Integrand size = 43, antiderivative size = 507 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {(a-b) \sqrt {a+b} \left (12 a^2 B-24 b^2 B+a b (27 A-56 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{12 b d}+\frac {\sqrt {a+b} \left (a b (27 A+72 B-56 C)+8 b^2 (3 A-3 B+C)+6 a^2 (A+2 (B+6 C))\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{12 d}-\frac {\sqrt {a+b} \left (15 A b^2+20 a b B+4 a^2 (A+2 C)\right ) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 d}+\frac {(5 A b+4 a B) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{4 d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}-\frac {b (21 A b+12 a B-8 b C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{12 d} \]
1/4*(5*A*b+4*B*a)*(a+b*sec(d*x+c))^(3/2)*sin(d*x+c)/d+1/2*A*cos(d*x+c)*(a+ b*sec(d*x+c))^(5/2)*sin(d*x+c)/d+1/12*(a-b)*(12*B*a^2-24*B*b^2+a*b*(27*A-5 6*C))*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b) )^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a- b))^(1/2)/b/d+1/12*(a*b*(27*A+72*B-56*C)+8*b^2*(3*A-3*B+C)+6*a^2*(A+2*B+12 *C))*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b)) ^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b ))^(1/2)/d-1/4*(15*A*b^2+20*B*a*b+4*a^2*(A+2*C))*cot(d*x+c)*EllipticPi((a+ b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*( b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/d-1/12*b*(21 *A*b+12*B*a-8*C*b)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(4887\) vs. \(2(507)=1014\).
Time = 36.53 (sec) , antiderivative size = 4887, normalized size of antiderivative = 9.64 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \]
Integrate[Cos[c + d*x]^2*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
((Cos[c + d*x]^2*(a + b*Sec[c + d*x])^(5/2)*((4*b*(3*b*B + 7*a*C)*Sin[c + d*x])/3 + (a^2*A*Sin[2*(c + d*x)])/2 + (4*b^2*C*Tan[c + d*x])/3))/(d*(b + a*Cos[c + d*x])^2) + (Sqrt[Cos[c + d*x]*Sec[(c + d*x)/2]^2]*((a^3*A)/(Sqrt [b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (6*a*A*b^2)/(Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (6*a^2*b*B)/(Sqrt[b + a*Cos[c + d*x]]*Sqrt[Se c[c + d*x]]) - (2*b^3*B)/(Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + ( 2*a^3*C)/(Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (14*a*b^2*C)/(3*S qrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (11*a^2*A*b*Sqrt[Sec[c + d*x ]])/(4*Sqrt[b + a*Cos[c + d*x]]) + (2*A*b^3*Sqrt[Sec[c + d*x]])/Sqrt[b + a *Cos[c + d*x]] + (a^3*B*Sqrt[Sec[c + d*x]])/Sqrt[b + a*Cos[c + d*x]] + (4* a*b^2*B*Sqrt[Sec[c + d*x]])/Sqrt[b + a*Cos[c + d*x]] + (4*a^2*b*C*Sqrt[Sec [c + d*x]])/(3*Sqrt[b + a*Cos[c + d*x]]) + (2*b^3*C*Sqrt[Sec[c + d*x]])/(3 *Sqrt[b + a*Cos[c + d*x]]) + (9*a^2*A*b*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x] ])/(4*Sqrt[b + a*Cos[c + d*x]]) + (a^3*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x ]])/Sqrt[b + a*Cos[c + d*x]] - (2*a*b^2*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d* x]])/Sqrt[b + a*Cos[c + d*x]] - (14*a^2*b*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(3*Sqrt[b + a*Cos[c + d*x]]))*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]] *(a + b*Sec[c + d*x])^(5/2)*((a + b)*(12*a^2*B - 24*b^2*B + a*b*(27*A - 56 *C))*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sec[(c + d*x)/2] ^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] - b*(a + b)*...
Time = 2.37 (sec) , antiderivative size = 513, normalized size of antiderivative = 1.01, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {3042, 4582, 27, 3042, 4582, 27, 3042, 4544, 27, 3042, 4546, 3042, 4409, 3042, 4271, 4319, 4492}
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.
| \(\displaystyle \int \cos ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 4582 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{2} \cos (c+d x) (a+b \sec (c+d x))^{3/2} \left (-b (3 A-4 C) \sec ^2(c+d x)+2 (2 b B+a (A+2 C)) \sec (c+d x)+5 A b+4 a B\right )dx+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \int \cos (c+d x) (a+b \sec (c+d x))^{3/2} \left (-b (3 A-4 C) \sec ^2(c+d x)+2 (2 b B+a (A+2 C)) \sec (c+d x)+5 A b+4 a B\right )dx+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (-b (3 A-4 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 (2 b B+a (A+2 C)) \csc \left (c+d x+\frac {\pi }{2}\right )+5 A b+4 a B\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 4582 |
| \(\displaystyle \frac {1}{4} \left (\int \frac {1}{2} \sqrt {a+b \sec (c+d x)} \left (4 (A+2 C) a^2+20 b B a+15 A b^2-b (21 A b-8 C b+12 a B) \sec ^2(c+d x)+2 b (4 b B-a (A-8 C)) \sec (c+d x)\right )dx+\frac {(4 a B+5 A b) \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \sqrt {a+b \sec (c+d x)} \left (4 (A+2 C) a^2+20 b B a+15 A b^2-b (21 A b-8 C b+12 a B) \sec ^2(c+d x)-2 b (a A-4 b B-8 a C) \sec (c+d x)\right )dx+\frac {(4 a B+5 A b) \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (4 (A+2 C) a^2+20 b B a+15 A b^2-b (21 A b-8 C b+12 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2-2 b (a A-4 b B-8 a C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {(4 a B+5 A b) \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 4544 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {2}{3} \int \frac {-b \left (12 B a^2+b (27 A-56 C) a-24 b^2 B\right ) \sec ^2(c+d x)+2 b \left (3 (A+12 C) a^2+36 b B a+4 b^2 (3 A+C)\right ) \sec (c+d x)+3 a \left (4 (A+2 C) a^2+20 b B a+15 A b^2\right )}{2 \sqrt {a+b \sec (c+d x)}}dx-\frac {2 b \tan (c+d x) (12 a B+21 A b-8 b C) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {(4 a B+5 A b) \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {1}{3} \int \frac {-b \left (12 B a^2+b (27 A-56 C) a-24 b^2 B\right ) \sec ^2(c+d x)+2 b \left (3 (A+12 C) a^2+36 b B a+4 b^2 (3 A+C)\right ) \sec (c+d x)+3 a \left (4 (A+2 C) a^2+20 b B a+15 A b^2\right )}{\sqrt {a+b \sec (c+d x)}}dx-\frac {2 b \tan (c+d x) (12 a B+21 A b-8 b C) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {(4 a B+5 A b) \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {1}{3} \int \frac {-b \left (12 B a^2+b (27 A-56 C) a-24 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 b \left (3 (A+12 C) a^2+36 b B a+4 b^2 (3 A+C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+3 a \left (4 (A+2 C) a^2+20 b B a+15 A b^2\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b \tan (c+d x) (12 a B+21 A b-8 b C) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {(4 a B+5 A b) \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 4546 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {1}{3} \left (\int \frac {3 a \left (4 (A+2 C) a^2+20 b B a+15 A b^2\right )+\left (b \left (12 B a^2+b (27 A-56 C) a-24 b^2 B\right )+2 b \left (3 (A+12 C) a^2+36 b B a+4 b^2 (3 A+C)\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-b \left (12 a^2 B+a b (27 A-56 C)-24 b^2 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\frac {2 b \tan (c+d x) (12 a B+21 A b-8 b C) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {(4 a B+5 A b) \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {1}{3} \left (\int \frac {3 a \left (4 (A+2 C) a^2+20 b B a+15 A b^2\right )+\left (b \left (12 B a^2+b (27 A-56 C) a-24 b^2 B\right )+2 b \left (3 (A+12 C) a^2+36 b B a+4 b^2 (3 A+C)\right )\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b \left (12 a^2 B+a b (27 A-56 C)-24 b^2 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 b \tan (c+d x) (12 a B+21 A b-8 b C) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {(4 a B+5 A b) \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 4409 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {1}{3} \left (-b \left (12 a^2 B+a b (27 A-56 C)-24 b^2 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a \left (4 a^2 (A+2 C)+20 a b B+15 A b^2\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx+b \left (6 a^2 (A+2 (B+6 C))+a b (27 A+72 B-56 C)+8 b^2 (3 A-3 B+C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\frac {2 b \tan (c+d x) (12 a B+21 A b-8 b C) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {(4 a B+5 A b) \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {1}{3} \left (3 a \left (4 a^2 (A+2 C)+20 a b B+15 A b^2\right ) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (6 a^2 (A+2 (B+6 C))+a b (27 A+72 B-56 C)+8 b^2 (3 A-3 B+C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b \left (12 a^2 B+a b (27 A-56 C)-24 b^2 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 b \tan (c+d x) (12 a B+21 A b-8 b C) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {(4 a B+5 A b) \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 4271 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {1}{3} \left (b \left (6 a^2 (A+2 (B+6 C))+a b (27 A+72 B-56 C)+8 b^2 (3 A-3 B+C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b \left (12 a^2 B+a b (27 A-56 C)-24 b^2 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 \sqrt {a+b} \cot (c+d x) \left (4 a^2 (A+2 C)+20 a b B+15 A b^2\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )-\frac {2 b \tan (c+d x) (12 a B+21 A b-8 b C) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {(4 a B+5 A b) \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {1}{3} \left (-b \left (12 a^2 B+a b (27 A-56 C)-24 b^2 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sqrt {a+b} \cot (c+d x) \left (6 a^2 (A+2 (B+6 C))+a b (27 A+72 B-56 C)+8 b^2 (3 A-3 B+C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {6 \sqrt {a+b} \cot (c+d x) \left (4 a^2 (A+2 C)+20 a b B+15 A b^2\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )-\frac {2 b \tan (c+d x) (12 a B+21 A b-8 b C) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {(4 a B+5 A b) \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {1}{3} \left (\frac {2 \sqrt {a+b} \cot (c+d x) \left (6 a^2 (A+2 (B+6 C))+a b (27 A+72 B-56 C)+8 b^2 (3 A-3 B+C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}+\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (12 a^2 B+a b (27 A-56 C)-24 b^2 B\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}-\frac {6 \sqrt {a+b} \cot (c+d x) \left (4 a^2 (A+2 C)+20 a b B+15 A b^2\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )-\frac {2 b \tan (c+d x) (12 a B+21 A b-8 b C) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {(4 a B+5 A b) \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{5/2}}{2 d}\) |
(A*Cos[c + d*x]*(a + b*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(2*d) + (((5*A*b + 4*a*B)*(a + b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/d + (((2*(a - b)*Sqrt[a + b]*(12*a^2*B - 24*b^2*B + a*b*(27*A - 56*C))*Cot[c + d*x]*EllipticE[ArcS in[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Se c[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d) + (2* Sqrt[a + b]*(a*b*(27*A + 72*B - 56*C) + 8*b^2*(3*A - 3*B + C) + 6*a^2*(A + 2*(B + 6*C)))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt [a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b* (1 + Sec[c + d*x]))/(a - b))])/d - (6*Sqrt[a + b]*(15*A*b^2 + 20*a*b*B + 4 *a^2*(A + 2*C))*Cot[c + d*x]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b )]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/d)/3 - (2*b*(21*A*b + 12*a*B - 8*b*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*d))/2)/4
3.10.54.3.1 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[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) *((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt [a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ .) + (a_)], x_Symbol] :> Simp[c Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[d Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Simp[(-C)*Cot [e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(m + 1) Int[( a + b*Csc[e + f*x])^(m - 1)*Simp[a*A*(m + 1) + ((A*b + a*B)*(m + 1) + b*C*m )*Csc[e + f*x] + (b*B*(m + 1) + a*C*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[ {a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Int[(A + (B - C )*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Simp[C Int[Csc[e + f*x]*(( 1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A , B, C}, x] && NeQ[a^2 - b^2, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d* Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Cs c[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a , b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(6318\) vs. \(2(461)=922\).
Time = 53.96 (sec) , antiderivative size = 6319, normalized size of antiderivative = 12.46
int(cos(d*x+c)^2*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, method=_RETURNVERBOSE)
\[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{2} \,d x } \]
integrate(cos(d*x+c)^2*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="fricas")
integral((C*b^2*cos(d*x + c)^2*sec(d*x + c)^4 + (2*C*a*b + B*b^2)*cos(d*x + c)^2*sec(d*x + c)^3 + A*a^2*cos(d*x + c)^2 + (C*a^2 + 2*B*a*b + A*b^2)*c os(d*x + c)^2*sec(d*x + c)^2 + (B*a^2 + 2*A*a*b)*cos(d*x + c)^2*sec(d*x + c))*sqrt(b*sec(d*x + c) + a), x)
Timed out. \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
\[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{2} \,d x } \]
integrate(cos(d*x+c)^2*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="maxima")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(5/ 2)*cos(d*x + c)^2, x)
\[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{2} \,d x } \]
integrate(cos(d*x+c)^2*(a+b*sec(d*x+c))^(5/2)*(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)^(5/ 2)*cos(d*x + c)^2, x)
Timed out. \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^2\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]