\(\int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx\) [495]

Optimal result

Integrand size = 35, antiderivative size = 130 \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {4 a (4 A+5 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (4 A+5 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}} \] Output:

4/15*a*(4*A+5*B)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/15 
*a*(4*A+5*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/5*a*A* 
sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.60 \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {2 \sqrt {a (1+\cos (c+d x))} (7 A+5 B+(4 A+5 B) \cos (c+d x)+(4 A+5 B) \cos (2 (c+d x))) \sec ^{\frac {5}{2}}(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )}{15 d} \] Input:

Integrate[Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x])*Sec[c + d*x]^(7/2) 
,x]
 

Output:

(2*Sqrt[a*(1 + Cos[c + d*x])]*(7*A + 5*B + (4*A + 5*B)*Cos[c + d*x] + (4*A 
 + 5*B)*Cos[2*(c + d*x)])*Sec[c + d*x]^(5/2)*Tan[(c + d*x)/2])/(15*d)
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {3042, 3440, 3042, 3459, 3042, 3251, 3042, 3250}

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.

\(\displaystyle \int \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a} (A+B \cos (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{7/2} \sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\)

\(\Big \downarrow \) 3440

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\cos (c+d x) a+a} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3459

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3251

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3250

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

Input:

Int[Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x])*Sec[c + d*x]^(7/2),x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*a*A*Sin[c + d*x])/(5*d*Cos[c + d 
*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) + ((4*A + 5*B)*((2*a*Sin[c + d*x])/(3* 
d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + (4*a*Sin[c + d*x])/(3*d*S 
qrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])))/5)
 

Defintions of rubi rules used

rule 3042

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

rule 3250

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(3/2), x_Symbol] :> Simp[-2*b^2*(Cos[e + f*x]/(f*(b*c + a*d)*Sq 
rt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])), x] /; FreeQ[{a, b, c, d, 
 e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3251

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((c + d*Sin[e 
+ f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] + Sim 
p[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2)))   Int[Sqrt[a + b*Sin[e 
+ f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x 
] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, 
-1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]
 

rule 3440

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* 
(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim 
p[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p   Int[(a + b*Sin[e + f*x])^m*((c + 
d*Sin[e + f*x])^n/(g*Sin[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g 
, m, n, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && I 
ntegerQ[n])
 

rule 3459

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( 
f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1) 
*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b* 
c - 2*a*d*(n + 1)))/(2*d*(n + 1)*(b*c + a*d))   Int[Sqrt[a + b*Sin[e + f*x] 
]*(c + d*Sin[e + f*x])^(n + 1), 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] && LtQ[n, 
-1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(287\) vs. \(2(112)=224\).

Time = 16.44 (sec) , antiderivative size = 288, normalized size of antiderivative = 2.22

method	result	size
default	\(\sqrt {2}\, \left (\frac {2 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (32 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-24 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+7\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}}{15 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}-\frac {2 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (32 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-24 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+7\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}}{15 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}+\frac {4 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (12 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-9 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}}{5 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}\right )\)	\(288\)
parts	\(\frac {2 A \sqrt {2}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (32 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-24 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+7\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}}{15 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}+B \sqrt {2}\, \left (-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (32 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-24 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+7\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}}{15 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (12 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-9 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}}{5 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}\right )\)	\(291\)

Input:

int((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x,method=_RET 
URNVERBOSE)
 

Output:

2^(1/2)*(2/15*A/d*tan(1/2*d*x+1/2*c)*(32*cos(1/2*d*x+1/2*c)^4-24*cos(1/2*d 
*x+1/2*c)^2+7)*(a*cos(1/2*d*x+1/2*c)^2)^(1/2)*(1/(2*cos(1/2*d*x+1/2*c)^2-1 
))^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^2-2/15*B/d*tan(1/2*d*x+1/2*c)*(32*cos( 
1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)^2+7)*(a*cos(1/2*d*x+1/2*c)^2)^(1/2) 
*(1/(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^2+4/5*B/d 
*tan(1/2*d*x+1/2*c)*(12*cos(1/2*d*x+1/2*c)^4-9*cos(1/2*d*x+1/2*c)^2+2)*(a* 
cos(1/2*d*x+1/2*c)^2)^(1/2)*(1/(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/(2*cos(1/ 
2*d*x+1/2*c)^2-1)^2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.66 \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {2 \, {\left (2 \, {\left (4 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (4 \, A + 5 \, B\right )} \cos \left (d x + c\right ) + 3 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )} \sqrt {\cos \left (d x + c\right )}} \] Input:

integrate((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x, algo 
rithm="fricas")
 

Output:

2/15*(2*(4*A + 5*B)*cos(d*x + c)^2 + (4*A + 5*B)*cos(d*x + c) + 3*A)*sqrt( 
a*cos(d*x + c) + a)*sin(d*x + c)/((d*cos(d*x + c)^3 + d*cos(d*x + c)^2)*sq 
rt(cos(d*x + c)))
 

Sympy [F(-1)]

Timed out. \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))**(1/2)*(A+B*cos(d*x+c))*sec(d*x+c)**(7/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 475 vs. \(2 (112) = 224\).

Time = 0.22 (sec) , antiderivative size = 475, normalized size of antiderivative = 3.65 \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx =\text {Too large to display} \] Input:

integrate((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x, algo 
rithm="maxima")
 

Output:

2/15*(A*(15*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 25*sqrt(2)*s 
qrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 17*sqrt(2)*sqrt(a)*sin(d*x + 
c)^5/(cos(d*x + c) + 1)^5 - 7*sqrt(2)*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) 
 + 1)^7)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^3/((sin(d*x + c)/(cos(d 
*x + c) + 1) + 1)^(7/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(3*si 
n(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 
+ sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 1)) + 5*B*(3*sqrt(2)*sqrt(a)*sin(d 
*x + c)/(cos(d*x + c) + 1) - 7*sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c 
) + 1)^3 + 5*sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - sqrt(2) 
*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)*(sin(d*x + c)^2/(cos(d*x + c 
) + 1)^2 + 1)^3/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(-sin(d*x + c 
)/(cos(d*x + c) + 1) + 1)^(7/2)*(3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3 
*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + sin(d*x + c)^6/(cos(d*x + c) + 1)^6 
 + 1)))/d
 

Giac [F(-1)]

Timed out. \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x, algo 
rithm="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 1.77 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.51 \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {4\,\sqrt {a\,\left (\cos \left (c+d\,x\right )+1\right )}\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (14\,A\,\sin \left (c+d\,x\right )+10\,B\,\sin \left (c+d\,x\right )+8\,A\,\sin \left (2\,c+2\,d\,x\right )+18\,A\,\sin \left (3\,c+3\,d\,x\right )+4\,A\,\sin \left (4\,c+4\,d\,x\right )+4\,A\,\sin \left (5\,c+5\,d\,x\right )+10\,B\,\sin \left (2\,c+2\,d\,x\right )+15\,B\,\sin \left (3\,c+3\,d\,x\right )+5\,B\,\sin \left (4\,c+4\,d\,x\right )+5\,B\,\sin \left (5\,c+5\,d\,x\right )\right )}{15\,d\,\left (10\,\cos \left (c+d\,x\right )+8\,\cos \left (2\,c+2\,d\,x\right )+5\,\cos \left (3\,c+3\,d\,x\right )+2\,\cos \left (4\,c+4\,d\,x\right )+\cos \left (5\,c+5\,d\,x\right )+6\right )} \] Input:

int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(7/2)*(a + a*cos(c + d*x))^(1/2) 
,x)
 

Output:

(4*(a*(cos(c + d*x) + 1))^(1/2)*(1/cos(c + d*x))^(1/2)*(14*A*sin(c + d*x) 
+ 10*B*sin(c + d*x) + 8*A*sin(2*c + 2*d*x) + 18*A*sin(3*c + 3*d*x) + 4*A*s 
in(4*c + 4*d*x) + 4*A*sin(5*c + 5*d*x) + 10*B*sin(2*c + 2*d*x) + 15*B*sin( 
3*c + 3*d*x) + 5*B*sin(4*c + 4*d*x) + 5*B*sin(5*c + 5*d*x)))/(15*d*(10*cos 
(c + d*x) + 8*cos(2*c + 2*d*x) + 5*cos(3*c + 3*d*x) + 2*cos(4*c + 4*d*x) + 
 cos(5*c + 5*d*x) + 6))
 

Reduce [F]

\[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\sqrt {a}\, \left (\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{3}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}d x \right ) a \right ) \] Input:

int((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x)
 

Output:

sqrt(a)*(int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec(c 
+ d*x)**3,x)*b + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*sec(c + d*x 
)**3,x)*a)