Integrand size = 35, antiderivative size = 203 \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {4 a b (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (7 b^2 (A+3 C)+a^2 (5 A+7 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {8 a A b \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (4 A b^2+a^2 (5 A+7 C)\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 a b (3 A+5 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)} \]
-4/5*a*b*(3*A+5*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipti cE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/21*(7*b^2*(A+3*C)+a^2*(5*A+7*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))/d+8/35*a*A*b*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/21*(4*A*b^2+a^2*(5*A+7* C))*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/7*A*(a+b*cos(d*x+c))^2*sin(d*x+c)/d/co s(d*x+c)^(7/2)+4/5*a*b*(3*A+5*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)
Time = 2.81 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.98 \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {-84 a b (3 A+5 C) \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 \left (7 b^2 (A+3 C)+a^2 (5 A+7 C)\right ) \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+84 a A b \sin (c+d x)+252 a A b \cos ^2(c+d x) \sin (c+d x)+420 a b C \cos ^2(c+d x) \sin (c+d x)+25 a^2 A \sin (2 (c+d x))+35 A b^2 \sin (2 (c+d x))+35 a^2 C \sin (2 (c+d x))+30 a^2 A \tan (c+d x)}{105 d \cos ^{\frac {5}{2}}(c+d x)} \]
(-84*a*b*(3*A + 5*C)*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2] + 10*(7* b^2*(A + 3*C) + a^2*(5*A + 7*C))*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 84*a*A*b*Sin[c + d*x] + 252*a*A*b*Cos[c + d*x]^2*Sin[c + d*x] + 420* a*b*C*Cos[c + d*x]^2*Sin[c + d*x] + 25*a^2*A*Sin[2*(c + d*x)] + 35*A*b^2*S in[2*(c + d*x)] + 35*a^2*C*Sin[2*(c + d*x)] + 30*a^2*A*Tan[c + d*x])/(105* d*Cos[c + d*x]^(5/2))
Time = 1.19 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.01, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.457, Rules used = {3042, 3527, 27, 3042, 3510, 27, 3042, 3500, 27, 3042, 3227, 3042, 3116, 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.
\(\displaystyle \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx\) |
\(\Big \downarrow \) 3527 |
\(\displaystyle \frac {2}{7} \int \frac {(a+b \cos (c+d x)) \left (b (A+7 C) \cos ^2(c+d x)+a (5 A+7 C) \cos (c+d x)+4 A b\right )}{2 \cos ^{\frac {7}{2}}(c+d x)}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \int \frac {(a+b \cos (c+d x)) \left (b (A+7 C) \cos ^2(c+d x)+a (5 A+7 C) \cos (c+d x)+4 A b\right )}{\cos ^{\frac {7}{2}}(c+d x)}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (b (A+7 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (5 A+7 C) \sin \left (c+d x+\frac {\pi }{2}\right )+4 A b\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3510 |
\(\displaystyle \frac {1}{7} \left (\frac {8 a A b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {2}{5} \int -\frac {5 b^2 (A+7 C) \cos ^2(c+d x)+14 a b (3 A+5 C) \cos (c+d x)+5 \left ((5 A+7 C) a^2+4 A b^2\right )}{2 \cos ^{\frac {5}{2}}(c+d x)}dx\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \int \frac {5 b^2 (A+7 C) \cos ^2(c+d x)+14 a b (3 A+5 C) \cos (c+d x)+5 \left ((5 A+7 C) a^2+4 A b^2\right )}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {8 a A b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \int \frac {5 b^2 (A+7 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+14 a b (3 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )+5 \left ((5 A+7 C) a^2+4 A b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {8 a A b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3500 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {2}{3} \int \frac {42 a b (3 A+5 C)+5 \left ((5 A+7 C) a^2+7 b^2 (A+3 C)\right ) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x)}dx+\frac {10 \left (a^2 (5 A+7 C)+4 A b^2\right ) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {8 a A b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {42 a b (3 A+5 C)+5 \left ((5 A+7 C) a^2+7 b^2 (A+3 C)\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {10 \left (a^2 (5 A+7 C)+4 A b^2\right ) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {8 a A b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {42 a b (3 A+5 C)+5 \left ((5 A+7 C) a^2+7 b^2 (A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {10 \left (a^2 (5 A+7 C)+4 A b^2\right ) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {8 a A b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (5 \left (a^2 (5 A+7 C)+7 b^2 (A+3 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+42 a b (3 A+5 C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx\right )+\frac {10 \left (a^2 (5 A+7 C)+4 A b^2\right ) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {8 a A b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (5 \left (a^2 (5 A+7 C)+7 b^2 (A+3 C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+42 a b (3 A+5 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\right )+\frac {10 \left (a^2 (5 A+7 C)+4 A b^2\right ) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {8 a A b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (5 \left (a^2 (5 A+7 C)+7 b^2 (A+3 C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+42 a b (3 A+5 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )\right )+\frac {10 \left (a^2 (5 A+7 C)+4 A b^2\right ) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {8 a A b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (5 \left (a^2 (5 A+7 C)+7 b^2 (A+3 C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+42 a b (3 A+5 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )+\frac {10 \left (a^2 (5 A+7 C)+4 A b^2\right ) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {8 a A b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (5 \left (a^2 (5 A+7 C)+7 b^2 (A+3 C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+42 a b (3 A+5 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {10 \left (a^2 (5 A+7 C)+4 A b^2\right ) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {8 a A b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {10 \left (a^2 (5 A+7 C)+4 A b^2\right ) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (\frac {10 \left (a^2 (5 A+7 C)+7 b^2 (A+3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+42 a b (3 A+5 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )\right )+\frac {8 a A b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
(2*A*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)) + ((8*a *A*b*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + ((10*(4*A*b^2 + a^2*(5*A + 7 *C))*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((10*(7*b^2*(A + 3*C) + a^2* (5*A + 7*C))*EllipticF[(c + d*x)/2, 2])/d + 42*a*b*(3*A + 5*C)*((-2*Ellipt icE[(c + d*x)/2, 2])/d + (2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])))/3)/5)/7
3.7.88.3.1 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[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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[(-(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. \(902\) vs. \(2(235)=470\).
Time = 29.69 (sec) , antiderivative size = 903, normalized size of antiderivative = 4.45
method | result | size |
default | \(\text {Expression too large to display}\) | \(903\) |
parts | \(\text {Expression too large to display}\) | \(1036\) |
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*b^2*C*(sin(1 /2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(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*A*a^2*(-1/56*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2 *c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^4-5/42*cos(1/2*d*x+1/2*c)*(-2*sin( 1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+ 5/21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*si n(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)))+4/5*A*a*b/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*si n(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)*(sin(1/2*d*x+1/2*c)^2)^(1 /2)*EllipticE(cos(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)*(sin(1 /2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(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)+4*C*a*b/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)-(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...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.12 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.42 \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {-42 i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} a b \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 42 i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} a b \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 5 \, \sqrt {2} {\left (i \, {\left (5 \, A + 7 \, C\right )} a^{2} + 7 i \, {\left (A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 \, \sqrt {2} {\left (-i \, {\left (5 \, A + 7 \, C\right )} a^{2} - 7 i \, {\left (A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (42 \, {\left (3 \, A + 5 \, C\right )} a b \cos \left (d x + c\right )^{3} + 42 \, A a b \cos \left (d x + c\right ) + 15 \, A a^{2} + 5 \, {\left ({\left (5 \, A + 7 \, C\right )} a^{2} + 7 \, A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{4}} \]
1/105*(-42*I*sqrt(2)*(3*A + 5*C)*a*b*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 42*I*sqrt(2) *(3*A + 5*C)*a*b*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse (-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 5*sqrt(2)*(I*(5*A + 7*C)*a^2 + 7 *I*(A + 3*C)*b^2)*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*sqrt(2)*(-I*(5*A + 7*C)*a^2 - 7*I*(A + 3*C)*b^2)*cos( d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*( 42*(3*A + 5*C)*a*b*cos(d*x + c)^3 + 42*A*a*b*cos(d*x + c) + 15*A*a^2 + 5*( (5*A + 7*C)*a^2 + 7*A*b^2)*cos(d*x + c)^2)*sqrt(cos(d*x + c))*sin(d*x + c) )/(d*cos(d*x + c)^4)
Timed out. \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \]
\[ \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \]
Time = 4.36 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {30\,A\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+70\,A\,b^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+84\,A\,a\,b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{105\,d\,{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {2\,C\,b^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {4\,C\,a\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]
(30*A*a^2*sin(c + d*x)*hypergeom([-7/4, 1/2], -3/4, cos(c + d*x)^2) + 70*A *b^2*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^ 2) + 84*A*a*b*cos(c + d*x)*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2))/(105*d*cos(c + d*x)^(7/2)*(1 - cos(c + d*x)^2)^(1/2)) + (2*C*b ^2*ellipticF(c/2 + (d*x)/2, 2))/d + (2*C*a^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d*cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2 )) + (4*C*a*b*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))