3.128 \(\int (a \cosh ^3(x))^{5/2} \, dx\)

Optimal. Leaf size=121 \[ \frac {26}{165} a^2 \sinh (x) \cosh ^3(x) \sqrt {a \cosh ^3(x)}+\frac {78}{385} a^2 \sinh (x) \cosh (x) \sqrt {a \cosh ^3(x)}+\frac {26}{77} a^2 \tanh (x) \sqrt {a \cosh ^3(x)}-\frac {26 i a^2 F\left (\left .\frac {i x}{2}\right |2\right ) \sqrt {a \cosh ^3(x)}}{77 \cosh ^{\frac {3}{2}}(x)}+\frac {2}{15} a^2 \sinh (x) \cosh ^5(x) \sqrt {a \cosh ^3(x)} \]

[Out]

-26/77*I*a^2*(cosh(1/2*x)^2)^(1/2)/cosh(1/2*x)*EllipticF(I*sinh(1/2*x),2^(1/2))*(a*cosh(x)^3)^(1/2)/cosh(x)^(3
/2)+78/385*a^2*cosh(x)*sinh(x)*(a*cosh(x)^3)^(1/2)+26/165*a^2*cosh(x)^3*sinh(x)*(a*cosh(x)^3)^(1/2)+2/15*a^2*c
osh(x)^5*sinh(x)*(a*cosh(x)^3)^(1/2)+26/77*a^2*(a*cosh(x)^3)^(1/2)*tanh(x)

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Rubi [A]  time = 0.05, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3207, 2635, 2641} \[ \frac {2}{15} a^2 \sinh (x) \cosh ^5(x) \sqrt {a \cosh ^3(x)}+\frac {26}{165} a^2 \sinh (x) \cosh ^3(x) \sqrt {a \cosh ^3(x)}+\frac {78}{385} a^2 \sinh (x) \cosh (x) \sqrt {a \cosh ^3(x)}+\frac {26}{77} a^2 \tanh (x) \sqrt {a \cosh ^3(x)}-\frac {26 i a^2 F\left (\left .\frac {i x}{2}\right |2\right ) \sqrt {a \cosh ^3(x)}}{77 \cosh ^{\frac {3}{2}}(x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-26*I)/77)*a^2*Sqrt[a*Cosh[x]^3]*EllipticF[(I/2)*x, 2])/Cosh[x]^(3/2) + (78*a^2*Cosh[x]*Sqrt[a*Cosh[x]^3]*S
inh[x])/385 + (26*a^2*Cosh[x]^3*Sqrt[a*Cosh[x]^3]*Sinh[x])/165 + (2*a^2*Cosh[x]^5*Sqrt[a*Cosh[x]^3]*Sinh[x])/1
5 + (26*a^2*Sqrt[a*Cosh[x]^3]*Tanh[x])/77

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 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]

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

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Mathematica [A]  time = 0.12, size = 65, normalized size = 0.54 \[ \frac {a \left (a \cosh ^3(x)\right )^{3/2} \left ((15465 \sinh (x)+3657 \sinh (3 x)+749 \sinh (5 x)+77 \sinh (7 x)) \sqrt {\cosh (x)}-12480 i F\left (\left .\frac {i x}{2}\right |2\right )\right )}{36960 \cosh ^{\frac {9}{2}}(x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(a*Cosh[x]^3)^(3/2)*((-12480*I)*EllipticF[(I/2)*x, 2] + Sqrt[Cosh[x]]*(15465*Sinh[x] + 3657*Sinh[3*x] + 749
*Sinh[5*x] + 77*Sinh[7*x])))/(36960*Cosh[x]^(9/2))

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fricas [F]  time = 1.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {a \cosh \relax (x)^{3}} a^{2} \cosh \relax (x)^{6}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \cosh \relax (x)^{3}\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*cosh(x)^3)^(5/2), x)

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maple [F]  time = 0.19, size = 0, normalized size = 0.00 \[ \int \left (a \left (\cosh ^{3}\relax (x )\right )\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \cosh \relax (x)^{3}\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*cosh(x)^3)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a\,{\mathrm {cosh}\relax (x)}^3\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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