\(\int \frac {1}{\sqrt {a \sinh ^3(x)}} \, dx\) [149]

Optimal result

Integrand size = 10, antiderivative size = 60 \[ \int \frac {1}{\sqrt {a \sinh ^3(x)}} \, dx=-\frac {2 \cosh (x) \sinh (x)}{\sqrt {a \sinh ^3(x)}}+\frac {2 i E\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right ) \sinh ^2(x)}{\sqrt {i \sinh (x)} \sqrt {a \sinh ^3(x)}} \] Output:

-2*cosh(x)*sinh(x)/(a*sinh(x)^3)^(1/2)+2*I*EllipticE(cos(1/4*Pi+1/2*I*x),2 
^(1/2))*sinh(x)^2/(I*sinh(x))^(1/2)/(a*sinh(x)^3)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\sqrt {a \sinh ^3(x)}} \, dx=-\frac {2 \left (\cosh (x)-E\left (\left .\frac {1}{4} (\pi -2 i x)\right |2\right ) \sqrt {i \sinh (x)}\right ) \sinh (x)}{\sqrt {a \sinh ^3(x)}} \] Input:

Integrate[1/Sqrt[a*Sinh[x]^3],x]
 

Output:

(-2*(Cosh[x] - EllipticE[(Pi - (2*I)*x)/4, 2]*Sqrt[I*Sinh[x]])*Sinh[x])/Sq 
rt[a*Sinh[x]^3]
 

Rubi [A] (verified)

Time = 0.35 (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

\(\displaystyle \frac {\sinh ^{\frac {3}{2}}(x) \left (-\frac {2 \cosh (x)}{\sqrt {\sinh (x)}}+\frac {2 i \sqrt {\sinh (x)} E\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right )}{\sqrt {i \sinh (x)}}\right )}{\sqrt {a \sinh ^3(x)}}\)

Input:

Int[1/Sqrt[a*Sinh[x]^3],x]
 

Output:

(((-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]
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3116

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]
 

rule 3119

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]
 

rule 3121

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]
 

rule 3686

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]])
 

Maple [F]

\[\int \frac {1}{\sqrt {a \sinh \left (x \right )^{3}}}d x\]

Input:

int(1/(a*sinh(x)^3)^(1/2),x)
 

Output:

int(1/(a*sinh(x)^3)^(1/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.40 \[ \int \frac {1}{\sqrt {a \sinh ^3(x)}} \, dx=-\frac {4 \, {\left (\sqrt {\frac {1}{2}} {\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} - 1\right )} \sqrt {a} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cosh \left (x\right ) + \sinh \left (x\right )\right )\right ) + {\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2}\right )} \sqrt {a \sinh \left (x\right )}\right )}}{a \cosh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) \sinh \left (x\right ) + a \sinh \left (x\right )^{2} - a} \] Input:

integrate(1/(a*sinh(x)^3)^(1/2),x, algorithm="fricas")
 

Output:

-4*(sqrt(1/2)*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)*sqrt(a)*weie 
rstrassZeta(4, 0, weierstrassPInverse(4, 0, cosh(x) + sinh(x))) + (cosh(x) 
^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)*sqrt(a*sinh(x)))/(a*cosh(x)^2 + 2*a*co 
sh(x)*sinh(x) + a*sinh(x)^2 - a)
 

Sympy [F]

\[ \int \frac {1}{\sqrt {a \sinh ^3(x)}} \, dx=\int \frac {1}{\sqrt {a \sinh ^{3}{\left (x \right )}}}\, dx \] Input:

integrate(1/(a*sinh(x)**3)**(1/2),x)
 

Output:

Integral(1/sqrt(a*sinh(x)**3), x)
 

Maxima [F]

\[ \int \frac {1}{\sqrt {a \sinh ^3(x)}} \, dx=\int { \frac {1}{\sqrt {a \sinh \left (x\right )^{3}}} \,d x } \] Input:

integrate(1/(a*sinh(x)^3)^(1/2),x, algorithm="maxima")
 

Output:

integrate(1/sqrt(a*sinh(x)^3), x)
 

Giac [F]

\[ \int \frac {1}{\sqrt {a \sinh ^3(x)}} \, dx=\int { \frac {1}{\sqrt {a \sinh \left (x\right )^{3}}} \,d x } \] Input:

integrate(1/(a*sinh(x)^3)^(1/2),x, algorithm="giac")
 

Output:

integrate(1/sqrt(a*sinh(x)^3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a \sinh ^3(x)}} \, dx=\int \frac {1}{\sqrt {a\,{\mathrm {sinh}\left (x\right )}^3}} \,d x \] Input:

int(1/(a*sinh(x)^3)^(1/2),x)
 

Output:

int(1/(a*sinh(x)^3)^(1/2), x)
 

Reduce [F]

\[ \int \frac {1}{\sqrt {a \sinh ^3(x)}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sinh \left (x \right )}}{\sinh \left (x \right )^{2}}d x \right )}{a} \] Input:

int(1/(a*sinh(x)^3)^(1/2),x)
 

Output:

(sqrt(a)*int(sqrt(sinh(x))/sinh(x)**2,x))/a