Integrand size = 18, antiderivative size = 103 \[ \int \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx=\frac {1}{2} x \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )}+\frac {e^{-a} x \left (c x^n\right )^{-2/n} \csc ^{-1}\left (e^a \left (c x^n\right )^{2/n}\right ) \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )}}{2 \sqrt {1-e^{-2 a} \left (c x^n\right )^{-4/n}}} \]
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)
Time = 0.19 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.72 \[ \int \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx=\frac {1}{2} x \left (1-\frac {\arctan \left (\sqrt {-1+e^{2 a} \left (c x^n\right )^{4/n}}\right )}{\sqrt {-1+e^{2 a} \left (c x^n\right )^{4/n}}}\right ) \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \]
(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
Time = 0.37 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6051, 6059, 868, 773, 247, 223}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx\) |
\(\Big \downarrow \) 6051 |
\(\displaystyle \frac {x \left (c x^n\right )^{-1/n} \int \left (c x^n\right )^{\frac {1}{n}-1} \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )}d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 6059 |
\(\displaystyle \frac {x \left (c x^n\right )^{-2/n} \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \int \left (c x^n\right )^{\frac {2}{n}-1} \sqrt {1-e^{-2 a} \left (c x^n\right )^{-4/n}}d\left (c x^n\right )}{n \sqrt {1-e^{-2 a} \left (c x^n\right )^{-4/n}}}\) |
\(\Big \downarrow \) 868 |
\(\displaystyle \frac {x \left (c x^n\right )^{-2/n} \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \int \sqrt {1-\frac {e^{-2 a} x^{-2 n}}{c^2}}d\left (c x^n\right )^{2/n}}{2 \sqrt {1-e^{-2 a} \left (c x^n\right )^{-4/n}}}\) |
\(\Big \downarrow \) 773 |
\(\displaystyle -\frac {x \left (c x^n\right )^{-2/n} \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \int \frac {x^{-2 n} \sqrt {1-c^2 e^{-2 a} x^{2 n}}}{c^2}d\frac {x^{-n}}{c}}{2 \sqrt {1-e^{-2 a} \left (c x^n\right )^{-4/n}}}\) |
\(\Big \downarrow \) 247 |
\(\displaystyle -\frac {x \left (c x^n\right )^{-2/n} \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \left (-e^{-2 a} \int \frac {1}{\sqrt {1-c^2 e^{-2 a} x^{2 n}}}d\frac {x^{-n}}{c}-\frac {x^{-n} \sqrt {1-e^{-2 a} c^2 x^{2 n}}}{c}\right )}{2 \sqrt {1-e^{-2 a} \left (c x^n\right )^{-4/n}}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle -\frac {x \left (c x^n\right )^{-2/n} \left (-e^{-a} \arcsin \left (\frac {e^{-a} x^{-n}}{c}\right )-\frac {x^{-n} \sqrt {1-e^{-2 a} c^2 x^{2 n}}}{c}\right ) \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )}}{2 \sqrt {1-e^{-2 a} \left (c x^n\right )^{-4/n}}}\) |
-1/2*(x*(-(Sqrt[1 - (c^2*x^(2*n))/E^(2*a)]/(c*x^n)) - ArcSin[1/(c*E^a*x^n) ]/E^a)*Sqrt[Sinh[a + (2*Log[c*x^n])/n]])/((c*x^n)^(2/n)*Sqrt[1 - 1/(E^(2*a )*(c*x^n)^(4/n))])
3.3.86.3.1 Defintions of rubi rules used
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]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
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] && !IntegerQ[p]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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]
Int[Sinh[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> S imp[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] )
Int[((e_.)*(x_))^(m_.)*Sinh[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_), x_Symbol] :> Simp[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] /; FreeQ[ {a, b, d, e, m, p}, x] && !IntegerQ[p]
\[\int \sqrt {\sinh \left (a +\frac {2 \ln \left (c \,x^{n}\right )}{n}\right )}d x\]
Time = 0.27 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.14 \[ \int \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx=\frac {1}{4} \, {\left (2 \, \sqrt {\frac {1}{2}} x \sqrt {\frac {x^{4} e^{\left (\frac {2 \, {\left (a n + 2 \, \log \left (c\right )\right )}}{n}\right )} - 1}{x^{2}}} e^{\left (\frac {a n + 2 \, \log \left (c\right )}{2 \, n}\right )} - \sqrt {2} \arctan \left (\sqrt {2} \sqrt {\frac {1}{2}} x \sqrt {\frac {x^{4} e^{\left (\frac {2 \, {\left (a n + 2 \, \log \left (c\right )\right )}}{n}\right )} - 1}{x^{2}}}\right ) e^{\left (\frac {a n + 2 \, \log \left (c\right )}{2 \, n}\right )}\right )} e^{\left (-\frac {a n + 2 \, \log \left (c\right )}{n}\right )} \]
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(sqrt(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)
\[ \int \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx=\int \sqrt {\sinh {\left (a + \frac {2 \log {\left (c x^{n} \right )}}{n} \right )}}\, dx \]
\[ \int \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx=\int { \sqrt {\sinh \left (a + \frac {2 \, \log \left (c x^{n}\right )}{n}\right )} \,d x } \]
Timed out. \[ \int \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx=\text {Timed out} \]
Timed out. \[ \int \sqrt {\sinh \left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx=\int \sqrt {\mathrm {sinh}\left (a+\frac {2\,\ln \left (c\,x^n\right )}{n}\right )} \,d x \]