Integrand size = 41, antiderivative size = 194 \[ \int \frac {(a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {2 (5 a A+3 b B+3 a C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 (7 A b+7 a B+5 b C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 b C \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 (b B+a C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (7 A b+7 a B+5 b C) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}} \]
2/7*b*C*sin(d*x+c)/d/sec(d*x+c)^(5/2)+2/5*(B*b+C*a)*sin(d*x+c)/d/sec(d*x+c )^(3/2)+2/21*(7*A*b+7*B*a+5*C*b)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2/5*(5*A*a+ 3*B*b+3*C*a)*(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*(7*A*b+7 *B*a+5*C*b)*(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 = 1.97 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.72 \[ \int \frac {(a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {\sqrt {\sec (c+d x)} \left (84 (5 a A+3 b B+3 a C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 (7 A b+7 a B+5 b C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+(70 A b+70 a B+65 b C+42 (b B+a C) \cos (c+d x)+15 b C \cos (2 (c+d x))) \sin (2 (c+d x))\right )}{210 d} \]
Integrate[((a + b*Cos[c + d*x])*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/S qrt[Sec[c + d*x]],x]
(Sqrt[Sec[c + d*x]]*(84*(5*a*A + 3*b*B + 3*a*C)*Sqrt[Cos[c + d*x]]*Ellipti cE[(c + d*x)/2, 2] + 20*(7*A*b + 7*a*B + 5*b*C)*Sqrt[Cos[c + d*x]]*Ellipti cF[(c + d*x)/2, 2] + (70*A*b + 70*a*B + 65*b*C + 42*(b*B + a*C)*Cos[c + d* x] + 15*b*C*Cos[2*(c + d*x)])*Sin[2*(c + d*x)]))/(210*d)
Time = 0.95 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.89, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.366, Rules used = {3042, 4709, 3042, 3512, 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.
\(\displaystyle \int \frac {(a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )}{\sqrt {\sec (c+d x)}}dx\) |
\(\Big \downarrow \) 4709 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)} (a+b \cos (c+d x)) \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )dx\) |
\(\Big \downarrow \) 3512 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2}{7} \int \frac {1}{2} \sqrt {\cos (c+d x)} \left (7 (b B+a C) \cos ^2(c+d x)+(7 A b+5 C b+7 a B) \cos (c+d x)+7 a A\right )dx+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \int \sqrt {\cos (c+d x)} \left (7 (b B+a C) \cos ^2(c+d x)+(7 A b+5 C b+7 a B) \cos (c+d x)+7 a A\right )dx+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (7 (b B+a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+(7 A b+5 C b+7 a B) \sin \left (c+d x+\frac {\pi }{2}\right )+7 a A\right )dx+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {2}{5} \int \frac {1}{2} \sqrt {\cos (c+d x)} (7 (5 a A+3 b B+3 a C)+5 (7 A b+5 C b+7 a B) \cos (c+d x))dx+\frac {14 (a C+b B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \int \sqrt {\cos (c+d x)} (7 (5 a A+3 b B+3 a C)+5 (7 A b+5 C b+7 a B) \cos (c+d x))dx+\frac {14 (a C+b B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (7 (5 a A+3 b B+3 a C)+5 (7 A b+5 C b+7 a B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {14 (a C+b B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (5 (7 a B+7 A b+5 b C) \int \cos ^{\frac {3}{2}}(c+d x)dx+7 (5 a A+3 a C+3 b B) \int \sqrt {\cos (c+d x)}dx\right )+\frac {14 (a C+b B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (7 (5 a A+3 a C+3 b B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 (7 a B+7 A b+5 b C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx\right )+\frac {14 (a C+b B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (7 (5 a A+3 a C+3 b B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 (7 a B+7 A b+5 b C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {14 (a C+b B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (7 (5 a A+3 a C+3 b B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 (7 a B+7 A b+5 b C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {14 (a C+b B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (5 (7 a B+7 A b+5 b C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {14 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (5 a A+3 a C+3 b B)}{d}\right )+\frac {14 (a C+b B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (\frac {14 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (5 a A+3 a C+3 b B)}{d}+5 (7 a B+7 A b+5 b C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {14 (a C+b B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )\) |
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*(b*B + a*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + ((1 4*(5*a*A + 3*b*B + 3*a*C)*EllipticE[(c + d*x)/2, 2])/d + 5*(7*A*b + 7*a*B + 5*b*C)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d*x]]*Sin[ c + d*x])/(3*d)))/5)/7)
3.15.56.3.1 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_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[(a + b*Si n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 , x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, 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]
Leaf count of result is larger than twice the leaf count of optimal. \(514\) vs. \(2(222)=444\).
Time = 8.50 (sec) , antiderivative size = 515, normalized size of antiderivative = 2.65
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 B b -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 +140 B a +168 B 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 -70 B a -42 B 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 +35 B 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}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-63 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}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b +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}\) | \(515\) |
parts | \(\text {Expression too large to display}\) | \(729\) |
int((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x,me thod=_RETURNVERBOSE)
-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*B*b-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+140*B*a+168*B*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-70*B*a-42*B*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)*Ell ipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a+35*B*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))-63*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)*EllipticE(c os(1/2*d*x+1/2*c),2^(1/2))*b+25*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)*EllipticF(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 = 0.11 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.20 \[ \int \frac {(a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=-\frac {5 \, \sqrt {2} {\left (7 i \, B a + i \, {\left (7 \, A + 5 \, C\right )} b\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-7 i \, B a - i \, {\left (7 \, A + 5 \, C\right )} b\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, \sqrt {2} {\left (-i \, {\left (5 \, A + 3 \, C\right )} a - 3 i \, B b\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 \, \sqrt {2} {\left (i \, {\left (5 \, A + 3 \, C\right )} a + 3 i \, B b\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (15 \, C b \cos \left (d x + c\right )^{3} + 21 \, {\left (C a + B b\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (7 \, B a + {\left (7 \, A + 5 \, C\right )} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{105 \, d} \]
integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2 ),x, algorithm="fricas")
-1/105*(5*sqrt(2)*(7*I*B*a + I*(7*A + 5*C)*b)*weierstrassPInverse(-4, 0, c os(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-7*I*B*a - I*(7*A + 5*C)*b)*wei erstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21*sqrt(2)*(-I*(5 *A + 3*C)*a - 3*I*B*b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, c os(d*x + c) + I*sin(d*x + c))) + 21*sqrt(2)*(I*(5*A + 3*C)*a + 3*I*B*b)*we ierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(15*C*b*cos(d*x + c)^3 + 21*(C*a + B*b)*cos(d*x + c)^2 + 5*(7*B* a + (7*A + 5*C)*b)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/d
\[ \int \frac {(a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {\left (a + b \cos {\left (c + d x \right )}\right ) \left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right )}{\sqrt {\sec {\left (c + d x \right )}}}\, dx \]
Integral((a + b*cos(c + d*x))*(A + B*cos(c + d*x) + C*cos(c + d*x)**2)/sqr t(sec(c + d*x)), x)
\[ \int \frac {(a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \]
integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2 ),x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)/sqr t(sec(d*x + c)), x)
\[ \int \frac {(a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \]
integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2 ),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)/sqr t(sec(d*x + c)), x)
Timed out. \[ \int \frac {(a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {\left (a+b\,\cos \left (c+d\,x\right )\right )\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]