Integrand size = 33, antiderivative size = 174 \[ \int \frac {(a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {2 a (5 A+3 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 b (7 A+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 b C \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a C \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 b (7 A+5 C) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}} \]
2/7*b*C*sin(d*x+c)/d/sec(d*x+c)^(5/2)+2/5*a*C*sin(d*x+c)/d/sec(d*x+c)^(3/2 )+2/21*b*(7*A+5*C)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2/5*a*(5*A+3*C)*(cos(1/2* d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2 ))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+2/21*b*(7*A+5*C)*(cos(1/2*d*x+1/2*c )^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d* x+c)^(1/2)*sec(d*x+c)^(1/2)/d
Time = 1.61 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.69 \[ \int \frac {(a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {\sqrt {\sec (c+d x)} \left (84 a (5 A+3 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 b (7 A+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+(70 A b+65 b C+42 a C \cos (c+d x)+15 b C \cos (2 (c+d x))) \sin (2 (c+d x))\right )}{210 d} \]
(Sqrt[Sec[c + d*x]]*(84*a*(5*A + 3*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d* x)/2, 2] + 20*b*(7*A + 5*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (70*A*b + 65*b*C + 42*a*C*Cos[c + d*x] + 15*b*C*Cos[2*(c + d*x)])*Sin[2*( c + d*x)]))/(210*d)
Time = 0.88 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.91, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4709, 3042, 3513, 27, 3042, 3502, 27, 3042, 3227, 3042, 3115, 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.
\(\displaystyle \int \frac {(a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \cos (c+d x)) \left (A+C \cos (c+d x)^2\right )}{\sqrt {\sec (c+d x)}}dx\) |
\(\Big \downarrow \) 4709 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)} (a+b \cos (c+d x)) \left (C \cos ^2(c+d x)+A\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx\) |
\(\Big \downarrow \) 3513 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2}{7} \int \frac {1}{2} \sqrt {\cos (c+d x)} \left (7 a C \cos ^2(c+d x)+b (7 A+5 C) \cos (c+d x)+7 a A\right )dx+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \int \sqrt {\cos (c+d x)} \left (7 a C \cos ^2(c+d x)+b (7 A+5 C) \cos (c+d x)+7 a A\right )dx+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (7 a C \sin \left (c+d x+\frac {\pi }{2}\right )^2+b (7 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )+7 a A\right )dx+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {2}{5} \int \frac {1}{2} \sqrt {\cos (c+d x)} (7 a (5 A+3 C)+5 b (7 A+5 C) \cos (c+d x))dx+\frac {14 a C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \int \sqrt {\cos (c+d x)} (7 a (5 A+3 C)+5 b (7 A+5 C) \cos (c+d x))dx+\frac {14 a C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (7 a (5 A+3 C)+5 b (7 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {14 a C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (7 a (5 A+3 C) \int \sqrt {\cos (c+d x)}dx+5 b (7 A+5 C) \int \cos ^{\frac {3}{2}}(c+d x)dx\right )+\frac {14 a C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (7 a (5 A+3 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 b (7 A+5 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx\right )+\frac {14 a C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (7 a (5 A+3 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 b (7 A+5 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {14 a C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (7 a (5 A+3 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 b (7 A+5 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {14 a C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (5 b (7 A+5 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {14 a (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {14 a C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (\frac {14 a (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+5 b (7 A+5 C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {14 a C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*b*C*Cos[c + d*x]^(5/2)*Sin[c + d *x])/(7*d) + ((14*a*C*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + ((14*a*(5*A + 3*C)*EllipticE[(c + d*x)/2, 2])/d + 5*b*(7*A + 5*C)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)))/5)/7)
3.14.65.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
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]
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]
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]
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]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[ (-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 3) )), x] + Simp[1/(b*(m + 3)) Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c *(m + 3) + b*d*(C*(m + 2) + A*(m + 3))*Sin[e + f*x] - (2*a*C*d - b*c*C*(m + 3))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Time = 7.84 (sec) , antiderivative size = 401, normalized size of antiderivative = 2.30
method | result | size |
default | \(-\frac {2 \sqrt {\left (-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (240 C b \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-168 C a -360 C b \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (140 A b +168 C a +280 C b \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-70 A b -42 C a -80 C b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+35 A b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-105 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a +25 C b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-63 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a \right )}{105 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(401\) |
parts | \(\text {Expression too large to display}\) | \(719\) |
-2/105*((-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*C*b*c os(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+(-168*C*a-360*C*b)*sin(1/2*d*x+1/2* c)^6*cos(1/2*d*x+1/2*c)+(140*A*b+168*C*a+280*C*b)*sin(1/2*d*x+1/2*c)^4*cos (1/2*d*x+1/2*c)+(-70*A*b-42*C*a-80*C*b)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1 /2*c)+35*A*b*(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))-105*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))*a+2 5*C*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipt icF(cos(1/2*d*x+1/2*c),2^(1/2))-63*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))*a)/(-2*sin( 1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(-1+2*cos( 1/2*d*x+1/2*c)^2)^(1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.11 \[ \int \frac {(a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {-5 i \, \sqrt {2} {\left (7 \, A + 5 \, C\right )} b {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} {\left (7 \, A + 5 \, C\right )} b {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 i \, \sqrt {2} {\left (5 \, A + 3 \, C\right )} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 i \, \sqrt {2} {\left (5 \, A + 3 \, C\right )} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (15 \, C b \cos \left (d x + c\right )^{3} + 21 \, C a \cos \left (d x + c\right )^{2} + 5 \, {\left (7 \, A + 5 \, C\right )} b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{105 \, d} \]
1/105*(-5*I*sqrt(2)*(7*A + 5*C)*b*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*(7*A + 5*C)*b*weierstrassPInverse(-4, 0, c os(d*x + c) - I*sin(d*x + c)) + 21*I*sqrt(2)*(5*A + 3*C)*a*weierstrassZeta (-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 21*I* sqrt(2)*(5*A + 3*C)*a*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, co s(d*x + c) - I*sin(d*x + c))) + 2*(15*C*b*cos(d*x + c)^3 + 21*C*a*cos(d*x + c)^2 + 5*(7*A + 5*C)*b*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/d
\[ \int \frac {(a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {\left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \left (a + b \cos {\left (c + d x \right )}\right )}{\sqrt {\sec {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {(a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \]
\[ \int \frac {(a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {(a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,\left (a+b\,\cos \left (c+d\,x\right )\right )}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]