Integrand size = 35, antiderivative size = 166 \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 \left (5 a^2 (A-C)-b^2 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a b (3 A+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}-\frac {4 a b (3 A-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}-\frac {2 b^2 (5 A-C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \] Output:
-2/5*(5*a^2*(A-C)-b^2*(5*A+3*C))*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+4 /3*a*b*(3*A+C)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d-4/3*a*b*(3*A-C)*co s(d*x+c)^(1/2)*sin(d*x+c)/d-2/5*b^2*(5*A-C)*cos(d*x+c)^(3/2)*sin(d*x+c)/d+ 2*A*(a+b*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(1/2)
Time = 2.50 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.72 \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {-6 \left (5 a^2 (A-C)-b^2 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 a b (3 A+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {\left (30 a^2 A+3 b^2 C+20 a b C \cos (c+d x)+3 b^2 C \cos (2 (c+d x))\right ) \sin (c+d x)}{\sqrt {\cos (c+d x)}}}{15 d} \] Input:
Integrate[((a + b*Cos[c + d*x])^2*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(3/ 2),x]
Output:
(-6*(5*a^2*(A - C) - b^2*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2] + 20*a*b*( 3*A + C)*EllipticF[(c + d*x)/2, 2] + ((30*a^2*A + 3*b^2*C + 20*a*b*C*Cos[c + d*x] + 3*b^2*C*Cos[2*(c + d*x)])*Sin[c + d*x])/Sqrt[Cos[c + d*x]])/(15* d)
Time = 1.03 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.04, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3527, 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.
| \(\displaystyle \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3527 |
| \(\displaystyle 2 \int \frac {(a+b \cos (c+d x)) \left (-b (5 A-C) \cos ^2(c+d x)-a (A-C) \cos (c+d x)+4 A b\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {(a+b \cos (c+d x)) \left (-b (5 A-C) \cos ^2(c+d x)-a (A-C) \cos (c+d x)+4 A b\right )}{\sqrt {\cos (c+d x)}}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (-b (5 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )^2-a (A-C) \sin \left (c+d x+\frac {\pi }{2}\right )+4 A b\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3512 |
| \(\displaystyle \frac {2}{5} \int \frac {-10 a b (3 A-C) \cos ^2(c+d x)-\left (5 a^2 (A-C)-b^2 (5 A+3 C)\right ) \cos (c+d x)+20 a A b}{2 \sqrt {\cos (c+d x)}}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (5 A-C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {-10 a b (3 A-C) \cos ^2(c+d x)-\left (5 a^2 (A-C)-b^2 (5 A+3 C)\right ) \cos (c+d x)+20 a A b}{\sqrt {\cos (c+d x)}}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (5 A-C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {-10 a b (3 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (b^2 (5 A+3 C)-5 a^2 (A-C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+20 a A b}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (5 A-C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {10 a b (3 A+C)-3 \left (5 a^2 (A-C)-b^2 (5 A+3 C)\right ) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx-\frac {20 a b (3 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (5 A-C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {10 a b (3 A+C)-3 \left (5 a^2 (A-C)-b^2 (5 A+3 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx-\frac {20 a b (3 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (5 A-C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {10 a b (3 A+C)-3 \left (5 a^2 (A-C)-b^2 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {20 a b (3 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (5 A-C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (10 a b (3 A+C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 \left (5 a^2 (A-C)-b^2 (5 A+3 C)\right ) \int \sqrt {\cos (c+d x)}dx\right )-\frac {20 a b (3 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (5 A-C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (10 a b (3 A+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 \left (5 a^2 (A-C)-b^2 (5 A+3 C)\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-\frac {20 a b (3 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (5 A-C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (10 a b (3 A+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 \left (5 a^2 (A-C)-b^2 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {20 a b (3 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (5 A-C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {20 a b (3 A+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 \left (5 a^2 (A-C)-b^2 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {20 a b (3 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (5 A-C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
Input:
Int[((a + b*Cos[c + d*x])^2*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(3/2),x]
Output:
(-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*(5*a^2*(A - C) - b^2*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2])/d + (20*a*b*(3*A + C)*Ellipti cF[(c + d*x)/2, 2])/d)/3 - (20*a*b*(3*A - C)*Sqrt[Cos[c + d*x]]*Sin[c + d* x])/(3*d))/5
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_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + 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*c*m + a*d*(n + 1)) - (A*d*(a*d*(n + 2) - b *c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*( A*d^2*(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, 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]
Leaf count of result is larger than twice the leaf count of optimal. \(480\) vs. \(2(155)=310\).
Time = 17.55 (sec) , antiderivative size = 481, normalized size of antiderivative = 2.90
| method | result | size |
| default | \(\frac {\frac {16 C \,b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{5}-\frac {16 C b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a}{3}-\frac {16 C \,b^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}+4 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{2}-4 a A b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-2 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}+2 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}+\frac {8 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a b}{3}+\frac {4 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b^{2}}{5}-\frac {4 a b C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{3}+2 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}+\frac {6 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}}{5}}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(481\) |
| parts | \(\text {Expression too large to display}\) | \(736\) |
Input:
int((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2),x,method=_RETUR NVERBOSE)
Output:
2/15*(24*C*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-40*C*b*sin(1/2*d*x+ 1/2*c)^4*cos(1/2*d*x+1/2*c)*a-24*C*b^2*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/ 2*c)+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+20*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a*b+6*C*cos(1/2*d* x+1/2*c)*sin(1/2*d*x+1/2*c)^2*b^2-10*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(co s(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)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.47 \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {-10 i \, \sqrt {2} {\left (3 \, A + C\right )} a b \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 10 i \, \sqrt {2} {\left (3 \, A + C\right )} a b \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (5 i \, {\left (A - C\right )} a^{2} - i \, {\left (5 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (-5 i \, {\left (A - C\right )} a^{2} + i \, {\left (5 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (3 \, C b^{2} \cos \left (d x + c\right )^{2} + 10 \, C a b \cos \left (d x + c\right ) + 15 \, A a^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )} \] Input:
integrate((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2),x, algori thm="fricas")
Output:
1/15*(-10*I*sqrt(2)*(3*A + C)*a*b*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 10*I*sqrt(2)*(3*A + C)*a*b*cos(d*x + c)*w eierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*sqrt(2)*(5*I* (A - C)*a^2 - I*(5*A + 3*C)*b^2)*cos(d*x + c)*weierstrassZeta(-4, 0, weier strassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*sqrt(2)*(-5*I*(A - C)*a^2 + I*(5*A + 3*C)*b^2)*cos(d*x + c)*weierstrassZeta(-4, 0, weierst rassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(3*C*b^2*cos(d*x + c)^2 + 10*C*a*b*cos(d*x + c) + 15*A*a^2)*sqrt(cos(d*x + c))*sin(d*x + c)) /(d*cos(d*x + c))
Timed out. \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**2*(A+C*cos(d*x+c)**2)/cos(d*x+c)**(3/2),x)
Output:
Timed out
\[ \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2),x, algori thm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^2/cos(d*x + c)^(3/2) , x)
\[ \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2),x, algori thm="giac")
Output:
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^2/cos(d*x + c)^(3/2) , x)
Time = 1.17 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2\,A\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,a\,b\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {4\,A\,a\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,b^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:
int(((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^2)/cos(c + d*x)^(3/2),x)
Output:
(2*A*b^2*ellipticE(c/2 + (d*x)/2, 2))/d + (2*C*a^2*ellipticE(c/2 + (d*x)/2 , 2))/d + (2*C*a*b*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c /2 + (d*x)/2, 2))/3))/d + (4*A*a*b*ellipticF(c/2 + (d*x)/2, 2))/d + (2*A*a ^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d* x)^(1/2)*(sin(c + d*x)^2)^(1/2)) - (2*C*b^2*cos(c + d*x)^(7/2)*sin(c + d*x )*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2) )
\[ \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=2 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) a^{2} b +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a^{3}+\left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) a^{2} c +\left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) a \,b^{2}+2 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )d x \right ) a b c +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}d x \right ) b^{2} c \] Input:
int((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2),x)
Output:
2*int(sqrt(cos(c + d*x))/cos(c + d*x),x)*a**2*b + int(sqrt(cos(c + d*x))/c os(c + d*x)**2,x)*a**3 + int(sqrt(cos(c + d*x)),x)*a**2*c + int(sqrt(cos(c + d*x)),x)*a*b**2 + 2*int(sqrt(cos(c + d*x))*cos(c + d*x),x)*a*b*c + int( sqrt(cos(c + d*x))*cos(c + d*x)**2,x)*b**2*c