3.122 \(\int (a \cosh ^2(x))^{5/2} \, dx\)

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

[Out]

4/15*a*(a*cosh(x)^2)^(3/2)*tanh(x)+1/5*(a*cosh(x)^2)^(5/2)*tanh(x)+8/15*a^2*(a*cosh(x)^2)^(1/2)*tanh(x)

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Rubi [A]  time = 0.04, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3203, 3207, 2637} \[ \frac {8}{15} a^2 \tanh (x) \sqrt {a \cosh ^2(x)}+\frac {1}{5} \tanh (x) \left (a \cosh ^2(x)\right )^{5/2}+\frac {4}{15} a \tanh (x) \left (a \cosh ^2(x)\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^2*Sqrt[a*Cosh[x]^2]*Tanh[x])/15 + (4*a*(a*Cosh[x]^2)^(3/2)*Tanh[x])/15 + ((a*Cosh[x]^2)^(5/2)*Tanh[x])/5

Rule 2637

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

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
]])

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 36, normalized size = 0.68 \[ \frac {1}{240} a^2 (150 \sinh (x)+25 \sinh (3 x)+3 \sinh (5 x)) \text {sech}(x) \sqrt {a \cosh ^2(x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.86, size = 501, normalized size = 9.45 \[ \frac {{\left (30 \, a^{2} \cosh \relax (x) e^{x} \sinh \relax (x)^{9} + 3 \, a^{2} e^{x} \sinh \relax (x)^{10} + 5 \, {\left (27 \, a^{2} \cosh \relax (x)^{2} + 5 \, a^{2}\right )} e^{x} \sinh \relax (x)^{8} + 40 \, {\left (9 \, a^{2} \cosh \relax (x)^{3} + 5 \, a^{2} \cosh \relax (x)\right )} e^{x} \sinh \relax (x)^{7} + 10 \, {\left (63 \, a^{2} \cosh \relax (x)^{4} + 70 \, a^{2} \cosh \relax (x)^{2} + 15 \, a^{2}\right )} e^{x} \sinh \relax (x)^{6} + 4 \, {\left (189 \, a^{2} \cosh \relax (x)^{5} + 350 \, a^{2} \cosh \relax (x)^{3} + 225 \, a^{2} \cosh \relax (x)\right )} e^{x} \sinh \relax (x)^{5} + 10 \, {\left (63 \, a^{2} \cosh \relax (x)^{6} + 175 \, a^{2} \cosh \relax (x)^{4} + 225 \, a^{2} \cosh \relax (x)^{2} - 15 \, a^{2}\right )} e^{x} \sinh \relax (x)^{4} + 40 \, {\left (9 \, a^{2} \cosh \relax (x)^{7} + 35 \, a^{2} \cosh \relax (x)^{5} + 75 \, a^{2} \cosh \relax (x)^{3} - 15 \, a^{2} \cosh \relax (x)\right )} e^{x} \sinh \relax (x)^{3} + 5 \, {\left (27 \, a^{2} \cosh \relax (x)^{8} + 140 \, a^{2} \cosh \relax (x)^{6} + 450 \, a^{2} \cosh \relax (x)^{4} - 180 \, a^{2} \cosh \relax (x)^{2} - 5 \, a^{2}\right )} e^{x} \sinh \relax (x)^{2} + 10 \, {\left (3 \, a^{2} \cosh \relax (x)^{9} + 20 \, a^{2} \cosh \relax (x)^{7} + 90 \, a^{2} \cosh \relax (x)^{5} - 60 \, a^{2} \cosh \relax (x)^{3} - 5 \, a^{2} \cosh \relax (x)\right )} e^{x} \sinh \relax (x) + {\left (3 \, a^{2} \cosh \relax (x)^{10} + 25 \, a^{2} \cosh \relax (x)^{8} + 150 \, a^{2} \cosh \relax (x)^{6} - 150 \, a^{2} \cosh \relax (x)^{4} - 25 \, a^{2} \cosh \relax (x)^{2} - 3 \, a^{2}\right )} e^{x}\right )} \sqrt {a e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} + a} e^{\left (-x\right )}}{480 \, {\left (\cosh \relax (x)^{5} e^{\left (2 \, x\right )} + {\left (e^{\left (2 \, x\right )} + 1\right )} \sinh \relax (x)^{5} + \cosh \relax (x)^{5} + 5 \, {\left (\cosh \relax (x) e^{\left (2 \, x\right )} + \cosh \relax (x)\right )} \sinh \relax (x)^{4} + 10 \, {\left (\cosh \relax (x)^{2} e^{\left (2 \, x\right )} + \cosh \relax (x)^{2}\right )} \sinh \relax (x)^{3} + 10 \, {\left (\cosh \relax (x)^{3} e^{\left (2 \, x\right )} + \cosh \relax (x)^{3}\right )} \sinh \relax (x)^{2} + 5 \, {\left (\cosh \relax (x)^{4} e^{\left (2 \, x\right )} + \cosh \relax (x)^{4}\right )} \sinh \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cosh(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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giac [A]  time = 0.12, size = 61, normalized size = 1.15 \[ \frac {1}{480} \, {\left (3 \, a^{2} e^{\left (5 \, x\right )} + 25 \, a^{2} e^{\left (3 \, x\right )} + 150 \, a^{2} e^{x} - {\left (150 \, a^{2} e^{\left (4 \, x\right )} + 25 \, a^{2} e^{\left (2 \, x\right )} + 3 \, a^{2}\right )} e^{\left (-5 \, x\right )}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.18, size = 32, normalized size = 0.60 \[ \frac {a^{3} \cosh \relax (x ) \sinh \relax (x ) \left (3 \left (\cosh ^{4}\relax (x )\right )+4 \left (\cosh ^{2}\relax (x )\right )+8\right )}{15 \sqrt {a \left (\cosh ^{2}\relax (x )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 53, normalized size = 1.00 \[ \frac {1}{160} \, a^{\frac {5}{2}} e^{\left (5 \, x\right )} + \frac {5}{96} \, a^{\frac {5}{2}} e^{\left (3 \, x\right )} - \frac {5}{16} \, a^{\frac {5}{2}} e^{\left (-x\right )} - \frac {5}{96} \, a^{\frac {5}{2}} e^{\left (-3 \, x\right )} - \frac {1}{160} \, a^{\frac {5}{2}} e^{\left (-5 \, x\right )} + \frac {5}{16} \, a^{\frac {5}{2}} e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cosh(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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\left (a\,{\mathrm {cosh}\relax (x)}^2\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a*cosh(x)^2)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cosh(x)**2)**(5/2),x)

[Out]

Timed out

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