\(\int \sqrt {a \sinh ^3(x)} \, dx\) [148]

Optimal result

Integrand size = 10, antiderivative size = 62 \[ \int \sqrt {a \sinh ^3(x)} \, dx=\frac {2}{3} \coth (x) \sqrt {a \sinh ^3(x)}-\frac {2}{3} i \text {csch}^2(x) \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {i x}{2},2\right ) \sqrt {i \sinh (x)} \sqrt {a \sinh ^3(x)} \]

[Out]

2/3*coth(x)*(a*sinh(x)^3)^(1/2)-2/3*I*csch(x)^2*(sin(1/4*Pi+1/2*I*x)^2)^(1/2)/sin(1/4*Pi+1/2*I*x)*EllipticF(co
s(1/4*Pi+1/2*I*x),2^(1/2))*(I*sinh(x))^(1/2)*(a*sinh(x)^3)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 62, 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, 2715, 2721, 2720} \[ \int \sqrt {a \sinh ^3(x)} \, dx=\frac {2}{3} \coth (x) \sqrt {a \sinh ^3(x)}-\frac {2}{3} i \sqrt {i \sinh (x)} \text {csch}^2(x) \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {i x}{2},2\right ) \sqrt {a \sinh ^3(x)} \]

[In]

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

[Out]

(2*Coth[x]*Sqrt[a*Sinh[x]^3])/3 - ((2*I)/3)*Csch[x]^2*EllipticF[Pi/4 - (I/2)*x, 2]*Sqrt[I*Sinh[x]]*Sqrt[a*Sinh
[x]^3]

Rule 2715

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

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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 {\sqrt {a \sinh ^3(x)} \int \sinh ^{\frac {3}{2}}(x) \, dx}{\sinh ^{\frac {3}{2}}(x)} \\ & = \frac {2}{3} \coth (x) \sqrt {a \sinh ^3(x)}-\frac {\sqrt {a \sinh ^3(x)} \int \frac {1}{\sqrt {\sinh (x)}} \, dx}{3 \sinh ^{\frac {3}{2}}(x)} \\ & = \frac {2}{3} \coth (x) \sqrt {a \sinh ^3(x)}-\frac {1}{3} \left (\text {csch}^2(x) \sqrt {i \sinh (x)} \sqrt {a \sinh ^3(x)}\right ) \int \frac {1}{\sqrt {i \sinh (x)}} \, dx \\ & = \frac {2}{3} \coth (x) \sqrt {a \sinh ^3(x)}-\frac {2}{3} i \text {csch}^2(x) \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {i x}{2},2\right ) \sqrt {i \sinh (x)} \sqrt {a \sinh ^3(x)} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.97 \[ \int \sqrt {a \sinh ^3(x)} \, dx=\frac {2}{3} \sqrt {a \sinh ^3(x)} \left (\coth (x)-\sqrt {2} \text {csch}^2(x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\cosh (2 x)+\sinh (2 x)\right ) \sqrt {-\sinh (x) (\cosh (x)+\sinh (x))}\right ) \]

[In]

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

[Out]

(2*Sqrt[a*Sinh[x]^3]*(Coth[x] - Sqrt[2]*Csch[x]^2*Hypergeometric2F1[1/4, 1/2, 5/4, Cosh[2*x] + Sinh[2*x]]*Sqrt
[-(Sinh[x]*(Cosh[x] + Sinh[x]))]))/3

Maple [F]

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

[In]

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

[Out]

int((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 = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.97 \[ \int \sqrt {a \sinh ^3(x)} \, dx=-\frac {2 \, {\left (\sqrt {2} \cosh \left (x\right ) + \sqrt {2} \sinh \left (x\right )\right )} \sqrt {a} {\rm weierstrassPInverse}\left (4, 0, \cosh \left (x\right ) + \sinh \left (x\right )\right ) - {\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 1\right )} \sqrt {a \sinh \left (x\right )}}{3 \, {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}} \]

[In]

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

[Out]

-1/3*(2*(sqrt(2)*cosh(x) + sqrt(2)*sinh(x))*sqrt(a)*weierstrassPInverse(4, 0, cosh(x) + sinh(x)) - (cosh(x)^2
+ 2*cosh(x)*sinh(x) + sinh(x)^2 + 1)*sqrt(a*sinh(x)))/(cosh(x) + sinh(x))

Sympy [F]

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

[In]

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

[Out]

Integral(sqrt(a*sinh(x)**3), x)

Maxima [F]

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

[In]

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

[Out]

integrate(sqrt(a*sinh(x)^3), x)

Giac [F]

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

[In]

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

[Out]

integrate(sqrt(a*sinh(x)^3), x)

Mupad [F(-1)]

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

[In]

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

[Out]

int((a*sinh(x)^3)^(1/2), x)