\(\int \frac {(a+a \sec (c+d x))^2 (A+C \sec ^2(c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) [1097]

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

Optimal result

Integrand size = 35, antiderivative size = 230 \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {16 a^2 (3 A+2 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {4 a^2 (7 A+5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 a^2 (21 A+19 C) \sin (c+d x)}{105 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4 a^2 (7 A+5 C) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {16 a^2 (3 A+2 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {8 C \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \cos ^{\frac {7}{2}}(c+d x)} \] Output:

-16/15*a^2*(3*A+2*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+4/21*a^2*(7*A 
+5*C)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+2/105*a^2*(21*A+19*C)*sin(d 
*x+c)/d/cos(d*x+c)^(5/2)+4/21*a^2*(7*A+5*C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+ 
16/15*a^2*(3*A+2*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)+2/9*C*(a+a*cos(d*x+c))^2 
*sin(d*x+c)/d/cos(d*x+c)^(9/2)+8/63*C*(a^2+a^2*cos(d*x+c))*sin(d*x+c)/d/co 
s(d*x+c)^(7/2)
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

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

Time = 11.68 (sec) , antiderivative size = 1137, normalized size of antiderivative = 4.94 \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(Cos[c + d*x]^(9/2)*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec 
[c + d*x]^2)*((8*(3*A + 2*C)*Csc[c]*Sec[c])/(15*d) + (C*Sec[c]*Sec[c + d*x 
]^5*Sin[d*x])/(9*d) + (Sec[c]*Sec[c + d*x]^4*(7*C*Sin[c] + 18*C*Sin[d*x])) 
/(63*d) + (2*Sec[c]*Sec[c + d*x]*(35*A*Sin[c] + 25*C*Sin[c] + 84*A*Sin[d*x 
] + 56*C*Sin[d*x]))/(105*d) + (Sec[c]*Sec[c + d*x]^3*(90*C*Sin[c] + 63*A*S 
in[d*x] + 112*C*Sin[d*x]))/(315*d) + (Sec[c]*Sec[c + d*x]^2*(63*A*Sin[c] + 
 112*C*Sin[c] + 210*A*Sin[d*x] + 150*C*Sin[d*x]))/(315*d)))/(A + 2*C + A*C 
os[2*c + 2*d*x]) - (2*A*Cos[c + d*x]^4*Csc[c]*HypergeometricPFQ[{1/4, 1/2} 
, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + 
d*x])^2*(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]]]])/(3*d*(A + 2*C + A*Cos[2*c + 2*d 
*x])*Sqrt[1 + Cot[c]^2]) - (10*C*Cos[c + d*x]^4*Csc[c]*HypergeometricPFQ[{ 
1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*(a + a 
*Sec[c + d*x])^2*(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 - Arc 
Tan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(21*d*(A + 2*C + A*Cos 
[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) + (4*A*Cos[c + d*x]^4*Csc[c]*Sec[c/2 + 
(d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2)*((HypergeometricP 
FQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[T...
 

Rubi [A] (verified)

Time = 1.44 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.543, Rules used = {3042, 4602, 3042, 3523, 27, 3042, 3454, 27, 3042, 3447, 3042, 3500, 3042, 3227, 3042, 3116, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4602

\(\displaystyle \int \frac {(a \cos (c+d x)+a)^2 \left (A \cos ^2(c+d x)+C\right )}{\cos ^{\frac {11}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A \sin \left (c+d x+\frac {\pi }{2}\right )^2+C\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{11/2}}dx\)

\(\Big \downarrow \) 3523

\(\displaystyle \frac {2 \int \frac {(\cos (c+d x) a+a)^2 (4 a C+3 a (3 A+C) \cos (c+d x))}{2 \cos ^{\frac {9}{2}}(c+d x)}dx}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(\cos (c+d x) a+a)^2 (4 a C+3 a (3 A+C) \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)}dx}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (4 a C+3 a (3 A+C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3454

\(\displaystyle \frac {\frac {2}{7} \int \frac {3 (\cos (c+d x) a+a) \left ((21 A+19 C) a^2+(21 A+11 C) \cos (c+d x) a^2\right )}{2 \cos ^{\frac {7}{2}}(c+d x)}dx+\frac {8 C \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3}{7} \int \frac {(\cos (c+d x) a+a) \left ((21 A+19 C) a^2+(21 A+11 C) \cos (c+d x) a^2\right )}{\cos ^{\frac {7}{2}}(c+d x)}dx+\frac {8 C \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{7} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((21 A+19 C) a^2+(21 A+11 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {8 C \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3447

\(\displaystyle \frac {\frac {3}{7} \int \frac {(21 A+11 C) \cos ^2(c+d x) a^3+(21 A+19 C) a^3+\left ((21 A+11 C) a^3+(21 A+19 C) a^3\right ) \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x)}dx+\frac {8 C \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{7} \int \frac {(21 A+11 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^3+(21 A+19 C) a^3+\left ((21 A+11 C) a^3+(21 A+19 C) a^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {8 C \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3500

\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \int \frac {15 (7 A+5 C) a^3+28 (3 A+2 C) \cos (c+d x) a^3}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a^3 (21 A+19 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {8 C \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \int \frac {15 (7 A+5 C) a^3+28 (3 A+2 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a^3 (21 A+19 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {8 C \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \left (15 a^3 (7 A+5 C) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)}dx+28 a^3 (3 A+2 C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx\right )+\frac {2 a^3 (21 A+19 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {8 C \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \left (15 a^3 (7 A+5 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+28 a^3 (3 A+2 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\right )+\frac {2 a^3 (21 A+19 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {8 C \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3116

\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \left (15 a^3 (7 A+5 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+28 a^3 (3 A+2 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )\right )+\frac {2 a^3 (21 A+19 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {8 C \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \left (15 a^3 (7 A+5 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+28 a^3 (3 A+2 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )+\frac {2 a^3 (21 A+19 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {8 C \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \left (15 a^3 (7 A+5 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+28 a^3 (3 A+2 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 a^3 (21 A+19 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {8 C \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {3}{7} \left (\frac {2 a^3 (21 A+19 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \left (15 a^3 (7 A+5 C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+28 a^3 (3 A+2 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )\right )+\frac {8 C \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

Input:

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

Output:

(2*C*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(9*d*Cos[c + d*x]^(9/2)) + ((8*C 
*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)) + (3*((2* 
a^3*(21*A + 19*C)*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (2*(15*a^3*(7*A 
 + 5*C)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sin[c + d*x])/(3*d*Cos[c 
 + d*x]^(3/2))) + 28*a^3*(3*A + 2*C)*((-2*EllipticE[(c + d*x)/2, 2])/d + ( 
2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]))))/5))/7)/(9*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 3116
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1))   I 
nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && 
 IntegerQ[2*n]
 

rule 3119
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3227
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 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 3500
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 
 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* 
(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A 
*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
 

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. \(1140\) vs. \(2(209)=418\).

Time = 8.99 (sec) , antiderivative size = 1141, normalized size of antiderivative = 4.96

method result size
default \(\text {Expression too large to display}\) \(1141\)

Input:

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

Output:

-8*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(1/2*A*(- 
1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2 
)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+1/4*A/sin(1/2*d*x+1/2*c)^2/(2*si 
n(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2) 
*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2))+1/4 
*C*(-1/144*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^ 
2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^5-7/180*cos(1/2*d*x+1/2*c)*(-2*sin(1/2 
*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^3-14/ 
15*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1)*s 
in(1/2*d*x+1/2*c)^2)^(1/2)+7/15*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(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)* 
EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-7/15*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(- 
2*cos(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)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/ 
2*c),2^(1/2))))+1/2*C*(-1/56*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+s 
in(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^4-5/42*cos(1/2*d*x+1 
/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.21 \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (15 i \, \sqrt {2} {\left (7 \, A + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 15 i \, \sqrt {2} {\left (7 \, A + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 84 i \, \sqrt {2} {\left (3 \, A + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 84 i \, \sqrt {2} {\left (3 \, A + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (168 \, {\left (3 \, A + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 30 \, {\left (7 \, A + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 7 \, {\left (9 \, A + 16 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 90 \, C a^{2} \cos \left (d x + c\right ) + 35 \, C a^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{315 \, d \cos \left (d x + c\right )^{5}} \] Input:

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

Output:

-2/315*(15*I*sqrt(2)*(7*A + 5*C)*a^2*cos(d*x + c)^5*weierstrassPInverse(-4 
, 0, cos(d*x + c) + I*sin(d*x + c)) - 15*I*sqrt(2)*(7*A + 5*C)*a^2*cos(d*x 
 + c)^5*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 84*I*s 
qrt(2)*(3*A + 2*C)*a^2*cos(d*x + c)^5*weierstrassZeta(-4, 0, weierstrassPI 
nverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 84*I*sqrt(2)*(3*A + 2*C)*a 
^2*cos(d*x + c)^5*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d* 
x + c) - I*sin(d*x + c))) - (168*(3*A + 2*C)*a^2*cos(d*x + c)^4 + 30*(7*A 
+ 5*C)*a^2*cos(d*x + c)^3 + 7*(9*A + 16*C)*a^2*cos(d*x + c)^2 + 90*C*a^2*c 
os(d*x + c) + 35*C*a^2)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5 
)
 

Sympy [F]

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

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

Output:

a**2*(Integral(A/cos(c + d*x)**(3/2), x) + Integral(2*A*sec(c + d*x)/cos(c 
 + d*x)**(3/2), x) + Integral(A*sec(c + d*x)**2/cos(c + d*x)**(3/2), x) + 
Integral(C*sec(c + d*x)**2/cos(c + d*x)**(3/2), x) + Integral(2*C*sec(c + 
d*x)**3/cos(c + d*x)**(3/2), x) + Integral(C*sec(c + d*x)**4/cos(c + d*x)* 
*(3/2), x))
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 15.63 (sec) , antiderivative size = 482, normalized size of antiderivative = 2.10 \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {4\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )\,\left (\frac {7\,A\,a^2\,\sin \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {4\,C\,a^2\,\sin \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {3\,C\,a^2\,\sin \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}\right )}{21\,d}-\frac {8\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {7}{4};\ {\cos \left (c+d\,x\right )}^2\right )\,\left (\frac {9\,A\,a^2\,\sin \left (c+d\,x\right )}{\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {16\,C\,a^2\,\sin \left (c+d\,x\right )}{\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {5\,C\,a^2\,\sin \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}\right )}{135\,d}+\frac {2\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )\,\left (\frac {81\,A\,a^2\,\sin \left (c+d\,x\right )}{\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {9\,A\,a^2\,\sin \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {64\,C\,a^2\,\sin \left (c+d\,x\right )}{\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {21\,C\,a^2\,\sin \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {5\,C\,a^2\,\sin \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{9/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}\right )}{45\,d}+\frac {16\,C\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {5}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{21\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}} \] Input:

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

Output:

(4*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2)*((7*A*a^2*sin(c + d*x))/(co 
s(c + d*x)^(3/2)*(1 - cos(c + d*x)^2)^(1/2)) + (4*C*a^2*sin(c + d*x))/(cos 
(c + d*x)^(3/2)*(1 - cos(c + d*x)^2)^(1/2)) + (3*C*a^2*sin(c + d*x))/(cos( 
c + d*x)^(7/2)*(1 - cos(c + d*x)^2)^(1/2))))/(21*d) - (8*hypergeom([-1/4, 
1/2], 7/4, cos(c + d*x)^2)*((9*A*a^2*sin(c + d*x))/(cos(c + d*x)^(1/2)*(1 
- cos(c + d*x)^2)^(1/2)) + (16*C*a^2*sin(c + d*x))/(cos(c + d*x)^(1/2)*(1 
- cos(c + d*x)^2)^(1/2)) + (5*C*a^2*sin(c + d*x))/(cos(c + d*x)^(5/2)*(1 - 
 cos(c + d*x)^2)^(1/2))))/(135*d) + (2*hypergeom([-1/4, 1/2], 3/4, cos(c + 
 d*x)^2)*((81*A*a^2*sin(c + d*x))/(cos(c + d*x)^(1/2)*(1 - cos(c + d*x)^2) 
^(1/2)) + (9*A*a^2*sin(c + d*x))/(cos(c + d*x)^(5/2)*(1 - cos(c + d*x)^2)^ 
(1/2)) + (64*C*a^2*sin(c + d*x))/(cos(c + d*x)^(1/2)*(1 - cos(c + d*x)^2)^ 
(1/2)) + (21*C*a^2*sin(c + d*x))/(cos(c + d*x)^(5/2)*(1 - cos(c + d*x)^2)^ 
(1/2)) + (5*C*a^2*sin(c + d*x))/(cos(c + d*x)^(9/2)*(1 - cos(c + d*x)^2)^( 
1/2))))/(45*d) + (16*C*a^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 5/4, cos(c 
+ d*x)^2))/(21*d*cos(c + d*x)^(3/2)*(1 - cos(c + d*x)^2)^(1/2))
 

Reduce [F]

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

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

Output:

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