\(\int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\) [205]

Optimal result

Integrand size = 25, antiderivative size = 153 \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 a \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {12 a \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {16 a \sin (c+d x)}{35 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {32 a \sin (c+d x)}{35 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \]

[Out]

2/7*a*sin(d*x+c)/d/cos(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(1/2)+12/35*a*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a+a*cos(d*x+
c))^(1/2)+16/35*a*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2)+32/35*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 = 153, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2851, 2850} \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {16 a \sin (c+d x)}{35 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {12 a \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {2 a \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {32 a \sin (c+d x)}{35 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}} \]

[In]

Int[Sqrt[a + a*Cos[c + d*x]]/Cos[c + d*x]^(9/2),x]

[Out]

(2*a*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)*Sqrt[a + a*Cos[c + d*x]]) + (12*a*Sin[c + d*x])/(35*d*Cos[c + d*x]^
(5/2)*Sqrt[a + a*Cos[c + d*x]]) + (16*a*Sin[c + d*x])/(35*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + (32
*a*Sin[c + d*x])/(35*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)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {6}{7} \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 a \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {12 a \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {24}{35} \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 a \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {12 a \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {16 a \sin (c+d x)}{35 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {16}{35} \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 a \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {12 a \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {16 a \sin (c+d x)}{35 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {32 a \sin (c+d x)}{35 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.46 \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 \sqrt {a (1+\cos (c+d x))} (9+18 \cos (c+d x)+4 \cos (2 (c+d x))+4 \cos (3 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{35 d \cos ^{\frac {7}{2}}(c+d x)} \]

[In]

Integrate[Sqrt[a + a*Cos[c + d*x]]/Cos[c + d*x]^(9/2),x]

[Out]

(2*Sqrt[a*(1 + Cos[c + d*x])]*(9 + 18*Cos[c + d*x] + 4*Cos[2*(c + d*x)] + 4*Cos[3*(c + d*x)])*Tan[(c + d*x)/2]
)/(35*d*Cos[c + d*x]^(7/2))

Maple [A] (verified)

Time = 4.55 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.47

method	result	size
default	\(\frac {2 \sin \left (d x +c \right ) \left (16 \left (\cos ^{3}\left (d x +c \right )\right )+8 \left (\cos ^{2}\left (d x +c \right )\right )+6 \cos \left (d x +c \right )+5\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{35 d \left (1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {7}{2}}}\)	\(72\)

[In]

int((a+cos(d*x+c)*a)^(1/2)/cos(d*x+c)^(9/2),x,method=_RETURNVERBOSE)

[Out]

2/35/d*sin(d*x+c)*(16*cos(d*x+c)^3+8*cos(d*x+c)^2+6*cos(d*x+c)+5)*(a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))/cos(
d*x+c)^(7/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.53 \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 \, {\left (16 \, \cos \left (d x + c\right )^{3} + 8 \, \cos \left (d x + c\right )^{2} + 6 \, \cos \left (d x + c\right ) + 5\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{35 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]

[In]

integrate((a+a*cos(d*x+c))^(1/2)/cos(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

2/35*(16*cos(d*x + c)^3 + 8*cos(d*x + c)^2 + 6*cos(d*x + c) + 5)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*s
in(d*x + c)/(d*cos(d*x + c)^5 + d*cos(d*x + c)^4)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))**(1/2)/cos(d*x+c)**(9/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (129) = 258\).

Time = 0.37 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.85 \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 \, {\left (\frac {35 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {70 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {84 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {58 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {9 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{4}}{35 \, d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {9}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {9}{2}} {\left (\frac {4 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {\sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + 1\right )}} \]

[In]

integrate((a+a*cos(d*x+c))^(1/2)/cos(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

2/35*(35*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 70*sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1
)^3 + 84*sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 58*sqrt(2)*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c)
+ 1)^7 + 9*sqrt(2)*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^4/(
d*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(4*sin(d*x + c)^2/(
cos(d*x + c) + 1)^2 + 6*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + sin(d*x
+ c)^8/(cos(d*x + c) + 1)^8 + 1))

Giac [A] (verification not implemented)

none

Time = 0.55 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {4 \, \sqrt {2} {\left ({\left ({\left ({\left (7 \, {\left (5 \, {\left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 10\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 267\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 3684\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1869\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 350\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 35\right )} \sqrt {a} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )}{35 \, {\left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1\right )}^{\frac {7}{2}} d} \]

[In]

integrate((a+a*cos(d*x+c))^(1/2)/cos(d*x+c)^(9/2),x, algorithm="giac")

[Out]

4/35*sqrt(2)*((((7*(5*(tan(1/4*d*x + 1/4*c)^2 - 10)*tan(1/4*d*x + 1/4*c)^2 + 267)*tan(1/4*d*x + 1/4*c)^2 - 368
4)*tan(1/4*d*x + 1/4*c)^2 + 1869)*tan(1/4*d*x + 1/4*c)^2 - 350)*tan(1/4*d*x + 1/4*c)^2 + 35)*sqrt(a)*sgn(cos(1
/2*d*x + 1/2*c))*tan(1/4*d*x + 1/4*c)/((tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1)^(7/2)*d)

Mupad [B] (verification not implemented)

Time = 19.82 (sec) , antiderivative size = 415, normalized size of antiderivative = 2.71 \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {\sqrt {a+a\,\left (\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}\right )}\,\left (\frac {32{}\mathrm {i}}{35\,d}+\frac {{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,16{}\mathrm {i}}{5\,d}-\frac {{\mathrm {e}}^{c\,5{}\mathrm {i}+d\,x\,5{}\mathrm {i}}\,16{}\mathrm {i}}{5\,d}-\frac {{\mathrm {e}}^{c\,7{}\mathrm {i}+d\,x\,7{}\mathrm {i}}\,32{}\mathrm {i}}{35\,d}\right )}{\sqrt {\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}+{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}+3\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}+3\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}+3\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}+3\,{\mathrm {e}}^{c\,5{}\mathrm {i}+d\,x\,5{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}+{\mathrm {e}}^{c\,6{}\mathrm {i}+d\,x\,6{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}+{\mathrm {e}}^{c\,7{}\mathrm {i}+d\,x\,7{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}} \]

[In]

int((a + a*cos(c + d*x))^(1/2)/cos(c + d*x)^(9/2),x)

[Out]

((a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(32i/(35*d) + (exp(c*2i + d*x*2i)*16i)/(5*d) -
(exp(c*5i + d*x*5i)*16i)/(5*d) - (exp(c*7i + d*x*7i)*32i)/(35*d)))/((exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1
i)/2)^(1/2) + exp(c*1i + d*x*1i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2) + 3*exp(c*2i + d*x*2i)*
(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2) + 3*exp(c*3i + d*x*3i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1
i + d*x*1i)/2)^(1/2) + 3*exp(c*4i + d*x*4i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2) + 3*exp(c*5i
+ d*x*5i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2) + exp(c*6i + d*x*6i)*(exp(- c*1i - d*x*1i)/2
+ exp(c*1i + d*x*1i)/2)^(1/2) + exp(c*7i + d*x*7i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2))