\(\int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3 (A+C \sec ^2(c+d x)) \, dx\) [1102]

Optimal result

Integrand size = 35, antiderivative size = 211 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {4 a^3 (9 A-5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^3 (3 A+5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {8 a^3 (3 A-10 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 C \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}+\frac {2 (3 A-35 C) \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d} \] Output:

4/5*a^3*(9*A-5*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+4/3*a^3*(3*A+5*C 
)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+8/15*a^3*(3*A-10*C)*cos(d*x+c)^ 
(1/2)*sin(d*x+c)/d+2/3*C*(a+a*cos(d*x+c))^3*sin(d*x+c)/d/cos(d*x+c)^(3/2)+ 
4*C*(a^2+a^2*cos(d*x+c))^2*sin(d*x+c)/a/d/cos(d*x+c)^(1/2)+2/15*(3*A-35*C) 
*cos(d*x+c)^(1/2)*(a^3+a^3*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 = 10.01 (sec) , antiderivative size = 1089, normalized size of antiderivative = 5.16 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

Integrate[Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2) 
,x]
 

Output:

(Cos[c + d*x]^(11/2)*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + C*Se 
c[c + d*x]^2)*(-1/20*((18*A - 25*C + 18*A*Cos[2*c] + 5*C*Cos[2*c])*Csc[c]* 
Sec[c])/d + (A*Cos[d*x]*Sin[c])/(2*d) + (A*Cos[2*d*x]*Sin[2*c])/(20*d) + ( 
A*Cos[c]*Sin[d*x])/(2*d) + (C*Sec[c]*Sec[c + d*x]^2*Sin[d*x])/(6*d) + (Sec 
[c]*Sec[c + d*x]*(C*Sin[c] + 9*C*Sin[d*x]))/(6*d) + (A*Cos[2*c]*Sin[2*d*x] 
)/(20*d)))/(A + 2*C + A*Cos[2*c + 2*d*x]) - (A*Cos[c + d*x]^5*Csc[c]*Hyper 
geometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d* 
x)/2]^6*(a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^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[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(A + 
2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (5*C*Cos[c + d*x]^5*Csc[c] 
*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 
 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2)*Sec[d*x - ArcT 
an[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 + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (9*A*Cos[c + d*x]^ 
5*Csc[c]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2 
)*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Si 
n[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[ 
1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sq...
 

Rubi [A] (verified)

Time = 1.53 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.05, number of steps used = 21, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 4602, 3042, 3523, 27, 3042, 3454, 27, 3042, 3455, 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.

\(\displaystyle \int \cos ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^3 \left (A+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (c+d x)^{5/2} (a \sec (c+d x)+a)^3 \left (A+C \sec (c+d x)^2\right )dx\)

\(\Big \downarrow \) 4602

\(\displaystyle \int \frac {(a \cos (c+d x)+a)^3 \left (A \cos ^2(c+d x)+C\right )}{\cos ^{\frac {5}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3523

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

\(\displaystyle \frac {2 \int \frac {(\cos (c+d x) a+a)^2 \left ((3 A+25 C) a^2+(3 A-35 C) \cos (c+d x) a^2\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(\cos (c+d x) a+a)^2 \left ((3 A+25 C) a^2+(3 A-35 C) \cos (c+d x) a^2\right )}{\sqrt {\cos (c+d x)}}dx+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left ((3 A+25 C) a^2+(3 A-35 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3455

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {6}{5} \left (\frac {2}{3} \int \frac {5 (3 A+5 C) a^4+3 (9 A-5 C) \cos (c+d x) a^4}{2 \sqrt {\cos (c+d x)}}dx+\frac {4 a^4 (3 A-10 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (3 A-35 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {6}{5} \left (\frac {1}{3} \int \frac {5 (3 A+5 C) a^4+3 (9 A-5 C) \cos (c+d x) a^4}{\sqrt {\cos (c+d x)}}dx+\frac {4 a^4 (3 A-10 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (3 A-35 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {6}{5} \left (\frac {1}{3} \int \frac {5 (3 A+5 C) a^4+3 (9 A-5 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^4}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {4 a^4 (3 A-10 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (3 A-35 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {6}{5} \left (\frac {1}{3} \left (5 a^4 (3 A+5 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 a^4 (9 A-5 C) \int \sqrt {\cos (c+d x)}dx\right )+\frac {4 a^4 (3 A-10 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (3 A-35 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {6}{5} \left (\frac {1}{3} \left (5 a^4 (3 A+5 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a^4 (9 A-5 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {4 a^4 (3 A-10 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (3 A-35 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {6}{5} \left (\frac {1}{3} \left (5 a^4 (3 A+5 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 a^4 (9 A-5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {4 a^4 (3 A-10 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (3 A-35 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {2 (3 A-35 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {6}{5} \left (\frac {4 a^4 (3 A-10 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}+\frac {1}{3} \left (\frac {10 a^4 (3 A+5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 a^4 (9 A-5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

Input:

Int[Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2),x]
 

Output:

(2*C*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((12* 
C*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + (2*(3* 
A - 35*C)*Sqrt[Cos[c + d*x]]*(a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/(5*d) 
+ (6*(((6*a^4*(9*A - 5*C)*EllipticE[(c + d*x)/2, 2])/d + (10*a^4*(3*A + 5* 
C)*EllipticF[(c + d*x)/2, 2])/d)/3 + (4*a^4*(3*A - 10*C)*Sqrt[Cos[c + d*x] 
]*Sin[c + d*x])/(3*d)))/5)/(3*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 3523

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
Simp[(-(c^2*C + 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*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(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, 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 4602

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x 
_)])^(m_.)*((A_.) + (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 + A*Cos 
[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] &&  !IntegerQ[n] 
 && IntegerQ[m]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(703\) vs. \(2(196)=392\).

Time = 158.31 (sec) , antiderivative size = 704, normalized size of antiderivative = 3.34

method	result	size
default	\(\text {Expression too large to display}\)	\(704\)

Input:

int(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x,method=_RETUR 
NVERBOSE)
 

Output:

-4/15*(24*A*(-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)^8-96*A*(-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)^6*cos(1/2*d*x+1/2*c)+6*(-2*sin(1/2*d*x+1 
/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(13*A+15*C)*cos(1/2*d*x+1/2*c)*sin(1/2 
*d*x+1/2*c)^4-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(9*A+ 
25*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-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)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1 
/2*c)^2)^(1/2)*(15*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-27*A*EllipticE( 
cos(1/2*d*x+1/2*c),2^(1/2))+25*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+15* 
C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*sin(1/2*d*x+1/2*c)^2+15*A*(-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)^(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))-27 
*A*(-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)^(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))+25*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) 
*EllipticF(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)+15*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)^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^3/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2* 
c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^(3/2)/sin(1/2*d*x+1/2*c)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.17 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} {\left (9 \, A - 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} {\left (9 \, A - 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (3 \, A a^{3} \cos \left (d x + c\right )^{3} + 15 \, A a^{3} \cos \left (d x + c\right )^{2} + 45 \, C a^{3} \cos \left (d x + c\right ) + 5 \, C a^{3}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{15 \, d \cos \left (d x + c\right )^{2}} \] Input:

integrate(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algori 
thm="fricas")
 

Output:

-2/15*(5*I*sqrt(2)*(3*A + 5*C)*a^3*cos(d*x + c)^2*weierstrassPInverse(-4, 
0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*(3*A + 5*C)*a^3*cos(d*x + 
c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*I*sqrt( 
2)*(9*A - 5*C)*a^3*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInver 
se(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*I*sqrt(2)*(9*A - 5*C)*a^3*co 
s(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c 
) - I*sin(d*x + c))) - (3*A*a^3*cos(d*x + c)^3 + 15*A*a^3*cos(d*x + c)^2 + 
 45*C*a^3*cos(d*x + c) + 5*C*a^3)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos( 
d*x + c)^2)
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**(5/2)*(a+a*sec(d*x+c))**3*(A+C*sec(d*x+c)**2),x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algori 
thm="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algori 
thm="giac")
 

Output:

integrate((C*sec(d*x + c)^2 + A)*(a*sec(d*x + c) + a)^3*cos(d*x + c)^(5/2) 
, x)
 

Mupad [B] (verification not implemented)

Time = 14.55 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.12 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,\left (C\,a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+3\,C\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{d}+\frac {6\,A\,a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,A\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,a^3\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{d}-\frac {2\,A\,a^3\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {6\,C\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

int(cos(c + d*x)^(5/2)*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^3,x)
 

Output:

(2*(C*a^3*ellipticE(c/2 + (d*x)/2, 2) + 3*C*a^3*ellipticF(c/2 + (d*x)/2, 2 
)))/d + (6*A*a^3*ellipticE(c/2 + (d*x)/2, 2))/d + (4*A*a^3*ellipticF(c/2 + 
 (d*x)/2, 2))/d + (2*A*a^3*cos(c + d*x)^(1/2)*sin(c + d*x))/d - (2*A*a^3*c 
os(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)) + (6*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)) + 
(2*C*a^3*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d*co 
s(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2))
 

Reduce [F]

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

int(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x)
 

Output:

a**3*(int(sqrt(cos(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**5,x)*c + 3*int( 
sqrt(cos(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**4,x)*c + int(sqrt(cos(c + 
 d*x))*cos(c + d*x)**2*sec(c + d*x)**3,x)*a + 3*int(sqrt(cos(c + d*x))*cos 
(c + d*x)**2*sec(c + d*x)**3,x)*c + 3*int(sqrt(cos(c + d*x))*cos(c + d*x)* 
*2*sec(c + d*x)**2,x)*a + int(sqrt(cos(c + d*x))*cos(c + d*x)**2*sec(c + d 
*x)**2,x)*c + 3*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,x)*a)