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

Optimal. Leaf size=53 \[ \frac{8}{15} a^2 \coth (x) \sqrt{a \sinh ^2(x)}+\frac{1}{5} \coth (x) \left (a \sinh ^2(x)\right )^{5/2}-\frac{4}{15} a \coth (x) \left (a \sinh ^2(x)\right )^{3/2} \]

[Out]

(8*a^2*Coth[x]*Sqrt[a*Sinh[x]^2])/15 - (4*a*Coth[x]*(a*Sinh[x]^2)^(3/2))/15 + (Coth[x]*(a*Sinh[x]^2)^(5/2))/5

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Rubi [A]  time = 0.0371284, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {3203, 3207, 2638} \[ \frac{8}{15} a^2 \coth (x) \sqrt{a \sinh ^2(x)}+\frac{1}{5} \coth (x) \left (a \sinh ^2(x)\right )^{5/2}-\frac{4}{15} a \coth (x) \left (a \sinh ^2(x)\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^2*Coth[x]*Sqrt[a*Sinh[x]^2])/15 - (4*a*Coth[x]*(a*Sinh[x]^2)^(3/2))/15 + (Coth[x]*(a*Sinh[x]^2)^(5/2))/5

Rule 3203

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> -Simp[(Cot[e + f*x]*(b*Sin[e + f*x]^2)^p)/(2*f*p), x]
+ Dist[(b*(2*p - 1))/(2*p), Int[(b*Sin[e + f*x]^2)^(p - 1), x], x] /; FreeQ[{b, e, f}, x] &&  !IntegerQ[p] &&
GtQ[p, 1]

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 2638

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

Rubi steps

\begin{align*} \int \left (a \sinh ^2(x)\right )^{5/2} \, dx &=\frac{1}{5} \coth (x) \left (a \sinh ^2(x)\right )^{5/2}-\frac{1}{5} (4 a) \int \left (a \sinh ^2(x)\right )^{3/2} \, dx\\ &=-\frac{4}{15} a \coth (x) \left (a \sinh ^2(x)\right )^{3/2}+\frac{1}{5} \coth (x) \left (a \sinh ^2(x)\right )^{5/2}+\frac{1}{15} \left (8 a^2\right ) \int \sqrt{a \sinh ^2(x)} \, dx\\ &=-\frac{4}{15} a \coth (x) \left (a \sinh ^2(x)\right )^{3/2}+\frac{1}{5} \coth (x) \left (a \sinh ^2(x)\right )^{5/2}+\frac{1}{15} \left (8 a^2 \text{csch}(x) \sqrt{a \sinh ^2(x)}\right ) \int \sinh (x) \, dx\\ &=\frac{8}{15} a^2 \coth (x) \sqrt{a \sinh ^2(x)}-\frac{4}{15} a \coth (x) \left (a \sinh ^2(x)\right )^{3/2}+\frac{1}{5} \coth (x) \left (a \sinh ^2(x)\right )^{5/2}\\ \end{align*}

Mathematica [A]  time = 0.0408078, size = 36, normalized size = 0.68 \[ \frac{1}{240} a^2 (150 \cosh (x)-25 \cosh (3 x)+3 \cosh (5 x)) \text{csch}(x) \sqrt{a \sinh ^2(x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(150*Cosh[x] - 25*Cosh[3*x] + 3*Cosh[5*x])*Csch[x]*Sqrt[a*Sinh[x]^2])/240

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Maple [A]  time = 0.045, size = 32, normalized size = 0.6 \begin{align*}{\frac{{a}^{3}\sinh \left ( x \right ) \cosh \left ( x \right ) \left ( 3\, \left ( \sinh \left ( x \right ) \right ) ^{4}-4\, \left ( \sinh \left ( x \right ) \right ) ^{2}+8 \right ) }{15}{\frac{1}{\sqrt{a \left ( \sinh \left ( x \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*a^3*sinh(x)*cosh(x)*(3*sinh(x)^4-4*sinh(x)^2+8)/(a*sinh(x)^2)^(1/2)

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Maxima [A]  time = 1.81499, size = 72, normalized size = 1.36 \begin{align*} -\frac{1}{160} \, a^{\frac{5}{2}} e^{\left (5 \, x\right )} + \frac{5}{96} \, a^{\frac{5}{2}} e^{\left (3 \, x\right )} - \frac{5}{16} \, a^{\frac{5}{2}} e^{\left (-x\right )} + \frac{5}{96} \, a^{\frac{5}{2}} e^{\left (-3 \, x\right )} - \frac{1}{160} \, a^{\frac{5}{2}} e^{\left (-5 \, x\right )} - \frac{5}{16} \, a^{\frac{5}{2}} e^{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/160*a^(5/2)*e^(5*x) + 5/96*a^(5/2)*e^(3*x) - 5/16*a^(5/2)*e^(-x) + 5/96*a^(5/2)*e^(-3*x) - 1/160*a^(5/2)*e^
(-5*x) - 5/16*a^(5/2)*e^x

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Fricas [B]  time = 1.78509, size = 1462, normalized size = 27.58 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(30*a^2*cosh(x)*e^x*sinh(x)^9 + 3*a^2*e^x*sinh(x)^10 + 5*(27*a^2*cosh(x)^2 - 5*a^2)*e^x*sinh(x)^8 + 40*(
9*a^2*cosh(x)^3 - 5*a^2*cosh(x))*e^x*sinh(x)^7 + 10*(63*a^2*cosh(x)^4 - 70*a^2*cosh(x)^2 + 15*a^2)*e^x*sinh(x)
^6 + 4*(189*a^2*cosh(x)^5 - 350*a^2*cosh(x)^3 + 225*a^2*cosh(x))*e^x*sinh(x)^5 + 10*(63*a^2*cosh(x)^6 - 175*a^
2*cosh(x)^4 + 225*a^2*cosh(x)^2 + 15*a^2)*e^x*sinh(x)^4 + 40*(9*a^2*cosh(x)^7 - 35*a^2*cosh(x)^5 + 75*a^2*cosh
(x)^3 + 15*a^2*cosh(x))*e^x*sinh(x)^3 + 5*(27*a^2*cosh(x)^8 - 140*a^2*cosh(x)^6 + 450*a^2*cosh(x)^4 + 180*a^2*
cosh(x)^2 - 5*a^2)*e^x*sinh(x)^2 + 10*(3*a^2*cosh(x)^9 - 20*a^2*cosh(x)^7 + 90*a^2*cosh(x)^5 + 60*a^2*cosh(x)^
3 - 5*a^2*cosh(x))*e^x*sinh(x) + (3*a^2*cosh(x)^10 - 25*a^2*cosh(x)^8 + 150*a^2*cosh(x)^6 + 150*a^2*cosh(x)^4
- 25*a^2*cosh(x)^2 + 3*a^2)*e^x)*sqrt(a*e^(4*x) - 2*a*e^(2*x) + a)*e^(-x)/(cosh(x)^5*e^(2*x) + (e^(2*x) - 1)*s
inh(x)^5 - cosh(x)^5 + 5*(cosh(x)*e^(2*x) - cosh(x))*sinh(x)^4 + 10*(cosh(x)^2*e^(2*x) - cosh(x)^2)*sinh(x)^3
+ 10*(cosh(x)^3*e^(2*x) - cosh(x)^3)*sinh(x)^2 + 5*(cosh(x)^4*e^(2*x) - cosh(x)^4)*sinh(x))

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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)**2)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.31344, size = 162, normalized size = 3.06 \begin{align*} \frac{1}{480} \,{\left (3 \, a^{2} e^{\left (5 \, x\right )} \mathrm{sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right ) - 25 \, a^{2} e^{\left (3 \, x\right )} \mathrm{sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right ) + 150 \, a^{2} e^{x} \mathrm{sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right ) +{\left (150 \, a^{2} e^{\left (4 \, x\right )} \mathrm{sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right ) - 25 \, a^{2} e^{\left (2 \, x\right )} \mathrm{sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right ) + 3 \, a^{2} \mathrm{sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right )\right )} e^{\left (-5 \, x\right )}\right )} \sqrt{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(3*a^2*e^(5*x)*sgn(e^(3*x) - e^x) - 25*a^2*e^(3*x)*sgn(e^(3*x) - e^x) + 150*a^2*e^x*sgn(e^(3*x) - e^x) +
(150*a^2*e^(4*x)*sgn(e^(3*x) - e^x) - 25*a^2*e^(2*x)*sgn(e^(3*x) - e^x) + 3*a^2*sgn(e^(3*x) - e^x))*e^(-5*x))
*sqrt(a)