3.15 \(\int (a \cosh (x))^{7/2} \, dx\)

Optimal. Leaf size=65 \[ -\frac {10 i a^4 \sqrt {\cosh (x)} F\left (\left .\frac {i x}{2}\right |2\right )}{21 \sqrt {a \cosh (x)}}+\frac {10}{21} a^3 \sinh (x) \sqrt {a \cosh (x)}+\frac {2}{7} a \sinh (x) (a \cosh (x))^{5/2} \]

[Out]

2/7*a*(a*cosh(x))^(5/2)*sinh(x)-10/21*I*a^4*(cosh(1/2*x)^2)^(1/2)/cosh(1/2*x)*EllipticF(I*sinh(1/2*x),2^(1/2))
*cosh(x)^(1/2)/(a*cosh(x))^(1/2)+10/21*a^3*sinh(x)*(a*cosh(x))^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2635, 2642, 2641} \[ \frac {10}{21} a^3 \sinh (x) \sqrt {a \cosh (x)}-\frac {10 i a^4 \sqrt {\cosh (x)} F\left (\left .\frac {i x}{2}\right |2\right )}{21 \sqrt {a \cosh (x)}}+\frac {2}{7} a \sinh (x) (a \cosh (x))^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cosh[x])^(7/2),x]

[Out]

(((-10*I)/21)*a^4*Sqrt[Cosh[x]]*EllipticF[(I/2)*x, 2])/Sqrt[a*Cosh[x]] + (10*a^3*Sqrt[a*Cosh[x]]*Sinh[x])/21 +
(2*a*(a*Cosh[x])^(5/2)*Sinh[x])/7

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 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rubi steps

\begin {align*} \int (a \cosh (x))^{7/2} \, dx &=\frac {2}{7} a (a \cosh (x))^{5/2} \sinh (x)+\frac {1}{7} \left (5 a^2\right ) \int (a \cosh (x))^{3/2} \, dx\\ &=\frac {10}{21} a^3 \sqrt {a \cosh (x)} \sinh (x)+\frac {2}{7} a (a \cosh (x))^{5/2} \sinh (x)+\frac {1}{21} \left (5 a^4\right ) \int \frac {1}{\sqrt {a \cosh (x)}} \, dx\\ &=\frac {10}{21} a^3 \sqrt {a \cosh (x)} \sinh (x)+\frac {2}{7} a (a \cosh (x))^{5/2} \sinh (x)+\frac {\left (5 a^4 \sqrt {\cosh (x)}\right ) \int \frac {1}{\sqrt {\cosh (x)}} \, dx}{21 \sqrt {a \cosh (x)}}\\ &=-\frac {10 i a^4 \sqrt {\cosh (x)} F\left (\left .\frac {i x}{2}\right |2\right )}{21 \sqrt {a \cosh (x)}}+\frac {10}{21} a^3 \sqrt {a \cosh (x)} \sinh (x)+\frac {2}{7} a (a \cosh (x))^{5/2} \sinh (x)\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 53, normalized size = 0.82 \[ \frac {a^3 \sqrt {a \cosh (x)} \left ((23 \sinh (x)+3 \sinh (3 x)) \sqrt {\cosh (x)}-20 i F\left (\left .\frac {i x}{2}\right |2\right )\right )}{42 \sqrt {\cosh (x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cosh[x])^(7/2),x]

[Out]

(a^3*Sqrt[a*Cosh[x]]*((-20*I)*EllipticF[(I/2)*x, 2] + Sqrt[Cosh[x]]*(23*Sinh[x] + 3*Sinh[3*x])))/(42*Sqrt[Cosh
[x]])

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {a \cosh \relax (x)} a^{3} \cosh \relax (x)^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cosh(x))^(7/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \cosh \relax (x)\right )^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cosh(x))^(7/2),x, algorithm="giac")

[Out]

integrate((a*cosh(x))^(7/2), x)

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maple [B]  time = 0.35, size = 145, normalized size = 2.23 \[ \frac {\sqrt {a \left (2 \left (\cosh ^{2}\left (\frac {x}{2}\right )\right )-1\right ) \left (\sinh ^{2}\left (\frac {x}{2}\right )\right )}\, a^{4} \left (96 \left (\cosh ^{9}\left (\frac {x}{2}\right )\right )-240 \left (\cosh ^{7}\left (\frac {x}{2}\right )\right )+256 \left (\cosh ^{5}\left (\frac {x}{2}\right )\right )+5 \sqrt {2}\, \sqrt {-2 \left (\cosh ^{2}\left (\frac {x}{2}\right )\right )+1}\, \sqrt {-\left (\sinh ^{2}\left (\frac {x}{2}\right )\right )}\, \EllipticF \left (\sqrt {2}\, \cosh \left (\frac {x}{2}\right ), \frac {\sqrt {2}}{2}\right )-144 \left (\cosh ^{3}\left (\frac {x}{2}\right )\right )+32 \cosh \left (\frac {x}{2}\right )\right )}{21 \sqrt {a \left (2 \left (\sinh ^{4}\left (\frac {x}{2}\right )\right )+\sinh ^{2}\left (\frac {x}{2}\right )\right )}\, \sinh \left (\frac {x}{2}\right ) \sqrt {a \left (2 \left (\cosh ^{2}\left (\frac {x}{2}\right )\right )-1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cosh(x))^(7/2),x)

[Out]

1/21*(a*(2*cosh(1/2*x)^2-1)*sinh(1/2*x)^2)^(1/2)*a^4*(96*cosh(1/2*x)^9-240*cosh(1/2*x)^7+256*cosh(1/2*x)^5+5*2
^(1/2)*(-2*cosh(1/2*x)^2+1)^(1/2)*(-sinh(1/2*x)^2)^(1/2)*EllipticF(2^(1/2)*cosh(1/2*x),1/2*2^(1/2))-144*cosh(1
/2*x)^3+32*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 = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \cosh \relax (x)\right )^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cosh(x))^(7/2),x, algorithm="maxima")

[Out]

integrate((a*cosh(x))^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\left (a\,\mathrm {cosh}\relax (x)\right )}^{7/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cosh(x))^(7/2),x)

[Out]

int((a*cosh(x))^(7/2), 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))**(7/2),x)

[Out]

Timed out

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