\(\int \frac {(a+a \cos (c+d x))^3 (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) [452]

Optimal result
Mathematica [C] (warning: unable to verify)
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 = 43, antiderivative size = 230 \[ \int \frac {(a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {4 a^3 (9 A+5 B-5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^3 (3 A+5 (B+C)) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}-\frac {4 a^3 (21 A+20 B+5 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A (a+a \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (6 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{15 a d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (33 A+35 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}} \] Output:

-4/5*a^3*(9*A+5*B-5*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+4/3*a^3*(3* 
A+5*B+5*C)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d-4/15*a^3*(21*A+20*B+5* 
C)*cos(d*x+c)^(1/2)*sin(d*x+c)/d+2/5*A*(a+a*cos(d*x+c))^3*sin(d*x+c)/d/cos 
(d*x+c)^(5/2)+2/15*(6*A+5*B)*(a^2+a^2*cos(d*x+c))^2*sin(d*x+c)/a/d/cos(d*x 
+c)^(3/2)+2/15*(33*A+35*B+15*C)*(a^3+a^3*cos(d*x+c))*sin(d*x+c)/d/cos(d*x+ 
c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

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

Time = 11.45 (sec) , antiderivative size = 1298, normalized size of antiderivative = 5.64 \[ \int \frac {(a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:

Integrate[((a + a*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)) 
/Cos[c + d*x]^(7/2),x]
 

Output:

Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(-1/40*((-3 
6*A - 25*B + 5*C + 5*B*Cos[2*c] + 15*C*Cos[2*c])*Csc[c]*Sec[c])/d + (C*Cos 
[d*x]*Sin[c])/(12*d) + (C*Cos[c]*Sin[d*x])/(12*d) + (A*Sec[c]*Sec[c + d*x] 
^3*Sin[d*x])/(20*d) + (Sec[c]*Sec[c + d*x]^2*(3*A*Sin[c] + 15*A*Sin[d*x] + 
 5*B*Sin[d*x]))/(60*d) + (Sec[c]*Sec[c + d*x]*(15*A*Sin[c] + 5*B*Sin[c] + 
54*A*Sin[d*x] + 45*B*Sin[d*x] + 15*C*Sin[d*x]))/(60*d)) - (A*(a + a*Cos[c 
+ d*x])^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot 
[c]]]^2]*Sec[c/2 + (d*x)/2]^6*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[Cot[c] 
]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(2*d*Sqrt[1 + Cot[c]^2]) - (5*B* 
(a + a*Cos[c + d*x])^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x 
 - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*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[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(6*d*Sqrt[1 + Cot[c 
]^2]) - (5*C*(a + a*Cos[c + d*x])^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, { 
5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*Sec[d*x - ArcTan[C 
ot[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]]]])/(6*d*S 
qrt[1 + Cot[c]^2]) + (9*A*(a + a*Cos[c + d*x])^3*Csc[c]*Sec[c/2 + (d*x)/2] 
^6*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2...
 

Rubi [A] (verified)

Time = 1.66 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.05, number of steps used = 19, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.442, Rules used = {3042, 3522, 27, 3042, 3454, 27, 3042, 3454, 27, 3042, 3447, 3042, 3502, 27, 3042, 3227, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3522

\(\displaystyle \frac {2 \int \frac {(\cos (c+d x) a+a)^3 (a (6 A+5 B)-a (3 A-5 C) \cos (c+d x))}{2 \cos ^{\frac {5}{2}}(c+d x)}dx}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(\cos (c+d x) a+a)^3 (a (6 A+5 B)-a (3 A-5 C) \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)}dx}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (a (6 A+5 B)-a (3 A-5 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3454

\(\displaystyle \frac {\frac {2}{3} \int \frac {(\cos (c+d x) a+a)^2 \left (a^2 (33 A+35 B+15 C)-3 a^2 (9 A+5 B-5 C) \cos (c+d x)\right )}{2 \cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 (6 A+5 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} \int \frac {(\cos (c+d x) a+a)^2 \left (a^2 (33 A+35 B+15 C)-3 a^2 (9 A+5 B-5 C) \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 (6 A+5 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a^2 (33 A+35 B+15 C)-3 a^2 (9 A+5 B-5 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 (6 A+5 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3454

\(\displaystyle \frac {\frac {1}{3} \left (2 \int \frac {3 (\cos (c+d x) a+a) \left (a^3 (12 A+15 B+10 C)-a^3 (21 A+20 B+5 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx+\frac {2 (33 A+35 B+15 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (6 A+5 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} \left (6 \int \frac {(\cos (c+d x) a+a) \left (a^3 (12 A+15 B+10 C)-a^3 (21 A+20 B+5 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx+\frac {2 (33 A+35 B+15 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (6 A+5 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (6 \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a^3 (12 A+15 B+10 C)-a^3 (21 A+20 B+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (33 A+35 B+15 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (6 A+5 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3447

\(\displaystyle \frac {\frac {1}{3} \left (6 \int \frac {-\left ((21 A+20 B+5 C) \cos ^2(c+d x) a^4\right )+(12 A+15 B+10 C) a^4+\left (a^4 (12 A+15 B+10 C)-a^4 (21 A+20 B+5 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {2 (33 A+35 B+15 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (6 A+5 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (6 \int \frac {-\left ((21 A+20 B+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^4\right )+(12 A+15 B+10 C) a^4+\left (a^4 (12 A+15 B+10 C)-a^4 (21 A+20 B+5 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (33 A+35 B+15 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (6 A+5 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {1}{3} \left (6 \left (\frac {2}{3} \int \frac {5 a^4 (3 A+5 (B+C))-3 a^4 (9 A+5 B-5 C) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx-\frac {2 a^4 (21 A+20 B+5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (33 A+35 B+15 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (6 A+5 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} \left (6 \left (\frac {1}{3} \int \frac {5 a^4 (3 A+5 (B+C))-3 a^4 (9 A+5 B-5 C) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx-\frac {2 a^4 (21 A+20 B+5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (33 A+35 B+15 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (6 A+5 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (6 \left (\frac {1}{3} \int \frac {5 a^4 (3 A+5 (B+C))-3 a^4 (9 A+5 B-5 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a^4 (21 A+20 B+5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (33 A+35 B+15 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (6 A+5 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {1}{3} \left (6 \left (\frac {1}{3} \left (5 a^4 (3 A+5 (B+C)) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 a^4 (9 A+5 B-5 C) \int \sqrt {\cos (c+d x)}dx\right )-\frac {2 a^4 (21 A+20 B+5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (33 A+35 B+15 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (6 A+5 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (6 \left (\frac {1}{3} \left (5 a^4 (3 A+5 (B+C)) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 a^4 (9 A+5 B-5 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-\frac {2 a^4 (21 A+20 B+5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (33 A+35 B+15 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (6 A+5 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {1}{3} \left (6 \left (\frac {1}{3} \left (5 a^4 (3 A+5 (B+C)) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a^4 (9 A+5 B-5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {2 a^4 (21 A+20 B+5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (33 A+35 B+15 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (6 A+5 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {1}{3} \left (\frac {2 (33 A+35 B+15 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d \sqrt {\cos (c+d x)}}+6 \left (\frac {1}{3} \left (\frac {10 a^4 (3 A+5 (B+C)) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 a^4 (9 A+5 B-5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {2 a^4 (21 A+20 B+5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {2 (6 A+5 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

Input:

Int[((a + a*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c 
 + d*x]^(7/2),x]
 

Output:

(2*A*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + ((2*( 
6*A + 5*B)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2 
)) + ((2*(33*A + 35*B + 15*C)*(a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/(d*Sq 
rt[Cos[c + d*x]]) + 6*(((-6*a^4*(9*A + 5*B - 5*C)*EllipticE[(c + d*x)/2, 2 
])/d + (10*a^4*(3*A + 5*(B + C))*EllipticF[(c + d*x)/2, 2])/d)/3 - (2*a^4* 
(21*A + 20*B + 5*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)))/3)/(5*a)
 

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 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 3447
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a 
 + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), 
 x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
 

rule 3454
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[(-b^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ 
e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + 
 a*d))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp 
[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B 
*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{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] 
&& GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 
])
 

rule 3502
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

rule 3522
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[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - 
d^2))), x] + Simp[1/(b*d*(n + 1)*(c^2 - d^2))   Int[(a + b*Sin[e + f*x])^m* 
(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C - B*d)*( 
a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2* 
(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, 
 x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ 
[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1146\) vs. \(2(214)=428\).

Time = 10.31 (sec) , antiderivative size = 1147, normalized size of antiderivative = 4.99

method result size
parts \(\text {Expression too large to display}\) \(1147\)
default \(\text {Expression too large to display}\) \(1328\)

Input:

int((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, 
method=_RETURNVERBOSE)
 

Output:

-2/3*(3*A*a^3+B*a^3)*(-2*(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))*sin(1/2*d*x+1/2*c)^2-2* 
sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si 
n(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))*((-1+2* 
cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4 
+sin(1/2*d*x+1/2*c)^2)^(1/2)/(-1+2*cos(1/2*d*x+1/2*c)^2)^(3/2)/sin(1/2*d*x 
+1/2*c)/d+2*(B*a^3+3*C*a^3)*((-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c 
)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)* 
EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))/(-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)/(-1+2*cos(1/2*d*x+1/2*c)^2)^(1/2)/d+2* 
(A*a^3+3*B*a^3+3*C*a^3)/d*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))-2*(3*A*a^ 
3+3*B*a^3+C*a^3)*(-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)* 
cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si 
n(1/2*d*x+1/2*c)^2-1)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2) 
^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-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)/(-1+2*cos(1/2*d*x+1/2*c)^2)^(1 
/2)/d-2/3*C*a^3*((-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*( 
4*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d 
*x+1/2*c)+(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*El 
lipticF(cos(1/2*d*x+1/2*c),2^(1/2)))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.18 \[ \int \frac {(a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (3 \, A + 5 \, B + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (3 \, A + 5 \, B + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (9 \, A + 5 \, B - 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (9 \, A + 5 \, B - 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (5 \, C a^{3} \cos \left (d x + c\right )^{3} + 3 \, {\left (18 \, A + 15 \, B + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 5 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + 3 \, A a^{3}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{15 \, d \cos \left (d x + c\right )^{3}} \] Input:

integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7 
/2),x, algorithm="fricas")
 

Output:

-2/15*(5*I*sqrt(2)*(3*A + 5*B + 5*C)*a^3*cos(d*x + c)^3*weierstrassPInvers 
e(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*(3*A + 5*B + 5*C)*a^ 
3*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) 
 + 3*I*sqrt(2)*(9*A + 5*B - 5*C)*a^3*cos(d*x + c)^3*weierstrassZeta(-4, 0, 
 weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*I*sqrt(2)* 
(9*A + 5*B - 5*C)*a^3*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPIn 
verse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (5*C*a^3*cos(d*x + c)^3 + 3 
*(18*A + 15*B + 5*C)*a^3*cos(d*x + c)^2 + 5*(3*A + B)*a^3*cos(d*x + c) + 3 
*A*a^3)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))**3*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)* 
*(7/2),x)
 

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \frac {(a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7 
/2),x, algorithm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^3/c 
os(d*x + c)^(7/2), x)
 

Giac [F]

\[ \int \frac {(a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7 
/2),x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^3/c 
os(d*x + c)^(7/2), x)
 

Mupad [B] (verification not implemented)

Time = 1.68 (sec) , antiderivative size = 408, normalized size of antiderivative = 1.77 \[ \int \frac {(a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2\,\left (B\,a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+3\,B\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{d}+\frac {C\,a^3\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,A\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,C\,a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,C\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,A\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,A\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,A\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {6\,B\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,B\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

int(((a + a*cos(c + d*x))^3*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c 
 + d*x)^(7/2),x)
 

Output:

(2*(B*a^3*ellipticE(c/2 + (d*x)/2, 2) + 3*B*a^3*ellipticF(c/2 + (d*x)/2, 2 
)))/d + (C*a^3*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + 
 (d*x)/2, 2))/3))/d + (2*A*a^3*ellipticF(c/2 + (d*x)/2, 2))/d + (6*C*a^3*e 
llipticE(c/2 + (d*x)/2, 2))/d + (6*C*a^3*ellipticF(c/2 + (d*x)/2, 2))/d + 
(6*A*a^3*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos( 
c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (2*A*a^3*sin(c + d*x)*hypergeom([ 
-3/4, 1/2], 1/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^( 
1/2)) + (2*A*a^3*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) 
)/(5*d*cos(c + d*x)^(5/2)*(sin(c + d*x)^2)^(1/2)) + (6*B*a^3*sin(c + d*x)* 
hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c 
+ d*x)^2)^(1/2)) + (2*B*a^3*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c 
 + d*x)^2))/(3*d*cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2)) + (2*C*a^3*sin 
(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/ 
2)*(sin(c + d*x)^2)^(1/2))
 

Reduce [F]

\[ \int \frac {(a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=a^{3} \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) a +3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) b +3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}}d x \right ) a +3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) b +3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a +3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) b +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) c +\left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) b +3 \left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) c +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )d x \right ) c \right ) \] Input:

int((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x)
 

Output:

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