Integrand size = 43, antiderivative size = 274 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \left (5 a^3 B-15 a b^2 B+15 a^2 b (A-C)-b^3 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (9 a^2 b B+b^3 B+3 a b^2 (3 A+C)+a^3 (A+3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 b \left (15 A b^2+35 a b B+24 a^2 C+9 b^2 C\right ) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 a^2 (5 a A-5 b B-9 a C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 (5 b B+6 a C) (b+a \cos (c+d x))^2 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (b+a \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]
2/5*(5*B*a^3-15*B*a*b^2+15*a^2*b*(A-C)-b^3*(5*A+3*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*(9 *B*a^2*b+B*b^3+3*a*b^2*(3*A+C)+a^3*(A+3*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/c os(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/15*(5*B*b+6*C* a)*(b+a*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/5*C*(b+a*cos(d*x+c)) ^3*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/15*b*(15*A*b^2+35*B*a*b+24*C*a^2+9*C*b^ 2)*sin(d*x+c)/d/cos(d*x+c)^(1/2)+2/15*a^2*(5*A*a-5*B*b-9*C*a)*sin(d*x+c)*c os(d*x+c)^(1/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 17.00 (sec) , antiderivative size = 3163, normalized size of antiderivative = 11.54 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \]
Integrate[Cos[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
(Cos[c + d*x]^(11/2)*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((-2*(15*a^2*A*b - 10*A*b^3 + 5*a^3*B - 30*a*b^2*B - 30*a^2*b*C - 6*b^3*C + 15*a^2*A*b*Cos[2*c] + 5*a^3*B*Cos[2*c])*Csc[c]*Sec[c])/(5*d) + (4*a^3*A*Cos[d*x]*Sin[c])/(3*d) + (4*a^3*A*Cos[c]*Sin[d*x])/(3*d) + (4*b^ 3*C*Sec[c]*Sec[c + d*x]^3*Sin[d*x])/(5*d) + (4*Sec[c]*Sec[c + d*x]^2*(3*b^ 3*C*Sin[c] + 5*b^3*B*Sin[d*x] + 15*a*b^2*C*Sin[d*x]))/(15*d) + (4*Sec[c]*S ec[c + d*x]*(5*b^3*B*Sin[c] + 15*a*b^2*C*Sin[c] + 15*A*b^3*Sin[d*x] + 45*a *b^2*B*Sin[d*x] + 45*a^2*b*C*Sin[d*x] + 9*b^3*C*Sin[d*x]))/(15*d)))/((b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) - (4* a^3*A*Cos[c + d*x]^5*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqr t[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d* x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (12*a*A*b^2*Cos[c + d*x]^5* Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*( a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - Ar cTan[Cot[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]]]])/ (d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d...
Time = 1.80 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.01, number of steps used = 19, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.442, Rules used = {3042, 4600, 3042, 3526, 27, 3042, 3526, 27, 3042, 3510, 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.
| \(\displaystyle \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (c+d x)^{3/2} (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )dx\) |
| \(\Big \downarrow \) 4600 |
| \(\displaystyle \int \frac {(a \cos (c+d x)+b)^3 \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right )}{\cos ^{\frac {7}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^3 \left (A \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+C\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle \frac {2}{5} \int \frac {(b+a \cos (c+d x))^2 \left (a (5 A-3 C) \cos ^2(c+d x)+(5 A b+3 C b+5 a B) \cos (c+d x)+5 b B+6 a C\right )}{2 \cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {(b+a \cos (c+d x))^2 \left (a (5 A-3 C) \cos ^2(c+d x)+(5 A b+3 C b+5 a B) \cos (c+d x)+5 b B+6 a C\right )}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (a (5 A-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+(5 A b+3 C b+5 a B) \sin \left (c+d x+\frac {\pi }{2}\right )+5 b B+6 a C\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {(b+a \cos (c+d x)) \left (24 C a^2+3 (5 a A-5 b B-9 a C) \cos ^2(c+d x) a+35 b B a+15 A b^2+9 b^2 C+\left (15 B a^2+6 b (5 A+C) a+5 b^2 B\right ) \cos (c+d x)\right )}{2 \cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 (6 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {(b+a \cos (c+d x)) \left (24 C a^2+3 (5 a A-5 b B-9 a C) \cos ^2(c+d x) a+35 b B a+15 A b^2+9 b^2 C+\left (15 B a^2+6 b (5 A+C) a+5 b^2 B\right ) \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 (6 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {\left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (24 C a^2+3 (5 a A-5 b B-9 a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a+35 b B a+15 A b^2+9 b^2 C+\left (15 B a^2+6 b (5 A+C) a+5 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 (6 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3510 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {2 b \sin (c+d x) \left (24 a^2 C+35 a b B+15 A b^2+9 b^2 C\right )}{d \sqrt {\cos (c+d x)}}-2 \int -\frac {24 C a^3+3 (5 a A-5 b B-9 a C) \cos ^2(c+d x) a^2+50 b B a^2+15 b^2 (3 A+C) a+5 b^3 B+3 \left (5 B a^3+15 b (A-C) a^2-15 b^2 B a-b^3 (5 A+3 C)\right ) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx\right )+\frac {2 (6 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {24 C a^3+3 (5 a A-5 b B-9 a C) \cos ^2(c+d x) a^2+50 b B a^2+15 b^2 (3 A+C) a+5 b^3 B+3 \left (5 B a^3+15 b (A-C) a^2-15 b^2 B a-b^3 (5 A+3 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {2 b \sin (c+d x) \left (24 a^2 C+35 a b B+15 A b^2+9 b^2 C\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (6 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {24 C a^3+3 (5 a A-5 b B-9 a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^2+50 b B a^2+15 b^2 (3 A+C) a+5 b^3 B+3 \left (5 B a^3+15 b (A-C) a^2-15 b^2 B a-b^3 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sin (c+d x) \left (24 a^2 C+35 a b B+15 A b^2+9 b^2 C\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (6 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {2}{3} \int \frac {3 \left (5 \left ((A+3 C) a^3+9 b B a^2+3 b^2 (3 A+C) a+b^3 B\right )+3 \left (5 B a^3+15 b (A-C) a^2-15 b^2 B a-b^3 (5 A+3 C)\right ) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {2 b \sin (c+d x) \left (24 a^2 C+35 a b B+15 A b^2+9 b^2 C\right )}{d \sqrt {\cos (c+d x)}}+\frac {2 a^2 \sin (c+d x) \sqrt {\cos (c+d x)} (5 a A-9 a C-5 b B)}{d}\right )+\frac {2 (6 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {5 \left ((A+3 C) a^3+9 b B a^2+3 b^2 (3 A+C) a+b^3 B\right )+3 \left (5 B a^3+15 b (A-C) a^2-15 b^2 B a-b^3 (5 A+3 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {2 b \sin (c+d x) \left (24 a^2 C+35 a b B+15 A b^2+9 b^2 C\right )}{d \sqrt {\cos (c+d x)}}+\frac {2 a^2 \sin (c+d x) \sqrt {\cos (c+d x)} (5 a A-9 a C-5 b B)}{d}\right )+\frac {2 (6 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {5 \left ((A+3 C) a^3+9 b B a^2+3 b^2 (3 A+C) a+b^3 B\right )+3 \left (5 B a^3+15 b (A-C) a^2-15 b^2 B a-b^3 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sin (c+d x) \left (24 a^2 C+35 a b B+15 A b^2+9 b^2 C\right )}{d \sqrt {\cos (c+d x)}}+\frac {2 a^2 \sin (c+d x) \sqrt {\cos (c+d x)} (5 a A-9 a C-5 b B)}{d}\right )+\frac {2 (6 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 \left (5 a^3 B+15 a^2 b (A-C)-15 a b^2 B-b^3 (5 A+3 C)\right ) \int \sqrt {\cos (c+d x)}dx+\frac {2 b \sin (c+d x) \left (24 a^2 C+35 a b B+15 A b^2+9 b^2 C\right )}{d \sqrt {\cos (c+d x)}}+\frac {2 a^2 \sin (c+d x) \sqrt {\cos (c+d x)} (5 a A-9 a C-5 b B)}{d}\right )+\frac {2 (6 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 \left (5 a^3 B+15 a^2 b (A-C)-15 a b^2 B-b^3 (5 A+3 C)\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 b \sin (c+d x) \left (24 a^2 C+35 a b B+15 A b^2+9 b^2 C\right )}{d \sqrt {\cos (c+d x)}}+\frac {2 a^2 \sin (c+d x) \sqrt {\cos (c+d x)} (5 a A-9 a C-5 b B)}{d}\right )+\frac {2 (6 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sin (c+d x) \left (24 a^2 C+35 a b B+15 A b^2+9 b^2 C\right )}{d \sqrt {\cos (c+d x)}}+\frac {2 a^2 \sin (c+d x) \sqrt {\cos (c+d x)} (5 a A-9 a C-5 b B)}{d}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^3 B+15 a^2 b (A-C)-15 a b^2 B-b^3 (5 A+3 C)\right )}{d}\right )+\frac {2 (6 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {2 b \sin (c+d x) \left (24 a^2 C+35 a b B+15 A b^2+9 b^2 C\right )}{d \sqrt {\cos (c+d x)}}+\frac {2 a^2 \sin (c+d x) \sqrt {\cos (c+d x)} (5 a A-9 a C-5 b B)}{d}+\frac {10 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right )}{d}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^3 B+15 a^2 b (A-C)-15 a b^2 B-b^3 (5 A+3 C)\right )}{d}\right )+\frac {2 (6 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
(2*C*(b + a*Cos[c + d*x])^3*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + ((2*( 5*b*B + 6*a*C)*(b + a*Cos[c + d*x])^2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2 )) + ((6*(5*a^3*B - 15*a*b^2*B + 15*a^2*b*(A - C) - b^3*(5*A + 3*C))*Ellip ticE[(c + d*x)/2, 2])/d + (10*(9*a^2*b*B + b^3*B + 3*a*b^2*(3*A + C) + a^3 *(A + 3*C))*EllipticF[(c + d*x)/2, 2])/d + (2*b*(15*A*b^2 + 35*a*b*B + 24* a^2*C + 9*b^2*C)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + (2*a^2*(5*a*A - 5* b*B - 9*a*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/d)/3)/5
3.14.9.3.1 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[(-(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_.) + (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]
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. \(1391\) vs. \(2(306)=612\).
Time = 4.24 (sec) , antiderivative size = 1392, normalized size of antiderivative = 5.08
int(cos(d*x+c)^(3/2)*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, method=_RETURNVERBOSE)
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A*a^3*(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^3*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*B*a^3*(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))-6*A*a^2*b*(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))+6*a*A*b^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)/(-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))+6*B*a^2*b*(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) )+4/3*A*a^3*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^ 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)^ 2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(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)^(1/ 2))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2/5*C*b^3/(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(...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.15 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.39 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {5 \, \sqrt {2} {\left (i \, {\left (A + 3 \, C\right )} a^{3} + 9 i \, B a^{2} b + 3 i \, {\left (3 \, A + C\right )} a b^{2} + i \, B b^{3}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, {\left (A + 3 \, C\right )} a^{3} - 9 i \, B a^{2} b - 3 i \, {\left (3 \, A + C\right )} a b^{2} - i \, B b^{3}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (-5 i \, B a^{3} - 15 i \, {\left (A - C\right )} a^{2} b + 15 i \, B a b^{2} + i \, {\left (5 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (5 i \, B a^{3} + 15 i \, {\left (A - C\right )} a^{2} b - 15 i \, B a b^{2} - i \, {\left (5 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (5 \, A a^{3} \cos \left (d x + c\right )^{3} + 3 \, C b^{3} + 3 \, {\left (15 \, C a^{2} b + 15 \, B a b^{2} + {\left (5 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{3}} \]
integrate(cos(d*x+c)^(3/2)*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="fricas")
-1/15*(5*sqrt(2)*(I*(A + 3*C)*a^3 + 9*I*B*a^2*b + 3*I*(3*A + C)*a*b^2 + I* B*b^3)*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I*(A + 3*C)*a^3 - 9*I*B*a^2*b - 3*I*(3*A + C)*a*b^2 - I*B*b^3)*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d* x + c)) + 3*sqrt(2)*(-5*I*B*a^3 - 15*I*(A - C)*a^2*b + 15*I*B*a*b^2 + I*(5 *A + 3*C)*b^3)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(- 4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(5*I*B*a^3 + 15*I*(A - C )*a^2*b - 15*I*B*a*b^2 - I*(5*A + 3*C)*b^3)*cos(d*x + c)^3*weierstrassZeta (-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(5* A*a^3*cos(d*x + c)^3 + 3*C*b^3 + 3*(15*C*a^2*b + 15*B*a*b^2 + (5*A + 3*C)* b^3)*cos(d*x + c)^2 + 5*(3*C*a*b^2 + B*b^3)*cos(d*x + c))*sqrt(cos(d*x + c ))*sin(d*x + c))/(d*cos(d*x + c)^3)
Timed out. \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Timed out. \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
integrate(cos(d*x+c)^(3/2)*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="maxima")
\[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac {3}{2}} \,d x } \]
integrate(cos(d*x+c)^(3/2)*(a+b*sec(d*x+c))^3*(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)*(b*sec(d*x + c) + a)^3*c os(d*x + c)^(3/2), x)
Time = 22.47 (sec) , antiderivative size = 414, normalized size of antiderivative = 1.51 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,\left (B\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\,a^3+3\,B\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\,a^2\right )}{d}+\frac {A\,a^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 {2\,C\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,A\,a^2\,b\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,A\,a\,b^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,b^3\,\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\,b^3\,\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\,b^3\,\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 {6\,B\,a\,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}}+\frac {6\,C\,a^2\,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 {2\,C\,a\,b^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 )}{d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]
(2*(B*a^3*ellipticE(c/2 + (d*x)/2, 2) + 3*B*a^2*b*ellipticF(c/2 + (d*x)/2, 2)))/d + (A*a^3*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d + (2*C*a^3*ellipticF(c/2 + (d*x)/2, 2))/d + (6*A*a^2 *b*ellipticE(c/2 + (d*x)/2, 2))/d + (6*A*a*b^2*ellipticF(c/2 + (d*x)/2, 2) )/d + (2*A*b^3*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*b^3*sin(c + d*x)*hyper geom([-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*b^3*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)) + (6*B*a*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)) + (6*C*a^2*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)) + (2 *C*a*b^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(d*cos( c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2))