\(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{a+a \cos (c+d x)} \, dx\) [456]

Optimal result

Integrand size = 43, antiderivative size = 210 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=-\frac {3 (5 A-7 B+7 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d}+\frac {5 (7 A-7 B+9 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 a d}+\frac {5 (7 A-7 B+9 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 a d}-\frac {(5 A-7 B+7 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 a d}+\frac {(7 A-7 B+9 C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 a d}-\frac {(A-B+C) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))} \] Output:

-3/5*(5*A-7*B+7*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a/d+5/21*(7*A-7*B 
+9*C)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a/d+5/21*(7*A-7*B+9*C)*cos(d* 
x+c)^(1/2)*sin(d*x+c)/a/d-1/5*(5*A-7*B+7*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/a/ 
d+1/7*(7*A-7*B+9*C)*cos(d*x+c)^(5/2)*sin(d*x+c)/a/d-(A-B+C)*cos(d*x+c)^(7/ 
2)*sin(d*x+c)/d/(a+a*cos(d*x+c))
 

Mathematica [C] (warning: unable to verify)

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

Time = 9.07 (sec) , antiderivative size = 1398, normalized size of antiderivative = 6.66 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx =\text {Too large to display} \] Input:

Integrate[(Cos[c + d*x]^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a 
+ a*Cos[c + d*x]),x]
 

Output:

(Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*((2*(5*A - 5*B + 5*C + 10*A*Cos[c 
] - 16*B*Cos[c] + 16*C*Cos[c])*Csc[c])/(5*d) + ((28*A - 28*B + 51*C)*Cos[d 
*x]*Sin[c])/(21*d) + (2*(B - C)*Cos[2*d*x]*Sin[2*c])/(5*d) + (C*Cos[3*d*x] 
*Sin[3*c])/(7*d) + (2*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/2] - B*Sin[ 
(d*x)/2] + C*Sin[(d*x)/2]))/d + ((28*A - 28*B + 51*C)*Cos[c]*Sin[d*x])/(21 
*d) + (2*(B - C)*Cos[2*c]*Sin[2*d*x])/(5*d) + (C*Cos[3*c]*Sin[3*d*x])/(7*d 
)))/(a + a*Cos[c + d*x]) - (5*A*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*Hypergeometr 
icPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - A 
rcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^ 
2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]]) 
/(3*d*(a + a*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]) + (5*B*Cos[c/2 + (d*x)/2]^2 
*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2 
]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*S 
qrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d 
*x - ArcTan[Cot[c]]]])/(3*d*(a + a*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]) - (15 
*C*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[ 
d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d 
*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Co 
t[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(7*d*(a + a*Cos[c + d*x])*Sq 
rt[1 + Cot[c]^2]) + (3*A*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*Sec[c/2]*((Hyper...
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.88, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.279, Rules used = {3042, 3520, 27, 3042, 3227, 3042, 3115, 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.

\(\displaystyle \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a \cos (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{a \sin \left (c+d x+\frac {\pi }{2}\right )+a}dx\)

\(\Big \downarrow \) 3520

\(\displaystyle \frac {\int -\frac {1}{2} \cos ^{\frac {5}{2}}(c+d x) (a (5 A-7 B+7 C)-a (7 A-7 B+9 C) \cos (c+d x))dx}{a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

\(\displaystyle -\frac {a (5 A-7 B+7 C) \left (\frac {3}{5} \int \sqrt {\cos (c+d x)}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )-a (7 A-7 B+9 C) \left (\frac {5}{7} \int \cos ^{\frac {3}{2}}(c+d x)dx+\frac {2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )}{2 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

Int[(Cos[c + d*x]^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + a*Co 
s[c + d*x]),x]
 

Output:

-(((A - B + C)*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]))) 
- (a*(5*A - 7*B + 7*C)*((6*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*Cos[c + d 
*x]^(3/2)*Sin[c + d*x])/(5*d)) - a*(7*A - 7*B + 9*C)*((2*Cos[c + d*x]^(5/2 
)*Sin[c + d*x])/(7*d) + (5*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sqrt[ 
Cos[c + d*x]]*Sin[c + d*x])/(3*d)))/7))/(2*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 3520

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*A - b*B + a*C)*Cos[e + f*x]*(a + b* 
Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x 
] + Simp[1/(b*(b*c - a*d)*(2*m + 1))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c 
+ d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) + B*(b*c*m + a 
*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(m + n + 2) + C*(b*c 
*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, 
d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c 
^2 - d^2, 0] && LtQ[m, -2^(-1)]
 

Maple [A] (verified)

Time = 12.35 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.62

method	result	size
default	\(-\frac {\sqrt {\left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (175 A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+315 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-175 B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-441 B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+225 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+441 C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-480 C \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (336 B +864 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-280 A -392 B -888 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (630 A -210 B +930 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-245 A +161 B -321 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{105 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(341\)

Input:

int(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c)),x,me 
thod=_RETURNVERBOSE)
 

Output:

-1/105*((-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(cos(1/2*d 
*x+1/2*c)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1 
75*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+315*A*EllipticE(cos(1/2*d*x+1/2 
*c),2^(1/2))-175*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-441*B*EllipticE(c 
os(1/2*d*x+1/2*c),2^(1/2))+225*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+441 
*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-480*C*sin(1/2*d*x+1/2*c)^10+(336 
*B+864*C)*sin(1/2*d*x+1/2*c)^8+(-280*A-392*B-888*C)*sin(1/2*d*x+1/2*c)^6+( 
630*A-210*B+930*C)*sin(1/2*d*x+1/2*c)^4+(-245*A+161*B-321*C)*sin(1/2*d*x+1 
/2*c)^2)/a/cos(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^ 
2)^(1/2)/sin(1/2*d*x+1/2*c)/(-1+2*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.50 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\frac {2 \, {\left (30 \, C \cos \left (d x + c\right )^{3} + 6 \, {\left (7 \, B - 2 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (35 \, A - 14 \, B + 39 \, C\right )} \cos \left (d x + c\right ) + 175 \, A - 175 \, B + 225 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 25 \, {\left (\sqrt {2} {\left (7 i \, A - 7 i \, B + 9 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (7 i \, A - 7 i \, B + 9 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 25 \, {\left (\sqrt {2} {\left (-7 i \, A + 7 i \, B - 9 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-7 i \, A + 7 i \, B - 9 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 63 \, {\left (\sqrt {2} {\left (5 i \, A - 7 i \, B + 7 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (5 i \, A - 7 i \, B + 7 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 63 \, {\left (\sqrt {2} {\left (-5 i \, A + 7 i \, B - 7 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-5 i \, A + 7 i \, B - 7 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{210 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:

integrate(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c) 
),x, algorithm="fricas")
 

Output:

1/210*(2*(30*C*cos(d*x + c)^3 + 6*(7*B - 2*C)*cos(d*x + c)^2 + 2*(35*A - 1 
4*B + 39*C)*cos(d*x + c) + 175*A - 175*B + 225*C)*sqrt(cos(d*x + c))*sin(d 
*x + c) - 25*(sqrt(2)*(7*I*A - 7*I*B + 9*I*C)*cos(d*x + c) + sqrt(2)*(7*I* 
A - 7*I*B + 9*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + 
c)) - 25*(sqrt(2)*(-7*I*A + 7*I*B - 9*I*C)*cos(d*x + c) + sqrt(2)*(-7*I*A 
+ 7*I*B - 9*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c) 
) - 63*(sqrt(2)*(5*I*A - 7*I*B + 7*I*C)*cos(d*x + c) + sqrt(2)*(5*I*A - 7* 
I*B + 7*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + 
c) + I*sin(d*x + c))) - 63*(sqrt(2)*(-5*I*A + 7*I*B - 7*I*C)*cos(d*x + c) 
+ sqrt(2)*(-5*I*A + 7*I*B - 7*I*C))*weierstrassZeta(-4, 0, weierstrassPInv 
erse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a*d*cos(d*x + c) + a*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+a*cos(d*x+ 
c)),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{a \cos \left (d x + c\right ) + a} \,d x } \] Input:

integrate(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c) 
),x, algorithm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)^(5/2)/(a*co 
s(d*x + c) + a), x)
 

Giac [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{a \cos \left (d x + c\right ) + a} \,d x } \] Input:

integrate(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c) 
),x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)^(5/2)/(a*co 
s(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{a+a\,\cos \left (c+d\,x\right )} \,d x \] Input:

int((cos(c + d*x)^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + a*co 
s(c + d*x)),x)
 

Output:

int((cos(c + d*x)^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + a*co 
s(c + d*x)), x)
 

Reduce [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\frac {\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{4}}{\cos \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{3}}{\cos \left (d x +c \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}}{\cos \left (d x +c \right )+1}d x \right ) a}{a} \] Input:

int(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c)),x)
 

Output:

(int((sqrt(cos(c + d*x))*cos(c + d*x)**4)/(cos(c + d*x) + 1),x)*c + int((s 
qrt(cos(c + d*x))*cos(c + d*x)**3)/(cos(c + d*x) + 1),x)*b + int((sqrt(cos 
(c + d*x))*cos(c + d*x)**2)/(cos(c + d*x) + 1),x)*a)/a