\(\int \cos ^4(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx\) [126]

Optimal result

Integrand size = 33, antiderivative size = 203 \[ \int \cos ^4(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {5 \sqrt {a} (7 A+8 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{64 d}+\frac {5 a (7 A+8 B) \sin (c+d x)}{64 d \sqrt {a+a \sec (c+d x)}}+\frac {5 a (7 A+8 B) \cos (c+d x) \sin (c+d x)}{96 d \sqrt {a+a \sec (c+d x)}}+\frac {a (7 A+8 B) \cos ^2(c+d x) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {a A \cos ^3(c+d x) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}} \] Output:

5/64*a^(1/2)*(7*A+8*B)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d 
+5/64*a*(7*A+8*B)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+5/96*a*(7*A+8*B)*cos 
(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/24*a*(7*A+8*B)*cos(d*x+c)^2* 
sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/4*a*A*cos(d*x+c)^3*sin(d*x+c)/d/(a+a 
*sec(d*x+c))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.13 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.34 \[ \int \cos ^4(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {2 \left (B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},4,\frac {3}{2},1-\sec (c+d x)\right )+A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},5,\frac {3}{2},1-\sec (c+d x)\right )\right ) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{d} \] Input:

Integrate[Cos[c + d*x]^4*Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x]),x]
 

Output:

(2*(B*Hypergeometric2F1[1/2, 4, 3/2, 1 - Sec[c + d*x]] + A*Hypergeometric2 
F1[1/2, 5, 3/2, 1 - Sec[c + d*x]])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x 
)/2])/d
 

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.94, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4503, 3042, 4292, 3042, 4292, 3042, 4292, 3042, 4261, 216}

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.

\(\displaystyle \int \cos ^4(c+d x) \sqrt {a \sec (c+d x)+a} (A+B \sec (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\)

\(\Big \downarrow \) 4503

\(\displaystyle \frac {1}{8} (7 A+8 B) \int \cos ^3(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{8} (7 A+8 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4292

\(\displaystyle \frac {1}{8} (7 A+8 B) \left (\frac {5}{6} \int \cos ^2(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{8} (7 A+8 B) \left (\frac {5}{6} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4292

\(\displaystyle \frac {1}{8} (7 A+8 B) \left (\frac {5}{6} \left (\frac {3}{4} \int \cos (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{8} (7 A+8 B) \left (\frac {5}{6} \left (\frac {3}{4} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4292

\(\displaystyle \frac {1}{8} (7 A+8 B) \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{8} (7 A+8 B) \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {1}{8} (7 A+8 B) \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{8} (7 A+8 B) \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}\)

Input:

Int[Cos[c + d*x]^4*Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x]),x]
 

Output:

(a*A*Cos[c + d*x]^3*Sin[c + d*x])/(4*d*Sqrt[a + a*Sec[c + d*x]]) + ((7*A + 
 8*B)*((a*Cos[c + d*x]^2*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (5 
*((a*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]]) + (3*((Sqrt 
[a]*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (a*Sin[c 
+ d*x])/(d*Sqrt[a + a*Sec[c + d*x]])))/4))/6))/8
 

Defintions of rubi rules used

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4261

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(b/d) 
  Subst[Int[1/(a + x^2), x], x, b*(Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
 

rule 4292

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)], x_Symbol] :> Simp[a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n*Sqrt[a 
+ b*Csc[e + f*x]])), x] + Simp[a*((2*n + 1)/(2*b*d*n))   Int[Sqrt[a + b*Csc 
[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && 
 EqQ[a^2 - b^2, 0] && LtQ[n, -2^(-1)] && IntegerQ[2*n]
 

rule 4503

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*b^2*Co 
t[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp 
[(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n)   Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[ 
e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a 
*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && LtQ[n, 0]
 

Maple [A] (verified)

Time = 1.64 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.37

method	result	size
default	\(\frac {\left (\left (-105 \cos \left (d x +c \right )-105\right ) A \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )+\left (-120 \cos \left (d x +c \right )-120\right ) B \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (48 \cos \left (d x +c \right )^{3}+56 \cos \left (d x +c \right )^{2}+70 \cos \left (d x +c \right )+105\right ) A +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (64 \cos \left (d x +c \right )^{2}+80 \cos \left (d x +c \right )+120\right ) B \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{192 d \left (1+\cos \left (d x +c \right )\right )}\)	\(279\)

Input:

int(cos(d*x+c)^4*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x,method=_RETURNV 
ERBOSE)
 

Output:

1/192/d*((-105*cos(d*x+c)-105)*A*(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan 
h(2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(cot(d*x+c)^2-2*csc(d*x+c)*cot(d*x+c)+c 
sc(d*x+c)^2-1)^(1/2))+(-120*cos(d*x+c)-120)*B*(-cos(d*x+c)/(1+cos(d*x+c))) 
^(1/2)*arctanh(2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(cot(d*x+c)^2-2*csc(d*x+c) 
*cot(d*x+c)+csc(d*x+c)^2-1)^(1/2))+sin(d*x+c)*cos(d*x+c)*(48*cos(d*x+c)^3+ 
56*cos(d*x+c)^2+70*cos(d*x+c)+105)*A+sin(d*x+c)*cos(d*x+c)*(64*cos(d*x+c)^ 
2+80*cos(d*x+c)+120)*B)*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 380, normalized size of antiderivative = 1.87 \[ \int \cos ^4(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\left [\frac {15 \, {\left ({\left (7 \, A + 8 \, B\right )} \cos \left (d x + c\right ) + 7 \, A + 8 \, B\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (48 \, A \cos \left (d x + c\right )^{4} + 8 \, {\left (7 \, A + 8 \, B\right )} \cos \left (d x + c\right )^{3} + 10 \, {\left (7 \, A + 8 \, B\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (7 \, A + 8 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{384 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {15 \, {\left ({\left (7 \, A + 8 \, B\right )} \cos \left (d x + c\right ) + 7 \, A + 8 \, B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (48 \, A \cos \left (d x + c\right )^{4} + 8 \, {\left (7 \, A + 8 \, B\right )} \cos \left (d x + c\right )^{3} + 10 \, {\left (7 \, A + 8 \, B\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (7 \, A + 8 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{192 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \] Input:

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x, algorith 
m="fricas")
 

Output:

[1/384*(15*((7*A + 8*B)*cos(d*x + c) + 7*A + 8*B)*sqrt(-a)*log((2*a*cos(d* 
x + c)^2 - 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c) 
*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*(48*A*cos(d*x 
+ c)^4 + 8*(7*A + 8*B)*cos(d*x + c)^3 + 10*(7*A + 8*B)*cos(d*x + c)^2 + 15 
*(7*A + 8*B)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x 
 + c))/(d*cos(d*x + c) + d), -1/192*(15*((7*A + 8*B)*cos(d*x + c) + 7*A + 
8*B)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/( 
sqrt(a)*sin(d*x + c))) - (48*A*cos(d*x + c)^4 + 8*(7*A + 8*B)*cos(d*x + c) 
^3 + 10*(7*A + 8*B)*cos(d*x + c)^2 + 15*(7*A + 8*B)*cos(d*x + c))*sqrt((a* 
cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d)]
 

Sympy [F(-1)]

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

integrate(cos(d*x+c)**4*(a+a*sec(d*x+c))**(1/2)*(A+B*sec(d*x+c)),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 8561 vs. \(2 (179) = 358\).

Time = 0.61 (sec) , antiderivative size = 8561, normalized size of antiderivative = 42.17 \[ \int \cos ^4(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x, algorith 
m="maxima")
 

Output:

1/768*(8*(4*(cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + sin( 
2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + 2*cos(2/3*arctan2(sin 
(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)^(3/4)*(cos(3/2*arctan2(sin(2/3*arct 
an2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c) 
, cos(3*d*x + 3*c))) + 1))*sin(3*d*x + 3*c) - (cos(3*d*x + 3*c) - 1)*sin(3 
/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*a 
rctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)))*sqrt(a) + 6*(cos(2/3*ar 
ctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + sin(2/3*arctan2(sin(3*d*x + 
 3*c), cos(3*d*x + 3*c)))^2 + 2*cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d* 
x + 3*c))) + 1)^(1/4)*((sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c) 
)) + 5*sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))))*cos(1/2*arcta 
n2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(s 
in(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)) - (cos(2/3*arctan2(sin(3*d*x + 3 
*c), cos(3*d*x + 3*c))) + 3*cos(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 
3*c))) - 4)*sin(1/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 
3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)))*sqrt(a 
) + 15*sqrt(a)*(arctan2(-(cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3* 
c)))^2 + sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + 2*cos(2/ 
3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)^(1/4)*(cos(1/2*arctan2 
(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1080 vs. \(2 (179) = 358\).

Time = 0.71 (sec) , antiderivative size = 1080, normalized size of antiderivative = 5.32 \[ \int \cos ^4(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x, algorith 
m="giac")
 

Output:

-1/384*(15*(7*A*sqrt(-a)*sgn(cos(d*x + c)) + 8*B*sqrt(-a)*sgn(cos(d*x + c) 
))*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 
 + a))^2 - a*(2*sqrt(2) + 3))) - 15*(7*A*sqrt(-a)*sgn(cos(d*x + c)) + 8*B* 
sqrt(-a)*sgn(cos(d*x + c)))*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt( 
-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3))) - 4*sqrt(2)*(279*( 
sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^14*A* 
sqrt(-a)*a*sgn(cos(d*x + c)) - 504*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(- 
a*tan(1/2*d*x + 1/2*c)^2 + a))^14*B*sqrt(-a)*a*sgn(cos(d*x + c)) + 285*(sq 
rt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^12*A*sq 
rt(-a)*a^2*sgn(cos(d*x + c)) + 5976*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt( 
-a*tan(1/2*d*x + 1/2*c)^2 + a))^12*B*sqrt(-a)*a^2*sgn(cos(d*x + c)) - 4605 
*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10* 
A*sqrt(-a)*a^3*sgn(cos(d*x + c)) - 31320*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - 
sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*B*sqrt(-a)*a^3*sgn(cos(d*x + c)) + 
 37281*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a 
))^8*A*sqrt(-a)*a^4*sgn(cos(d*x + c)) + 90168*(sqrt(-a)*tan(1/2*d*x + 1/2* 
c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*B*sqrt(-a)*a^4*sgn(cos(d*x + c 
)) - 35643*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 
 + a))^6*A*sqrt(-a)*a^5*sgn(cos(d*x + c)) - 66024*(sqrt(-a)*tan(1/2*d*x + 
1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*B*sqrt(-a)*a^5*sgn(cos(...
 

Mupad [F(-1)]

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

int(cos(c + d*x)^4*(A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(1/2),x)
 

Output:

int(cos(c + d*x)^4*(A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(1/2), x)
 

Reduce [F]

\[ \int \cos ^4(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\sqrt {a}\, \left (\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4}d x \right ) a \right ) \] Input:

int(cos(d*x+c)^4*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x)
 

Output:

sqrt(a)*(int(sqrt(sec(c + d*x) + 1)*cos(c + d*x)**4*sec(c + d*x),x)*b + in 
t(sqrt(sec(c + d*x) + 1)*cos(c + d*x)**4,x)*a)