Integrand size = 43, antiderivative size = 305 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {2 \left (18 a A b+9 a^2 B+7 b^2 B+14 a b C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 \left (110 a b B+11 a^2 (7 A+5 C)+5 b^2 (11 A+9 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{231 d}+\frac {2 \left (110 a b B+11 a^2 (7 A+5 C)+5 b^2 (11 A+9 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{231 d}+\frac {2 \left (18 a A b+9 a^2 B+7 b^2 B+14 a b C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 \left (11 A b^2+22 a b B+4 a^2 C+9 b^2 C\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{77 d}+\frac {2 b (11 b B+4 a C) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 C \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{11 d} \]
2/15*(18*A*a*b+9*B*a^2+7*B*b^2+14*C*a*b)*(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))/d+2/231*(110*B*a*b+11 *a^2*(7*A+5*C)+5*b^2*(11*A+9*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+2/45*(18*A*a*b+9*B*a^2+7*B* b^2+14*C*a*b)*cos(d*x+c)^(3/2)*sin(d*x+c)/d+2/77*(11*A*b^2+22*B*a*b+4*C*a^ 2+9*C*b^2)*cos(d*x+c)^(5/2)*sin(d*x+c)/d+2/99*b*(11*B*b+4*C*a)*cos(d*x+c)^ (7/2)*sin(d*x+c)/d+2/11*C*cos(d*x+c)^(5/2)*(a+b*cos(d*x+c))^2*sin(d*x+c)/d +2/231*(110*B*a*b+11*a^2*(7*A+5*C)+5*b^2*(11*A+9*C))*sin(d*x+c)*cos(d*x+c) ^(1/2)/d
Time = 3.21 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.78 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {154 \left (9 a^2 B+7 b^2 B+2 a b (9 A+7 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 \left (110 a b B+11 a^2 (7 A+5 C)+5 b^2 (11 A+9 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {1}{12} \sqrt {\cos (c+d x)} \left (154 \left (72 a A b+36 a^2 B+43 b^2 B+86 a b C\right ) \cos (c+d x)+5 \left (3432 a b B+132 a^2 (14 A+13 C)+3 b^2 (572 A+531 C)+36 \left (11 A b^2+22 a b B+11 a^2 C+16 b^2 C\right ) \cos (2 (c+d x))+154 b (b B+2 a C) \cos (3 (c+d x))+63 b^2 C \cos (4 (c+d x))\right )\right ) \sin (c+d x)}{1155 d} \]
Integrate[Cos[c + d*x]^(3/2)*(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]
(154*(9*a^2*B + 7*b^2*B + 2*a*b*(9*A + 7*C))*EllipticE[(c + d*x)/2, 2] + 1 0*(110*a*b*B + 11*a^2*(7*A + 5*C) + 5*b^2*(11*A + 9*C))*EllipticF[(c + d*x )/2, 2] + (Sqrt[Cos[c + d*x]]*(154*(72*a*A*b + 36*a^2*B + 43*b^2*B + 86*a* b*C)*Cos[c + d*x] + 5*(3432*a*b*B + 132*a^2*(14*A + 13*C) + 3*b^2*(572*A + 531*C) + 36*(11*A*b^2 + 22*a*b*B + 11*a^2*C + 16*b^2*C)*Cos[2*(c + d*x)] + 154*b*(b*B + 2*a*C)*Cos[3*(c + d*x)] + 63*b^2*C*Cos[4*(c + d*x)]))*Sin[c + d*x])/12)/(1155*d)
Time = 1.35 (sec) , antiderivative size = 273, normalized size of antiderivative = 0.90, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.372, Rules used = {3042, 3528, 27, 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.
\(\displaystyle \int \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 3528 |
\(\displaystyle \frac {2}{11} \int \frac {1}{2} \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x)) \left ((11 b B+4 a C) \cos ^2(c+d x)+(11 A b+9 C b+11 a B) \cos (c+d x)+a (11 A+5 C)\right )dx+\frac {2 C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2}{11 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{11} \int \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x)) \left ((11 b B+4 a C) \cos ^2(c+d x)+(11 A b+9 C b+11 a B) \cos (c+d x)+a (11 A+5 C)\right )dx+\frac {2 C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2}{11 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{11} \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left ((11 b B+4 a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+(11 A b+9 C b+11 a B) \sin \left (c+d x+\frac {\pi }{2}\right )+a (11 A+5 C)\right )dx+\frac {2 C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2}{11 d}\) |
\(\Big \downarrow \) 3512 |
\(\displaystyle \frac {1}{11} \left (\frac {2}{9} \int \frac {1}{2} \cos ^{\frac {3}{2}}(c+d x) \left (9 (11 A+5 C) a^2+9 \left (4 C a^2+22 b B a+11 A b^2+9 b^2 C\right ) \cos ^2(c+d x)+11 \left (9 B a^2+18 A b a+14 b C a+7 b^2 B\right ) \cos (c+d x)\right )dx+\frac {2 b (4 a C+11 b B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}\right )+\frac {2 C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2}{11 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{11} \left (\frac {1}{9} \int \cos ^{\frac {3}{2}}(c+d x) \left (9 (11 A+5 C) a^2+9 \left (4 C a^2+22 b B a+11 A b^2+9 b^2 C\right ) \cos ^2(c+d x)+11 \left (9 B a^2+18 A b a+14 b C a+7 b^2 B\right ) \cos (c+d x)\right )dx+\frac {2 b (4 a C+11 b B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}\right )+\frac {2 C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2}{11 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{11} \left (\frac {1}{9} \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (9 (11 A+5 C) a^2+9 \left (4 C a^2+22 b B a+11 A b^2+9 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+11 \left (9 B a^2+18 A b a+14 b C a+7 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 b (4 a C+11 b B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}\right )+\frac {2 C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2}{11 d}\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {2}{7} \int \frac {1}{2} \cos ^{\frac {3}{2}}(c+d x) \left (9 \left (11 (7 A+5 C) a^2+110 b B a+5 b^2 (11 A+9 C)\right )+77 \left (9 B a^2+18 A b a+14 b C a+7 b^2 B\right ) \cos (c+d x)\right )dx+\frac {18 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (4 a^2 C+22 a b B+11 A b^2+9 b^2 C\right )}{7 d}\right )+\frac {2 b (4 a C+11 b B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}\right )+\frac {2 C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2}{11 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {1}{7} \int \cos ^{\frac {3}{2}}(c+d x) \left (9 \left (11 (7 A+5 C) a^2+110 b B a+5 b^2 (11 A+9 C)\right )+77 \left (9 B a^2+18 A b a+14 b C a+7 b^2 B\right ) \cos (c+d x)\right )dx+\frac {18 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (4 a^2 C+22 a b B+11 A b^2+9 b^2 C\right )}{7 d}\right )+\frac {2 b (4 a C+11 b B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}\right )+\frac {2 C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2}{11 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {1}{7} \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (9 \left (11 (7 A+5 C) a^2+110 b B a+5 b^2 (11 A+9 C)\right )+77 \left (9 B a^2+18 A b a+14 b C a+7 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {18 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (4 a^2 C+22 a b B+11 A b^2+9 b^2 C\right )}{7 d}\right )+\frac {2 b (4 a C+11 b B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}\right )+\frac {2 C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2}{11 d}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {1}{7} \left (9 \left (11 a^2 (7 A+5 C)+110 a b B+5 b^2 (11 A+9 C)\right ) \int \cos ^{\frac {3}{2}}(c+d x)dx+77 \left (9 a^2 B+18 a A b+14 a b C+7 b^2 B\right ) \int \cos ^{\frac {5}{2}}(c+d x)dx\right )+\frac {18 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (4 a^2 C+22 a b B+11 A b^2+9 b^2 C\right )}{7 d}\right )+\frac {2 b (4 a C+11 b B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}\right )+\frac {2 C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2}{11 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {1}{7} \left (9 \left (11 a^2 (7 A+5 C)+110 a b B+5 b^2 (11 A+9 C)\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx+77 \left (9 a^2 B+18 a A b+14 a b C+7 b^2 B\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx\right )+\frac {18 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (4 a^2 C+22 a b B+11 A b^2+9 b^2 C\right )}{7 d}\right )+\frac {2 b (4 a C+11 b B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}\right )+\frac {2 C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2}{11 d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {1}{7} \left (77 \left (9 a^2 B+18 a A b+14 a b C+7 b^2 B\right ) \left (\frac {3}{5} \int \sqrt {\cos (c+d x)}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+9 \left (11 a^2 (7 A+5 C)+110 a b B+5 b^2 (11 A+9 C)\right ) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {18 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (4 a^2 C+22 a b B+11 A b^2+9 b^2 C\right )}{7 d}\right )+\frac {2 b (4 a C+11 b B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}\right )+\frac {2 C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2}{11 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {1}{7} \left (77 \left (9 a^2 B+18 a A b+14 a b C+7 b^2 B\right ) \left (\frac {3}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+9 \left (11 a^2 (7 A+5 C)+110 a b B+5 b^2 (11 A+9 C)\right ) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {18 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (4 a^2 C+22 a b B+11 A b^2+9 b^2 C\right )}{7 d}\right )+\frac {2 b (4 a C+11 b B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}\right )+\frac {2 C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2}{11 d}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {1}{7} \left (9 \left (11 a^2 (7 A+5 C)+110 a b B+5 b^2 (11 A+9 C)\right ) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+77 \left (9 a^2 B+18 a A b+14 a b C+7 b^2 B\right ) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\right )+\frac {18 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (4 a^2 C+22 a b B+11 A b^2+9 b^2 C\right )}{7 d}\right )+\frac {2 b (4 a C+11 b B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}\right )+\frac {2 C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2}{11 d}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {18 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (4 a^2 C+22 a b B+11 A b^2+9 b^2 C\right )}{7 d}+\frac {1}{7} \left (77 \left (9 a^2 B+18 a A b+14 a b C+7 b^2 B\right ) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+9 \left (11 a^2 (7 A+5 C)+110 a b B+5 b^2 (11 A+9 C)\right ) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )\right )+\frac {2 b (4 a C+11 b B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}\right )+\frac {2 C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2}{11 d}\) |
(2*C*Cos[c + d*x]^(5/2)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(11*d) + ((2* b*(11*b*B + 4*a*C)*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(9*d) + ((18*(11*A*b^2 + 22*a*b*B + 4*a^2*C + 9*b^2*C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(7*d) + (9*(110*a*b*B + 11*a^2*(7*A + 5*C) + 5*b^2*(11*A + 9*C))*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)) + 77*(18* a*A*b + 9*a^2*B + 7*b^2*B + 14*a*b*C)*((6*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d)))/7)/9)/11
3.11.72.3.1 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[((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)*Cos[e + f*x]*(a + b*Sin[e + f*x ])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a *c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} , x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Leaf count of result is larger than twice the leaf count of optimal. \(862\) vs. \(2(333)=666\).
Time = 15.55 (sec) , antiderivative size = 863, normalized size of antiderivative = 2.83
method | result | size |
default | \(\text {Expression too large to display}\) | \(863\) |
parts | \(\text {Expression too large to display}\) | \(1060\) |
int(cos(d*x+c)^(3/2)*(a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, method=_RETURNVERBOSE)
-2/3465*((-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(20160*C* b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^12+(-12320*B*b^2-24640*C*a*b-504 00*C*b^2)*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+(7920*A*b^2+15840*B*a*b +24640*B*b^2+7920*C*a^2+49280*C*a*b+56880*C*b^2)*sin(1/2*d*x+1/2*c)^8*cos( 1/2*d*x+1/2*c)+(-11088*A*a*b-11880*A*b^2-5544*B*a^2-23760*B*a*b-22792*B*b^ 2-11880*C*a^2-45584*C*a*b-34920*C*b^2)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/ 2*c)+(4620*A*a^2+11088*A*a*b+9240*A*b^2+5544*B*a^2+18480*B*a*b+10472*B*b^2 +9240*C*a^2+20944*C*a*b+13860*C*b^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2* c)+(-2310*A*a^2-2772*A*a*b-2640*A*b^2-1386*B*a^2-5280*B*a*b-1848*B*b^2-264 0*C*a^2-3696*C*a*b-2790*C*b^2)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+115 5*A*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)*Elli pticF(cos(1/2*d*x+1/2*c),2^(1/2))+825*A*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-4158 *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*b+1650*B*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))-2079 *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^2-1617*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si n(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+825* a^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)*Ell...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.15 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.17 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (315 \, C b^{2} \cos \left (d x + c\right )^{4} + 385 \, {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )^{3} + 165 \, {\left (7 \, A + 5 \, C\right )} a^{2} + 1650 \, B a b + 75 \, {\left (11 \, A + 9 \, C\right )} b^{2} + 45 \, {\left (11 \, C a^{2} + 22 \, B a b + {\left (11 \, A + 9 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} + 77 \, {\left (9 \, B a^{2} + 2 \, {\left (9 \, A + 7 \, C\right )} a b + 7 \, B b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 15 \, \sqrt {2} {\left (11 i \, {\left (7 \, A + 5 \, C\right )} a^{2} + 110 i \, B a b + 5 i \, {\left (11 \, A + 9 \, C\right )} b^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 15 \, \sqrt {2} {\left (-11 i \, {\left (7 \, A + 5 \, C\right )} a^{2} - 110 i \, B a b - 5 i \, {\left (11 \, A + 9 \, C\right )} b^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 231 \, \sqrt {2} {\left (-9 i \, B a^{2} - 2 i \, {\left (9 \, A + 7 \, C\right )} a b - 7 i \, B b^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 231 \, \sqrt {2} {\left (9 i \, B a^{2} + 2 i \, {\left (9 \, A + 7 \, C\right )} a b + 7 i \, B b^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{3465 \, d} \]
integrate(cos(d*x+c)^(3/2)*(a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c) ^2),x, algorithm="fricas")
1/3465*(2*(315*C*b^2*cos(d*x + c)^4 + 385*(2*C*a*b + B*b^2)*cos(d*x + c)^3 + 165*(7*A + 5*C)*a^2 + 1650*B*a*b + 75*(11*A + 9*C)*b^2 + 45*(11*C*a^2 + 22*B*a*b + (11*A + 9*C)*b^2)*cos(d*x + c)^2 + 77*(9*B*a^2 + 2*(9*A + 7*C) *a*b + 7*B*b^2)*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 15*sqrt(2) *(11*I*(7*A + 5*C)*a^2 + 110*I*B*a*b + 5*I*(11*A + 9*C)*b^2)*weierstrassPI nverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 15*sqrt(2)*(-11*I*(7*A + 5* C)*a^2 - 110*I*B*a*b - 5*I*(11*A + 9*C)*b^2)*weierstrassPInverse(-4, 0, co s(d*x + c) - I*sin(d*x + c)) - 231*sqrt(2)*(-9*I*B*a^2 - 2*I*(9*A + 7*C)*a *b - 7*I*B*b^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 231*sqrt(2)*(9*I*B*a^2 + 2*I*(9*A + 7*C)*a*b + 7 *I*B*b^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/d
Timed out. \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \]
\[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {3}{2}} \,d x } \]
integrate(cos(d*x+c)^(3/2)*(a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c) ^2),x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^2*c os(d*x + c)^(3/2), x)
\[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {3}{2}} \,d x } \]
integrate(cos(d*x+c)^(3/2)*(a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c) ^2),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^2*c os(d*x + c)^(3/2), x)
Time = 3.72 (sec) , antiderivative size = 401, normalized size of antiderivative = 1.31 \[ \int \cos ^{\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=\frac {2\,A\,a^2\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}-\frac {2\,B\,a^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}}-\frac {2\,A\,b^2\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,a^2\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,B\,b^2\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{11\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,b^2\,{\cos \left (c+d\,x\right )}^{13/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {13}{4};\ \frac {17}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{13\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {4\,A\,a\,b\,{\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}}-\frac {4\,B\,a\,b\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {4\,C\,a\,b\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{11\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]
(2*A*a^2*(cos(c + d*x)^(1/2)*sin(c + d*x) + ellipticF(c/2 + (d*x)/2, 2)))/ (3*d) - (2*B*a^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)) - (2*A*b^2*cos(c + d*x)^( 9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(9*d*(sin(c + d*x)^2)^(1/2)) - (2*C*a^2*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/ 2, 9/4], 13/4, cos(c + d*x)^2))/(9*d*(sin(c + d*x)^2)^(1/2)) - (2*B*b^2*co s(c + d*x)^(11/2)*sin(c + d*x)*hypergeom([1/2, 11/4], 15/4, cos(c + d*x)^2 ))/(11*d*(sin(c + d*x)^2)^(1/2)) - (2*C*b^2*cos(c + d*x)^(13/2)*sin(c + d* x)*hypergeom([1/2, 13/4], 17/4, cos(c + d*x)^2))/(13*d*(sin(c + d*x)^2)^(1 /2)) - (4*A*a*b*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)) - (4*B*a*b*cos(c + d*x)^(9 /2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(9*d*(sin(c + d*x)^2)^(1/2)) - (4*C*a*b*cos(c + d*x)^(11/2)*sin(c + d*x)*hypergeom([1/ 2, 11/4], 15/4, cos(c + d*x)^2))/(11*d*(sin(c + d*x)^2)^(1/2))