3.268 \(\int \frac{\sqrt{a-a \cos (c+d x)}}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=118 \[ -\frac{8 a \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a-a \cos (c+d x)}}+\frac{2 a \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a-a \cos (c+d x)}}+\frac{16 a \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a-a \cos (c+d x)}} \]

[Out]

(2*a*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)*Sqrt[a - a*Cos[c + d*x]]) - (8*a*Sin[c + d*x])/(15*d*Cos[c + d*x]^(
3/2)*Sqrt[a - a*Cos[c + d*x]]) + (16*a*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]*Sqrt[a - a*Cos[c + d*x]])

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Rubi [A]  time = 0.172602, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {2772, 2771} \[ -\frac{8 a \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a-a \cos (c+d x)}}+\frac{2 a \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a-a \cos (c+d x)}}+\frac{16 a \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a-a \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a - a*Cos[c + d*x]]/Cos[c + d*x]^(7/2),x]

[Out]

(2*a*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)*Sqrt[a - a*Cos[c + d*x]]) - (8*a*Sin[c + d*x])/(15*d*Cos[c + d*x]^(
3/2)*Sqrt[a - a*Cos[c + d*x]]) + (16*a*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]*Sqrt[a - a*Cos[c + d*x]])

Rule 2772

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[((b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x]
+ Dist[((2*n + 3)*(b*c - a*d))/(2*b*(n + 1)*(c^2 - d^2)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

Rule 2771

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Sim
p[(-2*b^2*Cos[e + f*x])/(f*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), x] /; FreeQ[{a, b,
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps

\begin{align*} \int \frac{\sqrt{a-a \cos (c+d x)}}{\cos ^{\frac{7}{2}}(c+d x)} \, dx &=\frac{2 a \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a-a \cos (c+d x)}}-\frac{4}{5} \int \frac{\sqrt{a-a \cos (c+d x)}}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a-a \cos (c+d x)}}-\frac{8 a \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a-a \cos (c+d x)}}+\frac{8}{15} \int \frac{\sqrt{a-a \cos (c+d x)}}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a-a \cos (c+d x)}}-\frac{8 a \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a-a \cos (c+d x)}}+\frac{16 a \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a-a \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.147521, size = 62, normalized size = 0.53 \[ \frac{2 (-4 \cos (c+d x)+4 \cos (2 (c+d x))+7) \cot \left (\frac{1}{2} (c+d x)\right ) \sqrt{a-a \cos (c+d x)}}{15 d \cos ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a - a*Cos[c + d*x]]/Cos[c + d*x]^(7/2),x]

[Out]

(2*Sqrt[a - a*Cos[c + d*x]]*(7 - 4*Cos[c + d*x] + 4*Cos[2*(c + d*x)])*Cot[(c + d*x)/2])/(15*d*Cos[c + d*x]^(5/
2))

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Maple [A]  time = 0.34, size = 66, normalized size = 0.6 \begin{align*} -{\frac{\sqrt{2} \left ( 8\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-4\,\cos \left ( dx+c \right ) +3 \right ) \sin \left ( dx+c \right ) }{15\,d \left ( -1+\cos \left ( dx+c \right ) \right ) }\sqrt{-2\,a \left ( -1+\cos \left ( dx+c \right ) \right ) } \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a-cos(d*x+c)*a)^(1/2)/cos(d*x+c)^(7/2),x)

[Out]

-1/15/d*2^(1/2)*(8*cos(d*x+c)^2-4*cos(d*x+c)+3)*(-2*a*(-1+cos(d*x+c)))^(1/2)*sin(d*x+c)/(-1+cos(d*x+c))/cos(d*
x+c)^(5/2)

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Maxima [B]  time = 1.54642, size = 298, normalized size = 2.53 \begin{align*} \frac{2 \,{\left (7 \, \sqrt{2} \sqrt{a} - \frac{17 \, \sqrt{2} \sqrt{a} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{25 \, \sqrt{2} \sqrt{a} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{15 \, \sqrt{2} \sqrt{a} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )}{\left (\frac{\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{3}}{15 \, d{\left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac{7}{2}}{\left (-\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac{7}{2}}{\left (\frac{3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{\sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*cos(d*x+c))^(1/2)/cos(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

2/15*(7*sqrt(2)*sqrt(a) - 17*sqrt(2)*sqrt(a)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 25*sqrt(2)*sqrt(a)*sin(d*x
+ c)^4/(cos(d*x + c) + 1)^4 - 15*sqrt(2)*sqrt(a)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6)*(sin(d*x + c)^2/(cos(d*x
+ c) + 1)^2 + 1)^3/(d*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2
)*(3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + sin(d*x + c)^6/(cos(d*x + c
) + 1)^6 + 1))

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Fricas [A]  time = 1.87124, size = 169, normalized size = 1.43 \begin{align*} \frac{2 \,{\left (8 \, \cos \left (d x + c\right )^{3} + 4 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 3\right )} \sqrt{-a \cos \left (d x + c\right ) + a}}{15 \, d \cos \left (d x + c\right )^{\frac{5}{2}} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*cos(d*x+c))^(1/2)/cos(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

2/15*(8*cos(d*x + c)^3 + 4*cos(d*x + c)^2 - cos(d*x + c) + 3)*sqrt(-a*cos(d*x + c) + a)/(d*cos(d*x + c)^(5/2)*
sin(d*x + c))

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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-a*cos(d*x+c))**(1/2)/cos(d*x+c)**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 2.20359, size = 219, normalized size = 1.86 \begin{align*} -\frac{\sqrt{2}{\left (2 \, a^{2}{\left (\frac{4 \, \sqrt{2}}{\sqrt{a}{\left | a \right |}} - \frac{15 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{2} + 20 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )} a + 12 \, a^{2}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}{\left | a \right |}}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right ) + \frac{\sqrt{2}{\left (7 \, \sqrt{2} a^{2} - 8 \, a^{2}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}{\sqrt{a}{\left | a \right |}}\right )}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*cos(d*x+c))^(1/2)/cos(d*x+c)^(7/2),x, algorithm="giac")

[Out]

-1/15*sqrt(2)*(2*a^2*(4*sqrt(2)/(sqrt(a)*abs(a)) - (15*(a*tan(1/2*d*x + 1/2*c)^2 - a)^2 + 20*(a*tan(1/2*d*x +
1/2*c)^2 - a)*a + 12*a^2)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*abs(a)))*sgn(t
an(1/2*d*x + 1/2*c)) + sqrt(2)*(7*sqrt(2)*a^2 - 8*a^2)*sgn(tan(1/2*d*x + 1/2*c))/(sqrt(a)*abs(a)))/d