Integrand size = 43, antiderivative size = 386 \[ \int \frac {(a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 \left (20 a^3 b B-20 a b^3 B+30 a^2 b^2 (A-C)-b^4 (5 A+3 C)+a^4 (3 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (a^4 B+18 a^2 b^2 B+b^4 B+4 a b^3 (3 A+C)+4 a^3 b (A+3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}-\frac {2 b \left (105 a^2 b B-5 b^3 B+4 a b^2 (33 A-5 C)+6 a^3 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}-\frac {2 b^2 \left (50 a b B+b^2 (59 A-3 C)+3 a^2 (3 A+5 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 \left (16 A b^2+15 a b B+a^2 (3 A+5 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 (8 A b+5 a B) (a+b \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]
-2/5*(20*B*a^3*b-20*B*a*b^3+30*a^2*b^2*(A-C)-b^4*(5*A+3*C)+a^4*(3*A+5*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))/d+2/3*(B*a^4+18*B*a^2*b^2+B*b^4+4*a*b^3*(3*A+C)+4*a^3*b*(A+3*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-2/15*b^2*(50*B*a*b+b^2*(59*A-3*C)+3*a^2*(3*A+5*C))*cos(d* x+c)^(3/2)*sin(d*x+c)/d+2/15*(8*A*b+5*B*a)*(a+b*cos(d*x+c))^3*sin(d*x+c)/d /cos(d*x+c)^(3/2)+2/5*A*(a+b*cos(d*x+c))^4*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2 /5*(16*A*b^2+15*B*a*b+a^2*(3*A+5*C))*(a+b*cos(d*x+c))^2*sin(d*x+c)/d/cos(d *x+c)^(1/2)-2/15*b*(105*B*a^2*b-5*B*b^3+4*a*b^2*(33*A-5*C)+6*a^3*(3*A+5*C) )*sin(d*x+c)*cos(d*x+c)^(1/2)/d
Time = 6.69 (sec) , antiderivative size = 316, normalized size of antiderivative = 0.82 \[ \int \frac {(a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \left (-9 a^4 A-90 a^2 A b^2+15 A b^4-60 a^3 b B+60 a b^3 B-15 a^4 C+90 a^2 b^2 C+9 b^4 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 \left (20 a^3 A b+60 a A b^3+5 a^4 B+90 a^2 b^2 B+5 b^4 B+60 a^3 b C+20 a b^3 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{15 d}+\frac {\sqrt {\cos (c+d x)} \left (\frac {2}{3} b^3 (b B+4 a C) \sin (c+d x)+\frac {2}{3} \sec ^2(c+d x) \left (4 a^3 A b \sin (c+d x)+a^4 B \sin (c+d x)\right )+\frac {2}{5} \sec (c+d x) \left (3 a^4 A \sin (c+d x)+30 a^2 A b^2 \sin (c+d x)+20 a^3 b B \sin (c+d x)+5 a^4 C \sin (c+d x)\right )+\frac {1}{5} b^4 C \sin (2 (c+d x))+\frac {2}{5} a^4 A \sec ^2(c+d x) \tan (c+d x)\right )}{d} \]
Integrate[((a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)) /Cos[c + d*x]^(7/2),x]
(2*(-9*a^4*A - 90*a^2*A*b^2 + 15*A*b^4 - 60*a^3*b*B + 60*a*b^3*B - 15*a^4* C + 90*a^2*b^2*C + 9*b^4*C)*EllipticE[(c + d*x)/2, 2] + 2*(20*a^3*A*b + 60 *a*A*b^3 + 5*a^4*B + 90*a^2*b^2*B + 5*b^4*B + 60*a^3*b*C + 20*a*b^3*C)*Ell ipticF[(c + d*x)/2, 2])/(15*d) + (Sqrt[Cos[c + d*x]]*((2*b^3*(b*B + 4*a*C) *Sin[c + d*x])/3 + (2*Sec[c + d*x]^2*(4*a^3*A*b*Sin[c + d*x] + a^4*B*Sin[c + d*x]))/3 + (2*Sec[c + d*x]*(3*a^4*A*Sin[c + d*x] + 30*a^2*A*b^2*Sin[c + d*x] + 20*a^3*b*B*Sin[c + d*x] + 5*a^4*C*Sin[c + d*x]))/5 + (b^4*C*Sin[2* (c + d*x)])/5 + (2*a^4*A*Sec[c + d*x]^2*Tan[c + d*x])/5))/d
Time = 2.59 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.465, Rules used = {3042, 3526, 27, 3042, 3526, 27, 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.
\(\displaystyle \int \frac {(a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^4 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
\(\Big \downarrow \) 3526 |
\(\displaystyle \frac {2}{5} \int \frac {(a+b \cos (c+d x))^3 \left (-5 b (A-C) \cos ^2(c+d x)+(3 a A+5 b B+5 a C) \cos (c+d x)+8 A b+5 a B\right )}{2 \cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {(a+b \cos (c+d x))^3 \left (-5 b (A-C) \cos ^2(c+d x)+(3 a A+5 b B+5 a C) \cos (c+d x)+8 A b+5 a B\right )}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (-5 b (A-C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+(3 a A+5 b B+5 a C) \sin \left (c+d x+\frac {\pi }{2}\right )+8 A b+5 a B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3526 |
\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {(a+b \cos (c+d x))^2 \left (-5 b (11 A b-3 C b+5 a B) \cos ^2(c+d x)+\left (5 B a^2+2 b (A+15 C) a+15 b^2 B\right ) \cos (c+d x)+3 \left ((3 A+5 C) a^2+15 b B a+16 A b^2\right )\right )}{2 \cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 (5 a B+8 A b) \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {(a+b \cos (c+d x))^2 \left (-5 b (11 A b-3 C b+5 a B) \cos ^2(c+d x)+\left (5 B a^2+2 b (A+15 C) a+15 b^2 B\right ) \cos (c+d x)+3 \left ((3 A+5 C) a^2+15 b B a+16 A b^2\right )\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 (5 a B+8 A b) \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (-5 b (11 A b-3 C b+5 a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (5 B a^2+2 b (A+15 C) a+15 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left ((3 A+5 C) a^2+15 b B a+16 A b^2\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 (5 a B+8 A b) \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3526 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (2 \int \frac {(a+b \cos (c+d x)) \left (5 B a^3+2 b (19 A+45 C) a^2+195 b^2 B a+192 A b^3-5 b \left (3 (3 A+5 C) a^2+50 b B a+b^2 (59 A-3 C)\right ) \cos ^2(c+d x)-\left (3 (3 A+5 C) a^3+65 b B a^2+b^2 (101 A-45 C) a-15 b^3 B\right ) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)+15 a b B+16 A b^2\right ) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (5 a B+8 A b) \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {(a+b \cos (c+d x)) \left (5 B a^3+(38 A b+90 C b) a^2+195 b^2 B a+192 A b^3-5 b \left (3 (3 A+5 C) a^2+50 b B a+b^2 (59 A-3 C)\right ) \cos ^2(c+d x)-\left (3 (3 A+5 C) a^3+65 b B a^2+b^2 (101 A-45 C) a-15 b^3 B\right ) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx+\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)+15 a b B+16 A b^2\right ) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (5 a B+8 A b) \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (5 B a^3+(38 A b+90 C b) a^2+195 b^2 B a+192 A b^3-5 b \left (3 (3 A+5 C) a^2+50 b B a+b^2 (59 A-3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (-3 (3 A+5 C) a^3-65 b B a^2-b^2 (101 A-45 C) a+15 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)+15 a b B+16 A b^2\right ) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (5 a B+8 A b) \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3512 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {2}{5} \int \frac {5 \left (-3 b \left (6 (3 A+5 C) a^3+105 b B a^2+4 b^2 (33 A-5 C) a-5 b^3 B\right ) \cos ^2(c+d x)-3 \left ((3 A+5 C) a^4+20 b B a^3+30 b^2 (A-C) a^2-20 b^3 B a-b^4 (5 A+3 C)\right ) \cos (c+d x)+a \left (5 B a^3+(38 A b+90 C b) a^2+195 b^2 B a+192 A b^3\right )\right )}{2 \sqrt {\cos (c+d x)}}dx-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (3 a^2 (3 A+5 C)+50 a b B+b^2 (59 A-3 C)\right )}{d}+\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)+15 a b B+16 A b^2\right ) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (5 a B+8 A b) \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {-3 b \left (6 (3 A+5 C) a^3+105 b B a^2+4 b^2 (33 A-5 C) a-5 b^3 B\right ) \cos ^2(c+d x)-3 \left ((3 A+5 C) a^4+20 b B a^3+30 b^2 (A-C) a^2-20 b^3 B a-b^4 (5 A+3 C)\right ) \cos (c+d x)+a \left (5 B a^3+(38 A b+90 C b) a^2+195 b^2 B a+192 A b^3\right )}{\sqrt {\cos (c+d x)}}dx-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (3 a^2 (3 A+5 C)+50 a b B+b^2 (59 A-3 C)\right )}{d}+\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)+15 a b B+16 A b^2\right ) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (5 a B+8 A b) \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {-3 b \left (6 (3 A+5 C) a^3+105 b B a^2+4 b^2 (33 A-5 C) a-5 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-3 \left ((3 A+5 C) a^4+20 b B a^3+30 b^2 (A-C) a^2-20 b^3 B a-b^4 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a \left (5 B a^3+(38 A b+90 C b) a^2+195 b^2 B a+192 A b^3\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (3 a^2 (3 A+5 C)+50 a b B+b^2 (59 A-3 C)\right )}{d}+\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)+15 a b B+16 A b^2\right ) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (5 a B+8 A b) \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {2}{3} \int \frac {3 \left (5 \left (B a^4+4 b (A+3 C) a^3+18 b^2 B a^2+4 b^3 (3 A+C) a+b^4 B\right )-3 \left ((3 A+5 C) a^4+20 b B a^3+30 b^2 (A-C) a^2-20 b^3 B a-b^4 (5 A+3 C)\right ) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (3 a^2 (3 A+5 C)+50 a b B+b^2 (59 A-3 C)\right )}{d}+\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)+15 a b B+16 A b^2\right ) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b \sin (c+d x) \sqrt {\cos (c+d x)} \left (6 a^3 (3 A+5 C)+105 a^2 b B+4 a b^2 (33 A-5 C)-5 b^3 B\right )}{d}\right )+\frac {2 (5 a B+8 A b) \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {5 \left (B a^4+4 b (A+3 C) a^3+18 b^2 B a^2+4 b^3 (3 A+C) a+b^4 B\right )-3 \left ((3 A+5 C) a^4+20 b B a^3+30 b^2 (A-C) a^2-20 b^3 B a-b^4 (5 A+3 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (3 a^2 (3 A+5 C)+50 a b B+b^2 (59 A-3 C)\right )}{d}+\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)+15 a b B+16 A b^2\right ) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b \sin (c+d x) \sqrt {\cos (c+d x)} \left (6 a^3 (3 A+5 C)+105 a^2 b B+4 a b^2 (33 A-5 C)-5 b^3 B\right )}{d}\right )+\frac {2 (5 a B+8 A b) \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {5 \left (B a^4+4 b (A+3 C) a^3+18 b^2 B a^2+4 b^3 (3 A+C) a+b^4 B\right )-3 \left ((3 A+5 C) a^4+20 b B a^3+30 b^2 (A-C) a^2-20 b^3 B a-b^4 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (3 a^2 (3 A+5 C)+50 a b B+b^2 (59 A-3 C)\right )}{d}+\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)+15 a b B+16 A b^2\right ) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b \sin (c+d x) \sqrt {\cos (c+d x)} \left (6 a^3 (3 A+5 C)+105 a^2 b B+4 a b^2 (33 A-5 C)-5 b^3 B\right )}{d}\right )+\frac {2 (5 a B+8 A b) \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (a^4 B+4 a^3 b (A+3 C)+18 a^2 b^2 B+4 a b^3 (3 A+C)+b^4 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 \left (a^4 (3 A+5 C)+20 a^3 b B+30 a^2 b^2 (A-C)-20 a b^3 B-b^4 (5 A+3 C)\right ) \int \sqrt {\cos (c+d x)}dx-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (3 a^2 (3 A+5 C)+50 a b B+b^2 (59 A-3 C)\right )}{d}+\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)+15 a b B+16 A b^2\right ) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b \sin (c+d x) \sqrt {\cos (c+d x)} \left (6 a^3 (3 A+5 C)+105 a^2 b B+4 a b^2 (33 A-5 C)-5 b^3 B\right )}{d}\right )+\frac {2 (5 a B+8 A b) \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (a^4 B+4 a^3 b (A+3 C)+18 a^2 b^2 B+4 a b^3 (3 A+C)+b^4 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 \left (a^4 (3 A+5 C)+20 a^3 b B+30 a^2 b^2 (A-C)-20 a b^3 B-b^4 (5 A+3 C)\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (3 a^2 (3 A+5 C)+50 a b B+b^2 (59 A-3 C)\right )}{d}+\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)+15 a b B+16 A b^2\right ) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b \sin (c+d x) \sqrt {\cos (c+d x)} \left (6 a^3 (3 A+5 C)+105 a^2 b B+4 a b^2 (33 A-5 C)-5 b^3 B\right )}{d}\right )+\frac {2 (5 a B+8 A b) \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (a^4 B+4 a^3 b (A+3 C)+18 a^2 b^2 B+4 a b^3 (3 A+C)+b^4 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (3 a^2 (3 A+5 C)+50 a b B+b^2 (59 A-3 C)\right )}{d}+\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)+15 a b B+16 A b^2\right ) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b \sin (c+d x) \sqrt {\cos (c+d x)} \left (6 a^3 (3 A+5 C)+105 a^2 b B+4 a b^2 (33 A-5 C)-5 b^3 B\right )}{d}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^4 (3 A+5 C)+20 a^3 b B+30 a^2 b^2 (A-C)-20 a b^3 B-b^4 (5 A+3 C)\right )}{d}\right )+\frac {2 (5 a B+8 A b) \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (3 a^2 (3 A+5 C)+50 a b B+b^2 (59 A-3 C)\right )}{d}+\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)+15 a b B+16 A b^2\right ) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b \sin (c+d x) \sqrt {\cos (c+d x)} \left (6 a^3 (3 A+5 C)+105 a^2 b B+4 a b^2 (33 A-5 C)-5 b^3 B\right )}{d}+\frac {10 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^4 B+4 a^3 b (A+3 C)+18 a^2 b^2 B+4 a b^3 (3 A+C)+b^4 B\right )}{d}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^4 (3 A+5 C)+20 a^3 b B+30 a^2 b^2 (A-C)-20 a b^3 B-b^4 (5 A+3 C)\right )}{d}\right )+\frac {2 (5 a B+8 A b) \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
(2*A*(a + b*Cos[c + d*x])^4*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + ((2*( 8*A*b + 5*a*B)*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2 )) + ((-6*(20*a^3*b*B - 20*a*b^3*B + 30*a^2*b^2*(A - C) - b^4*(5*A + 3*C) + a^4*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/d + (10*(a^4*B + 18*a^2*b^2* B + b^4*B + 4*a*b^3*(3*A + C) + 4*a^3*b*(A + 3*C))*EllipticF[(c + d*x)/2, 2])/d - (2*b*(105*a^2*b*B - 5*b^3*B + 4*a*b^2*(33*A - 5*C) + 6*a^3*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/d - (2*b^2*(50*a*b*B + b^2*(59*A - 3*C) + 3*a^2*(3*A + 5*C))*Cos[c + d*x]^(3/2)*Sin[c + d*x])/d + (6*(16*A*b^ 2 + 15*a*b*B + a^2*(3*A + 5*C))*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(d*Sq rt[Cos[c + d*x]]))/3)/5
3.11.91.3.1 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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 5.63 (sec) , antiderivative size = 1382, normalized size of antiderivative = 3.58
method | result | size |
parts | \(\text {Expression too large to display}\) | \(1382\) |
default | \(\text {Expression too large to display}\) | \(1857\) |
int((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, method=_RETURNVERBOSE)
-2/3*(4*A*a^3*b+B*a^4)*(-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))*sin(1/2*d*x+1/2*c)^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))*((-1+ 2*cos(1/2*d*x+1/2*c)^2)*sin(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)/(-1+2*cos(1/2*d*x+1/2*c)^2)^(3/2)/sin(1/2*d *x+1/2*c)/d-2/3*(B*b^4+4*C*a*b^3)*((-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x +1/2*c)^2)^(1/2)*(4*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-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)*EllipticF(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)/sin(1/2*d*x+1/2*c)/(-1+2*cos(1/2*d*x+1/2 *c)^2)^(1/2)/d+2*(A*b^4+4*B*a*b^3+6*C*a^2*b^2)*((-1+2*cos(1/2*d*x+1/2*c)^2 )*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(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)/sin(1/2*d*x+1/2*c)/(-1+2*cos(1/2*d*x+1 /2*c)^2)^(1/2)/d+2*(4*A*a*b^3+6*B*a^2*b^2+4*C*a^3*b)/d*InverseJacobiAM(1/2 *d*x+1/2*c,2^(1/2))-2*(6*A*a^2*b^2+4*B*a^3*b+C*a^4)*(-2*(-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)*sin(1/2*d*x+1/2*c)^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)*(-2*sin(1/2 *d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.16 (sec) , antiderivative size = 455, normalized size of antiderivative = 1.18 \[ \int \frac {(a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {5 \, \sqrt {2} {\left (i \, B a^{4} + 4 i \, {\left (A + 3 \, C\right )} a^{3} b + 18 i \, B a^{2} b^{2} + 4 i \, {\left (3 \, A + C\right )} a b^{3} + i \, B b^{4}\right )} \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 ) + 5 \, \sqrt {2} {\left (-i \, B a^{4} - 4 i \, {\left (A + 3 \, C\right )} a^{3} b - 18 i \, B a^{2} b^{2} - 4 i \, {\left (3 \, A + C\right )} a b^{3} - i \, B b^{4}\right )} \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^{4} + 20 i \, B a^{3} b + 30 i \, {\left (A - C\right )} a^{2} b^{2} - 20 i \, B a b^{3} - i \, {\left (5 \, A + 3 \, C\right )} b^{4}\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^{4} - 20 i \, B a^{3} b - 30 i \, {\left (A - C\right )} a^{2} b^{2} + 20 i \, B a b^{3} + i \, {\left (5 \, A + 3 \, C\right )} b^{4}\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 (3 \, C b^{4} \cos \left (d x + c\right )^{4} + 3 \, A a^{4} + 5 \, {\left (4 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left ({\left (3 \, A + 5 \, C\right )} a^{4} + 20 \, B a^{3} b + 30 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{3}} \]
integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7 /2),x, algorithm="fricas")
-1/15*(5*sqrt(2)*(I*B*a^4 + 4*I*(A + 3*C)*a^3*b + 18*I*B*a^2*b^2 + 4*I*(3* A + C)*a*b^3 + I*B*b^4)*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I*B*a^4 - 4*I*(A + 3*C)*a^3*b - 18*I* B*a^2*b^2 - 4*I*(3*A + C)*a*b^3 - I*B*b^4)*cos(d*x + c)^3*weierstrassPInve rse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt(2)*(I*(3*A + 5*C)*a^4 + 20*I*B*a^3*b + 30*I*(A - C)*a^2*b^2 - 20*I*B*a*b^3 - I*(5*A + 3*C)*b^4)*c os(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^4 - 20*I*B*a^3*b - 30* I*(A - C)*a^2*b^2 + 20*I*B*a*b^3 + I*(5*A + 3*C)*b^4)*cos(d*x + c)^3*weier strassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c) )) - 2*(3*C*b^4*cos(d*x + c)^4 + 3*A*a^4 + 5*(4*C*a*b^3 + B*b^4)*cos(d*x + c)^3 + 3*((3*A + 5*C)*a^4 + 20*B*a^3*b + 30*A*a^2*b^2)*cos(d*x + c)^2 + 5 *(B*a^4 + 4*A*a^3*b)*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos (d*x + c)^3)
Timed out. \[ \int \frac {(a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{4}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7 /2),x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^4/c os(d*x + c)^(7/2), x)
\[ \int \frac {(a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{4}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7 /2),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^4/c os(d*x + c)^(7/2), x)
Time = 7.13 (sec) , antiderivative size = 524, normalized size of antiderivative = 1.36 \[ \int \frac {(a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2\,\left (B\,b^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+12\,B\,a\,b^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+B\,b^4\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+18\,B\,a^2\,b^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,A\,b^4\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {8\,A\,a\,b^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {8\,C\,a^3\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,C\,a\,b^3\,\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 {12\,C\,a^2\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,a^4\,\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 )}{5\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,B\,a^4\,\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 {2\,C\,a^4\,\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^4\,{\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 {8\,A\,a^3\,b\,\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 {8\,B\,a^3\,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}}+\frac {12\,A\,a^2\,b^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}} \]
(2*(B*b^4*ellipticF(c/2 + (d*x)/2, 2) + 12*B*a*b^3*ellipticE(c/2 + (d*x)/2 , 2) + B*b^4*cos(c + d*x)^(1/2)*sin(c + d*x) + 18*B*a^2*b^2*ellipticF(c/2 + (d*x)/2, 2)))/(3*d) + (2*A*b^4*ellipticE(c/2 + (d*x)/2, 2))/d + (8*A*a*b ^3*ellipticF(c/2 + (d*x)/2, 2))/d + (8*C*a^3*b*ellipticF(c/2 + (d*x)/2, 2) )/d + (4*C*a*b^3*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d + (12*C*a^2*b^2*ellipticE(c/2 + (d*x)/2, 2))/d + (2* A*a^4*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2))/(5*d*cos( c + d*x)^(5/2)*(sin(c + d*x)^2)^(1/2)) + (2*B*a^4*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)) + (2*C*a^4*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^4*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)) + (8*A*a^3*b*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, co s(c + d*x)^2))/(3*d*cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2)) + (8*B*a^3* 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)) + (12*A*a^2*b^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))