\(\int \sqrt {\cos (c+d x)} (a+a \sec (c+d x)) (A+C \sec ^2(c+d x)) \, dx\) [1087]

Optimal result

Integrand size = 33, antiderivative size = 95 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a (3 A+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a C \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a C \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \] Output:

2*a*(A-C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/3*a*(3*A+C)*InverseJac 
obiAM(1/2*d*x+1/2*c,2^(1/2))/d+2/3*a*C*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2*a*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 = 6.74 (sec) , antiderivative size = 817, normalized size of antiderivative = 8.60 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

a*(Sqrt[Cos[c + d*x]]*(1 + Cos[c + d*x])*Sec[c/2 + (d*x)/2]^2*(-1/2*((A - 
2*C + A*Cos[2*c])*Csc[c]*Sec[c])/d + (C*Sec[c]*Sec[c + d*x]^2*Sin[d*x])/(3 
*d) + (Sec[c]*Sec[c + d*x]*(C*Sin[c] + 3*C*Sin[d*x]))/(3*d)) - (A*(1 + Cos 
[c + d*x])*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Co 
t[c]]]^2]*Sec[c/2 + (d*x)/2]^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*Sqrt[1 + Cot[c]^2]) - (C*(1 
+ Cos[c + d*x])*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcT 
an[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^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*Sqrt[1 + Cot[c]^2]) - 
 (A*(1 + Cos[c + d*x])*Csc[c]*Sec[c/2 + (d*x)/2]^2*((HypergeometricPFQ[{-1 
/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*T 
an[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]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + T 
an[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Co 
s[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2) 
)/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(2*d) + (C*( 
1 + Cos[c + d*x])*Csc[c]*Sec[c/2 + (d*x)/2]^2*((HypergeometricPFQ[{-1/2, - 
1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan...
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.03, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3042, 4602, 3042, 3511, 27, 3042, 3500, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4602

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3511

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int \frac {3 a A \cos ^2(c+d x)+a (3 A+C) \cos (c+d x)+3 a C}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a C \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

(2*a*C*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((6*a*(A - C)*EllipticE[(c 
 + d*x)/2, 2])/d + (2*a*(3*A + C)*EllipticF[(c + d*x)/2, 2])/d + (6*a*C*Si 
n[c + d*x])/(d*Sqrt[Cos[c + d*x]]))/3
 

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 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 3511

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

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. \(409\) vs. \(2(90)=180\).

Time = 3.48 (sec) , antiderivative size = 410, normalized size of antiderivative = 4.32

method	result	size
default	\(-\frac {4 \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a \left (\frac {A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \left (\operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{2 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}+\frac {C \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{6 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{2}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )}{2}+\frac {C \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\right )}{2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(410\)

Input:

int(cos(d*x+c)^(1/2)*(a+a*sec(d*x+c))*(A+C*sec(d*x+c)^2),x,method=_RETURNV 
ERBOSE)
 

Output:

-4*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a*(1/2*A*(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*(-1/6*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)))+1/2*C/sin(1/2*d*x+1/2*c)^2/(2*sin(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)))/sin(1/2*d*x+1/2*c)/(2*cos( 
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 = 198, normalized size of antiderivative = 2.08 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {-i \, \sqrt {2} {\left (3 \, A + C\right )} a \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} {\left (3 \, A + C\right )} a \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (A - C\right )} a \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (A - C\right )} a \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (3 \, C a \cos \left (d x + c\right ) + C a\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{2}} \] Input:

integrate(cos(d*x+c)^(1/2)*(a+a*sec(d*x+c))*(A+C*sec(d*x+c)^2),x, algorith 
m="fricas")
 

Output:

1/3*(-I*sqrt(2)*(3*A + C)*a*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos( 
d*x + c) + I*sin(d*x + c)) + I*sqrt(2)*(3*A + C)*a*cos(d*x + c)^2*weierstr 
assPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*I*sqrt(2)*(A - C)*a* 
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*cos(d*x + c)^2*weierstrassZ 
eta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2* 
(3*C*a*cos(d*x + c) + C*a)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c 
)^2)
 

Sympy [F]

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

integrate(cos(d*x+c)**(1/2)*(a+a*sec(d*x+c))*(A+C*sec(d*x+c)**2),x)
 

Output:

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

Maxima [F]

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

integrate(cos(d*x+c)^(1/2)*(a+a*sec(d*x+c))*(A+C*sec(d*x+c)^2),x, algorith 
m="maxima")
 

Output:

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

Giac [F]

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

integrate(cos(d*x+c)^(1/2)*(a+a*sec(d*x+c))*(A+C*sec(d*x+c)^2),x, algorith 
m="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 13.87 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.29 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,A\,a\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,a\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,a\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,a\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

int(cos(c + d*x)^(1/2)*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x)),x)
 

Output:

(2*A*a*ellipticE(c/2 + (d*x)/2, 2))/d + (2*A*a*ellipticF(c/2 + (d*x)/2, 2) 
)/d + (2*C*a*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*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))
 

Reduce [F]

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

int(cos(d*x+c)^(1/2)*(a+a*sec(d*x+c))*(A+C*sec(d*x+c)^2),x)
 

Output:

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