Integrand size = 10, antiderivative size = 60 \[ \int \frac {1}{\sqrt {a \sinh ^3(x)}} \, dx=-\frac {2 \cosh (x) \sinh (x)}{\sqrt {a \sinh ^3(x)}}+\frac {2 i E\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right ) \sinh ^2(x)}{\sqrt {i \sinh (x)} \sqrt {a \sinh ^3(x)}} \]
-2*cosh(x)*sinh(x)/(a*sinh(x)^3)^(1/2)+2*I*(sin(1/4*Pi+1/2*I*x)^2)^(1/2)/s in(1/4*Pi+1/2*I*x)*EllipticE(cos(1/4*Pi+1/2*I*x),2^(1/2))*sinh(x)^2/(I*sin h(x))^(1/2)/(a*sinh(x)^3)^(1/2)
Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\sqrt {a \sinh ^3(x)}} \, dx=-\frac {2 \left (\cosh (x)-E\left (\left .\frac {1}{4} (\pi -2 i x)\right |2\right ) \sqrt {i \sinh (x)}\right ) \sinh (x)}{\sqrt {a \sinh ^3(x)}} \]
Time = 0.33 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {3042, 3686, 3042, 3116, 3042, 3121, 3042, 3119}
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.
\(\displaystyle \int \frac {1}{\sqrt {a \sinh ^3(x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {i a \sin (i x)^3}}dx\) |
\(\Big \downarrow \) 3686 |
\(\displaystyle \frac {\sinh ^{\frac {3}{2}}(x) \int \frac {1}{\sinh ^{\frac {3}{2}}(x)}dx}{\sqrt {a \sinh ^3(x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sinh ^{\frac {3}{2}}(x) \int \frac {1}{(-i \sin (i x))^{3/2}}dx}{\sqrt {a \sinh ^3(x)}}\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle \frac {\sinh ^{\frac {3}{2}}(x) \left (\int \sqrt {\sinh (x)}dx-\frac {2 \cosh (x)}{\sqrt {\sinh (x)}}\right )}{\sqrt {a \sinh ^3(x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sinh ^{\frac {3}{2}}(x) \left (-\frac {2 \cosh (x)}{\sqrt {\sinh (x)}}+\int \sqrt {-i \sin (i x)}dx\right )}{\sqrt {a \sinh ^3(x)}}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {\sinh ^{\frac {3}{2}}(x) \left (-\frac {2 \cosh (x)}{\sqrt {\sinh (x)}}+\frac {\sqrt {\sinh (x)} \int \sqrt {i \sinh (x)}dx}{\sqrt {i \sinh (x)}}\right )}{\sqrt {a \sinh ^3(x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sinh ^{\frac {3}{2}}(x) \left (-\frac {2 \cosh (x)}{\sqrt {\sinh (x)}}+\frac {\sqrt {\sinh (x)} \int \sqrt {\sin (i x)}dx}{\sqrt {i \sinh (x)}}\right )}{\sqrt {a \sinh ^3(x)}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\sinh ^{\frac {3}{2}}(x) \left (-\frac {2 \cosh (x)}{\sqrt {\sinh (x)}}+\frac {2 i \sqrt {\sinh (x)} E\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right )}{\sqrt {i \sinh (x)}}\right )}{\sqrt {a \sinh ^3(x)}}\) |
(((-2*Cosh[x])/Sqrt[Sinh[x]] + ((2*I)*EllipticE[Pi/4 - (I/2)*x, 2]*Sqrt[Si nh[x]])/Sqrt[I*Sinh[x]])*Sinh[x]^(3/2))/Sqrt[a*Sinh[x]^3]
3.2.49.3.1 Defintions of rubi rules used
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2*n]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Sin[e + f*x]^ n)^FracPart[p]/(Sin[e + f*x]/ff)^(n*FracPart[p])) Int[ActivateTrig[u]*(Si n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] && !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / ; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
\[\int \frac {1}{\sqrt {a \sinh \left (x \right )^{3}}}d x\]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.62 \[ \int \frac {1}{\sqrt {a \sinh ^3(x)}} \, dx=-\frac {2 \, {\left ({\left (\sqrt {2} \cosh \left (x\right )^{2} + 2 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right ) + \sqrt {2} \sinh \left (x\right )^{2} - \sqrt {2}\right )} \sqrt {a} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cosh \left (x\right ) + \sinh \left (x\right )\right )\right ) + 2 \, {\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2}\right )} \sqrt {a \sinh \left (x\right )}\right )}}{a \cosh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) \sinh \left (x\right ) + a \sinh \left (x\right )^{2} - a} \]
-2*((sqrt(2)*cosh(x)^2 + 2*sqrt(2)*cosh(x)*sinh(x) + sqrt(2)*sinh(x)^2 - s qrt(2))*sqrt(a)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cosh(x) + sinh(x))) + 2*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)*sqrt(a*sinh(x))) /(a*cosh(x)^2 + 2*a*cosh(x)*sinh(x) + a*sinh(x)^2 - a)
\[ \int \frac {1}{\sqrt {a \sinh ^3(x)}} \, dx=\int \frac {1}{\sqrt {a \sinh ^{3}{\left (x \right )}}}\, dx \]
\[ \int \frac {1}{\sqrt {a \sinh ^3(x)}} \, dx=\int { \frac {1}{\sqrt {a \sinh \left (x\right )^{3}}} \,d x } \]
\[ \int \frac {1}{\sqrt {a \sinh ^3(x)}} \, dx=\int { \frac {1}{\sqrt {a \sinh \left (x\right )^{3}}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt {a \sinh ^3(x)}} \, dx=\int \frac {1}{\sqrt {a\,{\mathrm {sinh}\left (x\right )}^3}} \,d x \]