3.274 \(\int \frac{\sqrt{1-\cos (c+d x)}}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=112 \[ -\frac{8 \sin (c+d x)}{15 d \sqrt{1-\cos (c+d x)} \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \sin (c+d x)}{5 d \sqrt{1-\cos (c+d x)} \cos ^{\frac{5}{2}}(c+d x)}+\frac{16 \sin (c+d x)}{15 d \sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}} \]

[Out]

(2*Sin[c + d*x])/(5*d*Sqrt[1 - Cos[c + d*x]]*Cos[c + d*x]^(5/2)) - (8*Sin[c + d*x])/(15*d*Sqrt[1 - Cos[c + d*x
]]*Cos[c + d*x]^(3/2)) + (16*Sin[c + d*x])/(15*d*Sqrt[1 - Cos[c + d*x]]*Sqrt[Cos[c + d*x]])

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Rubi [A]  time = 0.131729, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {2772, 2771} \[ -\frac{8 \sin (c+d x)}{15 d \sqrt{1-\cos (c+d x)} \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \sin (c+d x)}{5 d \sqrt{1-\cos (c+d x)} \cos ^{\frac{5}{2}}(c+d x)}+\frac{16 \sin (c+d x)}{15 d \sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - Cos[c + d*x]]/Cos[c + d*x]^(7/2),x]

[Out]

(2*Sin[c + d*x])/(5*d*Sqrt[1 - Cos[c + d*x]]*Cos[c + d*x]^(5/2)) - (8*Sin[c + d*x])/(15*d*Sqrt[1 - Cos[c + d*x
]]*Cos[c + d*x]^(3/2)) + (16*Sin[c + d*x])/(15*d*Sqrt[1 - Cos[c + d*x]]*Sqrt[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{1-\cos (c+d x)}}{\cos ^{\frac{7}{2}}(c+d x)} \, dx &=\frac{2 \sin (c+d x)}{5 d \sqrt{1-\cos (c+d x)} \cos ^{\frac{5}{2}}(c+d x)}-\frac{4}{5} \int \frac{\sqrt{1-\cos (c+d x)}}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 \sin (c+d x)}{5 d \sqrt{1-\cos (c+d x)} \cos ^{\frac{5}{2}}(c+d x)}-\frac{8 \sin (c+d x)}{15 d \sqrt{1-\cos (c+d x)} \cos ^{\frac{3}{2}}(c+d x)}+\frac{8}{15} \int \frac{\sqrt{1-\cos (c+d x)}}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 \sin (c+d x)}{5 d \sqrt{1-\cos (c+d x)} \cos ^{\frac{5}{2}}(c+d x)}-\frac{8 \sin (c+d x)}{15 d \sqrt{1-\cos (c+d x)} \cos ^{\frac{3}{2}}(c+d x)}+\frac{16 \sin (c+d x)}{15 d \sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.117131, size = 61, normalized size = 0.54 \[ \frac{2 \sqrt{1-\cos (c+d x)} \left (8 \cos ^2(c+d x)-4 \cos (c+d x)+3\right ) \cot \left (\frac{1}{2} (c+d x)\right )}{15 d \cos ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 - Cos[c + d*x]]/Cos[c + d*x]^(7/2),x]

[Out]

(2*Sqrt[1 - Cos[c + d*x]]*(3 - 4*Cos[c + d*x] + 8*Cos[c + d*x]^2)*Cot[(c + d*x)/2])/(15*d*Cos[c + d*x]^(5/2))

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Maple [A]  time = 0.293, size = 65, 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{2-2\,\cos \left ( dx+c \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((1-cos(d*x+c))^(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)*(2-2*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 = 1.56646, size = 282, normalized size = 2.52 \begin{align*} \frac{2 \,{\left (7 \, \sqrt{2} - \frac{17 \, \sqrt{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{25 \, \sqrt{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{15 \, \sqrt{2} \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((1-cos(d*x+c))^(1/2)/cos(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

2/15*(7*sqrt(2) - 17*sqrt(2)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 25*sqrt(2)*sin(d*x + c)^4/(cos(d*x + c) + 1
)^4 - 15*sqrt(2)*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 = 1.83888, size = 166, normalized size = 1.48 \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{-\cos \left (d x + c\right ) + 1}}{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((1-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(-cos(d*x + c) + 1)/(d*cos(d*x + c)^(5/2)*si
n(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((1-cos(d*x+c))**(1/2)/cos(d*x+c)**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 2.16339, size = 154, normalized size = 1.38 \begin{align*} -\frac{\sqrt{2}{\left (\sqrt{2}{\left (7 \, \sqrt{2} - 8\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right ) + 2 \,{\left (4 \, \sqrt{2} - \frac{15 \,{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2} + 20 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2} \sqrt{-\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1}}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )\right )}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*sqrt(2)*(sqrt(2)*(7*sqrt(2) - 8)*sgn(tan(1/2*d*x + 1/2*c)) + 2*(4*sqrt(2) - (15*(tan(1/2*d*x + 1/2*c)^2
- 1)^2 + 20*tan(1/2*d*x + 1/2*c)^2 - 8)/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*sqrt(-tan(1/2*d*x + 1/2*c)^2 + 1)))*sg
n(tan(1/2*d*x + 1/2*c)))/d