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

Optimal. Leaf size=71 \[ \frac {14}{45} a \sinh (x) \sqrt {a \cosh ^3(x)}-\frac {14 i a E\left (\left .\frac {i x}{2}\right |2\right ) \sqrt {a \cosh ^3(x)}}{15 \cosh ^{\frac {3}{2}}(x)}+\frac {2}{9} a \sinh (x) \cosh ^2(x) \sqrt {a \cosh ^3(x)} \]

[Out]

-14/15*I*a*(cosh(1/2*x)^2)^(1/2)/cosh(1/2*x)*EllipticE(I*sinh(1/2*x),2^(1/2))*(a*cosh(x)^3)^(1/2)/cosh(x)^(3/2
)+14/45*a*sinh(x)*(a*cosh(x)^3)^(1/2)+2/9*a*cosh(x)^2*sinh(x)*(a*cosh(x)^3)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 71, 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 = {3207, 2635, 2639} \[ \frac {2}{9} a \sinh (x) \cosh ^2(x) \sqrt {a \cosh ^3(x)}+\frac {14}{45} a \sinh (x) \sqrt {a \cosh ^3(x)}-\frac {14 i a E\left (\left .\frac {i x}{2}\right |2\right ) \sqrt {a \cosh ^3(x)}}{15 \cosh ^{\frac {3}{2}}(x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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]

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 )^{3/2} \, dx &=\frac {\left (a \sqrt {a \cosh ^3(x)}\right ) \int \cosh ^{\frac {9}{2}}(x) \, dx}{\cosh ^{\frac {3}{2}}(x)}\\ &=\frac {2}{9} a \cosh ^2(x) \sqrt {a \cosh ^3(x)} \sinh (x)+\frac {\left (7 a \sqrt {a \cosh ^3(x)}\right ) \int \cosh ^{\frac {5}{2}}(x) \, dx}{9 \cosh ^{\frac {3}{2}}(x)}\\ &=\frac {14}{45} a \sqrt {a \cosh ^3(x)} \sinh (x)+\frac {2}{9} a \cosh ^2(x) \sqrt {a \cosh ^3(x)} \sinh (x)+\frac {\left (7 a \sqrt {a \cosh ^3(x)}\right ) \int \sqrt {\cosh (x)} \, dx}{15 \cosh ^{\frac {3}{2}}(x)}\\ &=-\frac {14 i a \sqrt {a \cosh ^3(x)} E\left (\left .\frac {i x}{2}\right |2\right )}{15 \cosh ^{\frac {3}{2}}(x)}+\frac {14}{45} a \sqrt {a \cosh ^3(x)} \sinh (x)+\frac {2}{9} a \cosh ^2(x) \sqrt {a \cosh ^3(x)} \sinh (x)\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 54, normalized size = 0.76 \[ \frac {\left (a \cosh ^3(x)\right )^{3/2} \left ((38 \sinh (2 x)+5 \sinh (4 x)) \sqrt {\cosh (x)}-168 i E\left (\left .\frac {i x}{2}\right |2\right )\right )}{180 \cosh ^{\frac {9}{2}}(x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*Cosh[x]^3)^(3/2)*((-168*I)*EllipticE[(I/2)*x, 2] + Sqrt[Cosh[x]]*(38*Sinh[2*x] + 5*Sinh[4*x])))/(180*Cosh[
x]^(9/2))

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fricas [F]  time = 1.98, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {a \cosh \relax (x)^{3}} a \cosh \relax (x)^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*cosh(x)^3)^(3/2), x)

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maple [F]  time = 0.22, size = 0, normalized size = 0.00 \[ \int \left (a \left (\cosh ^{3}\relax (x )\right )\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*cosh(x)^3)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a*cosh(x)^3)^(3/2), 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)**(3/2),x)

[Out]

Timed out

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