Integrand size = 41, antiderivative size = 505 \[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {(a-b) \sqrt {a+b} \left (70 a b B-a^2 (15 A-46 C)+6 b^2 (5 A+3 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b d}+\frac {\sqrt {a+b} \left (a^2 b (15 A+90 B-46 C)+30 a^3 C-2 b^3 (15 A-5 B+9 C)+2 a b^2 (45 A-35 B+17 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b d}-\frac {a \sqrt {a+b} (5 A b+2 a B) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}+\frac {A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{d}-\frac {b (15 a A-10 b B-16 a C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}-\frac {b (5 A-2 C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \]
A*(a+b*sec(d*x+c))^(5/2)*sin(d*x+c)/d-1/15*(a-b)*(70*B*a*b-a^2*(15*A-46*C) +6*b^2*(5*A+3*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+1/15*(a^2*b*(15*A+90*B-46*C)+30*a^3*C-2*b^3*(15* A-5*B+9*C)+2*a*b^2*(45*A-35*B+17*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)/b/d-a*(5*A*b+2*B*a)*cot(d*x+c)*El lipticPi((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-1/5*b*(5*A-2*C)*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/d-1/15*b*(15*A*a-10*B* b-16*C*a)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(1490\) vs. \(2(505)=1010\).
Time = 34.98 (sec) , antiderivative size = 1490, normalized size of antiderivative = 2.95 \[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \]
(2*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sqrt [(1 - Tan[(c + d*x)/2]^2)^(-1)]*(15*a^3*A*Tan[(c + d*x)/2] + 15*a^2*A*b*Ta n[(c + d*x)/2] - 30*a*A*b^2*Tan[(c + d*x)/2] - 30*A*b^3*Tan[(c + d*x)/2] - 70*a^2*b*B*Tan[(c + d*x)/2] - 70*a*b^2*B*Tan[(c + d*x)/2] - 46*a^3*C*Tan[ (c + d*x)/2] - 46*a^2*b*C*Tan[(c + d*x)/2] - 18*a*b^2*C*Tan[(c + d*x)/2] - 18*b^3*C*Tan[(c + d*x)/2] - 30*a^3*A*Tan[(c + d*x)/2]^3 + 60*a*A*b^2*Tan[ (c + d*x)/2]^3 + 140*a^2*b*B*Tan[(c + d*x)/2]^3 + 92*a^3*C*Tan[(c + d*x)/2 ]^3 + 36*a*b^2*C*Tan[(c + d*x)/2]^3 + 15*a^3*A*Tan[(c + d*x)/2]^5 - 15*a^2 *A*b*Tan[(c + d*x)/2]^5 - 30*a*A*b^2*Tan[(c + d*x)/2]^5 + 30*A*b^3*Tan[(c + d*x)/2]^5 - 70*a^2*b*B*Tan[(c + d*x)/2]^5 + 70*a*b^2*B*Tan[(c + d*x)/2]^ 5 - 46*a^3*C*Tan[(c + d*x)/2]^5 + 46*a^2*b*C*Tan[(c + d*x)/2]^5 - 18*a*b^2 *C*Tan[(c + d*x)/2]^5 + 18*b^3*C*Tan[(c + d*x)/2]^5 + 150*a^2*A*b*Elliptic Pi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2 ]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 60*a^3*B*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x )/2]^2)/(a + b)] + 150*a^2*A*b*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 60*a^3*B*Ellipt icPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*...
Time = 2.26 (sec) , antiderivative size = 516, normalized size of antiderivative = 1.02, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.415, Rules used = {3042, 4582, 27, 3042, 4544, 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.
| \(\displaystyle \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4582 |
| \(\displaystyle \int \frac {1}{2} (a+b \sec (c+d x))^{3/2} \left (-b (5 A-2 C) \sec ^2(c+d x)+2 (b B+a C) \sec (c+d x)+5 A b+2 a B\right )dx+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int (a+b \sec (c+d x))^{3/2} \left (-b (5 A-2 C) \sec ^2(c+d x)+2 (b B+a C) \sec (c+d x)+5 A b+2 a B\right )dx+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (-b (5 A-2 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 (b B+a C) \csc \left (c+d x+\frac {\pi }{2}\right )+5 A b+2 a B\right )dx+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\) |
| \(\Big \downarrow \) 4544 |
| \(\displaystyle \frac {1}{2} \left (\frac {2}{5} \int \frac {1}{2} \sqrt {a+b \sec (c+d x)} \left (-b (15 a A-10 b B-16 a C) \sec ^2(c+d x)+2 \left (5 C a^2+10 b B a+5 A b^2+3 b^2 C\right ) \sec (c+d x)+5 a (5 A b+2 a B)\right )dx-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{5} \int \sqrt {a+b \sec (c+d x)} \left (-b (15 a A-10 b B-16 a C) \sec ^2(c+d x)+2 \left (5 C a^2+10 b B a+5 A b^2+3 b^2 C\right ) \sec (c+d x)+5 a (5 A b+2 a B)\right )dx-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{5} \int \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (-b (15 a A-10 b B-16 a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 \left (5 C a^2+10 b B a+5 A b^2+3 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+5 a (5 A b+2 a B)\right )dx-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\) |
| \(\Big \downarrow \) 4544 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {2}{3} \int \frac {15 (5 A b+2 a B) a^2+b \left (-\left ((15 A-46 C) a^2\right )+70 b B a+6 b^2 (5 A+3 C)\right ) \sec ^2(c+d x)+2 \left (15 C a^3+45 b B a^2+b^2 (45 A+17 C) a+5 b^3 B\right ) \sec (c+d x)}{2 \sqrt {a+b \sec (c+d x)}}dx-\frac {2 b \tan (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {15 (5 A b+2 a B) a^2+b \left (-\left ((15 A-46 C) a^2\right )+70 b B a+6 b^2 (5 A+3 C)\right ) \sec ^2(c+d x)+2 \left (15 C a^3+45 b B a^2+b^2 (45 A+17 C) a+5 b^3 B\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-\frac {2 b \tan (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {15 (5 A b+2 a B) a^2+b \left (-\left ((15 A-46 C) a^2\right )+70 b B a+6 b^2 (5 A+3 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 \left (15 C a^3+45 b B a^2+b^2 (45 A+17 C) a+5 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b \tan (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\) |
| \(\Big \downarrow \) 4546 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (b \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx+\int \frac {15 (5 A b+2 a B) a^2+\left (2 \left (15 C a^3+45 b B a^2+b^2 (45 A+17 C) a+5 b^3 B\right )-b \left (-\left ((15 A-46 C) a^2\right )+70 b B a+6 b^2 (5 A+3 C)\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\frac {2 b \tan (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (b \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\int \frac {15 (5 A b+2 a B) a^2+\left (2 \left (15 C a^3+45 b B a^2+b^2 (45 A+17 C) a+5 b^3 B\right )-b \left (-\left ((15 A-46 C) a^2\right )+70 b B a+6 b^2 (5 A+3 C)\right )\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 b \tan (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\) |
| \(\Big \downarrow \) 4409 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (b \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+15 a^2 (2 a B+5 A b) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx+\left (30 a^3 C+a^2 b (15 A+90 B-46 C)+2 a b^2 (45 A-35 B+17 C)-2 b^3 (15 A-5 B+9 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\frac {2 b \tan (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (b \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+15 a^2 (2 a B+5 A b) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (30 a^3 C+a^2 b (15 A+90 B-46 C)+2 a b^2 (45 A-35 B+17 C)-2 b^3 (15 A-5 B+9 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 b \tan (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\) |
| \(\Big \downarrow \) 4271 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (b \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (30 a^3 C+a^2 b (15 A+90 B-46 C)+2 a b^2 (45 A-35 B+17 C)-2 b^3 (15 A-5 B+9 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {30 a \sqrt {a+b} (2 a B+5 A b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )-\frac {2 b \tan (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (b \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sqrt {a+b} \cot (c+d x) \left (30 a^3 C+a^2 b (15 A+90 B-46 C)+2 a b^2 (45 A-35 B+17 C)-2 b^3 (15 A-5 B+9 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {30 a \sqrt {a+b} (2 a B+5 A b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )-\frac {2 b \tan (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}+\frac {2 \sqrt {a+b} \cot (c+d x) \left (30 a^3 C+a^2 b (15 A+90 B-46 C)+2 a b^2 (45 A-35 B+17 C)-2 b^3 (15 A-5 B+9 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {30 a \sqrt {a+b} (2 a B+5 A b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )-\frac {2 b \tan (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\) |
(A*(a + b*Sec[c + d*x])^(5/2)*Sin[c + d*x])/d + ((-2*b*(5*A - 2*C)*(a + b* Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (((-2*(a - b)*Sqrt[a + b]*(70*a* b*B - a^2*(15*A - 46*C) + 6*b^2*(5*A + 3*C))*Cot[c + d*x]*EllipticE[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))])/(b*d) + (2*Sq rt[a + b]*(a^2*b*(15*A + 90*B - 46*C) + 30*a^3*C - 2*b^3*(15*A - 5*B + 9*C ) + 2*a*b^2*(45*A - 35*B + 17*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))])/(b*d) - (30*a*Sqrt[a + b ]*(5*A*b + 2*a*B)*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*(15*a*A - 10*b*B - 16*a*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*d))/5)/2
3.10.53.3.1 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. \(6347\) vs. \(2(464)=928\).
Time = 53.14 (sec) , antiderivative size = 6348, normalized size of antiderivative = 12.57
int(cos(d*x+c)*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,me thod=_RETURNVERBOSE)
\[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right ) \,d x } \]
integrate(cos(d*x+c)*(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)*sec(d*x + c)^4 + (2*C*a*b + B*b^2)*cos(d*x + c)*sec(d*x + c)^3 + A*a^2*cos(d*x + c) + (C*a^2 + 2*B*a*b + A*b^2)*cos(d*x + c)*sec(d*x + c)^2 + (B*a^2 + 2*A*a*b)*cos(d*x + c)*sec(d*x + c))*sqrt(b *sec(d*x + c) + a), x)
Timed out. \[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
\[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right ) \,d x } \]
integrate(cos(d*x+c)*(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), x)
\[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right ) \,d x } \]
integrate(cos(d*x+c)*(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), x)
Timed out. \[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \cos \left (c+d\,x\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]