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

Optimal. Leaf size=83 \[ \frac{2}{9} a \sinh ^2(x) \cosh (x) \sqrt{a \sinh ^3(x)}-\frac{14}{45} a \cosh (x) \sqrt{a \sinh ^3(x)}+\frac{14 i a \text{csch}(x) E\left (\left .\frac{\pi }{4}-\frac{i x}{2}\right |2\right ) \sqrt{a \sinh ^3(x)}}{15 \sqrt{i \sinh (x)}} \]

[Out]

(-14*a*Cosh[x]*Sqrt[a*Sinh[x]^3])/45 + (((14*I)/15)*a*Csch[x]*EllipticE[Pi/4 - (I/2)*x, 2]*Sqrt[a*Sinh[x]^3])/
Sqrt[I*Sinh[x]] + (2*a*Cosh[x]*Sinh[x]^2*Sqrt[a*Sinh[x]^3])/9

________________________________________________________________________________________

Rubi [A]  time = 0.0436829, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3207, 2635, 2640, 2639} \[ \frac{2}{9} a \sinh ^2(x) \cosh (x) \sqrt{a \sinh ^3(x)}-\frac{14}{45} a \cosh (x) \sqrt{a \sinh ^3(x)}+\frac{14 i a \text{csch}(x) E\left (\left .\frac{\pi }{4}-\frac{i x}{2}\right |2\right ) \sqrt{a \sinh ^3(x)}}{15 \sqrt{i \sinh (x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-14*a*Cosh[x]*Sqrt[a*Sinh[x]^3])/45 + (((14*I)/15)*a*Csch[x]*EllipticE[Pi/4 - (I/2)*x, 2]*Sqrt[a*Sinh[x]^3])/
Sqrt[I*Sinh[x]] + (2*a*Cosh[x]*Sinh[x]^2*Sqrt[a*Sinh[x]^3])/9

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 2640

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

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(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 )^{3/2} \, dx &=\frac{\left (a \sqrt{a \sinh ^3(x)}\right ) \int \sinh ^{\frac{9}{2}}(x) \, dx}{\sinh ^{\frac{3}{2}}(x)}\\ &=\frac{2}{9} a \cosh (x) \sinh ^2(x) \sqrt{a \sinh ^3(x)}-\frac{\left (7 a \sqrt{a \sinh ^3(x)}\right ) \int \sinh ^{\frac{5}{2}}(x) \, dx}{9 \sinh ^{\frac{3}{2}}(x)}\\ &=-\frac{14}{45} a \cosh (x) \sqrt{a \sinh ^3(x)}+\frac{2}{9} a \cosh (x) \sinh ^2(x) \sqrt{a \sinh ^3(x)}+\frac{\left (7 a \sqrt{a \sinh ^3(x)}\right ) \int \sqrt{\sinh (x)} \, dx}{15 \sinh ^{\frac{3}{2}}(x)}\\ &=-\frac{14}{45} a \cosh (x) \sqrt{a \sinh ^3(x)}+\frac{2}{9} a \cosh (x) \sinh ^2(x) \sqrt{a \sinh ^3(x)}+\frac{\left (7 a \text{csch}(x) \sqrt{a \sinh ^3(x)}\right ) \int \sqrt{i \sinh (x)} \, dx}{15 \sqrt{i \sinh (x)}}\\ &=-\frac{14}{45} a \cosh (x) \sqrt{a \sinh ^3(x)}+\frac{14 i a \text{csch}(x) E\left (\left .\frac{\pi }{4}-\frac{i x}{2}\right |2\right ) \sqrt{a \sinh ^3(x)}}{15 \sqrt{i \sinh (x)}}+\frac{2}{9} a \cosh (x) \sinh ^2(x) \sqrt{a \sinh ^3(x)}\\ \end{align*}

Mathematica [A]  time = 0.0644298, size = 57, normalized size = 0.69 \[ \frac{1}{180} a \text{csch}(x) \sqrt{a \sinh ^3(x)} \left (-38 \sinh (2 x)+5 \sinh (4 x)+168 \sqrt{i \sinh (x)} \text{csch}(x) E\left (\left .\frac{1}{4} (\pi -2 i x)\right |2\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Csch[x]*Sqrt[a*Sinh[x]^3]*(168*Csch[x]*EllipticE[(Pi - (2*I)*x)/4, 2]*Sqrt[I*Sinh[x]] - 38*Sinh[2*x] + 5*Si
nh[4*x]))/180

________________________________________________________________________________________

Maple [F]  time = 0.06, size = 0, normalized size = 0. \begin{align*} \int \left ( a \left ( \sinh \left ( x \right ) \right ) ^{3} \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sinh(x)^3)^(3/2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*sinh(x)**3)**(3/2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sinh(x)^3)^(3/2), x)