Integrand size = 41, antiderivative size = 502 \[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (45 a^3 b B+435 a b^3 B-10 a^4 C+21 b^4 (9 A+7 C)+3 a^2 b^2 (161 A+93 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^3 d}+\frac {2 (a-b) \sqrt {a+b} \left (10 a^3 C+15 a^2 b (21 A-3 B+11 C)-6 a b^2 (28 A-60 B+19 C)+3 b^3 (63 A-25 B+49 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^2 d}+\frac {2 \left (45 a^2 b B+75 b^3 B-10 a^3 C+6 a b^2 (28 A+19 C)\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b d}+\frac {2 \left (63 A b^2+45 a b B-10 a^2 C+49 b^2 C\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b d}+\frac {2 (9 b B-2 a C) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac {2 C (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d} \]
-2/315*(a-b)*(45*B*a^3*b+435*B*a*b^3-10*a^4*C+21*b^4*(9*A+7*C)+3*a^2*b^2*( 161*A+93*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^3/d+2/315*(a-b)*(10*a^3*C+15*a^2*b*(21*A-3*B+11*C)-6*a* b^2*(28*A-60*B+19*C)+3*b^3*(63*A-25*B+49*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^2/d+2/315*(63*A*b^2+45* B*a*b-10*C*a^2+49*C*b^2)*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/b/d+2/63*(9*B*b -2*C*a)*(a+b*sec(d*x+c))^(5/2)*tan(d*x+c)/b/d+2/9*C*(a+b*sec(d*x+c))^(7/2) *tan(d*x+c)/b/d+2/315*(45*B*a^2*b+75*B*b^3-10*a^3*C+6*a*b^2*(28*A+19*C))*( a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b/d
Leaf count is larger than twice the leaf count of optimal. \(4831\) vs. \(2(502)=1004\).
Time = 29.69 (sec) , antiderivative size = 4831, normalized size of antiderivative = 9.62 \[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \]
(Cos[c + d*x]^4*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((4*(483*a^2*A*b^2 + 189*A*b^4 + 45*a^3*b*B + 435*a*b^3*B - 10*a^ 4*C + 279*a^2*b^2*C + 147*b^4*C)*Sin[c + d*x])/(315*b^2) + (4*Sec[c + d*x] ^3*(9*b^2*B*Sin[c + d*x] + 19*a*b*C*Sin[c + d*x]))/63 + (4*Sec[c + d*x]^2* (63*A*b^2*Sin[c + d*x] + 135*a*b*B*Sin[c + d*x] + 75*a^2*C*Sin[c + d*x] + 49*b^2*C*Sin[c + d*x]))/315 + (4*Sec[c + d*x]*(231*a*A*b^2*Sin[c + d*x] + 135*a^2*b*B*Sin[c + d*x] + 75*b^3*B*Sin[c + d*x] + 5*a^3*C*Sin[c + d*x] + 163*a*b^2*C*Sin[c + d*x]))/(315*b) + (4*b^2*C*Sec[c + d*x]^3*Tan[c + d*x]) /9))/(d*(b + a*Cos[c + d*x])^2*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2 *d*x])) + (4*((-46*a^2*A*b)/(15*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x] ]) - (6*A*b^3)/(5*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (2*a^3*B) /(7*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (58*a*b^2*B)/(21*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (4*a^4*C)/(63*b*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (62*a^2*b*C)/(35*Sqrt[b + a*Cos[c + d*x]]*Sq rt[Sec[c + d*x]]) - (14*b^3*C)/(15*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d *x]]) - (16*a^3*A*Sqrt[Sec[c + d*x]])/(15*Sqrt[b + a*Cos[c + d*x]]) + (16* a*A*b^2*Sqrt[Sec[c + d*x]])/(15*Sqrt[b + a*Cos[c + d*x]]) - (2*a^4*B*Sqrt[ Sec[c + d*x]])/(7*b*Sqrt[b + a*Cos[c + d*x]]) - (4*a^2*b*B*Sqrt[Sec[c + d* x]])/(21*Sqrt[b + a*Cos[c + d*x]]) + (10*b^3*B*Sqrt[Sec[c + d*x]])/(21*Sqr t[b + a*Cos[c + d*x]]) - (248*a^3*C*Sqrt[Sec[c + d*x]])/(315*Sqrt[b + a...
Time = 2.17 (sec) , antiderivative size = 512, 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, 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.
| \(\displaystyle \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{5/2} (b (9 A+7 C)+(9 b B-2 a C) \sec (c+d x))dx}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sec (c+d x) (a+b \sec (c+d x))^{5/2} (b (9 A+7 C)+(9 b B-2 a C) \sec (c+d x))dx}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (b (9 A+7 C)+(9 b B-2 a C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {2}{7} \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (3 b (21 a A+15 b B+13 a C)+\left (-10 C a^2+45 b B a+63 A b^2+49 b^2 C\right ) \sec (c+d x)\right )dx+\frac {2 (9 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{7} \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (3 b (21 a A+15 b B+13 a C)+\left (-10 C a^2+45 b B a+63 A b^2+49 b^2 C\right ) \sec (c+d x)\right )dx+\frac {2 (9 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{7} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (3 b (21 a A+15 b B+13 a C)+\left (-10 C a^2+45 b B a+63 A b^2+49 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 (9 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {2}{5} \int \frac {3}{2} \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b \left (5 (21 A+11 C) a^2+120 b B a+7 b^2 (9 A+7 C)\right )+\left (-10 C a^3+45 b B a^2+6 b^2 (28 A+19 C) a+75 b^3 B\right ) \sec (c+d x)\right )dx+\frac {2 \tan (c+d x) \left (-10 a^2 C+45 a b B+63 A b^2+49 b^2 C\right ) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 (9 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b \left (5 (21 A+11 C) a^2+120 b B a+7 b^2 (9 A+7 C)\right )+\left (-10 C a^3+45 b B a^2+6 b^2 (28 A+19 C) a+75 b^3 B\right ) \sec (c+d x)\right )dx+\frac {2 \tan (c+d x) \left (-10 a^2 C+45 a b B+63 A b^2+49 b^2 C\right ) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 (9 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (b \left (5 (21 A+11 C) a^2+120 b B a+7 b^2 (9 A+7 C)\right )+\left (-10 C a^3+45 b B a^2+6 b^2 (28 A+19 C) a+75 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 \tan (c+d x) \left (-10 a^2 C+45 a b B+63 A b^2+49 b^2 C\right ) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 (9 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {2}{3} \int \frac {\sec (c+d x) \left (b \left (5 (63 A+31 C) a^3+405 b B a^2+3 b^2 (119 A+87 C) a+75 b^3 B\right )+\left (-10 C a^4+45 b B a^3+3 b^2 (161 A+93 C) a^2+435 b^3 B a+21 b^4 (9 A+7 C)\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {2 \tan (c+d x) \left (-10 a^3 C+45 a^2 b B+6 a b^2 (28 A+19 C)+75 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 \tan (c+d x) \left (-10 a^2 C+45 a b B+63 A b^2+49 b^2 C\right ) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 (9 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \int \frac {\sec (c+d x) \left (b \left (5 (63 A+31 C) a^3+405 b B a^2+3 b^2 (119 A+87 C) a+75 b^3 B\right )+\left (-10 C a^4+45 b B a^3+3 b^2 (161 A+93 C) a^2+435 b^3 B a+21 b^4 (9 A+7 C)\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx+\frac {2 \tan (c+d x) \left (-10 a^3 C+45 a^2 b B+6 a b^2 (28 A+19 C)+75 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 \tan (c+d x) \left (-10 a^2 C+45 a b B+63 A b^2+49 b^2 C\right ) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 (9 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (5 (63 A+31 C) a^3+405 b B a^2+3 b^2 (119 A+87 C) a+75 b^3 B\right )+\left (-10 C a^4+45 b B a^3+3 b^2 (161 A+93 C) a^2+435 b^3 B a+21 b^4 (9 A+7 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \tan (c+d x) \left (-10 a^3 C+45 a^2 b B+6 a b^2 (28 A+19 C)+75 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 \tan (c+d x) \left (-10 a^2 C+45 a b B+63 A b^2+49 b^2 C\right ) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 (9 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 4493 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left ((a-b) \left (10 a^3 C+15 a^2 b (21 A-3 B+11 C)-6 a b^2 (28 A-60 B+19 C)+3 b^3 (63 A-25 B+49 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+\left (-10 a^4 C+45 a^3 b B+3 a^2 b^2 (161 A+93 C)+435 a b^3 B+21 b^4 (9 A+7 C)\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 \tan (c+d x) \left (-10 a^3 C+45 a^2 b B+6 a b^2 (28 A+19 C)+75 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 \tan (c+d x) \left (-10 a^2 C+45 a b B+63 A b^2+49 b^2 C\right ) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 (9 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left ((a-b) \left (10 a^3 C+15 a^2 b (21 A-3 B+11 C)-6 a b^2 (28 A-60 B+19 C)+3 b^3 (63 A-25 B+49 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (-10 a^4 C+45 a^3 b B+3 a^2 b^2 (161 A+93 C)+435 a b^3 B+21 b^4 (9 A+7 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 \tan (c+d x) \left (-10 a^3 C+45 a^2 b B+6 a b^2 (28 A+19 C)+75 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 \tan (c+d x) \left (-10 a^2 C+45 a b B+63 A b^2+49 b^2 C\right ) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 (9 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left (\left (-10 a^4 C+45 a^3 b B+3 a^2 b^2 (161 A+93 C)+435 a b^3 B+21 b^4 (9 A+7 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (10 a^3 C+15 a^2 b (21 A-3 B+11 C)-6 a b^2 (28 A-60 B+19 C)+3 b^3 (63 A-25 B+49 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )+\frac {2 \tan (c+d x) \left (-10 a^3 C+45 a^2 b B+6 a b^2 (28 A+19 C)+75 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 \tan (c+d x) \left (-10 a^2 C+45 a b B+63 A b^2+49 b^2 C\right ) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 (9 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {2 \tan (c+d x) \left (-10 a^2 C+45 a b B+63 A b^2+49 b^2 C\right ) (a+b \sec (c+d x))^{3/2}}{5 d}+\frac {3}{5} \left (\frac {2 \tan (c+d x) \left (-10 a^3 C+45 a^2 b B+6 a b^2 (28 A+19 C)+75 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{3 d}+\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (10 a^3 C+15 a^2 b (21 A-3 B+11 C)-6 a b^2 (28 A-60 B+19 C)+3 b^3 (63 A-25 B+49 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-10 a^4 C+45 a^3 b B+3 a^2 b^2 (161 A+93 C)+435 a b^3 B+21 b^4 (9 A+7 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}\right )\right )\right )+\frac {2 (9 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
(2*C*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(9*b*d) + ((2*(9*b*B - 2*a*C )*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*d) + ((2*(63*A*b^2 + 45*a*b* B - 10*a^2*C + 49*b^2*C)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (3*(((-2*(a - b)*Sqrt[a + b]*(45*a^3*b*B + 435*a*b^3*B - 10*a^4*C + 21*b^4 *(9*A + 7*C) + 3*a^2*b^2*(161*A + 93*C))*Cot[c + d*x]*EllipticE[ArcSin[Sqr t[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]*(10*a^3*C + 15*a^2*b*(21*A - 3*B + 11*C) - 6*a*b^2*(28*A - 60*B + 19*C) + 3*b^3*(63*A - 25*B + 49*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))/3 + (2* (45*a^2*b*B + 75*b^3*B - 10*a^3*C + 6*a*b^2*(28*A + 19*C))*Sqrt[a + b*Sec[ c + d*x]]*Tan[c + d*x])/(3*d)))/5)/7)/(9*b)
3.10.51.3.1 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]
Leaf count of result is larger than twice the leaf count of optimal. \(7820\) vs. \(2(464)=928\).
Time = 48.53 (sec) , antiderivative size = 7821, normalized size of antiderivative = 15.58
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(7821\) |
| default | \(\text {Expression too large to display}\) | \(7915\) |
int(sec(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)
\[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right ) \,d x } \]
integrate(sec(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*sec(d*x + c)^5 + (2*C*a*b + B*b^2)*sec(d*x + c)^4 + A*a^2* sec(d*x + c) + (C*a^2 + 2*B*a*b + A*b^2)*sec(d*x + c)^3 + (B*a^2 + 2*A*a*b )*sec(d*x + c)^2)*sqrt(b*sec(d*x + c) + a), x)
\[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \]
Integral((a + b*sec(c + d*x))**(5/2)*(A + B*sec(c + d*x) + C*sec(c + d*x)* *2)*sec(c + d*x), x)
Timed out. \[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
integrate(sec(d*x+c)*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="maxima")
\[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right ) \,d x } \]
integrate(sec(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)*sec(d*x + c), x)
Timed out. \[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\cos \left (c+d\,x\right )} \,d x \]