Integrand size = 35, antiderivative size = 169 \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 \left (5 b^2 (A-C)+a^2 (3 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a b (A+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {8 a A b \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (4 A b^2+a^2 (3 A+5 C)\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)} \] Output:
-2/5*(5*b^2*(A-C)+a^2*(3*A+5*C))*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+4 /3*a*b*(A+3*C)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+8/15*a*A*b*sin(d*x +c)/d/cos(d*x+c)^(3/2)+2/5*(4*A*b^2+a^2*(3*A+5*C))*sin(d*x+c)/d/cos(d*x+c) ^(1/2)+2/5*A*(a+b*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(5/2)
Time = 2.40 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.93 \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {-6 \left (5 b^2 (A-C)+a^2 (3 A+5 C)\right ) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 a b (A+3 C) \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+20 a A b \sin (c+d x)+9 a^2 A \sin (2 (c+d x))+15 A b^2 \sin (2 (c+d x))+15 a^2 C \sin (2 (c+d x))+6 a^2 A \tan (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)} \] Input:
Integrate[((a + b*Cos[c + d*x])^2*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/ 2),x]
Output:
(-6*(5*b^2*(A - C) + a^2*(3*A + 5*C))*Cos[c + d*x]^(3/2)*EllipticE[(c + d* x)/2, 2] + 20*a*b*(A + 3*C)*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] + 20*a*A*b*Sin[c + d*x] + 9*a^2*A*Sin[2*(c + d*x)] + 15*A*b^2*Sin[2*(c + d* x)] + 15*a^2*C*Sin[2*(c + d*x)] + 6*a^2*A*Tan[c + d*x])/(15*d*Cos[c + d*x] ^(3/2))
Time = 1.05 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.02, 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, 3510, 27, 3042, 3500, 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.
| \(\displaystyle \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 3527 |
| \(\displaystyle \frac {2}{5} \int \frac {(a+b \cos (c+d x)) \left (-b (A-5 C) \cos ^2(c+d x)+a (3 A+5 C) \cos (c+d x)+4 A b\right )}{2 \cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {(a+b \cos (c+d x)) \left (-b (A-5 C) \cos ^2(c+d x)+a (3 A+5 C) \cos (c+d x)+4 A b\right )}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (-b (A-5 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (3 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )+4 A b\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3510 |
| \(\displaystyle \frac {1}{5} \left (\frac {8 a A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {2}{3} \int -\frac {-3 b^2 (A-5 C) \cos ^2(c+d x)+10 a b (A+3 C) \cos (c+d x)+3 \left ((3 A+5 C) a^2+4 A b^2\right )}{2 \cos ^{\frac {3}{2}}(c+d x)}dx\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {-3 b^2 (A-5 C) \cos ^2(c+d x)+10 a b (A+3 C) \cos (c+d x)+3 \left ((3 A+5 C) a^2+4 A b^2\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {8 a A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {-3 b^2 (A-5 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+10 a b (A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left ((3 A+5 C) a^2+4 A b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {8 a A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (2 \int \frac {10 a b (A+3 C)-3 \left ((3 A+5 C) a^2+5 b^2 (A-C)\right ) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx+\frac {6 \left (a^2 (3 A+5 C)+4 A b^2\right ) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {8 a A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {10 a b (A+3 C)-3 \left ((3 A+5 C) a^2+5 b^2 (A-C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {6 \left (a^2 (3 A+5 C)+4 A b^2\right ) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {8 a A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {10 a b (A+3 C)-3 \left ((3 A+5 C) a^2+5 b^2 (A-C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 \left (a^2 (3 A+5 C)+4 A b^2\right ) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {8 a A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (-3 \left (a^2 (3 A+5 C)+5 b^2 (A-C)\right ) \int \sqrt {\cos (c+d x)}dx+10 a b (A+3 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {6 \left (a^2 (3 A+5 C)+4 A b^2\right ) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {8 a A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (-3 \left (a^2 (3 A+5 C)+5 b^2 (A-C)\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+10 a b (A+3 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 \left (a^2 (3 A+5 C)+4 A b^2\right ) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {8 a A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (10 a b (A+3 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 \left (a^2 (3 A+5 C)+5 b^2 (A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {6 \left (a^2 (3 A+5 C)+4 A b^2\right ) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {8 a A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (-\frac {6 \left (a^2 (3 A+5 C)+5 b^2 (A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {6 \left (a^2 (3 A+5 C)+4 A b^2\right ) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {20 a b (A+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {8 a A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
Input:
Int[((a + b*Cos[c + d*x])^2*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/2),x]
Output:
(2*A*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + ((8*a *A*b*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((-6*(5*b^2*(A - C) + a^2*(3 *A + 5*C))*EllipticE[(c + d*x)/2, 2])/d + (20*a*b*(A + 3*C)*EllipticF[(c + d*x)/2, 2])/d + (6*(4*A*b^2 + a^2*(3*A + 5*C))*Sin[c + d*x])/(d*Sqrt[Cos[ c + d*x]]))/3)/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[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
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[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S imp[1/(b^2*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) ))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F reeQ[{a, b, c, d, e, f, A, B, C}, 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. \(885\) vs. \(2(156)=312\).
Time = 13.61 (sec) , antiderivative size = 886, normalized size of antiderivative = 5.24
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(886\) |
| parts | \(\text {Expression too large to display}\) | \(926\) |
Input:
int((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x,method=_RETUR NVERBOSE)
Output:
-(-(1-2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*(A*b^2+C*a^2) /sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+ sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(si n(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)))+2/5*A*a^2/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1 /2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(24*cos(1/2*d*x+1/2 *c)*sin(1/2*d*x+1/2*c)^6-12*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4 -24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*s in(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*(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)))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-2*C* b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-2*cos(1/2*d*x+1/2*c)^2)^(1/2)/(-2*sin( 1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c), 2^(1/2))+2*C*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-2*cos(1/2*d*x+1/2*c)^2)^( 1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1 /2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+4*a*b*C*(sin (1/2*d*x+1/2*c)^2)^(1/2)*(1-2*cos(1/2*d*x+1/2*c)^2)^(1/2)/(-2*sin(1/2*d*x+ 1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/...
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.57 \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {-10 i \, \sqrt {2} {\left (A + 3 \, C\right )} a b \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 10 i \, \sqrt {2} {\left (A + 3 \, C\right )} a b \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (i \, {\left (3 \, A + 5 \, C\right )} a^{2} + 5 i \, {\left (A - C\right )} b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (-i \, {\left (3 \, A + 5 \, C\right )} a^{2} - 5 i \, {\left (A - C\right )} b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (10 \, A a b \cos \left (d x + c\right ) + 3 \, A a^{2} + 3 \, {\left ({\left (3 \, A + 5 \, C\right )} a^{2} + 5 \, A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algori thm="fricas")
Output:
1/15*(-10*I*sqrt(2)*(A + 3*C)*a*b*cos(d*x + c)^3*weierstrassPInverse(-4, 0 , cos(d*x + c) + I*sin(d*x + c)) + 10*I*sqrt(2)*(A + 3*C)*a*b*cos(d*x + c) ^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*sqrt(2)*( I*(3*A + 5*C)*a^2 + 5*I*(A - C)*b^2)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*sqrt(2)*(- I*(3*A + 5*C)*a^2 - 5*I*(A - C)*b^2)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(10*A*a*b* cos(d*x + c) + 3*A*a^2 + 3*((3*A + 5*C)*a^2 + 5*A*b^2)*cos(d*x + c)^2)*sqr t(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)
Timed out. \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**2*(A+C*cos(d*x+c)**2)/cos(d*x+c)**(7/2),x)
Output:
Timed out
\[ \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algori thm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^2/cos(d*x + c)^(7/2) , x)
\[ \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algori thm="giac")
Output:
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^2/cos(d*x + c)^(7/2) , x)
Time = 1.74 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.18 \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {6\,A\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+30\,A\,b^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+20\,A\,a\,b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {2\,C\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,C\,a\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:
int(((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^2)/cos(c + d*x)^(7/2),x)
Output:
(6*A*a^2*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) + 30*A* b^2*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2 ) + 20*A*a*b*cos(c + d*x)*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(15*d*cos(c + d*x)^(5/2)*(1 - cos(c + d*x)^2)^(1/2)) + (2*C*b^2* ellipticE(c/2 + (d*x)/2, 2))/d + (4*C*a*b*ellipticF(c/2 + (d*x)/2, 2))/d + (2*C*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))
\[ \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=2 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) a b c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}}d x \right ) a^{3}+2 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) a^{2} b +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a^{2} c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a \,b^{2}+\left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) b^{2} c \] Input:
int((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x)
Output:
2*int(sqrt(cos(c + d*x))/cos(c + d*x),x)*a*b*c + int(sqrt(cos(c + d*x))/co s(c + d*x)**4,x)*a**3 + 2*int(sqrt(cos(c + d*x))/cos(c + d*x)**3,x)*a**2*b + int(sqrt(cos(c + d*x))/cos(c + d*x)**2,x)*a**2*c + int(sqrt(cos(c + d*x ))/cos(c + d*x)**2,x)*a*b**2 + int(sqrt(cos(c + d*x)),x)*b**2*c