqrt {2} \sqrt {\frac {1}{2}} x \sqrt {\frac {x^{4} e^{\left (\frac {2 \, {\left (a n + 2 \, \log \relax (c)\right )}}{n}\right )} - 1}{x^{2}}}\right ) e^{\left (\frac {a n + 2 \, \log \relax (c)}{2 \, n}\right )}\right )} e^{\left (-\frac {a n + 2 \, \log \relax (c)}{n}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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maple [F]  time = 0.27, size = 0, normalized size = 0.00 \[ \int \sqrt {\sinh }\left (a +\frac {2 \ln \left (c \,x^{n}\right )}{n}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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