3.146 \(\int (a \sinh ^3(x))^{5/2} \, dx\)

Optimal. Leaf size=135 \[ \frac{26}{77} i a^2 \sqrt{i \sinh (x)} \text{csch}^2(x) \text{EllipticF}\left (\frac{\pi }{4}-\frac{i x}{2},2\right ) \sqrt{a \sinh ^3(x)}+\frac{2}{15} a^2 \sinh ^5(x) \cosh (x) \sqrt{a \sinh ^3(x)}-\frac{26}{165} a^2 \sinh ^3(x) \cosh (x) \sqrt{a \sinh ^3(x)}+\frac{78}{385} a^2 \sinh (x) \cosh (x) \sqrt{a \sinh ^3(x)}-\frac{26}{77} a^2 \coth (x) \sqrt{a \sinh ^3(x)} \]

[Out]

(-26*a^2*Coth[x]*Sqrt[a*Sinh[x]^3])/77 + ((26*I)/77)*a^2*Csch[x]^2*EllipticF[Pi/4 - (I/2)*x, 2]*Sqrt[I*Sinh[x]
]*Sqrt[a*Sinh[x]^3] + (78*a^2*Cosh[x]*Sinh[x]*Sqrt[a*Sinh[x]^3])/385 - (26*a^2*Cosh[x]*Sinh[x]^3*Sqrt[a*Sinh[x
]^3])/165 + (2*a^2*Cosh[x]*Sinh[x]^5*Sqrt[a*Sinh[x]^3])/15

________________________________________________________________________________________

Rubi [A]  time = 0.0605775, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3207, 2635, 2642, 2641} \[ \frac{2}{15} a^2 \sinh ^5(x) \cosh (x) \sqrt{a \sinh ^3(x)}-\frac{26}{165} a^2 \sinh ^3(x) \cosh (x) \sqrt{a \sinh ^3(x)}+\frac{78}{385} a^2 \sinh (x) \cosh (x) \sqrt{a \sinh ^3(x)}-\frac{26}{77} a^2 \coth (x) \sqrt{a \sinh ^3(x)}+\frac{26}{77} i a^2 \sqrt{i \sinh (x)} \text{csch}^2(x) F\left (\left .\frac{\pi }{4}-\frac{i x}{2}\right |2\right ) \sqrt{a \sinh ^3(x)} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sinh[x]^3)^(5/2),x]

[Out]

(-26*a^2*Coth[x]*Sqrt[a*Sinh[x]^3])/77 + ((26*I)/77)*a^2*Csch[x]^2*EllipticF[Pi/4 - (I/2)*x, 2]*Sqrt[I*Sinh[x]
]*Sqrt[a*Sinh[x]^3] + (78*a^2*Cosh[x]*Sinh[x]*Sqrt[a*Sinh[x]^3])/385 - (26*a^2*Cosh[x]*Sinh[x]^3*Sqrt[a*Sinh[x
]^3])/165 + (2*a^2*Cosh[x]*Sinh[x]^5*Sqrt[a*Sinh[x]^3])/15

Rule 3207

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
]])

Rule 2635

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] && Integer
Q[2*n]

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \left (a \sinh ^3(x)\right )^{5/2} \, dx &=\frac{\left (a^2 \sqrt{a \sinh ^3(x)}\right ) \int \sinh ^{\frac{15}{2}}(x) \, dx}{\sinh ^{\frac{3}{2}}(x)}\\ &=\frac{2}{15} a^2 \cosh (x) \sinh ^5(x) \sqrt{a \sinh ^3(x)}-\frac{\left (13 a^2 \sqrt{a \sinh ^3(x)}\right ) \int \sinh ^{\frac{11}{2}}(x) \, dx}{15 \sinh ^{\frac{3}{2}}(x)}\\ &=-\frac{26}{165} a^2 \cosh (x) \sinh ^3(x) \sqrt{a \sinh ^3(x)}+\frac{2}{15} a^2 \cosh (x) \sinh ^5(x) \sqrt{a \sinh ^3(x)}+\frac{\left (39 a^2 \sqrt{a \sinh ^3(x)}\right ) \int \sinh ^{\frac{7}{2}}(x) \, dx}{55 \sinh ^{\frac{3}{2}}(x)}\\ &=\frac{78}{385} a^2 \cosh (x) \sinh (x) \sqrt{a \sinh ^3(x)}-\frac{26}{165} a^2 \cosh (x) \sinh ^3(x) \sqrt{a \sinh ^3(x)}+\frac{2}{15} a^2 \cosh (x) \sinh ^5(x) \sqrt{a \sinh ^3(x)}-\frac{\left (39 a^2 \sqrt{a \sinh ^3(x)}\right ) \int \sinh ^{\frac{3}{2}}(x) \, dx}{77 \sinh ^{\frac{3}{2}}(x)}\\ &=-\frac{26}{77} a^2 \coth (x) \sqrt{a \sinh ^3(x)}+\frac{78}{385} a^2 \cosh (x) \sinh (x) \sqrt{a \sinh ^3(x)}-\frac{26}{165} a^2 \cosh (x) \sinh ^3(x) \sqrt{a \sinh ^3(x)}+\frac{2}{15} a^2 \cosh (x) \sinh ^5(x) \sqrt{a \sinh ^3(x)}+\frac{\left (13 a^2 \sqrt{a \sinh ^3(x)}\right ) \int \frac{1}{\sqrt{\sinh (x)}} \, dx}{77 \sinh ^{\frac{3}{2}}(x)}\\ &=-\frac{26}{77} a^2 \coth (x) \sqrt{a \sinh ^3(x)}+\frac{78}{385} a^2 \cosh (x) \sinh (x) \sqrt{a \sinh ^3(x)}-\frac{26}{165} a^2 \cosh (x) \sinh ^3(x) \sqrt{a \sinh ^3(x)}+\frac{2}{15} a^2 \cosh (x) \sinh ^5(x) \sqrt{a \sinh ^3(x)}+\frac{1}{77} \left (13 a^2 \text{csch}^2(x) \sqrt{i \sinh (x)} \sqrt{a \sinh ^3(x)}\right ) \int \frac{1}{\sqrt{i \sinh (x)}} \, dx\\ &=-\frac{26}{77} a^2 \coth (x) \sqrt{a \sinh ^3(x)}+\frac{26}{77} i a^2 \text{csch}^2(x) F\left (\left .\frac{\pi }{4}-\frac{i x}{2}\right |2\right ) \sqrt{i \sinh (x)} \sqrt{a \sinh ^3(x)}+\frac{78}{385} a^2 \cosh (x) \sinh (x) \sqrt{a \sinh ^3(x)}-\frac{26}{165} a^2 \cosh (x) \sinh ^3(x) \sqrt{a \sinh ^3(x)}+\frac{2}{15} a^2 \cosh (x) \sinh ^5(x) \sqrt{a \sinh ^3(x)}\\ \end{align*}

Mathematica [A]  time = 0.182696, size = 67, normalized size = 0.5 \[ \frac{a^2 \text{csch}(x) \sqrt{a \sinh ^3(x)} \left (-\frac{12480 \text{EllipticF}\left (\frac{1}{4} (\pi -2 i x),2\right )}{\sqrt{i \sinh (x)}}-15465 \cosh (x)+3657 \cosh (3 x)-749 \cosh (5 x)+77 \cosh (7 x)\right )}{36960} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sinh[x]^3)^(5/2),x]

[Out]

(a^2*Csch[x]*(-15465*Cosh[x] + 3657*Cosh[3*x] - 749*Cosh[5*x] + 77*Cosh[7*x] - (12480*EllipticF[(Pi - (2*I)*x)
/4, 2])/Sqrt[I*Sinh[x]])*Sqrt[a*Sinh[x]^3])/36960

________________________________________________________________________________________

Maple [F]  time = 0.076, size = 0, normalized size = 0. \begin{align*} \int \left ( a \left ( \sinh \left ( x \right ) \right ) ^{3} \right ) ^{{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sinh(x)^3)^(5/2),x)

[Out]

int((a*sinh(x)^3)^(5/2),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \sinh \left (x\right )^{3}\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sinh(x)^3)^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sinh(x)^3)^(5/2), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{a \sinh \left (x\right )^{3}} a^{2} \sinh \left (x\right )^{6}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sinh(x)^3)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(a*sinh(x)^3)*a^2*sinh(x)^6, x)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sinh(x)**3)**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \sinh \left (x\right )^{3}\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sinh(x)^3)^(5/2),x, algorithm="giac")

[Out]

integrate((a*sinh(x)^3)^(5/2), x)