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

Optimal. Leaf size=48 \[ \frac{2}{5} a \sinh (x) (a \cosh (x))^{3/2}-\frac{6 i a^2 E\left (\left .\frac{i x}{2}\right |2\right ) \sqrt{a \cosh (x)}}{5 \sqrt{\cosh (x)}} \]

[Out]

(((-6*I)/5)*a^2*Sqrt[a*Cosh[x]]*EllipticE[(I/2)*x, 2])/Sqrt[Cosh[x]] + (2*a*(a*Cosh[x])^(3/2)*Sinh[x])/5

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Rubi [A]  time = 0.0252784, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {2635, 2640, 2639} \[ \frac{2}{5} a \sinh (x) (a \cosh (x))^{3/2}-\frac{6 i a^2 E\left (\left .\frac{i x}{2}\right |2\right ) \sqrt{a \cosh (x)}}{5 \sqrt{\cosh (x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-6*I)/5)*a^2*Sqrt[a*Cosh[x]]*EllipticE[(I/2)*x, 2])/Sqrt[Cosh[x]] + (2*a*(a*Cosh[x])^(3/2)*Sinh[x])/5

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 (a \cosh (x))^{5/2} \, dx &=\frac{2}{5} a (a \cosh (x))^{3/2} \sinh (x)+\frac{1}{5} \left (3 a^2\right ) \int \sqrt{a \cosh (x)} \, dx\\ &=\frac{2}{5} a (a \cosh (x))^{3/2} \sinh (x)+\frac{\left (3 a^2 \sqrt{a \cosh (x)}\right ) \int \sqrt{\cosh (x)} \, dx}{5 \sqrt{\cosh (x)}}\\ &=-\frac{6 i a^2 \sqrt{a \cosh (x)} E\left (\left .\frac{i x}{2}\right |2\right )}{5 \sqrt{\cosh (x)}}+\frac{2}{5} a (a \cosh (x))^{3/2} \sinh (x)\\ \end{align*}

Mathematica [A]  time = 0.0413884, size = 41, normalized size = 0.85 \[ \frac{2 (a \cosh (x))^{5/2} \left (\sinh (x) \cosh ^{\frac{3}{2}}(x)-3 i E\left (\left .\frac{i x}{2}\right |2\right )\right )}{5 \cosh ^{\frac{5}{2}}(x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a*Cosh[x])^(5/2)*((-3*I)*EllipticE[(I/2)*x, 2] + Cosh[x]^(3/2)*Sinh[x]))/(5*Cosh[x]^(5/2))

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Maple [B]  time = 0.043, size = 184, normalized size = 3.8 \begin{align*}{\frac{{a}^{3}}{5}\sqrt{a \left ( 2\, \left ( \cosh \left ( x/2 \right ) \right ) ^{2}-1 \right ) \left ( \sinh \left ({\frac{x}{2}} \right ) \right ) ^{2}} \left ( 16\,\cosh \left ( x/2 \right ) \left ( \sinh \left ( x/2 \right ) \right ) ^{6}+16\, \left ( \sinh \left ( x/2 \right ) \right ) ^{4}\cosh \left ( x/2 \right ) +3\,\sqrt{2}\sqrt{-2\, \left ( \sinh \left ( x/2 \right ) \right ) ^{2}-1}\sqrt{- \left ( \sinh \left ( x/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cosh \left ( x/2 \right ) \sqrt{2},1/2\,\sqrt{2} \right ) -6\,\sqrt{2}\sqrt{-2\, \left ( \sinh \left ( x/2 \right ) \right ) ^{2}-1}\sqrt{- \left ( \sinh \left ( x/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cosh \left ( x/2 \right ) \sqrt{2},1/2\,\sqrt{2} \right ) +4\, \left ( \sinh \left ( x/2 \right ) \right ) ^{2}\cosh \left ( x/2 \right ) \right ){\frac{1}{\sqrt{a \left ( 2\, \left ( \sinh \left ( x/2 \right ) \right ) ^{4}+ \left ( \sinh \left ({\frac{x}{2}} \right ) \right ) ^{2} \right ) }}} \left ( \sinh \left ({\frac{x}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{a \left ( 2\, \left ( \cosh \left ( x/2 \right ) \right ) ^{2}-1 \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(a*(2*cosh(1/2*x)^2-1)*sinh(1/2*x)^2)^(1/2)*a^3*(16*cosh(1/2*x)*sinh(1/2*x)^6+16*sinh(1/2*x)^4*cosh(1/2*x)
+3*2^(1/2)*(-2*sinh(1/2*x)^2-1)^(1/2)*(-sinh(1/2*x)^2)^(1/2)*EllipticF(cosh(1/2*x)*2^(1/2),1/2*2^(1/2))-6*2^(1
/2)*(-2*sinh(1/2*x)^2-1)^(1/2)*(-sinh(1/2*x)^2)^(1/2)*EllipticE(cosh(1/2*x)*2^(1/2),1/2*2^(1/2))+4*sinh(1/2*x)
^2*cosh(1/2*x))/(a*(2*sinh(1/2*x)^4+sinh(1/2*x)^2))^(1/2)/sinh(1/2*x)/(a*(2*cosh(1/2*x)^2-1))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \cosh \left (x\right )\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*cosh(x))^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{a \cosh \left (x\right )} a^{2} \cosh \left (x\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a*cosh(x))*a^2*cosh(x)^2, 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*cosh(x))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \cosh \left (x\right )\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*cosh(x))^(5/2), x)