Integrand size = 43, antiderivative size = 229 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {2 \left (10 a b B-5 a^2 (A-C)+b^2 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (3 a^2 B+b^2 B+2 a b (3 A+C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {2 b^2 (5 A-C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 b (6 a A-b B-2 a C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{d} \]
-2/5*b^2*(5*A-C)*sin(d*x+c)/d/sec(d*x+c)^(3/2)-2/3*b*(6*A*a-B*b-2*C*a)*sin (d*x+c)/d/sec(d*x+c)^(1/2)+2*A*(a+b*cos(d*x+c))^2*sin(d*x+c)*sec(d*x+c)^(1 /2)/d+2/5*(10*B*a*b-5*a^2*(A-C)+b^2*(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+2/3*(3*B*a^2+B*b^2+2*a*b*(3*A+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 = 3.22 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.72 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (12 \left (10 a b B-5 a^2 (A-C)+b^2 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 \left (3 a^2 B+b^2 B+2 a b (3 A+C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {2 \left (10 b (b B+2 a C) \cos (c+d x)+3 \left (10 a^2 A+b^2 C+b^2 C \cos (2 (c+d x))\right )\right ) \sin (c+d x)}{\sqrt {\cos (c+d x)}}\right )}{30 d} \]
Integrate[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*S ec[c + d*x]^(3/2),x]
(Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(12*(10*a*b*B - 5*a^2*(A - C) + b^2 *(5*A + 3*C))*EllipticE[(c + d*x)/2, 2] + 20*(3*a^2*B + b^2*B + 2*a*b*(3*A + C))*EllipticF[(c + d*x)/2, 2] + (2*(10*b*(b*B + 2*a*C)*Cos[c + d*x] + 3 *(10*a^2*A + b^2*C + b^2*C*Cos[2*(c + d*x)]))*Sin[c + d*x])/Sqrt[Cos[c + d *x]]))/(30*d)
Time = 1.34 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.94, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.372, Rules used = {3042, 4709, 3042, 3526, 27, 3042, 3512, 27, 3042, 3502, 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.
| \(\displaystyle \int \sec ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (c+d x)^{3/2} (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(a+b \cos (c+d x))^2 \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (2 \int \frac {(a+b \cos (c+d x)) \left (-b (5 A-C) \cos ^2(c+d x)+(b B-a (A-C)) \cos (c+d x)+4 A b+a B\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\int \frac {(a+b \cos (c+d x)) \left (-b (5 A-C) \cos ^2(c+d x)+(b B-a (A-C)) \cos (c+d x)+4 A b+a B\right )}{\sqrt {\cos (c+d x)}}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (-b (5 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+(b B-a (A-C)) \sin \left (c+d x+\frac {\pi }{2}\right )+4 A b+a B\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3512 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2}{5} \int \frac {-5 b (6 a A-b B-2 a C) \cos ^2(c+d x)+\left (-5 (A-C) a^2+10 b B a+b^2 (5 A+3 C)\right ) \cos (c+d x)+5 a (4 A b+a B)}{2 \sqrt {\cos (c+d x)}}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (5 A-C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \int \frac {-5 b (6 a A-b B-2 a C) \cos ^2(c+d x)+\left (-5 (A-C) a^2+10 b B a+b^2 (5 A+3 C)\right ) \cos (c+d x)+5 a (4 A b+a B)}{\sqrt {\cos (c+d x)}}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (5 A-C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \int \frac {-5 b (6 a A-b B-2 a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (-5 (A-C) a^2+10 b B a+b^2 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+5 a (4 A b+a B)}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (5 A-C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {2}{3} \int \frac {5 \left (3 B a^2+2 b (3 A+C) a+b^2 B\right )+3 \left (-5 (A-C) a^2+10 b B a+b^2 (5 A+3 C)\right ) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx-\frac {10 b \sin (c+d x) \sqrt {\cos (c+d x)} (6 a A-2 a C-b B)}{3 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (5 A-C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {5 \left (3 B a^2+2 b (3 A+C) a+b^2 B\right )+3 \left (-5 (A-C) a^2+10 b B a+b^2 (5 A+3 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx-\frac {10 b \sin (c+d x) \sqrt {\cos (c+d x)} (6 a A-2 a C-b B)}{3 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (5 A-C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {5 \left (3 B a^2+2 b (3 A+C) a+b^2 B\right )+3 \left (-5 (A-C) a^2+10 b B a+b^2 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {10 b \sin (c+d x) \sqrt {\cos (c+d x)} (6 a A-2 a C-b B)}{3 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (5 A-C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \left (5 \left (3 a^2 B+2 a b (3 A+C)+b^2 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 \left (-5 a^2 (A-C)+10 a b B+b^2 (5 A+3 C)\right ) \int \sqrt {\cos (c+d x)}dx\right )-\frac {10 b \sin (c+d x) \sqrt {\cos (c+d x)} (6 a A-2 a C-b B)}{3 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (5 A-C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \left (5 \left (3 a^2 B+2 a b (3 A+C)+b^2 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 \left (-5 a^2 (A-C)+10 a b B+b^2 (5 A+3 C)\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-\frac {10 b \sin (c+d x) \sqrt {\cos (c+d x)} (6 a A-2 a C-b B)}{3 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (5 A-C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \left (5 \left (3 a^2 B+2 a b (3 A+C)+b^2 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^2 (A-C)+10 a b B+b^2 (5 A+3 C)\right )}{d}\right )-\frac {10 b \sin (c+d x) \sqrt {\cos (c+d x)} (6 a A-2 a C-b B)}{3 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (5 A-C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \left (\frac {10 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (3 a^2 B+2 a b (3 A+C)+b^2 B\right )}{d}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^2 (A-C)+10 a b B+b^2 (5 A+3 C)\right )}{d}\right )-\frac {10 b \sin (c+d x) \sqrt {\cos (c+d x)} (6 a A-2 a C-b B)}{3 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (5 A-C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((-2*b^2*(5*A - C)*Cos[c + d*x]^(3/2 )*Sin[c + d*x])/(5*d) + (2*A*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(d*Sqrt[ Cos[c + d*x]]) + (((6*(10*a*b*B - 5*a^2*(A - C) + b^2*(5*A + 3*C))*Ellipti cE[(c + d*x)/2, 2])/d + (10*(3*a^2*B + b^2*B + 2*a*b*(3*A + C))*EllipticF[ (c + d*x)/2, 2])/d)/3 - (10*b*(6*a*A - b*B - 2*a*C)*Sqrt[Cos[c + d*x]]*Sin [c + d*x])/(3*d))/5)
3.15.62.3.1 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[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[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x ] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f *x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d , 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -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. \(678\) vs. \(2(259)=518\).
Time = 10.57 (sec) , antiderivative size = 679, normalized size of antiderivative = 2.97
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(679\) |
| parts | \(\text {Expression too large to display}\) | \(872\) |
int((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2),x, method=_RETURNVERBOSE)
2/15*(24*C*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-20*B*cos(1/2*d*x+1/ 2*c)*sin(1/2*d*x+1/2*c)^4*b^2-40*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4 *a*b-24*C*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+30*A*cos(1/2*d*x+1/2 *c)*sin(1/2*d*x+1/2*c)^2*a^2-30*a*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))-15*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+15*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))*b^2+10*B*cos(1/2*d *x+1/2*c)*sin(1/2*d*x+1/2*c)^2*b^2-15*B*a^2*(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))-5*B* b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elliptic F(cos(1/2*d*x+1/2*c),2^(1/2))+30*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(cos(1/2*d*x+1/2*c),2^(1/2))*a*b+20*C*b*si n(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*a+6*C*b^2*sin(1/2*d*x+1/2*c)^2*cos(1 /2*d*x+1/2*c)-10*a*b*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+15*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+9*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))*b^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 = 0.12 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.14 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=-\frac {5 \, \sqrt {2} {\left (3 i \, B a^{2} + 2 i \, {\left (3 \, A + C\right )} a b + i \, B b^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-3 i \, B a^{2} - 2 i \, {\left (3 \, A + C\right )} a b - i \, B b^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (5 i \, {\left (A - C\right )} a^{2} - 10 i \, B a b - i \, {\left (5 \, A + 3 \, C\right )} b^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-5 i \, {\left (A - C\right )} a^{2} + 10 i \, B a b + i \, {\left (5 \, A + 3 \, C\right )} b^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (3 \, C b^{2} \cos \left (d x + c\right )^{2} + 15 \, A a^{2} + 5 \, {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d} \]
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3 /2),x, algorithm="fricas")
-1/15*(5*sqrt(2)*(3*I*B*a^2 + 2*I*(3*A + C)*a*b + I*B*b^2)*weierstrassPInv erse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-3*I*B*a^2 - 2*I*( 3*A + C)*a*b - I*B*b^2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d* x + c)) + 3*sqrt(2)*(5*I*(A - C)*a^2 - 10*I*B*a*b - I*(5*A + 3*C)*b^2)*wei erstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(-5*I*(A - C)*a^2 + 10*I*B*a*b + I*(5*A + 3*C)*b^2)*weier strassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c) )) - 2*(3*C*b^2*cos(d*x + c)^2 + 15*A*a^2 + 5*(2*C*a*b + B*b^2)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/d
Timed out. \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out} \]
\[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \]
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3 /2),x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^2*s ec(d*x + c)^(3/2), x)
\[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \]
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3 /2),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^2*s ec(d*x + c)^(3/2), x)
Timed out. \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]