Integrand size = 35, antiderivative size = 534 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 a (a-b) \sqrt {a+b} \left (8 a^4 C+3 a^2 b^2 (33 A+17 C)+3 b^4 (319 A+247 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{693 b^4 d}-\frac {2 (a-b) \sqrt {a+b} \left (8 a^4 C+6 a^3 b C+15 b^4 (11 A+9 C)+3 a^2 b^2 (33 A+19 C)-6 a b^3 (132 A+101 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{693 b^3 d}+\frac {2 \left (8 a^4 C+15 b^4 (11 A+9 C)+3 a^2 b^2 (33 A+19 C)\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{693 b^2 d}+\frac {2 a \left (99 A b^2+8 a^2 C+67 b^2 C\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{693 b^2 d}+\frac {2 \left (8 a^2 C+9 b^2 (11 A+9 C)\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}-\frac {8 a C (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac {2 C \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d} \] Output:
-2/693*a*(a-b)*(a+b)^(1/2)*(8*a^4*C+3*a^2*b^2*(33*A+17*C)+3*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))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^4 /d-2/693*(a-b)*(a+b)^(1/2)*(8*a^4*C+6*a^3*b*C+15*b^4*(11*A+9*C)+3*a^2*b^2* (33*A+19*C)-6*a*b^3*(132*A+101*C))*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^( 1/2)/(a+b)^(1/2),((a+b)/(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+2/693*(8*a^4*C+15*b^4*(11*A+9*C)+3*a^2*b^ 2*(33*A+19*C))*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^2/d+2/693*a*(99*A*b^2+8 *C*a^2+67*C*b^2)*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/b^2/d+2/693*(8*C*a^2+9* b^2*(11*A+9*C))*(a+b*sec(d*x+c))^(5/2)*tan(d*x+c)/b^2/d-8/99*a*C*(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)*ta n(d*x+c)/b/d
Leaf count is larger than twice the leaf count of optimal. \(3989\) vs. \(2(534)=1068\).
Time = 24.17 (sec) , antiderivative size = 3989, normalized size of antiderivative = 7.47 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \] Input:
Integrate[Sec[c + d*x]^2*(a + b*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2) ,x]
Output:
(Cos[c + d*x]^4*(a + b*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2)*((4*a*(9 9*a^2*A*b^2 + 957*A*b^4 + 8*a^4*C + 51*a^2*b^2*C + 741*b^4*C)*Sin[c + d*x] )/(693*b^3) + (4*Sec[c + d*x]^3*(99*A*b^2*Sin[c + d*x] + 113*a^2*C*Sin[c + d*x] + 81*b^2*C*Sin[c + d*x]))/693 + (4*Sec[c + d*x]^2*(297*a*A*b^2*Sin[c + d*x] + 3*a^3*C*Sin[c + d*x] + 229*a*b^2*C*Sin[c + d*x]))/(693*b) + (4*S ec[c + d*x]*(297*a^2*A*b^2*Sin[c + d*x] + 165*A*b^4*Sin[c + d*x] - 4*a^4*C *Sin[c + d*x] + 205*a^2*b^2*C*Sin[c + d*x] + 135*b^4*C*Sin[c + d*x]))/(693 *b^2) + (92*a*b*C*Sec[c + d*x]^3*Tan[c + d*x])/99 + (4*b^2*C*Sec[c + d*x]^ 4*Tan[c + d*x])/11))/(d*(b + a*Cos[c + d*x])^2*(A + 2*C + A*Cos[2*c + 2*d* x])) - (4*((-2*a^3*A)/(7*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (5 8*a*A*b^2)/(21*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (34*a^3*C)/( 231*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (16*a^5*C)/(693*b^2*Sqr t[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (494*a*b^2*C)/(231*Sqrt[b + a* Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (2*a^4*A*Sqrt[Sec[c + d*x]])/(7*b*Sqrt [b + a*Cos[c + d*x]]) - (4*a^2*A*b*Sqrt[Sec[c + d*x]])/(21*Sqrt[b + a*Cos[ c + d*x]]) + (10*A*b^3*Sqrt[Sec[c + d*x]])/(21*Sqrt[b + a*Cos[c + d*x]]) - (16*a^6*C*Sqrt[Sec[c + d*x]])/(693*b^3*Sqrt[b + a*Cos[c + d*x]]) - (14*a^ 4*C*Sqrt[Sec[c + d*x]])/(99*b*Sqrt[b + a*Cos[c + d*x]]) - (52*a^2*b*C*Sqrt [Sec[c + d*x]])/(231*Sqrt[b + a*Cos[c + d*x]]) + (30*b^3*C*Sqrt[Sec[c + d* x]])/(77*Sqrt[b + a*Cos[c + d*x]]) - (2*a^4*A*Cos[2*(c + d*x)]*Sqrt[Sec...
Time = 2.54 (sec) , antiderivative size = 552, normalized size of antiderivative = 1.03, number of steps used = 20, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4581, 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.
| \(\displaystyle \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4581 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (-4 a C \sec ^2(c+d x)+b (11 A+9 C) \sec (c+d x)+2 a C\right )dx}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (-4 a C \sec ^2(c+d x)+b (11 A+9 C) \sec (c+d x)+2 a C\right )dx}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (-4 a C \csc \left (c+d x+\frac {\pi }{2}\right )^2+b (11 A+9 C) \csc \left (c+d x+\frac {\pi }{2}\right )+2 a C\right )dx}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle \frac {\frac {2 \int -\frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (10 a b C-\left (8 C a^2+9 b^2 (11 A+9 C)\right ) \sec (c+d x)\right )dx}{9 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (10 a b C-\left (8 C a^2+9 b^2 (11 A+9 C)\right ) \sec (c+d x)\right )dx}{9 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (10 a b C+\left (-8 C a^2-9 b^2 (11 A+9 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{9 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {-\frac {\frac {2}{7} \int -\frac {5}{2} \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (3 b \left (-2 C a^2+33 A b^2+27 b^2 C\right )+a \left (8 C a^2+99 A b^2+67 b^2 C\right ) \sec (c+d x)\right )dx-\frac {2 \left (8 a^2 C+9 b^2 (11 A+9 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {-\frac {5}{7} \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (3 b \left (-2 C a^2+33 A b^2+27 b^2 C\right )+a \left (8 C a^2+99 A b^2+67 b^2 C\right ) \sec (c+d x)\right )dx-\frac {2 \left (8 a^2 C+9 b^2 (11 A+9 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {-\frac {5}{7} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (3 b \left (-2 C a^2+33 A b^2+27 b^2 C\right )+a \left (8 C a^2+99 A b^2+67 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {2 \left (8 a^2 C+9 b^2 (11 A+9 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {-\frac {-\frac {5}{7} \left (\frac {2}{5} \int -\frac {3}{2} \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (2 a b \left (a^2 C-b^2 (132 A+101 C)\right )-\left (8 C a^4+3 b^2 (33 A+19 C) a^2+15 b^4 (11 A+9 C)\right ) \sec (c+d x)\right )dx+\frac {2 a \left (8 a^2 C+99 A b^2+67 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {2 \left (8 a^2 C+9 b^2 (11 A+9 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {-\frac {5}{7} \left (\frac {2 a \left (8 a^2 C+99 A b^2+67 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}-\frac {3}{5} \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (2 a b \left (a^2 C-b^2 (132 A+101 C)\right )-\left (8 C a^4+3 b^2 (33 A+19 C) a^2+15 b^4 (11 A+9 C)\right ) \sec (c+d x)\right )dx\right )-\frac {2 \left (8 a^2 C+9 b^2 (11 A+9 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {-\frac {5}{7} \left (\frac {2 a \left (8 a^2 C+99 A b^2+67 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}-\frac {3}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (2 a b \left (a^2 C-b^2 (132 A+101 C)\right )+\left (-8 C a^4-3 b^2 (33 A+19 C) a^2-15 b^4 (11 A+9 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\right )-\frac {2 \left (8 a^2 C+9 b^2 (11 A+9 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {-\frac {-\frac {5}{7} \left (\frac {2 a \left (8 a^2 C+99 A b^2+67 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}-\frac {3}{5} \left (\frac {2}{3} \int -\frac {\sec (c+d x) \left (b \left (2 C a^4+3 b^2 (297 A+221 C) a^2+15 b^4 (11 A+9 C)\right )+a \left (8 C a^4+3 b^2 (33 A+17 C) a^2+3 b^4 (319 A+247 C)\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx-\frac {2 \left (8 a^4 C+3 a^2 b^2 (33 A+19 C)+15 b^4 (11 A+9 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )\right )-\frac {2 \left (8 a^2 C+9 b^2 (11 A+9 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {-\frac {5}{7} \left (\frac {2 a \left (8 a^2 C+99 A b^2+67 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}-\frac {3}{5} \left (-\frac {1}{3} \int \frac {\sec (c+d x) \left (b \left (2 C a^4+3 b^2 (297 A+221 C) a^2+15 b^4 (11 A+9 C)\right )+a \left (8 C a^4+3 b^2 (33 A+17 C) a^2+3 b^4 (319 A+247 C)\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx-\frac {2 \left (8 a^4 C+3 a^2 b^2 (33 A+19 C)+15 b^4 (11 A+9 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )\right )-\frac {2 \left (8 a^2 C+9 b^2 (11 A+9 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {-\frac {5}{7} \left (\frac {2 a \left (8 a^2 C+99 A b^2+67 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}-\frac {3}{5} \left (-\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (2 C a^4+3 b^2 (297 A+221 C) a^2+15 b^4 (11 A+9 C)\right )+a \left (8 C a^4+3 b^2 (33 A+17 C) a^2+3 b^4 (319 A+247 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \left (8 a^4 C+3 a^2 b^2 (33 A+19 C)+15 b^4 (11 A+9 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )\right )-\frac {2 \left (8 a^2 C+9 b^2 (11 A+9 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 4493 |
| \(\displaystyle \frac {-\frac {-\frac {5}{7} \left (\frac {2 a \left (8 a^2 C+99 A b^2+67 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}-\frac {3}{5} \left (\frac {1}{3} \left ((a-b) \left (8 a^4 C+6 a^3 b C+3 a^2 b^2 (33 A+19 C)-6 a b^3 (132 A+101 C)+15 b^4 (11 A+9 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-a \left (8 a^4 C+3 a^2 b^2 (33 A+17 C)+3 b^4 (319 A+247 C)\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\frac {2 \left (8 a^4 C+3 a^2 b^2 (33 A+19 C)+15 b^4 (11 A+9 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )\right )-\frac {2 \left (8 a^2 C+9 b^2 (11 A+9 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {-\frac {5}{7} \left (\frac {2 a \left (8 a^2 C+99 A b^2+67 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}-\frac {3}{5} \left (\frac {1}{3} \left ((a-b) \left (8 a^4 C+6 a^3 b C+3 a^2 b^2 (33 A+19 C)-6 a b^3 (132 A+101 C)+15 b^4 (11 A+9 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-a \left (8 a^4 C+3 a^2 b^2 (33 A+17 C)+3 b^4 (319 A+247 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 \left (8 a^4 C+3 a^2 b^2 (33 A+19 C)+15 b^4 (11 A+9 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )\right )-\frac {2 \left (8 a^2 C+9 b^2 (11 A+9 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {-\frac {-\frac {5}{7} \left (\frac {2 a \left (8 a^2 C+99 A b^2+67 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}-\frac {3}{5} \left (\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} \left (8 a^4 C+6 a^3 b C+3 a^2 b^2 (33 A+19 C)-6 a b^3 (132 A+101 C)+15 b^4 (11 A+9 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-a \left (8 a^4 C+3 a^2 b^2 (33 A+17 C)+3 b^4 (319 A+247 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 \left (8 a^4 C+3 a^2 b^2 (33 A+19 C)+15 b^4 (11 A+9 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )\right )-\frac {2 \left (8 a^2 C+9 b^2 (11 A+9 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {-\frac {-\frac {2 \left (8 a^2 C+9 b^2 (11 A+9 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}-\frac {5}{7} \left (\frac {2 a \left (8 a^2 C+99 A b^2+67 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}-\frac {3}{5} \left (\frac {1}{3} \left (\frac {2 a (a-b) \sqrt {a+b} \left (8 a^4 C+3 a^2 b^2 (33 A+17 C)+3 b^4 (319 A+247 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}+\frac {2 (a-b) \sqrt {a+b} \left (8 a^4 C+6 a^3 b C+3 a^2 b^2 (33 A+19 C)-6 a b^3 (132 A+101 C)+15 b^4 (11 A+9 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )-\frac {2 \left (8 a^4 C+3 a^2 b^2 (33 A+19 C)+15 b^4 (11 A+9 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )\right )}{9 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
Input:
Int[Sec[c + d*x]^2*(a + b*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2),x]
Output:
(2*C*Sec[c + d*x]*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(11*b*d) + ((-8 *a*C*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(9*b*d) - ((-2*(8*a^2*C + 9* b^2*(11*A + 9*C))*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*d) - (5*((2* a*(99*A*b^2 + 8*a^2*C + 67*b^2*C)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x]) /(5*d) - (3*(((2*a*(a - b)*Sqrt[a + b]*(8*a^4*C + 3*a^2*b^2*(33*A + 17*C) + 3*b^4*(319*A + 247*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^2*d) + (2*(a - b)*Sqrt[a + b]* (8*a^4*C + 6*a^3*b*C + 15*b^4*(11*A + 9*C) + 3*a^2*b^2*(33*A + 19*C) - 6*a *b^3*(132*A + 101*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)]*Sqr t[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d))/3 - (2*(8*a^4*C + 15*b^4*(11* A + 9*C) + 3*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)
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_)]^2*(C_.))*(cs c[(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*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, 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. \(2617\) vs. \(2(492)=984\).
Time = 195.73 (sec) , antiderivative size = 2618, normalized size of antiderivative = 4.90
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(2618\) |
| parts | \(\text {Expression too large to display}\) | \(2621\) |
Input:
int(sec(d*x+c)^2*(a+b*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x,method=_RETUR NVERBOSE)
Output:
2/693/d/b^3*(a+b*sec(d*x+c))^(1/2)/(cos(d*x+c)^2*a+a*cos(d*x+c)+b*cos(d*x+ c)+b)*((205*cos(d*x+c)^3+256*cos(d*x+c)^2+116*cos(d*x+c)+116)*C*a^3*b^3*ta n(d*x+c)*sec(d*x+c)+(741*cos(d*x+c)^4+434*cos(d*x+c)^3+434*cos(d*x+c)^2+27 4*cos(d*x+c)+274)*C*a^2*b^4*tan(d*x+c)*sec(d*x+c)^2+(135*cos(d*x+c)^5+876* cos(d*x+c)^4+310*cos(d*x+c)^3+310*cos(d*x+c)^2+224*cos(d*x+c)+224)*C*b^5*a *tan(d*x+c)*sec(d*x+c)^3+33*(5*cos(d*x+c)^3+34*cos(d*x+c)^2+12*cos(d*x+c)+ 12)*A*b^5*a*tan(d*x+c)*sec(d*x+c)+165*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*A*(1/ (a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^( 1/2)*b^6*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+135*(-cos(d *x+c)^2-2*cos(d*x+c)-1)*C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)* (cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^6*EllipticF(-csc(d*x+c)+cot(d*x+c),((a -b)/(a+b))^(1/2))+8*(cos(d*x+c)^2+2*cos(d*x+c)+1)*C*(1/(a+b)*(b+a*cos(d*x+ c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^6*EllipticE( -csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(51*cos(d*x+c)^2-cos(d*x+c)-1) *a^4*b^2*C*tan(d*x+c)+33*(29*cos(d*x+c)^2+18*cos(d*x+c)+18)*a^2*A*b^4*tan( d*x+c)+99*A*a^4*b^2*cos(d*x+c)*sin(d*x+c)+9*(15*cos(d*x+c)^5+15*cos(d*x+c) ^4+9*cos(d*x+c)^3+9*cos(d*x+c)^2+7*cos(d*x+c)+7)*C*b^6*tan(d*x+c)*sec(d*x+ c)^4+33*(5*cos(d*x+c)^3+5*cos(d*x+c)^2+3*cos(d*x+c)+3)*A*b^6*tan(d*x+c)*se c(d*x+c)^2+99*sin(d*x+c)*(4+3*cos(d*x+c))*A*a^3*b^3+4*sin(d*x+c)*(1-cos(d* x+c))*a^5*C*b+51*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*C*(1/(a+b)*(b+a*cos(d*x...
\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{2} \,d x } \] Input:
integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algori thm="fricas")
Output:
integral((C*b^2*sec(d*x + c)^6 + 2*C*a*b*sec(d*x + c)^5 + 2*A*a*b*sec(d*x + c)^3 + A*a^2*sec(d*x + c)^2 + (C*a^2 + A*b^2)*sec(d*x + c)^4)*sqrt(b*sec (d*x + c) + a), x)
Timed out. \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**2*(a+b*sec(d*x+c))**(5/2)*(A+C*sec(d*x+c)**2),x)
Output:
Timed out
Timed out. \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algori thm="maxima")
Output:
Timed out
\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{2} \,d x } \] Input:
integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algori thm="giac")
Output:
integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c) + a)^(5/2)*sec(d*x + c)^2 , x)
Timed out. \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^2} \,d x \] Input:
int(((A + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^(5/2))/cos(c + d*x)^2,x)
Output:
int(((A + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^(5/2))/cos(c + d*x)^2, x)
\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{6}d x \right ) b^{2} c +2 \left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{5}d x \right ) a b c +\left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{4}d x \right ) a^{2} c +\left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{4}d x \right ) a \,b^{2}+2 \left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{3}d x \right ) a^{2} b +\left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{2}d x \right ) a^{3} \] Input:
int(sec(d*x+c)^2*(a+b*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x)
Output:
int(sqrt(sec(c + d*x)*b + a)*sec(c + d*x)**6,x)*b**2*c + 2*int(sqrt(sec(c + d*x)*b + a)*sec(c + d*x)**5,x)*a*b*c + int(sqrt(sec(c + d*x)*b + a)*sec( c + d*x)**4,x)*a**2*c + int(sqrt(sec(c + d*x)*b + a)*sec(c + d*x)**4,x)*a* b**2 + 2*int(sqrt(sec(c + d*x)*b + a)*sec(c + d*x)**3,x)*a**2*b + int(sqrt (sec(c + d*x)*b + a)*sec(c + d*x)**2,x)*a**3