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

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3286, 2716, 2721, 2719} \[ \int \frac {1}{\sqrt {a \sinh ^3(x)}} \, dx=-\frac {2 \sinh (x) \cosh (x)}{\sqrt {a \sinh ^3(x)}}+\frac {2 i \sinh ^2(x) E\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right )}{\sqrt {i \sinh (x)} \sqrt {a \sinh ^3(x)}} \]

[In]

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

[Out]

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

Rule 2716

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] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2719

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 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 3286

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[(b*ff^n)^IntPart[p]*((b*Sin[e + f*x]^n)^FracPart[p]/(Sin[e + f*x]/ff)^(n*FracPart[p])), Int[ActivateTrig[u]
*(Sin[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
]])

Rubi steps \begin{align*} \text {integral}& = \frac {\sinh ^{\frac {3}{2}}(x) \int \frac {1}{\sinh ^{\frac {3}{2}}(x)} \, dx}{\sqrt {a \sinh ^3(x)}} \\ & = -\frac {2 \cosh (x) \sinh (x)}{\sqrt {a \sinh ^3(x)}}+\frac {\sinh ^{\frac {3}{2}}(x) \int \sqrt {\sinh (x)} \, dx}{\sqrt {a \sinh ^3(x)}} \\ & = -\frac {2 \cosh (x) \sinh (x)}{\sqrt {a \sinh ^3(x)}}+\frac {\sinh ^2(x) \int \sqrt {i \sinh (x)} \, dx}{\sqrt {i \sinh (x)} \sqrt {a \sinh ^3(x)}} \\ & = -\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)}} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

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 \[ \int \frac {1}{\sqrt {a \sinh ^3(x)}} \, dx=-\frac {2 \, {\left ({\left (\sqrt {2} \cosh \left (x\right )^{2} + 2 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right ) + \sqrt {2} \sinh \left (x\right )^{2} - \sqrt {2}\right )} \sqrt {a} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cosh \left (x\right ) + \sinh \left (x\right )\right )\right ) + 2 \, {\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2}\right )} \sqrt {a \sinh \left (x\right )}\right )}}{a \cosh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) \sinh \left (x\right ) + a \sinh \left (x\right )^{2} - a} \]

[In]

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

[Out]

-2*((sqrt(2)*cosh(x)^2 + 2*sqrt(2)*cosh(x)*sinh(x) + sqrt(2)*sinh(x)^2 - sqrt(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)

Sympy [F]

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

[In]

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

[Out]

Integral(1/sqrt(a*sinh(x)**3), x)

Maxima [F]

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

[In]

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

[Out]

integrate(1/sqrt(a*sinh(x)^3), x)

Giac [F]

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

[In]

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

[Out]

integrate(1/sqrt(a*sinh(x)^3), x)

Mupad [F(-1)]

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

[In]

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

[Out]

int(1/(a*sinh(x)^3)^(1/2), x)