Integrand size = 33, antiderivative size = 566 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {2 (a-b) \sqrt {a+b} \left (110 a^4 A b-3069 a^2 A b^3-1617 A b^5-40 a^5 B-255 a^3 b^2 B-3705 a b^4 B\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3465 b^4 d}-\frac {2 (a-b) \sqrt {a+b} \left (6 a b^3 (209 A-505 B)-3 b^4 (539 A-225 B)-15 a^2 b^2 (121 A-19 B)+40 a^4 B-a^3 (110 A b-30 b B)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3465 b^3 d}-\frac {2 \left (110 a^3 A b-1254 a A b^3-40 a^4 B-285 a^2 b^2 B-675 b^4 B\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3465 b^2 d}-\frac {2 \left (110 a^2 A b-539 A b^3-40 a^3 B-335 a b^2 B\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{3465 b^2 d}-\frac {2 \left (22 a A b-8 a^2 B-81 b^2 B\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}+\frac {2 (11 A b-4 a B) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac {2 B \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d} \]
2/3465*(a-b)*(110*A*a^4*b-3069*A*a^2*b^3-1617*A*b^5-40*B*a^5-255*B*a^3*b^2 -3705*B*a*b^4)*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)*(6*a*b^3*(209*A-505*B)-3*b^4*(539*A- 225*B)-15*a^2*b^2*(121*A-19*B)+40*B*a^4-a^3*(110*A*b-30*B*b))*cot(d*x+c)*E llipticF((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*A*a^2*b-539*A*b^3-40*B*a^3-335*B*a*b^2)*(a+b*sec(d*x+c))^(3/2)*t an(d*x+c)/b^2/d-2/693*(22*A*a*b-8*B*a^2-81*B*b^2)*(a+b*sec(d*x+c))^(5/2)*t an(d*x+c)/b^2/d+2/99*(11*A*b-4*B*a)*(a+b*sec(d*x+c))^(7/2)*tan(d*x+c)/b^2/ d+2/11*B*sec(d*x+c)*(a+b*sec(d*x+c))^(7/2)*tan(d*x+c)/b/d-2/3465*(110*A*a^ 3*b-1254*A*a*b^3-40*B*a^4-285*B*a^2*b^2-675*B*b^4)*(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. \(4227\) vs. \(2(566)=1132\).
Time = 26.40 (sec) , antiderivative size = 4227, normalized size of antiderivative = 7.47 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Result too large to show} \]
(Cos[c + d*x]^2*(a + b*Sec[c + d*x])^(5/2)*((2*(-110*a^4*A*b + 3069*a^2*A* b^3 + 1617*A*b^5 + 40*a^5*B + 255*a^3*b^2*B + 3705*a*b^4*B)*Sin[c + d*x])/ (3465*b^3) + (2*Sec[c + d*x]^4*(11*A*b^2*Sin[c + d*x] + 23*a*b*B*Sin[c + d *x]))/99 + (2*Sec[c + d*x]^3*(209*a*A*b*Sin[c + d*x] + 113*a^2*B*Sin[c + d *x] + 81*b^2*B*Sin[c + d*x]))/693 + (2*Sec[c + d*x]^2*(825*a^2*A*b*Sin[c + d*x] + 539*A*b^3*Sin[c + d*x] + 15*a^3*B*Sin[c + d*x] + 1145*a*b^2*B*Sin[ c + d*x]))/(3465*b) + (2*Sec[c + d*x]*(55*a^3*A*b*Sin[c + d*x] + 1793*a*A* b^3*Sin[c + d*x] - 20*a^4*B*Sin[c + d*x] + 1025*a^2*b^2*B*Sin[c + d*x] + 6 75*b^4*B*Sin[c + d*x]))/(3465*b^2) + (2*b^2*B*Sec[c + d*x]^4*Tan[c + d*x]) /11))/(d*(b + a*Cos[c + d*x])^2) - (2*((2*a^4*A)/(63*b*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (31*a^2*A*b)/(35*Sqrt[b + a*Cos[c + d*x]]*Sqrt [Sec[c + d*x]]) - (7*A*b^3)/(15*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x] ]) - (17*a^3*B)/(231*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (8*a^5 *B)/(693*b^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (247*a*b^2*B)/ (231*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (124*a^3*A*Sqrt[Sec[c + d*x]])/(315*Sqrt[b + a*Cos[c + d*x]]) + (2*a^5*A*Sqrt[Sec[c + d*x]])/(63 *b^2*Sqrt[b + a*Cos[c + d*x]]) + (38*a*A*b^2*Sqrt[Sec[c + d*x]])/(105*Sqrt [b + a*Cos[c + d*x]]) - (8*a^6*B*Sqrt[Sec[c + d*x]])/(693*b^3*Sqrt[b + a*C os[c + d*x]]) - (7*a^4*B*Sqrt[Sec[c + d*x]])/(99*b*Sqrt[b + a*Cos[c + d*x] ]) - (26*a^2*b*B*Sqrt[Sec[c + d*x]])/(231*Sqrt[b + a*Cos[c + d*x]]) + (...
Time = 2.69 (sec) , antiderivative size = 584, normalized size of antiderivative = 1.03, number of steps used = 20, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.606, Rules used = {3042, 4521, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4521 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left ((11 A b-4 a B) \sec ^2(c+d x)+9 b B \sec (c+d x)+2 a B\right )dx}{11 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left ((11 A b-4 a B) \sec ^2(c+d x)+9 b B \sec (c+d x)+2 a B\right )dx}{11 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left ((11 A b-4 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+9 b B \csc \left (c+d x+\frac {\pi }{2}\right )+2 a B\right )dx}{11 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle \frac {\frac {2 \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (b (77 A b-10 a B)-\left (-8 B a^2+22 A b a-81 b^2 B\right ) \sec (c+d x)\right )dx}{9 b}+\frac {2 (11 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (b (77 A b-10 a B)-\left (-8 B a^2+22 A b a-81 b^2 B\right ) \sec (c+d x)\right )dx}{9 b}+\frac {2 (11 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (b (77 A b-10 a B)+\left (8 B a^2-22 A b a+81 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{9 b}+\frac {2 (11 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {\frac {2}{7} \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (3 b \left (-10 B a^2+143 A b a+135 b^2 B\right )-\left (-40 B a^3+110 A b a^2-335 b^2 B a-539 A b^3\right ) \sec (c+d x)\right )dx-\frac {2 \left (-8 a^2 B+22 a A b-81 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {1}{7} \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (3 b \left (-10 B a^2+143 A b a+135 b^2 B\right )-\left (-40 B a^3+110 A b a^2-335 b^2 B a-539 A b^3\right ) \sec (c+d x)\right )dx-\frac {2 \left (-8 a^2 B+22 a A b-81 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {1}{7} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (3 b \left (-10 B a^2+143 A b a+135 b^2 B\right )+\left (40 B a^3-110 A b a^2+335 b^2 B a+539 A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {2 \left (-8 a^2 B+22 a A b-81 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {2}{5} \int \frac {3}{2} \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b \left (-10 B a^3+605 A b a^2+1010 b^2 B a+539 A b^3\right )-\left (-40 B a^4+110 A b a^3-285 b^2 B a^2-1254 A b^3 a-675 b^4 B\right ) \sec (c+d x)\right )dx-\frac {2 \left (-40 a^3 B+110 a^2 A b-335 a b^2 B-539 A b^3\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {2 \left (-8 a^2 B+22 a A b-81 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b \left (-10 B a^3+605 A b a^2+1010 b^2 B a+539 A b^3\right )-\left (-40 B a^4+110 A b a^3-285 b^2 B a^2-1254 A b^3 a-675 b^4 B\right ) \sec (c+d x)\right )dx-\frac {2 \left (-40 a^3 B+110 a^2 A b-335 a b^2 B-539 A b^3\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {2 \left (-8 a^2 B+22 a A b-81 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (b \left (-10 B a^3+605 A b a^2+1010 b^2 B a+539 A b^3\right )+\left (40 B a^4-110 A b a^3+285 b^2 B a^2+1254 A b^3 a+675 b^4 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {2 \left (-40 a^3 B+110 a^2 A b-335 a b^2 B-539 A b^3\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {2 \left (-8 a^2 B+22 a A b-81 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {2}{3} \int \frac {\sec (c+d x) \left (b \left (10 B a^4+1705 A b a^3+3315 b^2 B a^2+2871 A b^3 a+675 b^4 B\right )-\left (-40 B a^5+110 A b a^4-255 b^2 B a^3-3069 A b^3 a^2-3705 b^4 B a-1617 A b^5\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx-\frac {2 \left (-40 a^4 B+110 a^3 A b-285 a^2 b^2 B-1254 a A b^3-675 b^4 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-40 a^3 B+110 a^2 A b-335 a b^2 B-539 A b^3\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {2 \left (-8 a^2 B+22 a A b-81 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \int \frac {\sec (c+d x) \left (b \left (10 B a^4+1705 A b a^3+3315 b^2 B a^2+2871 A b^3 a+675 b^4 B\right )-\left (-40 B a^5+110 A b a^4-255 b^2 B a^3-3069 A b^3 a^2-3705 b^4 B a-1617 A b^5\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx-\frac {2 \left (-40 a^4 B+110 a^3 A b-285 a^2 b^2 B-1254 a A b^3-675 b^4 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-40 a^3 B+110 a^2 A b-335 a b^2 B-539 A b^3\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {2 \left (-8 a^2 B+22 a A b-81 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (10 B a^4+1705 A b a^3+3315 b^2 B a^2+2871 A b^3 a+675 b^4 B\right )+\left (40 B a^5-110 A b a^4+255 b^2 B a^3+3069 A b^3 a^2+3705 b^4 B a+1617 A b^5\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \left (-40 a^4 B+110 a^3 A b-285 a^2 b^2 B-1254 a A b^3-675 b^4 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-40 a^3 B+110 a^2 A b-335 a b^2 B-539 A b^3\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {2 \left (-8 a^2 B+22 a A b-81 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 4493 |
| \(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left (-\left ((a-b) \left (40 a^4 B-a^3 (110 A b-30 b B)-15 a^2 b^2 (121 A-19 B)+6 a b^3 (209 A-505 B)-3 b^4 (539 A-225 B)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\left (-40 a^5 B+110 a^4 A b-255 a^3 b^2 B-3069 a^2 A b^3-3705 a b^4 B-1617 A b^5\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\frac {2 \left (-40 a^4 B+110 a^3 A b-285 a^2 b^2 B-1254 a A b^3-675 b^4 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-40 a^3 B+110 a^2 A b-335 a b^2 B-539 A b^3\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {2 \left (-8 a^2 B+22 a A b-81 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left (-\left ((a-b) \left (40 a^4 B-a^3 (110 A b-30 b B)-15 a^2 b^2 (121 A-19 B)+6 a b^3 (209 A-505 B)-3 b^4 (539 A-225 B)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\left (-40 a^5 B+110 a^4 A b-255 a^3 b^2 B-3069 a^2 A b^3-3705 a b^4 B-1617 A b^5\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 \left (-40 a^4 B+110 a^3 A b-285 a^2 b^2 B-1254 a A b^3-675 b^4 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-40 a^3 B+110 a^2 A b-335 a b^2 B-539 A b^3\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {2 \left (-8 a^2 B+22 a A b-81 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left (-\left (-40 a^5 B+110 a^4 A b-255 a^3 b^2 B-3069 a^2 A b^3-3705 a b^4 B-1617 A b^5\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \left (40 a^4 B-a^3 (110 A b-30 b B)-15 a^2 b^2 (121 A-19 B)+6 a b^3 (209 A-505 B)-3 b^4 (539 A-225 B)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )-\frac {2 \left (-40 a^4 B+110 a^3 A b-285 a^2 b^2 B-1254 a A b^3-675 b^4 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-40 a^3 B+110 a^2 A b-335 a b^2 B-539 A b^3\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {2 \left (-8 a^2 B+22 a A b-81 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} \left (-40 a^5 B+110 a^4 A b-255 a^3 b^2 B-3069 a^2 A b^3-3705 a b^4 B-1617 A b^5\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}-\frac {2 (a-b) \sqrt {a+b} \left (40 a^4 B-a^3 (110 A b-30 b B)-15 a^2 b^2 (121 A-19 B)+6 a b^3 (209 A-505 B)-3 b^4 (539 A-225 B)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )-\frac {2 \left (-40 a^4 B+110 a^3 A b-285 a^2 b^2 B-1254 a A b^3-675 b^4 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-40 a^3 B+110 a^2 A b-335 a b^2 B-539 A b^3\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {2 \left (-8 a^2 B+22 a A b-81 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
(2*B*Sec[c + d*x]*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(11*b*d) + ((2* (11*A*b - 4*a*B)*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(9*b*d) + ((-2*( 22*a*A*b - 8*a^2*B - 81*b^2*B)*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7 *d) + ((-2*(110*a^2*A*b - 539*A*b^3 - 40*a^3*B - 335*a*b^2*B)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (3*(((2*(a - b)*Sqrt[a + b]*(110*a^4* A*b - 3069*a^2*A*b^3 - 1617*A*b^5 - 40*a^5*B - 255*a^3*b^2*B - 3705*a*b^4* B)*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^2*d) - (2*(a - b)*Sqrt[a + b]*(6*a*b^3*(209*A - 505* B) - 3*b^4*(539*A - 225*B) - 15*a^2*b^2*(121*A - 19*B) + 40*a^4*B - a^3*(1 10*A*b - 30*b*B))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/S qrt[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))/3 - (2*(110*a^3*A*b - 1254*a*A*b^ 3 - 40*a^4*B - 285*a^2*b^2*B - 675*b^4*B)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*d)))/5)/7)/(9*b))/(11*b)
3.4.63.3.1 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_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*d^ 2*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 2)/(b*f* (m + n))), x] + Simp[d^2/(b*(m + n)) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 2)*Simp[a*B*(n - 2) + B*b*(m + n - 1)*Csc[e + f*x] + (A*b*(m + n) - a*B*(n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B , m}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && NeQ[m + n, 0] && !IGtQ[m, 1]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(6883\) vs. \(2(524)=1048\).
Time = 119.07 (sec) , antiderivative size = 6884, normalized size of antiderivative = 12.16
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(6884\) |
| default | \(\text {Expression too large to display}\) | \(6974\) |
\[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{3} \,d x } \]
integral((B*b^2*sec(d*x + c)^6 + A*a^2*sec(d*x + c)^3 + (2*B*a*b + A*b^2)* sec(d*x + c)^5 + (B*a^2 + 2*A*a*b)*sec(d*x + c)^4)*sqrt(b*sec(d*x + c) + a ), x)
Timed out. \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]
Timed out. \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]
\[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{3} \,d x } \]
Timed out. \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^3} \,d x \]