\(\int \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2 \, dx\) [568]

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

Optimal result

Integrand size = 23, antiderivative size = 160 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {2 \left (9 a^2+7 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {20 a b \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {20 a b \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 \left (9 a^2+7 b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {4 a b \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {2 b^2 \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9 d} \] Output:

2/15*(9*a^2+7*b^2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+20/21*a*b*Inver 
seJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+20/21*a*b*cos(d*x+c)^(1/2)*sin(d*x+c)/ 
d+2/45*(9*a^2+7*b^2)*cos(d*x+c)^(3/2)*sin(d*x+c)/d+4/7*a*b*cos(d*x+c)^(5/2 
)*sin(d*x+c)/d+2/9*b^2*cos(d*x+c)^(7/2)*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 1.32 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.71 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {84 \left (9 a^2+7 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+600 a b \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\sqrt {\cos (c+d x)} \left (7 \left (36 a^2+43 b^2\right ) \cos (c+d x)+5 b (156 a+36 a \cos (2 (c+d x))+7 b \cos (3 (c+d x)))\right ) \sin (c+d x)}{630 d} \] Input:

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

Output:

(84*(9*a^2 + 7*b^2)*EllipticE[(c + d*x)/2, 2] + 600*a*b*EllipticF[(c + d*x 
)/2, 2] + Sqrt[Cos[c + d*x]]*(7*(36*a^2 + 43*b^2)*Cos[c + d*x] + 5*b*(156* 
a + 36*a*Cos[2*(c + d*x)] + 7*b*Cos[3*(c + d*x)]))*Sin[c + d*x])/(630*d)
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.99, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 3268, 3042, 3115, 3042, 3115, 3042, 3120, 3493, 3042, 3115, 3042, 3119}

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.

\(\displaystyle \int \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2 \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3268

\(\displaystyle \int \cos ^{\frac {5}{2}}(c+d x) \left (a^2+b^2 \cos ^2(c+d x)\right )dx+2 a b \int \cos ^{\frac {7}{2}}(c+d x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a^2+b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \int \sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}dx\)

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a^2+b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \left (\frac {5}{7} \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a^2+b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \left (\frac {5}{7} \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a^2+b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \left (\frac {2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}+\frac {5}{7} \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )\)

\(\Big \downarrow \) 3493

\(\displaystyle \frac {1}{9} \left (9 a^2+7 b^2\right ) \int \cos ^{\frac {5}{2}}(c+d x)dx+2 a b \left (\frac {2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}+\frac {5}{7} \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {2 b^2 \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (9 a^2+7 b^2\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx+2 a b \left (\frac {2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}+\frac {5}{7} \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {2 b^2 \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {1}{9} \left (9 a^2+7 b^2\right ) \left (\frac {3}{5} \int \sqrt {\cos (c+d x)}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+2 a b \left (\frac {2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}+\frac {5}{7} \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {2 b^2 \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (9 a^2+7 b^2\right ) \left (\frac {3}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+2 a b \left (\frac {2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}+\frac {5}{7} \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {2 b^2 \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}\)

\(\Big \downarrow \) 3119

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

Input:

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

Output:

(2*b^2*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(9*d) + ((9*a^2 + 7*b^2)*((6*Ellip 
ticE[(c + d*x)/2, 2])/(5*d) + (2*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d)))/ 
9 + 2*a*b*((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)
 

Defintions of rubi rules used

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 3268
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)])^2, x_Symbol] :> Simp[2*c*(d/b)   Int[(b*Sin[e + f*x])^(m + 1), x], x] 
+ Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c, d, 
e, f, m}, x]
 

rule 3493
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*( 
x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f 
*(m + 2))), x] + Simp[(A*(m + 2) + C*(m + 1))/(m + 2)   Int[(b*Sin[e + f*x] 
)^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(397\) vs. \(2(143)=286\).

Time = 24.64 (sec) , antiderivative size = 398, normalized size of antiderivative = 2.49

method result size
default \(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-1120 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{10} b^{2}+\left (1440 a b +2240 b^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-504 a^{2}-2160 a b -2072 b^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (504 a^{2}+1680 a b +952 b^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-126 a^{2}-480 a b -168 b^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+150 a b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-189 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}-147 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}\right )}{315 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(398\)
parts \(-\frac {2 a^{2} \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 (-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\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}-\frac {2 b^{2} \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 (160 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}-480 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}+616 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-432 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+160 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-21 \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 {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-24 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{45 \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}-\frac {4 a b \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 (48 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}-120 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+128 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-72 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+5 \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 )+16 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{21 \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}\) \(620\)

Input:

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

Output:

-2/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-1120*cos( 
1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^10*b^2+(1440*a*b+2240*b^2)*sin(1/2*d*x+1 
/2*c)^8*cos(1/2*d*x+1/2*c)+(-504*a^2-2160*a*b-2072*b^2)*sin(1/2*d*x+1/2*c) 
^6*cos(1/2*d*x+1/2*c)+(504*a^2+1680*a*b+952*b^2)*sin(1/2*d*x+1/2*c)^4*cos( 
1/2*d*x+1/2*c)+(-126*a^2-480*a*b-168*b^2)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x 
+1/2*c)+150*a*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1 
/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-189*(sin(1/2*d*x+1/2*c)^2)^(1/2) 
*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^ 
2-147*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipt 
icE(cos(1/2*d*x+1/2*c),2^(1/2))*b^2)/(-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)/(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 = 0.10 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.22 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {-150 i \, \sqrt {2} a b {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 150 i \, \sqrt {2} a b {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (35 \, b^{2} \cos \left (d x + c\right )^{3} + 90 \, a b \cos \left (d x + c\right )^{2} + 150 \, a b + 7 \, {\left (9 \, a^{2} + 7 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 21 \, \sqrt {2} {\left (-9 i \, a^{2} - 7 i \, b^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 \, \sqrt {2} {\left (9 i \, a^{2} + 7 i \, b^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{315 \, d} \] Input:

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

Output:

1/315*(-150*I*sqrt(2)*a*b*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin( 
d*x + c)) + 150*I*sqrt(2)*a*b*weierstrassPInverse(-4, 0, cos(d*x + c) - I* 
sin(d*x + c)) + 2*(35*b^2*cos(d*x + c)^3 + 90*a*b*cos(d*x + c)^2 + 150*a*b 
 + 7*(9*a^2 + 7*b^2)*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 21*sq 
rt(2)*(-9*I*a^2 - 7*I*b^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 
0, cos(d*x + c) + I*sin(d*x + c))) - 21*sqrt(2)*(9*I*a^2 + 7*I*b^2)*weiers 
trassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) 
))/d
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

integrate((b*cos(d*x + c) + a)^2*cos(d*x + c)^(5/2), x)
 

Giac [F]

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

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

Output:

integrate((b*cos(d*x + c) + a)^2*cos(d*x + c)^(5/2), x)
 

Mupad [B] (verification not implemented)

Time = 45.05 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.84 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2 \, dx=-\frac {2\,a^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,b^2\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{11\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {4\,a\,b\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

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

Output:

- (2*a^2*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c 
 + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2)) - (2*b^2*cos(c + d*x)^(11/2)*sin( 
c + d*x)*hypergeom([1/2, 11/4], 15/4, cos(c + d*x)^2))/(11*d*(sin(c + d*x) 
^2)^(1/2)) - (4*a*b*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 
13/4, cos(c + d*x)^2))/(9*d*(sin(c + d*x)^2)^(1/2))
 

Reduce [F]

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

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

Output:

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