Integrand size = 10, antiderivative size = 62 \[ \int \sqrt {a \sinh ^3(x)} \, dx=\frac {2}{3} \coth (x) \sqrt {a \sinh ^3(x)}-\frac {2}{3} i \text {csch}^2(x) \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {i x}{2},2\right ) \sqrt {i \sinh (x)} \sqrt {a \sinh ^3(x)} \] Output:
2/3*coth(x)*(a*sinh(x)^3)^(1/2)+2/3*I*csch(x)^2*InverseJacobiAM(-1/4*Pi+1/ 2*I*x,2^(1/2))*(I*sinh(x))^(1/2)*(a*sinh(x)^3)^(1/2)
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 \[ \int \sqrt {a \sinh ^3(x)} \, dx=\frac {2}{3} \sqrt {a \sinh ^3(x)} \left (\coth (x)-\sqrt {2} \text {csch}^2(x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\cosh (2 x)+\sinh (2 x)\right ) \sqrt {-\sinh (x) (\cosh (x)+\sinh (x))}\right ) \] Input:
Integrate[Sqrt[a*Sinh[x]^3],x]
Output:
(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
Time = 0.35 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.08, 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, 3115, 3042, 3121, 3042, 3120}
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.
\(\displaystyle \int \sqrt {a \sinh ^3(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {i a \sin (i x)^3}dx\) |
\(\Big \downarrow \) 3686 |
\(\displaystyle \frac {\sqrt {a \sinh ^3(x)} \int \sinh ^{\frac {3}{2}}(x)dx}{\sinh ^{\frac {3}{2}}(x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {a \sinh ^3(x)} \int (-i \sin (i x))^{3/2}dx}{\sinh ^{\frac {3}{2}}(x)}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {\sqrt {a \sinh ^3(x)} \left (\frac {2}{3} \sqrt {\sinh (x)} \cosh (x)-\frac {1}{3} \int \frac {1}{\sqrt {\sinh (x)}}dx\right )}{\sinh ^{\frac {3}{2}}(x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {a \sinh ^3(x)} \left (\frac {2}{3} \sqrt {\sinh (x)} \cosh (x)-\frac {1}{3} \int \frac {1}{\sqrt {-i \sin (i x)}}dx\right )}{\sinh ^{\frac {3}{2}}(x)}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {\sqrt {a \sinh ^3(x)} \left (\frac {2}{3} \sqrt {\sinh (x)} \cosh (x)-\frac {\sqrt {i \sinh (x)} \int \frac {1}{\sqrt {i \sinh (x)}}dx}{3 \sqrt {\sinh (x)}}\right )}{\sinh ^{\frac {3}{2}}(x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {a \sinh ^3(x)} \left (\frac {2}{3} \sqrt {\sinh (x)} \cosh (x)-\frac {\sqrt {i \sinh (x)} \int \frac {1}{\sqrt {\sin (i x)}}dx}{3 \sqrt {\sinh (x)}}\right )}{\sinh ^{\frac {3}{2}}(x)}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\sqrt {a \sinh ^3(x)} \left (\frac {2}{3} \sqrt {\sinh (x)} \cosh (x)-\frac {2 i \sqrt {i \sinh (x)} \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {i x}{2},2\right )}{3 \sqrt {\sinh (x)}}\right )}{\sinh ^{\frac {3}{2}}(x)}\) |
Input:
Int[Sqrt[a*Sinh[x]^3],x]
Output:
(((((-2*I)/3)*EllipticF[Pi/4 - (I/2)*x, 2]*Sqrt[I*Sinh[x]])/Sqrt[Sinh[x]] + (2*Cosh[x]*Sqrt[Sinh[x]])/3)*Sqrt[a*Sinh[x]^3])/Sinh[x]^(3/2)
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] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
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]
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]])
\[\int \sqrt {a \sinh \left (x \right )^{3}}d x\]
Input:
int((a*sinh(x)^3)^(1/2),x)
Output:
int((a*sinh(x)^3)^(1/2),x)
Time = 0.10 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.89 \[ \int \sqrt {a \sinh ^3(x)} \, dx=-\frac {4 \, \sqrt {\frac {1}{2}} \sqrt {a} {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} {\rm weierstrassPInverse}\left (4, 0, \cosh \left (x\right ) + \sinh \left (x\right )\right ) - {\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 1\right )} \sqrt {a \sinh \left (x\right )}}{3 \, {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}} \] Input:
integrate((a*sinh(x)^3)^(1/2),x, algorithm="fricas")
Output:
-1/3*(4*sqrt(1/2)*sqrt(a)*(cosh(x) + sinh(x))*weierstrassPInverse(4, 0, co sh(x) + sinh(x)) - (cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 1)*sqrt(a* sinh(x)))/(cosh(x) + sinh(x))
\[ \int \sqrt {a \sinh ^3(x)} \, dx=\int \sqrt {a \sinh ^{3}{\left (x \right )}}\, dx \] Input:
integrate((a*sinh(x)**3)**(1/2),x)
Output:
Integral(sqrt(a*sinh(x)**3), x)
\[ \int \sqrt {a \sinh ^3(x)} \, dx=\int { \sqrt {a \sinh \left (x\right )^{3}} \,d x } \] Input:
integrate((a*sinh(x)^3)^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(a*sinh(x)^3), x)
\[ \int \sqrt {a \sinh ^3(x)} \, dx=\int { \sqrt {a \sinh \left (x\right )^{3}} \,d x } \] Input:
integrate((a*sinh(x)^3)^(1/2),x, algorithm="giac")
Output:
integrate(sqrt(a*sinh(x)^3), x)
Timed out. \[ \int \sqrt {a \sinh ^3(x)} \, dx=\int \sqrt {a\,{\mathrm {sinh}\left (x\right )}^3} \,d x \] Input:
int((a*sinh(x)^3)^(1/2),x)
Output:
int((a*sinh(x)^3)^(1/2), x)
\[ \int \sqrt {a \sinh ^3(x)} \, dx=\sqrt {a}\, \left (\int \sqrt {\sinh \left (x \right )}\, \sinh \left (x \right )d x \right ) \] Input:
int((a*sinh(x)^3)^(1/2),x)
Output:
sqrt(a)*int(sqrt(sinh(x))*sinh(x),x)