\(\int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [1213]

Optimal result

Integrand size = 43, antiderivative size = 267 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {56 a^4 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {8 a^4 (4 A+5 B+4 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {4 a^4 (A-25 B-41 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a (5 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (5 A+15 B+19 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}-\frac {4 (6 A+25 B+34 C) \sqrt {\cos (c+d x)} \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{15 d} \] Output:

56/5*a^4*(A-C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+8/3*a^4*(4*A+5*B+4* 
C)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+4/15*a^4*(A-25*B-41*C)*cos(d*x 
+c)^(1/2)*sin(d*x+c)/d+2/15*a*(5*B+8*C)*(a+a*cos(d*x+c))^3*sin(d*x+c)/d/co 
s(d*x+c)^(3/2)+2/5*C*(a+a*cos(d*x+c))^4*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/5* 
(5*A+15*B+19*C)*(a^2+a^2*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(1/2)-4/15* 
(6*A+25*B+34*C)*cos(d*x+c)^(1/2)*(a^4+a^4*cos(d*x+c))*sin(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

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

Time = 12.71 (sec) , antiderivative size = 1449, normalized size of antiderivative = 5.43 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

Integrate[Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + 
C*Sec[c + d*x]^2),x]
 

Output:

(Cos[c + d*x]^(13/2)*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Se 
c[c + d*x] + C*Sec[c + d*x]^2)*(-1/40*((23*A - 20*B - 61*C + 33*A*Cos[2*c] 
 + 20*B*Cos[2*c] + 5*C*Cos[2*c])*Csc[c]*Sec[c])/d + ((4*A + B)*Cos[d*x]*Si 
n[c])/(12*d) + (A*Cos[2*d*x]*Sin[2*c])/(40*d) + ((4*A + B)*Cos[c]*Sin[d*x] 
)/(12*d) + (C*Sec[c]*Sec[c + d*x]^3*Sin[d*x])/(20*d) + (Sec[c]*Sec[c + d*x 
]^2*(3*C*Sin[c] + 5*B*Sin[d*x] + 20*C*Sin[d*x]))/(60*d) + (Sec[c]*Sec[c + 
d*x]*(5*B*Sin[c] + 20*C*Sin[c] + 15*A*Sin[d*x] + 60*B*Sin[d*x] + 99*C*Sin[ 
d*x]))/(60*d) + (A*Cos[2*c]*Sin[2*d*x])/(40*d)))/(A + 2*C + 2*B*Cos[c + d* 
x] + A*Cos[2*c + 2*d*x]) - (4*A*Cos[c + d*x]^6*Csc[c]*HypergeometricPFQ[{1 
/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^8*(a + a* 
Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*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]]]])/(3*d*( 
A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (5* 
B*Cos[c + d*x]^6*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - Arc 
Tan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + 
 d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - Arc 
Tan[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 + 2*C + 2*B*Cos[c + d*x] + A 
*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (4*C*Cos[c + d*x]^6*Csc[c]*Hyp...
 

Rubi [A] (verified)

Time = 2.13 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.08, number of steps used = 23, number of rules used = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.535, Rules used = {3042, 4600, 3042, 3522, 27, 3042, 3454, 27, 3042, 3454, 27, 3042, 3455, 27, 3042, 3447, 3042, 3502, 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.

\(\displaystyle \int \cos ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (c+d x)^{5/2} (a \sec (c+d x)+a)^4 \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )dx\)

\(\Big \downarrow \) 4600

\(\displaystyle \int \frac {(a \cos (c+d x)+a)^4 \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right )}{\cos ^{\frac {7}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4 \left (A \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+C\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\)

\(\Big \downarrow \) 3522

\(\displaystyle \frac {2 \int \frac {(\cos (c+d x) a+a)^4 (a (5 B+8 C)+5 a (A-C) \cos (c+d x))}{2 \cos ^{\frac {5}{2}}(c+d x)}dx}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(\cos (c+d x) a+a)^4 (a (5 B+8 C)+5 a (A-C) \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)}dx}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^4 \left (a (5 B+8 C)+5 a (A-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3454

\(\displaystyle \frac {\frac {2}{3} \int \frac {(\cos (c+d x) a+a)^3 \left (3 (5 A+15 B+19 C) a^2+5 (3 A-5 B-11 C) \cos (c+d x) a^2\right )}{2 \cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} \int \frac {(\cos (c+d x) a+a)^3 \left (3 (5 A+15 B+19 C) a^2+5 (3 A-5 B-11 C) \cos (c+d x) a^2\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (3 (5 A+15 B+19 C) a^2+5 (3 A-5 B-11 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3454

\(\displaystyle \frac {\frac {1}{3} \left (2 \int \frac {5 (\cos (c+d x) a+a)^2 \left (a^3 (9 A+20 B+23 C)-a^3 (6 A+25 B+34 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} \left (10 \int \frac {(\cos (c+d x) a+a)^2 \left (a^3 (9 A+20 B+23 C)-a^3 (6 A+25 B+34 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (10 \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a^3 (9 A+20 B+23 C)-a^3 (6 A+25 B+34 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3455

\(\displaystyle \frac {\frac {1}{3} \left (10 \left (\frac {2}{5} \int \frac {3 (\cos (c+d x) a+a) \left ((13 A+25 B+27 C) a^4+(A-25 B-41 C) \cos (c+d x) a^4\right )}{2 \sqrt {\cos (c+d x)}}dx-\frac {2 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^5 \cos (c+d x)+a^5\right )}{5 d}\right )+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} \left (10 \left (\frac {3}{5} \int \frac {(\cos (c+d x) a+a) \left ((13 A+25 B+27 C) a^4+(A-25 B-41 C) \cos (c+d x) a^4\right )}{\sqrt {\cos (c+d x)}}dx-\frac {2 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^5 \cos (c+d x)+a^5\right )}{5 d}\right )+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (10 \left (\frac {3}{5} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((13 A+25 B+27 C) a^4+(A-25 B-41 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^4\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^5 \cos (c+d x)+a^5\right )}{5 d}\right )+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3447

\(\displaystyle \frac {\frac {1}{3} \left (10 \left (\frac {3}{5} \int \frac {(A-25 B-41 C) \cos ^2(c+d x) a^5+(13 A+25 B+27 C) a^5+\left ((A-25 B-41 C) a^5+(13 A+25 B+27 C) a^5\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx-\frac {2 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^5 \cos (c+d x)+a^5\right )}{5 d}\right )+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (10 \left (\frac {3}{5} \int \frac {(A-25 B-41 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^5+(13 A+25 B+27 C) a^5+\left ((A-25 B-41 C) a^5+(13 A+25 B+27 C) a^5\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^5 \cos (c+d x)+a^5\right )}{5 d}\right )+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {1}{3} \left (10 \left (\frac {3}{5} \left (\frac {2}{3} \int \frac {5 (4 A+5 B+4 C) a^5+21 (A-C) \cos (c+d x) a^5}{\sqrt {\cos (c+d x)}}dx+\frac {2 a^5 (A-25 B-41 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^5 \cos (c+d x)+a^5\right )}{5 d}\right )+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (10 \left (\frac {3}{5} \left (\frac {2}{3} \int \frac {5 (4 A+5 B+4 C) a^5+21 (A-C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^5}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^5 (A-25 B-41 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^5 \cos (c+d x)+a^5\right )}{5 d}\right )+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {1}{3} \left (10 \left (\frac {3}{5} \left (\frac {2}{3} \left (5 a^5 (4 A+5 B+4 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+21 a^5 (A-C) \int \sqrt {\cos (c+d x)}dx\right )+\frac {2 a^5 (A-25 B-41 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^5 \cos (c+d x)+a^5\right )}{5 d}\right )+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (10 \left (\frac {3}{5} \left (\frac {2}{3} \left (5 a^5 (4 A+5 B+4 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+21 a^5 (A-C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {2 a^5 (A-25 B-41 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^5 \cos (c+d x)+a^5\right )}{5 d}\right )+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {1}{3} \left (10 \left (\frac {3}{5} \left (\frac {2}{3} \left (5 a^5 (4 A+5 B+4 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {42 a^5 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 a^5 (A-25 B-41 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^5 \cos (c+d x)+a^5\right )}{5 d}\right )+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (10 \left (\frac {3}{5} \left (\frac {2 a^5 (A-25 B-41 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}+\frac {2}{3} \left (\frac {10 a^5 (4 A+5 B+4 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {42 a^5 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )-\frac {2 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^5 \cos (c+d x)+a^5\right )}{5 d}\right )+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

Input:

Int[Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[ 
c + d*x]^2),x]
 

Output:

(2*C*(a + a*Cos[c + d*x])^4*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + ((2*a 
^2*(5*B + 8*C)*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2 
)) + ((6*a^3*(5*A + 15*B + 19*C)*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(d*S 
qrt[Cos[c + d*x]]) + 10*((-2*(6*A + 25*B + 34*C)*Sqrt[Cos[c + d*x]]*(a^5 + 
 a^5*Cos[c + d*x])*Sin[c + d*x])/(5*d) + (3*((2*((42*a^5*(A - C)*EllipticE 
[(c + d*x)/2, 2])/d + (10*a^5*(4*A + 5*B + 4*C)*EllipticF[(c + d*x)/2, 2]) 
/d))/3 + (2*a^5*(A - 25*B - 41*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d))) 
/5))/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 3455

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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n 
 + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a + b*Sin[e + 
f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 
) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + 
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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])
 

rule 4600

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x 
_)])^(m_.)*((A_.) + (B_.)*sec[(e_.) + (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.) 
*(x_)]^2), x_Symbol] :> Simp[d^(m + 2)   Int[(b + a*Cos[e + f*x])^m*(d*Cos[ 
e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; Fr 
eeQ[{a, b, d, e, f, A, B, C, n}, x] &&  !IntegerQ[n] && IntegerQ[m]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1213\) vs. \(2(246)=492\).

Time = 498.95 (sec) , antiderivative size = 1214, normalized size of antiderivative = 4.55

method	result	size
default	\(\text {Expression too large to display}\)	\(1214\)

Input:

int(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, 
method=_RETURNVERBOSE)
 

Output:

-8/15*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^4/(8*sin 
(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/ 
2*d*x+1/2*c)^3*(-102*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+19*sin(1/2* 
d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*A+61*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2* 
c)*C-128*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+24*A*cos(1/2*d*x+1/2*c) 
*sin(1/2*d*x+1/2*c)^10+84*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2 
*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4 
-84*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/ 
2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+198*C*cos(1/2*d*x+1/2 
*c)*sin(1/2*d*x+1/2*c)^6-218*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-140 
*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+35*B*cos(1/2*d*x+1/2*c)*sin(1/2 
*d*x+1/2*c)^2-80*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2) 
^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-80*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)*EllipticF(cos(1 
/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-84*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)) 
*sin(1/2*d*x+1/2*c)^4+80*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1 
/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+ 
80*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)*Ellipti 
cF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+84*C*(2*sin(1/2*d*x...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.05 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (10 i \, \sqrt {2} {\left (4 \, A + 5 \, B + 4 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 10 i \, \sqrt {2} {\left (4 \, A + 5 \, B + 4 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 42 i \, \sqrt {2} {\left (A - C\right )} a^{4} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 42 i \, \sqrt {2} {\left (A - C\right )} a^{4} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (3 \, A a^{4} \cos \left (d x + c\right )^{4} + 5 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A + 20 \, B + 33 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 5 \, {\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) + 3 \, C a^{4}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{15 \, d \cos \left (d x + c\right )^{3}} \] Input:

integrate(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c) 
^2),x, algorithm="fricas")
 

Output:

-2/15*(10*I*sqrt(2)*(4*A + 5*B + 4*C)*a^4*cos(d*x + c)^3*weierstrassPInver 
se(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 10*I*sqrt(2)*(4*A + 5*B + 4*C)* 
a^4*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c 
)) - 42*I*sqrt(2)*(A - C)*a^4*cos(d*x + c)^3*weierstrassZeta(-4, 0, weiers 
trassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 42*I*sqrt(2)*(A - C 
)*a^4*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos 
(d*x + c) - I*sin(d*x + c))) - (3*A*a^4*cos(d*x + c)^4 + 5*(4*A + B)*a^4*c 
os(d*x + c)^3 + 3*(5*A + 20*B + 33*C)*a^4*cos(d*x + c)^2 + 5*(B + 4*C)*a^4 
*cos(d*x + c) + 3*C*a^4)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^ 
3)
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**(5/2)*(a+a*sec(d*x+c))**4*(A+B*sec(d*x+c)+C*sec(d*x+ 
c)**2),x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c) 
^2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c) 
^2),x, algorithm="giac")
 

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^4*c 
os(d*x + c)^(5/2), x)
 

Mupad [B] (verification not implemented)

Time = 16.85 (sec) , antiderivative size = 525, normalized size of antiderivative = 1.97 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

int(cos(c + d*x)^(5/2)*(a + a/cos(c + d*x))^4*(A + B/cos(c + d*x) + C/cos( 
c + d*x)^2),x)
 

Output:

(2*(12*B*a^4*ellipticE(c/2 + (d*x)/2, 2) + 19*B*a^4*ellipticF(c/2 + (d*x)/ 
2, 2) + B*a^4*cos(c + d*x)^(1/2)*sin(c + d*x)))/(3*d) + (2*(C*a^4*elliptic 
E(c/2 + (d*x)/2, 2) + 4*C*a^4*ellipticF(c/2 + (d*x)/2, 2)))/d + (2*((34*C* 
a^4*sin(c + d*x))/(cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (C*a^4*sin 
(c + d*x))/(cos(c + d*x)^(5/2)*(sin(c + d*x)^2)^(1/2)))*hypergeom([-1/4, 1 
/2], 3/4, cos(c + d*x)^2))/(5*d) + (4*A*a^4*((2*cos(c + d*x)^(1/2)*sin(c + 
 d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d + (12*A*a^4*ellipticE(c/2 
 + (d*x)/2, 2))/d + (8*A*a^4*ellipticF(c/2 + (d*x)/2, 2))/d + (2*A*a^4*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^4*cos(c + d*x)^(7/2)*sin(c + d*x)*hype 
rgeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2)) + (8 
*B*a^4*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^4*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)) + (8*C*a^4*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)) - (8*C*a^4*sin(c + d*x)*h 
ypergeom([-1/4, 1/2], 7/4, cos(c + d*x)^2))/(15*d*cos(c + d*x)^(1/2)*(sin( 
c + d*x)^2)^(1/2))
 

Reduce [F]

\[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a^{4} \left (\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{6}d x \right ) c +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{5}d x \right ) b +4 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{5}d x \right ) c +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{4}d x \right ) a +4 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{4}d x \right ) b +6 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{4}d x \right ) c +4 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3}d x \right ) a +6 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3}d x \right ) b +4 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3}d x \right ) c +6 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}d x \right ) a +4 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}d x \right ) b +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}d x \right ) c +4 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )d x \right ) b +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}d x \right ) a \right ) \] Input:

int(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
 

Output:

a**4*(int(sqrt(cos(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**6,x)*c + int(sq 
rt(cos(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**5,x)*b + 4*int(sqrt(cos(c + 
 d*x))*cos(c + d*x)**2*sec(c + d*x)**5,x)*c + int(sqrt(cos(c + d*x))*cos(c 
 + d*x)**2*sec(c + d*x)**4,x)*a + 4*int(sqrt(cos(c + d*x))*cos(c + d*x)**2 
*sec(c + d*x)**4,x)*b + 6*int(sqrt(cos(c + d*x))*cos(c + d*x)**2*sec(c + d 
*x)**4,x)*c + 4*int(sqrt(cos(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**3,x)* 
a + 6*int(sqrt(cos(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**3,x)*b + 4*int( 
sqrt(cos(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**3,x)*c + 6*int(sqrt(cos(c 
 + d*x))*cos(c + d*x)**2*sec(c + d*x)**2,x)*a + 4*int(sqrt(cos(c + d*x))*c 
os(c + d*x)**2*sec(c + d*x)**2,x)*b + int(sqrt(cos(c + d*x))*cos(c + d*x)* 
*2*sec(c + d*x)**2,x)*c + 4*int(sqrt(cos(c + d*x))*cos(c + d*x)**2*sec(c + 
 d*x),x)*a + int(sqrt(cos(c + d*x))*cos(c + d*x)**2*sec(c + d*x),x)*b + in 
t(sqrt(cos(c + d*x))*cos(c + d*x)**2,x)*a)