\(\int \frac {\sqrt {a-a \cos (c+d x)}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) [268]

Optimal result

Integrand size = 26, antiderivative size = 118 \[ \int \frac {\sqrt {a-a \cos (c+d x)}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}-\frac {8 a \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}+\frac {16 a \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}} \]

[Out]

2/5*a*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a-a*cos(d*x+c))^(1/2)-8/15*a*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a-a*cos(d*x+c
))^(1/2)+16/15*a*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a-a*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2851, 2850} \[ \int \frac {\sqrt {a-a \cos (c+d x)}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {8 a \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}+\frac {2 a \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}+\frac {16 a \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}} \]

[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 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)}{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] (verified)

Time = 0.40 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.53 \[ \int \frac {\sqrt {a-a \cos (c+d x)}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \sqrt {a-a \cos (c+d x)} (7-4 \cos (c+d x)+4 \cos (2 (c+d x))) \cot \left (\frac {1}{2} (c+d x)\right )}{15 d \cos ^{\frac {5}{2}}(c+d x)} \]

[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))

Maple [A] (verified)

Time = 5.45 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.53

method	result	size
default	\(\frac {2 \csc \left (d x +c \right ) \sqrt {-a \left (\cos \left (d x +c \right )-1\right )}\, \left (3+8 \left (\cos ^{3}\left (d x +c \right )\right )+4 \left (\cos ^{2}\left (d x +c \right )\right )-\cos \left (d x +c \right )\right )}{15 d \cos \left (d x +c \right )^{\frac {5}{2}}}\)	\(63\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.54 \[ \int \frac {\sqrt {a-a \cos (c+d x)}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \, {\left (8 \, \cos \left (d x + c\right )^{3} + 4 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 3\right )} \sqrt {-a \cos \left (d x + c\right ) + a}}{15 \, d \cos \left (d x + c\right )^{\frac {5}{2}} \sin \left (d x + c\right )} \]

[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))

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {a-a \cos (c+d x)}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (100) = 200\).

Time = 0.31 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.87 \[ \int \frac {\sqrt {a-a \cos (c+d x)}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \, {\left (7 \, \sqrt {2} \sqrt {a} - \frac {17 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {25 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {15 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{3}}{15 \, d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (\frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {\sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 1\right )}} \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.02 \[ \int \frac {\sqrt {a-a \cos (c+d x)}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 \, \sqrt {2} {\left ({\left ({\left ({\left ({\left (7 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 75\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 430\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 430\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 75\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 7\right )} \sqrt {a} \mathrm {sgn}\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{15 \, {\left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1\right )}^{\frac {5}{2}} d} \]

[In]

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

[Out]

-2/15*sqrt(2)*(((((7*tan(1/4*d*x + 1/4*c)^2 - 75)*tan(1/4*d*x + 1/4*c)^2 + 430)*tan(1/4*d*x + 1/4*c)^2 - 430)*
tan(1/4*d*x + 1/4*c)^2 + 75)*tan(1/4*d*x + 1/4*c)^2 - 7)*sqrt(a)*sgn(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)^(5/2)*d)

Mupad [B] (verification not implemented)

Time = 15.97 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.34 \[ \int \frac {\sqrt {a-a \cos (c+d x)}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {8\,\sqrt {2\,a\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\,\left (7\,\sin \left (c+d\,x\right )-4\,\sin \left (2\,c+2\,d\,x\right )+9\,\sin \left (3\,c+3\,d\,x\right )-2\,\sin \left (4\,c+4\,d\,x\right )+2\,\sin \left (5\,c+5\,d\,x\right )\right )}{15\,d\,\sqrt {1-2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\,\left (-16\,{\sin \left (c+d\,x\right )}^2-4\,{\sin \left (2\,c+2\,d\,x\right )}^2+20\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+10\,{\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}^2+2\,{\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}^2\right )} \]

[In]

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

[Out]

(8*(2*a*sin(c/2 + (d*x)/2)^2)^(1/2)*(7*sin(c + d*x) - 4*sin(2*c + 2*d*x) + 9*sin(3*c + 3*d*x) - 2*sin(4*c + 4*
d*x) + 2*sin(5*c + 5*d*x)))/(15*d*(1 - 2*sin(c/2 + (d*x)/2)^2)^(1/2)*(20*sin(c/2 + (d*x)/2)^2 - 4*sin(2*c + 2*
d*x)^2 + 10*sin((3*c)/2 + (3*d*x)/2)^2 + 2*sin((5*c)/2 + (5*d*x)/2)^2 - 16*sin(c + d*x)^2))