\(\int \frac {\cos ^{\frac {11}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx\) [189]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [C] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 23, antiderivative size = 207 \[ \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {231 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {21 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{2 a^3 d}-\frac {21 \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a^3 d}+\frac {77 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{10 a^3 d}-\frac {\cos ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {4 \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 a d (a+a \cos (c+d x))^2}-\frac {63 \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )} \] Output:

231/10*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/d-21/2*InverseJacobiAM(1/ 
2*d*x+1/2*c,2^(1/2))/a^3/d-21/2*cos(d*x+c)^(1/2)*sin(d*x+c)/a^3/d+77/10*co 
s(d*x+c)^(3/2)*sin(d*x+c)/a^3/d-1/5*cos(d*x+c)^(9/2)*sin(d*x+c)/d/(a+a*cos 
(d*x+c))^3-4/5*cos(d*x+c)^(7/2)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^2-63/10*co 
s(d*x+c)^(5/2)*sin(d*x+c)/d/(a^3+a^3*cos(d*x+c))
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 3.36 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\sqrt {\cos (c+d x)} \csc (c+d x) \left (-\left ((614+2995 \cos (c+d x)-766 \cos (2 (c+d x))-1139 \cos (3 (c+d x))+290 \cos (4 (c+d x))+127 \cos (5 (c+d x))-10 \cos (6 (c+d x))+\cos (7 (c+d x))) \csc ^4(c+d x)\right )+1680 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}+7040 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {7}{2},\frac {7}{4},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )}{160 a^3 d} \] Input:

Integrate[Cos[c + d*x]^(11/2)/(a + a*Cos[c + d*x])^3,x]
 

Output:

(Sqrt[Cos[c + d*x]]*Csc[c + d*x]*(-((614 + 2995*Cos[c + d*x] - 766*Cos[2*( 
c + d*x)] - 1139*Cos[3*(c + d*x)] + 290*Cos[4*(c + d*x)] + 127*Cos[5*(c + 
d*x)] - 10*Cos[6*(c + d*x)] + Cos[7*(c + d*x)])*Csc[c + d*x]^4) + 1680*Hyp 
ergeometric2F1[1/4, 1/2, 5/4, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2] + 7040* 
Cos[c + d*x]*Hypergeometric2F1[3/4, 7/2, 7/4, Cos[c + d*x]^2]*Sqrt[Sin[c + 
 d*x]^2]))/(160*a^3*d)
 

Rubi [A] (verified)

Time = 1.15 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.09, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.696, Rules used = {3042, 3244, 27, 3042, 3456, 27, 3042, 3456, 27, 3042, 3227, 3042, 3115, 3042, 3119, 3120}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{(a \cos (c+d x)+a)^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{11/2}}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\)

\(\Big \downarrow \) 3244

\(\displaystyle -\frac {\int \frac {3 \cos ^{\frac {7}{2}}(c+d x) (3 a-5 a \cos (c+d x))}{2 (\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {\sin (c+d x) \cos ^{\frac {9}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {3 \int \frac {\cos ^{\frac {7}{2}}(c+d x) (3 a-5 a \cos (c+d x))}{(\cos (c+d x) a+a)^2}dx}{10 a^2}-\frac {\sin (c+d x) \cos ^{\frac {9}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {3 \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (3 a-5 a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{10 a^2}-\frac {\sin (c+d x) \cos ^{\frac {9}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3456

\(\displaystyle -\frac {3 \left (\frac {\int \frac {7 \cos ^{\frac {5}{2}}(c+d x) \left (4 a^2-5 a^2 \cos (c+d x)\right )}{\cos (c+d x) a+a}dx}{3 a^2}+\frac {8 a \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{10 a^2}-\frac {\sin (c+d x) \cos ^{\frac {9}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {3 \left (\frac {7 \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (4 a^2-5 a^2 \cos (c+d x)\right )}{\cos (c+d x) a+a}dx}{3 a^2}+\frac {8 a \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{10 a^2}-\frac {\sin (c+d x) \cos ^{\frac {9}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {3 \left (\frac {7 \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (4 a^2-5 a^2 \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {8 a \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{10 a^2}-\frac {\sin (c+d x) \cos ^{\frac {9}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3456

\(\displaystyle -\frac {3 \left (\frac {7 \left (\frac {\int \frac {5}{2} \cos ^{\frac {3}{2}}(c+d x) \left (9 a^3-11 a^3 \cos (c+d x)\right )dx}{a^2}+\frac {9 a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\right )}{3 a^2}+\frac {8 a \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{10 a^2}-\frac {\sin (c+d x) \cos ^{\frac {9}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {3 \left (\frac {7 \left (\frac {5 \int \cos ^{\frac {3}{2}}(c+d x) \left (9 a^3-11 a^3 \cos (c+d x)\right )dx}{2 a^2}+\frac {9 a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\right )}{3 a^2}+\frac {8 a \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{10 a^2}-\frac {\sin (c+d x) \cos ^{\frac {9}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {3 \left (\frac {7 \left (\frac {5 \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (9 a^3-11 a^3 \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{2 a^2}+\frac {9 a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\right )}{3 a^2}+\frac {8 a \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{10 a^2}-\frac {\sin (c+d x) \cos ^{\frac {9}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3227

\(\displaystyle -\frac {3 \left (\frac {7 \left (\frac {5 \left (9 a^3 \int \cos ^{\frac {3}{2}}(c+d x)dx-11 a^3 \int \cos ^{\frac {5}{2}}(c+d x)dx\right )}{2 a^2}+\frac {9 a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\right )}{3 a^2}+\frac {8 a \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{10 a^2}-\frac {\sin (c+d x) \cos ^{\frac {9}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {3 \left (\frac {7 \left (\frac {5 \left (9 a^3 \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx-11 a^3 \int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx\right )}{2 a^2}+\frac {9 a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\right )}{3 a^2}+\frac {8 a \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{10 a^2}-\frac {\sin (c+d x) \cos ^{\frac {9}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3115

\(\displaystyle -\frac {3 \left (\frac {7 \left (\frac {5 \left (9 a^3 \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-11 a^3 \left (\frac {3}{5} \int \sqrt {\cos (c+d x)}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\right )}{2 a^2}+\frac {9 a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\right )}{3 a^2}+\frac {8 a \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{10 a^2}-\frac {\sin (c+d x) \cos ^{\frac {9}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {3 \left (\frac {7 \left (\frac {5 \left (9 a^3 \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-11 a^3 \left (\frac {3}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\right )}{2 a^2}+\frac {9 a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\right )}{3 a^2}+\frac {8 a \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{10 a^2}-\frac {\sin (c+d x) \cos ^{\frac {9}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3119

\(\displaystyle -\frac {3 \left (\frac {7 \left (\frac {5 \left (9 a^3 \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-11 a^3 \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\right )}{2 a^2}+\frac {9 a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\right )}{3 a^2}+\frac {8 a \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{10 a^2}-\frac {\sin (c+d x) \cos ^{\frac {9}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3120

\(\displaystyle -\frac {3 \left (\frac {7 \left (\frac {9 a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}+\frac {5 \left (9 a^3 \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-11 a^3 \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\right )}{2 a^2}\right )}{3 a^2}+\frac {8 a \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{10 a^2}-\frac {\sin (c+d x) \cos ^{\frac {9}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

Input:

Int[Cos[c + d*x]^(11/2)/(a + a*Cos[c + d*x])^3,x]
 

Output:

-1/5*(Cos[c + d*x]^(9/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^3) - (3*((8 
*a*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) + (7*((9* 
a^2*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])) + (5*(9*a^3* 
((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x]) 
/(3*d)) - 11*a^3*((6*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*Cos[c + d*x]^(3 
/2)*Sin[c + d*x])/(5*d))))/(2*a^2)))/(3*a^2)))/(10*a^2)
 

Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3115
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3119
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3227
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3244
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e 
+ f*x])^m*((c + d*Sin[e + f*x])^(n - 1)/(a*f*(2*m + 1))), x] + Simp[1/(a*b* 
(2*m + 1))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)* 
Simp[b*(c^2*(m + 1) + d^2*(n - 1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) 
 + b*c*(m + n))*Sin[e + f*x], x], 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[m, -1] 
&& GtQ[n, 1] && (IntegersQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
 

rule 3456
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( 
a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1))   Int[(a + b*Sin[e + f*x])^(m 
+ 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + 
 b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre 
eQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] & 
& NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (In 
tegerQ[2*n] || EqQ[c, 0])
 
Maple [A] (verified)

Time = 11.91 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.43

method result size
default \(-\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (64 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}-288 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-76 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-210 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-462 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+530 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-248 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+19 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}{20 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(296\)

Input:

int(cos(d*x+c)^(11/2)/(a+a*cos(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

-1/20*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(64*cos(1/2* 
d*x+1/2*c)^12-288*cos(1/2*d*x+1/2*c)^10-76*cos(1/2*d*x+1/2*c)^8-210*(sin(1 
/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2 
*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5-462*(sin(1/2*d*x+1/2*c)^2)^(1/2) 
*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*cos(1/2*d*x+1/2*c)^5*EllipticE(cos(1/2* 
d*x+1/2*c),2^(1/2))+530*cos(1/2*d*x+1/2*c)^6-248*cos(1/2*d*x+1/2*c)^4+19*c 
os(1/2*d*x+1/2*c)^2-1)/a^3/cos(1/2*d*x+1/2*c)^5/(-2*sin(1/2*d*x+1/2*c)^4+s 
in(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^( 
1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.76 \[ \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {2 \, {\left (4 \, \cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{3} - 147 \, \cos \left (d x + c\right )^{2} - 238 \, \cos \left (d x + c\right ) - 105\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 105 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{3} - 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 3 i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 105 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{3} + 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 3 i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 231 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{3} - 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 3 i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 231 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{3} + 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 3 i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{20 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:

integrate(cos(d*x+c)^(11/2)/(a+a*cos(d*x+c))^3,x, algorithm="fricas")
 

Output:

1/20*(2*(4*cos(d*x + c)^4 - 8*cos(d*x + c)^3 - 147*cos(d*x + c)^2 - 238*co 
s(d*x + c) - 105)*sqrt(cos(d*x + c))*sin(d*x + c) - 105*(-I*sqrt(2)*cos(d* 
x + c)^3 - 3*I*sqrt(2)*cos(d*x + c)^2 - 3*I*sqrt(2)*cos(d*x + c) - I*sqrt( 
2))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 105*(I*sqr 
t(2)*cos(d*x + c)^3 + 3*I*sqrt(2)*cos(d*x + c)^2 + 3*I*sqrt(2)*cos(d*x + c 
) + I*sqrt(2))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 
 231*(-I*sqrt(2)*cos(d*x + c)^3 - 3*I*sqrt(2)*cos(d*x + c)^2 - 3*I*sqrt(2) 
*cos(d*x + c) - I*sqrt(2))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 
0, cos(d*x + c) + I*sin(d*x + c))) - 231*(I*sqrt(2)*cos(d*x + c)^3 + 3*I*s 
qrt(2)*cos(d*x + c)^2 + 3*I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*weierstrassZ 
eta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^ 
3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d 
)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**(11/2)/(a+a*cos(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\int { \frac {\cos \left (d x + c\right )^{\frac {11}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:

integrate(cos(d*x+c)^(11/2)/(a+a*cos(d*x+c))^3,x, algorithm="maxima")
 

Output:

integrate(cos(d*x + c)^(11/2)/(a*cos(d*x + c) + a)^3, x)
 

Giac [F]

\[ \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\int { \frac {\cos \left (d x + c\right )^{\frac {11}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:

integrate(cos(d*x+c)^(11/2)/(a+a*cos(d*x+c))^3,x, algorithm="giac")
 

Output:

integrate(cos(d*x + c)^(11/2)/(a*cos(d*x + c) + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{11/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^3} \,d x \] Input:

int(cos(c + d*x)^(11/2)/(a + a*cos(c + d*x))^3,x)
 

Output:

int(cos(c + d*x)^(11/2)/(a + a*cos(c + d*x))^3, x)
 

Reduce [F]

\[ \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{5}}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x}{a^{3}} \] Input:

int(cos(d*x+c)^(11/2)/(a+a*cos(d*x+c))^3,x)
 

Output:

int((sqrt(cos(c + d*x))*cos(c + d*x)**5)/(cos(c + d*x)**3 + 3*cos(c + d*x) 
**2 + 3*cos(c + d*x) + 1),x)/a**3