\(\int \frac {\sqrt {a-a \cos (c+d x)}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) [267]

Optimal result

Integrand size = 26, antiderivative size = 79 \[ \int \frac {\sqrt {a-a \cos (c+d x)}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}-\frac {4 a \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2851, 2850} \[ \int \frac {\sqrt {a-a \cos (c+d x)}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}-\frac {4 a \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}} \]

[In]

Int[Sqrt[a - a*Cos[c + d*x]]/Cos[c + d*x]^(5/2),x]

[Out]

(2*a*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*Sqrt[a - a*Cos[c + d*x]]) - (4*a*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*
x]]*Sqrt[a - a*Cos[c + d*x]])

Rule 2850

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]

Rule 2851

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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}-\frac {2}{3} \int \frac {\sqrt {a-a \cos (c+d x)}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}-\frac {4 a \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.66 \[ \int \frac {\sqrt {a-a \cos (c+d x)}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 (-1+2 \cos (c+d x)) \sqrt {a-a \cos (c+d x)} \cot \left (\frac {1}{2} (c+d x)\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)} \]

[In]

Integrate[Sqrt[a - a*Cos[c + d*x]]/Cos[c + d*x]^(5/2),x]

[Out]

(-2*(-1 + 2*Cos[c + d*x])*Sqrt[a - a*Cos[c + d*x]]*Cot[(c + d*x)/2])/(3*d*Cos[c + d*x]^(3/2))

Maple [A] (verified)

Time = 5.58 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.65

method	result	size
default	\(-\frac {2 \csc \left (d x +c \right ) \sqrt {-a \left (\cos \left (d x +c \right )-1\right )}\, \left (-1+2 \left (\cos ^{2}\left (d x +c \right )\right )+\cos \left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{\frac {3}{2}}}\)	\(51\)

[In]

int((a-cos(d*x+c)*a)^(1/2)/cos(d*x+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/3/d*csc(d*x+c)*(-a*(cos(d*x+c)-1))^(1/2)*(-1+2*cos(d*x+c)^2+cos(d*x+c))/cos(d*x+c)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.66 \[ \int \frac {\sqrt {a-a \cos (c+d x)}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 \, \sqrt {-a \cos \left (d x + c\right ) + a} {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right ) - 1\right )}}{3 \, d \cos \left (d x + c\right )^{\frac {3}{2}} \sin \left (d x + c\right )} \]

[In]

integrate((a-a*cos(d*x+c))^(1/2)/cos(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

-2/3*sqrt(-a*cos(d*x + c) + a)*(2*cos(d*x + c)^2 + cos(d*x + c) - 1)/(d*cos(d*x + c)^(3/2)*sin(d*x + c))

Sympy [F]

\[ \int \frac {\sqrt {a-a \cos (c+d x)}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\sqrt {- a \left (\cos {\left (c + d x \right )} - 1\right )}}{\cos ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]

[In]

integrate((a-a*cos(d*x+c))**(1/2)/cos(d*x+c)**(5/2),x)

[Out]

Integral(sqrt(-a*(cos(c + d*x) - 1))/cos(c + d*x)**(5/2), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 174 vs. \(2 (67) = 134\).

Time = 0.31 (sec) , antiderivative size = 174, normalized size of antiderivative = 2.20 \[ \int \frac {\sqrt {a-a \cos (c+d x)}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (\sqrt {2} \sqrt {a} - \frac {4 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{2}}{3 \, d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (\frac {2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 1\right )}} \]

[In]

integrate((a-a*cos(d*x+c))^(1/2)/cos(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

-2/3*(sqrt(2)*sqrt(a) - 4*sqrt(2)*sqrt(a)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*sqrt(2)*sqrt(a)*sin(d*x + c)
^4/(cos(d*x + c) + 1)^4)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^2/(d*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^
(5/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^4/(co
s(d*x + c) + 1)^4 + 1))

Giac [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.14 \[ \int \frac {\sqrt {a-a \cos (c+d x)}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 \, \sqrt {2} {\left ({\left ({\left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 15\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 15\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 1\right )} \sqrt {a} \mathrm {sgn}\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{3 \, {\left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1\right )}^{\frac {3}{2}} d} \]

[In]

integrate((a-a*cos(d*x+c))^(1/2)/cos(d*x+c)^(5/2),x, algorithm="giac")

[Out]

2/3*sqrt(2)*(((tan(1/4*d*x + 1/4*c)^2 - 15)*tan(1/4*d*x + 1/4*c)^2 + 15)*tan(1/4*d*x + 1/4*c)^2 - 1)*sqrt(a)*s
gn(sin(1/2*d*x + 1/2*c))/((tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1)^(3/2)*d)

Mupad [B] (verification not implemented)

Time = 14.73 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.08 \[ \int \frac {\sqrt {a-a \cos (c+d x)}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {4\,\sqrt {-a\,\left (\cos \left (c+d\,x\right )-1\right )}\,\left (\sin \left (c+d\,x\right )-\sin \left (2\,c+2\,d\,x\right )+\sin \left (3\,c+3\,d\,x\right )\right )}{3\,d\,\sqrt {\cos \left (c+d\,x\right )}\,\left (3\,\cos \left (c+d\,x\right )-2\,\cos \left (2\,c+2\,d\,x\right )+\cos \left (3\,c+3\,d\,x\right )-2\right )} \]

[In]

int((a - a*cos(c + d*x))^(1/2)/cos(c + d*x)^(5/2),x)

[Out]

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