\(\int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+C \sec ^2(c+d x)) \, dx\) [1094]

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 = 35, antiderivative size = 154 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {4 a^2 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {8 a^2 (A+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a^2 (A-5 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {8 C \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}} \] Output:

4*a^2*(A-C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+8/3*a^2*(A+C)*InverseJ 
acobiAM(1/2*d*x+1/2*c,2^(1/2))/d+2/3*a^2*(A-5*C)*cos(d*x+c)^(1/2)*sin(d*x+ 
c)/d+2/3*C*(a+a*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(3/2)+8/3*C*(a^2+a^2 
*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 = 8.58 (sec) , antiderivative size = 1040, normalized size of antiderivative = 6.75 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

Integrate[Cos[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^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)*(-(((A - 2*C + A*Cos[2*c])*Csc[c]*Sec[c])/d) + (A*Cos[d*x]*Si 
n[c])/(3*d) + (A*Cos[c]*Sin[d*x])/(3*d) + (C*Sec[c]*Sec[c + d*x]^2*Sin[d*x 
])/(3*d) + (Sec[c]*Sec[c + d*x]*(C*Sin[c] + 6*C*Sin[d*x]))/(3*d)))/(A + 2* 
C + A*Cos[2*c + 2*d*x]) - (4*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*S 
ec[c + d*x])^2*(A + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - S 
in[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTa 
n[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]) - (4*C*Cos[c + d*x]^4*Csc[c]*Hypergeometri 
cPFQ[{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]]]*Sq 
rt[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]) - (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)*((Hypergeometr 
icPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[T 
an[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + Ar 
cTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*S 
qrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c...
 

Rubi [A] (verified)

Time = 1.18 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.08, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.486, Rules used = {3042, 4602, 3042, 3523, 27, 3042, 3454, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4602

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3523

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {3 \left (\frac {2}{3} \left (2 a^3 (A+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 a^3 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 a^3 (A-5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {8 C \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3120

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

Input:

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

Output:

(2*C*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((8*C 
*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + 3*((2*((6 
*a^3*(A - C)*EllipticE[(c + d*x)/2, 2])/d + (4*a^3*(A + C)*EllipticF[(c + 
d*x)/2, 2])/d))/3 + (2*a^3*(A - 5*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d 
)))/(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 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. \(650\) vs. \(2(143)=286\).

Time = 3.84 (sec) , antiderivative size = 651, normalized size of antiderivative = 4.23

method result size
default \(\frac {4 a^{2} \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (4 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-4 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+4 A \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-6 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-12 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+4 C \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+6 C \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) A -2 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+7 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) C -2 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 \left (4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(651\)

Input:

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

Output:

4/3*a^2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(4*sin(1 
/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)/sin(1/2*d*x+1/2*c)^3*(4*A*cos(1/ 
2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-4*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c 
)^4+4*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)*Elli 
pticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-6*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)^2-12*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c) 
^4+4*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)*Ellip 
ticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+6*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)^2+sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*A-2 
*A*(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))+3*A*(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))+7*sin(1/2*d*x+1 
/2*c)^2*cos(1/2*d*x+1/2*c)*C-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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*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)*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)/(2*co 
s(1/2*d*x+1/2*c)^2-1)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.42 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (2 i \, \sqrt {2} {\left (A + C\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 2 i \, \sqrt {2} {\left (A + C\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} {\left (A - C\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} {\left (A - C\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (A a^{2} \cos \left (d x + c\right )^{2} + 6 \, C a^{2} \cos \left (d x + c\right ) + C a^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{3 \, d \cos \left (d x + c\right )^{2}} \] Input:

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

Output:

-2/3*(2*I*sqrt(2)*(A + C)*a^2*cos(d*x + c)^2*weierstrassPInverse(-4, 0, co 
s(d*x + c) + I*sin(d*x + c)) - 2*I*sqrt(2)*(A + C)*a^2*cos(d*x + c)^2*weie 
rstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*I*sqrt(2)*(A - C 
)*a^2*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos 
(d*x + c) + I*sin(d*x + c))) + 3*I*sqrt(2)*(A - C)*a^2*cos(d*x + c)^2*weie 
rstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c 
))) - (A*a^2*cos(d*x + c)^2 + 6*C*a^2*cos(d*x + c) + C*a^2)*sqrt(cos(d*x + 
 c))*sin(d*x + c))/(d*cos(d*x + c)^2)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Mupad [B] (verification not implemented)

Time = 14.43 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.05 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,A\,a^2\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+6\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,C\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,C\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

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

Output:

(2*A*a^2*(cos(c + d*x)^(1/2)*sin(c + d*x) + 6*ellipticE(c/2 + (d*x)/2, 2) 
+ 4*ellipticF(c/2 + (d*x)/2, 2)))/(3*d) + (2*C*a^2*ellipticF(c/2 + (d*x)/2 
, 2))/d + (4*C*a^2*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^2*sin(c + d*x)*h 
ypergeom([-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))
 

Reduce [F]

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

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

Output:

a**2*(int(sqrt(cos(c + d*x))*cos(c + d*x)*sec(c + d*x)**4,x)*c + 2*int(sqr 
t(cos(c + d*x))*cos(c + d*x)*sec(c + d*x)**3,x)*c + int(sqrt(cos(c + d*x)) 
*cos(c + d*x)*sec(c + d*x)**2,x)*a + int(sqrt(cos(c + d*x))*cos(c + d*x)*s 
ec(c + d*x)**2,x)*c + 2*int(sqrt(cos(c + d*x))*cos(c + d*x)*sec(c + d*x),x 
)*a + int(sqrt(cos(c + d*x))*cos(c + d*x),x)*a)