Integrand size = 43, antiderivative size = 267 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {56 a^4 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {8 a^4 (4 A+5 B+4 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {4 a^4 (A-25 B-41 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a (5 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (5 A+15 B+19 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}-\frac {4 (6 A+25 B+34 C) \sqrt {\cos (c+d x)} \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{15 d} \]
56/5*a^4*(A-C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(s in(1/2*d*x+1/2*c),2^(1/2))/d+8/3*a^4*(4*A+5*B+4*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*a*(5 *B+8*C)*(a+a*cos(d*x+c))^3*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/5*C*(a+a*cos(d* x+c))^4*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/5*(5*A+15*B+19*C)*(a^2+a^2*cos(d*x +c))^2*sin(d*x+c)/d/cos(d*x+c)^(1/2)+4/15*a^4*(A-25*B-41*C)*sin(d*x+c)*cos (d*x+c)^(1/2)/d-4/15*(6*A+25*B+34*C)*(a^4+a^4*cos(d*x+c))*sin(d*x+c)*cos(d *x+c)^(1/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 13.58 (sec) , antiderivative size = 1449, normalized size of antiderivative = 5.43 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \]
Integrate[Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
(Cos[c + d*x]^(13/2)*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Se c[c + d*x] + C*Sec[c + d*x]^2)*(-1/40*((23*A - 20*B - 61*C + 33*A*Cos[2*c] + 20*B*Cos[2*c] + 5*C*Cos[2*c])*Csc[c]*Sec[c])/d + ((4*A + B)*Cos[d*x]*Si n[c])/(12*d) + (A*Cos[2*d*x]*Sin[2*c])/(40*d) + ((4*A + B)*Cos[c]*Sin[d*x] )/(12*d) + (C*Sec[c]*Sec[c + d*x]^3*Sin[d*x])/(20*d) + (Sec[c]*Sec[c + d*x ]^2*(3*C*Sin[c] + 5*B*Sin[d*x] + 20*C*Sin[d*x]))/(60*d) + (Sec[c]*Sec[c + d*x]*(5*B*Sin[c] + 20*C*Sin[c] + 15*A*Sin[d*x] + 60*B*Sin[d*x] + 99*C*Sin[ d*x]))/(60*d) + (A*Cos[2*c]*Sin[2*d*x])/(40*d)))/(A + 2*C + 2*B*Cos[c + d* x] + A*Cos[2*c + 2*d*x]) - (4*A*Cos[c + d*x]^6*Csc[c]*HypergeometricPFQ[{1 /4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^8*(a + a* Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[C ot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[ c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*( A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (5* B*Cos[c + d*x]^6*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - Arc Tan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - Arc Tan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])] *Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A *Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (4*C*Cos[c + d*x]^6*Csc[c]*Hyp...
Time = 2.11 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.08, number of steps used = 23, number of rules used = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.535, Rules used = {3042, 4600, 3042, 3522, 27, 3042, 3454, 27, 3042, 3454, 27, 3042, 3455, 27, 3042, 3447, 3042, 3502, 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.
\(\displaystyle \int \cos ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^{5/2} (a \sec (c+d x)+a)^4 \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )dx\) |
\(\Big \downarrow \) 4600 |
\(\displaystyle \int \frac {(a \cos (c+d x)+a)^4 \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right )}{\cos ^{\frac {7}{2}}(c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4 \left (A \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+C\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
\(\Big \downarrow \) 3522 |
\(\displaystyle \frac {2 \int \frac {(\cos (c+d x) a+a)^4 (a (5 B+8 C)+5 a (A-C) \cos (c+d x))}{2 \cos ^{\frac {5}{2}}(c+d x)}dx}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {(\cos (c+d x) a+a)^4 (a (5 B+8 C)+5 a (A-C) \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)}dx}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^4 \left (a (5 B+8 C)+5 a (A-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3454 |
\(\displaystyle \frac {\frac {2}{3} \int \frac {(\cos (c+d x) a+a)^3 \left (3 (5 A+15 B+19 C) a^2+5 (3 A-5 B-11 C) \cos (c+d x) a^2\right )}{2 \cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{3} \int \frac {(\cos (c+d x) a+a)^3 \left (3 (5 A+15 B+19 C) a^2+5 (3 A-5 B-11 C) \cos (c+d x) a^2\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (3 (5 A+15 B+19 C) a^2+5 (3 A-5 B-11 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3454 |
\(\displaystyle \frac {\frac {1}{3} \left (2 \int \frac {5 (\cos (c+d x) a+a)^2 \left (a^3 (9 A+20 B+23 C)-a^3 (6 A+25 B+34 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{3} \left (10 \int \frac {(\cos (c+d x) a+a)^2 \left (a^3 (9 A+20 B+23 C)-a^3 (6 A+25 B+34 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} \left (10 \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a^3 (9 A+20 B+23 C)-a^3 (6 A+25 B+34 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3455 |
\(\displaystyle \frac {\frac {1}{3} \left (10 \left (\frac {2}{5} \int \frac {3 (\cos (c+d x) a+a) \left ((13 A+25 B+27 C) a^4+(A-25 B-41 C) \cos (c+d x) a^4\right )}{2 \sqrt {\cos (c+d x)}}dx-\frac {2 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^5 \cos (c+d x)+a^5\right )}{5 d}\right )+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{3} \left (10 \left (\frac {3}{5} \int \frac {(\cos (c+d x) a+a) \left ((13 A+25 B+27 C) a^4+(A-25 B-41 C) \cos (c+d x) a^4\right )}{\sqrt {\cos (c+d x)}}dx-\frac {2 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^5 \cos (c+d x)+a^5\right )}{5 d}\right )+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} \left (10 \left (\frac {3}{5} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((13 A+25 B+27 C) a^4+(A-25 B-41 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^4\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^5 \cos (c+d x)+a^5\right )}{5 d}\right )+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3447 |
\(\displaystyle \frac {\frac {1}{3} \left (10 \left (\frac {3}{5} \int \frac {(A-25 B-41 C) \cos ^2(c+d x) a^5+(13 A+25 B+27 C) a^5+\left ((A-25 B-41 C) a^5+(13 A+25 B+27 C) a^5\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx-\frac {2 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^5 \cos (c+d x)+a^5\right )}{5 d}\right )+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} \left (10 \left (\frac {3}{5} \int \frac {(A-25 B-41 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^5+(13 A+25 B+27 C) a^5+\left ((A-25 B-41 C) a^5+(13 A+25 B+27 C) a^5\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^5 \cos (c+d x)+a^5\right )}{5 d}\right )+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {\frac {1}{3} \left (10 \left (\frac {3}{5} \left (\frac {2}{3} \int \frac {5 (4 A+5 B+4 C) a^5+21 (A-C) \cos (c+d x) a^5}{\sqrt {\cos (c+d x)}}dx+\frac {2 a^5 (A-25 B-41 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^5 \cos (c+d x)+a^5\right )}{5 d}\right )+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} \left (10 \left (\frac {3}{5} \left (\frac {2}{3} \int \frac {5 (4 A+5 B+4 C) a^5+21 (A-C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^5}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^5 (A-25 B-41 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^5 \cos (c+d x)+a^5\right )}{5 d}\right )+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {\frac {1}{3} \left (10 \left (\frac {3}{5} \left (\frac {2}{3} \left (5 a^5 (4 A+5 B+4 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+21 a^5 (A-C) \int \sqrt {\cos (c+d x)}dx\right )+\frac {2 a^5 (A-25 B-41 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^5 \cos (c+d x)+a^5\right )}{5 d}\right )+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} \left (10 \left (\frac {3}{5} \left (\frac {2}{3} \left (5 a^5 (4 A+5 B+4 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+21 a^5 (A-C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {2 a^5 (A-25 B-41 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^5 \cos (c+d x)+a^5\right )}{5 d}\right )+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {1}{3} \left (10 \left (\frac {3}{5} \left (\frac {2}{3} \left (5 a^5 (4 A+5 B+4 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {42 a^5 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 a^5 (A-25 B-41 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^5 \cos (c+d x)+a^5\right )}{5 d}\right )+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {2 a^2 (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (10 \left (\frac {3}{5} \left (\frac {2 a^5 (A-25 B-41 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}+\frac {2}{3} \left (\frac {10 a^5 (4 A+5 B+4 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {42 a^5 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )-\frac {2 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^5 \cos (c+d x)+a^5\right )}{5 d}\right )+\frac {6 a^3 (5 A+15 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
(2*C*(a + a*Cos[c + d*x])^4*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + ((2*a ^2*(5*B + 8*C)*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2 )) + ((6*a^3*(5*A + 15*B + 19*C)*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(d*S qrt[Cos[c + d*x]]) + 10*((-2*(6*A + 25*B + 34*C)*Sqrt[Cos[c + d*x]]*(a^5 + a^5*Cos[c + d*x])*Sin[c + d*x])/(5*d) + (3*((2*((42*a^5*(A - C)*EllipticE [(c + d*x)/2, 2])/d + (10*a^5*(4*A + 5*B + 4*C)*EllipticF[(c + d*x)/2, 2]) /d))/3 + (2*a^5*(A - 25*B - 41*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d))) /5))/3)/(5*a)
3.13.13.3.1 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_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-b^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp [a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B *(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 ])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 ) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1 ] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
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_)])^(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/(b*d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m* (c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C - B*d)*( a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2* (n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ [m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x _)])^(m_.)*((A_.) + (B_.)*sec[(e_.) + (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.) *(x_)]^2), x_Symbol] :> Simp[d^(m + 2) Int[(b + a*Cos[e + f*x])^m*(d*Cos[ e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; Fr eeQ[{a, b, d, e, f, A, B, C, n}, x] && !IntegerQ[n] && IntegerQ[m]
Leaf count of result is larger than twice the leaf count of optimal. \(1213\) vs. \(2(295)=590\).
Time = 647.27 (sec) , antiderivative size = 1214, normalized size of antiderivative = 4.55
int(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, method=_RETURNVERBOSE)
-8/15*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^4/(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)^3*(-84*C*(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)^4-84*A* (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))*(si n(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+84*C*(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)^2+186*A*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+84* A*(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)^4-25*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))-102*A*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+19*A*sin(1/2*d*x+1/2*c )^2*cos(1/2*d*x+1/2*c)-100*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/ 2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^ 4+100*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)) *(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+21*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))+198*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-218*C*cos(1/2*d* x+1/2*c)*sin(1/2*d*x+1/2*c)^4-20*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-20*C*(sin(...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.13 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.05 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (10 i \, \sqrt {2} {\left (4 \, A + 5 \, B + 4 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 10 i \, \sqrt {2} {\left (4 \, A + 5 \, B + 4 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 42 i \, \sqrt {2} {\left (A - C\right )} a^{4} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 42 i \, \sqrt {2} {\left (A - C\right )} a^{4} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (3 \, A a^{4} \cos \left (d x + c\right )^{4} + 5 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A + 20 \, B + 33 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 5 \, {\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) + 3 \, C a^{4}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{15 \, d \cos \left (d x + c\right )^{3}} \]
integrate(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="fricas")
-2/15*(10*I*sqrt(2)*(4*A + 5*B + 4*C)*a^4*cos(d*x + c)^3*weierstrassPInver se(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 10*I*sqrt(2)*(4*A + 5*B + 4*C)* a^4*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c )) - 42*I*sqrt(2)*(A - C)*a^4*cos(d*x + c)^3*weierstrassZeta(-4, 0, weiers trassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 42*I*sqrt(2)*(A - C )*a^4*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos (d*x + c) - I*sin(d*x + c))) - (3*A*a^4*cos(d*x + c)^4 + 5*(4*A + B)*a^4*c os(d*x + c)^3 + 3*(5*A + 20*B + 33*C)*a^4*cos(d*x + c)^2 + 5*(B + 4*C)*a^4 *cos(d*x + c) + 3*C*a^4)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^ 3)
Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
integrate(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="maxima")
\[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{4} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x } \]
integrate(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^4*c os(d*x + c)^(5/2), x)
Time = 22.20 (sec) , antiderivative size = 525, normalized size of antiderivative = 1.97 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,\left (12\,B\,a^4\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+19\,B\,a^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+B\,a^4\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )\right )}{3\,d}+\frac {2\,\left (C\,a^4\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+4\,C\,a^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{d}+\frac {2\,\left (\frac {34\,C\,a^4\,\sin \left (c+d\,x\right )}{\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {C\,a^4\,\sin \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d}+\frac {4\,A\,a^4\,\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\,A\,a^4\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {8\,A\,a^4\,\mathrm {F}\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 {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\,A\,a^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\,B\,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\,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 {8\,C\,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 {8\,C\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {7}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]
(2*(12*B*a^4*ellipticE(c/2 + (d*x)/2, 2) + 19*B*a^4*ellipticF(c/2 + (d*x)/ 2, 2) + B*a^4*cos(c + d*x)^(1/2)*sin(c + d*x)))/(3*d) + (2*(C*a^4*elliptic E(c/2 + (d*x)/2, 2) + 4*C*a^4*ellipticF(c/2 + (d*x)/2, 2)))/d + (2*((34*C* a^4*sin(c + d*x))/(cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (C*a^4*sin (c + d*x))/(cos(c + d*x)^(5/2)*(sin(c + d*x)^2)^(1/2)))*hypergeom([-1/4, 1 /2], 3/4, cos(c + d*x)^2))/(5*d) + (4*A*a^4*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d + (12*A*a^4*ellipticE(c/2 + (d*x)/2, 2))/d + (8*A*a^4*ellipticF(c/2 + (d*x)/2, 2))/d + (2*A*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*A*a^4*cos(c + d*x)^(7/2)*sin(c + d*x)*hype rgeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2)) + (8 *B*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*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)) + (8*C*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)) - (8*C*a^4*sin(c + d*x)*h ypergeom([-1/4, 1/2], 7/4, cos(c + d*x)^2))/(15*d*cos(c + d*x)^(1/2)*(sin( c + d*x)^2)^(1/2))