Integrand size = 43, antiderivative size = 610 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 (a-b) \sqrt {a+b} \left (110 a^4 b B-3069 a^2 b^3 B-1617 b^5 B-40 a^5 C-15 a^3 b^2 (33 A+17 C)-15 a b^4 (319 A+247 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3465 b^4 d}+\frac {2 (a-b) \sqrt {a+b} \left (10 a^3 b (11 B-3 C)-40 a^4 C-15 a^2 b^2 (33 A-121 B+19 C)-3 b^4 (275 A-539 B+225 C)+6 a b^3 (660 A-209 B+505 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3465 b^3 d}-\frac {2 \left (110 a^3 b B-1254 a b^3 B-40 a^4 C-75 b^4 (11 A+9 C)-15 a^2 b^2 (33 A+19 C)\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3465 b^2 d}-\frac {2 \left (110 a^2 b B-539 b^3 B-40 a^3 C-5 a b^2 (99 A+67 C)\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{3465 b^2 d}+\frac {2 \left (99 A b^2-22 a b B+8 a^2 C+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}+\frac {2 (11 b B-4 a C) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac {2 C \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d} \]
2/3465*(a-b)*(110*B*a^4*b-3069*B*a^2*b^3-1617*B*b^5-40*a^5*C-15*a^3*b^2*(3 3*A+17*C)-15*a*b^4*(319*A+247*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^4/d+2/3465*(a-b)*(10*a^3*b*(11*B-3 *C)-40*a^4*C-15*a^2*b^2*(33*A-121*B+19*C)-3*b^4*(275*A-539*B+225*C)+6*a*b^ 3*(660*A-209*B+505*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^3/d-2/3465*(110*B*a^2*b-539*B*b^3-40*a^3*C-5* a*b^2*(99*A+67*C))*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/b^2/d+2/693*(99*A*b^2 -22*B*a*b+8*C*a^2+81*C*b^2)*(a+b*sec(d*x+c))^(5/2)*tan(d*x+c)/b^2/d+2/99*( 11*B*b-4*C*a)*(a+b*sec(d*x+c))^(7/2)*tan(d*x+c)/b^2/d+2/11*C*sec(d*x+c)*(a +b*sec(d*x+c))^(7/2)*tan(d*x+c)/b/d-2/3465*(110*B*a^3*b-1254*B*a*b^3-40*a^ 4*C-75*b^4*(11*A+9*C)-15*a^2*b^2*(33*A+19*C))*(a+b*sec(d*x+c))^(1/2)*tan(d *x+c)/b^2/d
Leaf count is larger than twice the leaf count of optimal. \(5369\) vs. \(2(610)=1220\).
Time = 32.94 (sec) , antiderivative size = 5369, normalized size of antiderivative = 8.80 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \]
Integrate[Sec[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]
Time = 2.83 (sec) , antiderivative size = 627, normalized size of antiderivative = 1.03, number of steps used = 20, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.465, Rules used = {3042, 4580, 27, 3042, 4570, 27, 3042, 4490, 27, 3042, 4490, 27, 3042, 4490, 27, 3042, 4493, 3042, 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.
| \(\displaystyle \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4580 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left ((11 b B-4 a C) \sec ^2(c+d x)+b (11 A+9 C) \sec (c+d x)+2 a C\right )dx}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left ((11 b B-4 a C) \sec ^2(c+d x)+b (11 A+9 C) \sec (c+d x)+2 a C\right )dx}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left ((11 b B-4 a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+b (11 A+9 C) \csc \left (c+d x+\frac {\pi }{2}\right )+2 a C\right )dx}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle \frac {\frac {2 \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (b (77 b B-10 a C)+\left (8 C a^2-22 b B a+99 A b^2+81 b^2 C\right ) \sec (c+d x)\right )dx}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (b (77 b B-10 a C)+\left (8 C a^2-22 b B a+99 A b^2+81 b^2 C\right ) \sec (c+d x)\right )dx}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (b (77 b B-10 a C)+\left (8 C a^2-22 b B a+99 A b^2+81 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {\frac {2}{7} \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (3 b \left (-10 C a^2+143 b B a+165 A b^2+135 b^2 C\right )-\left (-40 C a^3+110 b B a^2-5 b^2 (99 A+67 C) a-539 b^3 B\right ) \sec (c+d x)\right )dx+\frac {2 \tan (c+d x) \left (8 a^2 C-22 a b B+99 A b^2+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {1}{7} \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (3 b \left (-10 C a^2+143 b B a+165 A b^2+135 b^2 C\right )-\left (-40 C a^3+110 b B a^2-5 b^2 (99 A+67 C) a-539 b^3 B\right ) \sec (c+d x)\right )dx+\frac {2 \tan (c+d x) \left (8 a^2 C-22 a b B+99 A b^2+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {1}{7} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (3 b \left (-10 C a^2+143 b B a+165 A b^2+135 b^2 C\right )+\left (40 C a^3-110 b B a^2+5 b^2 (99 A+67 C) a+539 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 \tan (c+d x) \left (8 a^2 C-22 a b B+99 A b^2+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {2}{5} \int \frac {3}{2} \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b \left (-10 C a^3+605 b B a^2+10 b^2 (132 A+101 C) a+539 b^3 B\right )-\left (-40 C a^4+110 b B a^3-15 b^2 (33 A+19 C) a^2-1254 b^3 B a-75 b^4 (11 A+9 C)\right ) \sec (c+d x)\right )dx-\frac {2 \tan (c+d x) \left (-40 a^3 C+110 a^2 b B-5 a b^2 (99 A+67 C)-539 b^3 B\right ) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 \tan (c+d x) \left (8 a^2 C-22 a b B+99 A b^2+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b \left (-10 C a^3+605 b B a^2+10 b^2 (132 A+101 C) a+539 b^3 B\right )-\left (-40 C a^4+110 b B a^3-15 b^2 (33 A+19 C) a^2-1254 b^3 B a-75 b^4 (11 A+9 C)\right ) \sec (c+d x)\right )dx-\frac {2 \tan (c+d x) \left (-40 a^3 C+110 a^2 b B-5 a b^2 (99 A+67 C)-539 b^3 B\right ) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 \tan (c+d x) \left (8 a^2 C-22 a b B+99 A b^2+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (b \left (-10 C a^3+605 b B a^2+10 b^2 (132 A+101 C) a+539 b^3 B\right )+\left (40 C a^4-110 b B a^3+15 b^2 (33 A+19 C) a^2+1254 b^3 B a+75 b^4 (11 A+9 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {2 \tan (c+d x) \left (-40 a^3 C+110 a^2 b B-5 a b^2 (99 A+67 C)-539 b^3 B\right ) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 \tan (c+d x) \left (8 a^2 C-22 a b B+99 A b^2+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {2}{3} \int \frac {\sec (c+d x) \left (b \left (10 C a^4+1705 b B a^3+15 b^2 (297 A+221 C) a^2+2871 b^3 B a+75 b^4 (11 A+9 C)\right )-\left (-40 C a^5+110 b B a^4-15 b^2 (33 A+17 C) a^3-3069 b^3 B a^2-15 b^4 (319 A+247 C) a-1617 b^5 B\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx-\frac {2 \tan (c+d x) \left (-40 a^4 C+110 a^3 b B-15 a^2 b^2 (33 A+19 C)-1254 a b^3 B-75 b^4 (11 A+9 C)\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \tan (c+d x) \left (-40 a^3 C+110 a^2 b B-5 a b^2 (99 A+67 C)-539 b^3 B\right ) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 \tan (c+d x) \left (8 a^2 C-22 a b B+99 A b^2+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \int \frac {\sec (c+d x) \left (b \left (10 C a^4+1705 b B a^3+15 b^2 (297 A+221 C) a^2+2871 b^3 B a+75 b^4 (11 A+9 C)\right )-\left (-40 C a^5+110 b B a^4-15 b^2 (33 A+17 C) a^3-3069 b^3 B a^2-15 b^4 (319 A+247 C) a-1617 b^5 B\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx-\frac {2 \tan (c+d x) \left (-40 a^4 C+110 a^3 b B-15 a^2 b^2 (33 A+19 C)-1254 a b^3 B-75 b^4 (11 A+9 C)\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \tan (c+d x) \left (-40 a^3 C+110 a^2 b B-5 a b^2 (99 A+67 C)-539 b^3 B\right ) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 \tan (c+d x) \left (8 a^2 C-22 a b B+99 A b^2+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (10 C a^4+1705 b B a^3+15 b^2 (297 A+221 C) a^2+2871 b^3 B a+75 b^4 (11 A+9 C)\right )+\left (40 C a^5-110 b B a^4+15 b^2 (33 A+17 C) a^3+3069 b^3 B a^2+15 b^4 (319 A+247 C) a+1617 b^5 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \tan (c+d x) \left (-40 a^4 C+110 a^3 b B-15 a^2 b^2 (33 A+19 C)-1254 a b^3 B-75 b^4 (11 A+9 C)\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \tan (c+d x) \left (-40 a^3 C+110 a^2 b B-5 a b^2 (99 A+67 C)-539 b^3 B\right ) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 \tan (c+d x) \left (8 a^2 C-22 a b B+99 A b^2+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 4493 |
| \(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left ((a-b) \left (-40 a^4 C+a^3 b (110 B-30 C)-15 a^2 b^2 (33 A-121 B+19 C)+6 a b^3 (660 A-209 B+505 C)-3 b^4 (275 A-539 B+225 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-\left (-40 a^5 C+110 a^4 b B-15 a^3 b^2 (33 A+17 C)-3069 a^2 b^3 B-15 a b^4 (319 A+247 C)-1617 b^5 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\frac {2 \tan (c+d x) \left (-40 a^4 C+110 a^3 b B-15 a^2 b^2 (33 A+19 C)-1254 a b^3 B-75 b^4 (11 A+9 C)\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \tan (c+d x) \left (-40 a^3 C+110 a^2 b B-5 a b^2 (99 A+67 C)-539 b^3 B\right ) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 \tan (c+d x) \left (8 a^2 C-22 a b B+99 A b^2+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left ((a-b) \left (-40 a^4 C+a^3 b (110 B-30 C)-15 a^2 b^2 (33 A-121 B+19 C)+6 a b^3 (660 A-209 B+505 C)-3 b^4 (275 A-539 B+225 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (-40 a^5 C+110 a^4 b B-15 a^3 b^2 (33 A+17 C)-3069 a^2 b^3 B-15 a b^4 (319 A+247 C)-1617 b^5 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 \tan (c+d x) \left (-40 a^4 C+110 a^3 b B-15 a^2 b^2 (33 A+19 C)-1254 a b^3 B-75 b^4 (11 A+9 C)\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \tan (c+d x) \left (-40 a^3 C+110 a^2 b B-5 a b^2 (99 A+67 C)-539 b^3 B\right ) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 \tan (c+d x) \left (8 a^2 C-22 a b B+99 A b^2+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-40 a^4 C+a^3 b (110 B-30 C)-15 a^2 b^2 (33 A-121 B+19 C)+6 a b^3 (660 A-209 B+505 C)-3 b^4 (275 A-539 B+225 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\left (-40 a^5 C+110 a^4 b B-15 a^3 b^2 (33 A+17 C)-3069 a^2 b^3 B-15 a b^4 (319 A+247 C)-1617 b^5 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 \tan (c+d x) \left (-40 a^4 C+110 a^3 b B-15 a^2 b^2 (33 A+19 C)-1254 a b^3 B-75 b^4 (11 A+9 C)\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \tan (c+d x) \left (-40 a^3 C+110 a^2 b B-5 a b^2 (99 A+67 C)-539 b^3 B\right ) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 \tan (c+d x) \left (8 a^2 C-22 a b B+99 A b^2+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {\frac {\frac {2 \tan (c+d x) \left (8 a^2 C-22 a b B+99 A b^2+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2}}{7 d}+\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-40 a^4 C+a^3 b (110 B-30 C)-15 a^2 b^2 (33 A-121 B+19 C)+6 a b^3 (660 A-209 B+505 C)-3 b^4 (275 A-539 B+225 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}+\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-40 a^5 C+110 a^4 b B-15 a^3 b^2 (33 A+17 C)-3069 a^2 b^3 B-15 a b^4 (319 A+247 C)-1617 b^5 B\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}\right )-\frac {2 \tan (c+d x) \left (-40 a^4 C+110 a^3 b B-15 a^2 b^2 (33 A+19 C)-1254 a b^3 B-75 b^4 (11 A+9 C)\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \tan (c+d x) \left (-40 a^3 C+110 a^2 b B-5 a b^2 (99 A+67 C)-539 b^3 B\right ) (a+b \sec (c+d x))^{3/2}}{5 d}\right )}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
(2*C*Sec[c + d*x]*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(11*b*d) + ((2* (11*b*B - 4*a*C)*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(9*b*d) + ((2*(9 9*A*b^2 - 22*a*b*B + 8*a^2*C + 81*b^2*C)*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*d) + ((-2*(110*a^2*b*B - 539*b^3*B - 40*a^3*C - 5*a*b^2*(99*A + 67*C))*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (3*(((2*(a - b)*S qrt[a + b]*(110*a^4*b*B - 3069*a^2*b^3*B - 1617*b^5*B - 40*a^5*C - 15*a^3* b^2*(33*A + 17*C) - 15*a*b^4*(319*A + 247*C))*Cot[c + d*x]*EllipticE[ArcSi n[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^2*d) + (2 *(a - b)*Sqrt[a + b]*(a^3*b*(110*B - 30*C) - 40*a^4*C - 15*a^2*b^2*(33*A - 121*B + 19*C) - 3*b^4*(275*A - 539*B + 225*C) + 6*a*b^3*(660*A - 209*B + 505*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 + Se c[c + d*x]))/(a - b))])/(b*d))/3 - (2*(110*a^3*b*B - 1254*a*b^3*B - 40*a^4 *C - 75*b^4*(11*A + 9*C) - 15*a^2*b^2*(33*A + 19*C))*Sqrt[a + b*Sec[c + d* x]]*Tan[c + d*x])/(3*d)))/5)/7)/(9*b))/(11*b)
3.10.50.3.1 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[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_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(m + 1) Int[Csc[e + f*x]* (a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1 ))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a* B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 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[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(A - B) Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[B 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} , x] && NeQ[a^2 - b^2, 0] && NeQ[A^2 - B^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e _.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S ymbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2) )), x] + Simp[1/(b*(m + 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[ b*A*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[ (e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x _Symbol] :> Simp[(-C)*Csc[e + f*x]*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m + 2) + A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B* (m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] & & NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(9226\) vs. \(2(568)=1136\).
Time = 134.14 (sec) , antiderivative size = 9227, normalized size of antiderivative = 15.13
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(9227\) |
| default | \(\text {Expression too large to display}\) | \(9339\) |
int(sec(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)
\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{2} \,d x } \]
integrate(sec(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*sec(d*x + c)^6 + (2*C*a*b + B*b^2)*sec(d*x + c)^5 + A*a^2* sec(d*x + c)^2 + (C*a^2 + 2*B*a*b + A*b^2)*sec(d*x + c)^4 + (B*a^2 + 2*A*a *b)*sec(d*x + c)^3)*sqrt(b*sec(d*x + c) + a), x)
Timed out. \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Timed out. \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
integrate(sec(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")
\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{2} \,d x } \]
integrate(sec(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)*sec(d*x + c)^2, x)
Timed out. \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\cos \left (c+d\,x\right )}^2} \,d x \]