Integrand size = 43, antiderivative size = 447 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=-\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 b^2 \left (a^2-b^2\right ) d}-\frac {\left (3 A b^4+3 a^3 b B-5 a b^3 B-a^2 b^2 (A-7 C)-5 a^4 C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^3 (a+b)^2 d}+\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]
1/3*(3*A*b^2-3*B*a*b+5*C*a^2-2*C*b^2)*sec(d*x+c)^(3/2)*sin(d*x+c)/b^2/(a^2 -b^2)/d-(A*b^2-a*(B*b-C*a))*sec(d*x+c)^(5/2)*sin(d*x+c)/b/(a^2-b^2)/d/(a+b *sec(d*x+c))+(3*B*a^2*b-2*B*b^3-a*b^2*(A-4*C)-5*a^3*C)*sin(d*x+c)*sec(d*x+ c)^(1/2)/b^3/(a^2-b^2)/d-(3*B*a^2*b-2*B*b^3-a*b^2*(A-4*C)-5*a^3*C)*(cos(1/ 2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1 /2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/b^3/(a^2-b^2)/d+1/3*(3*A*b^2-3*B*a* b+5*C*a^2-2*C*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipti cF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/b^2/(a^2- b^2)/d-(3*A*b^4+3*B*a^3*b-5*B*a*b^3-a^2*b^2*(A-7*C)-5*a^4*C)*(cos(1/2*d*x+ 1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b) ,2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/(a-b)/b^3/(a+b)^2/d
Leaf count is larger than twice the leaf count of optimal. \(926\) vs. \(2(447)=894\).
Time = 12.81 (sec) , antiderivative size = 926, normalized size of antiderivative = 2.07 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {(b+a \cos (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {2 \left (9 a^2 A b^2-12 A b^4-27 a^3 b B+30 a b^3 B+45 a^4 C-44 a^2 b^2 C-4 b^4 C\right ) \cos ^2(c+d x) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) (a+b \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 \left (12 a A b^3-24 a^2 b^2 B+12 b^4 B+40 a^3 b C-28 a b^3 C\right ) \cos ^2(c+d x) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) (a+b \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{a (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {\left (3 a^2 A b^2-9 a^3 b B+6 a b^3 B+15 a^4 C-12 a^2 b^2 C\right ) \cos (2 (c+d x)) (a+b \sec (c+d x)) \left (-4 a b+4 a b \sec ^2(c+d x)-4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-2 a (a-2 b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 a^2 \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-4 b^2 \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a^2 b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}\right )}{6 (a-b) b^3 (a+b) d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+b \sec (c+d x))^2}+\frac {(b+a \cos (c+d x))^2 \sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {2 \left (a A b^2-3 a^2 b B+2 b^3 B+5 a^3 C-4 a b^2 C\right ) \sin (c+d x)}{b^3 \left (-a^2+b^2\right )}-\frac {2 \left (a A b^2 \sin (c+d x)-a^2 b B \sin (c+d x)+a^3 C \sin (c+d x)\right )}{b^2 \left (-a^2+b^2\right ) (b+a \cos (c+d x))}+\frac {4 C \tan (c+d x)}{3 b^2}\right )}{d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+b \sec (c+d x))^2} \]
Integrate[(Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^2,x]
((b + a*Cos[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((2*(9*a^2 *A*b^2 - 12*A*b^4 - 27*a^3*b*B + 30*a*b^3*B + 45*a^4*C - 44*a^2*b^2*C - 4* b^4*C)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1] - Ellipti cPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1])*(a + b*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(b*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2 )) + (2*(12*a*A*b^3 - 24*a^2*b^2*B + 12*b^4*B + 40*a^3*b*C - 28*a*b^3*C)*C os[c + d*x]^2*EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*(a + b*Se c[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(a*(b + a*Cos[c + d*x]) *(1 - Cos[c + d*x]^2)) + ((3*a^2*A*b^2 - 9*a^3*b*B + 6*a*b^3*B + 15*a^4*C - 12*a^2*b^2*C)*Cos[2*(c + d*x)]*(a + b*Sec[c + d*x])*(-4*a*b + 4*a*b*Sec[ c + d*x]^2 - 4*a*b*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] - 2*a*(a - 2*b)*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + 2*a^2*Ellipti cPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Se c[c + d*x]^2] - 4*b^2*EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*S qrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2])*Sin[c + d*x])/(a^2*b*(b + a*Co s[c + d*x])*(1 - Cos[c + d*x]^2)*Sqrt[Sec[c + d*x]]*(2 - Sec[c + d*x]^2))) )/(6*(a - b)*b^3*(a + b)*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x ])*(a + b*Sec[c + d*x])^2) + ((b + a*Cos[c + d*x])^2*Sqrt[Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((2*(a*A*b^2 - 3*a^2*b*B + 2*b^3*...
Time = 3.32 (sec) , antiderivative size = 431, normalized size of antiderivative = 0.96, number of steps used = 21, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.488, Rules used = {3042, 4586, 27, 3042, 4590, 27, 3042, 4590, 27, 3042, 4594, 3042, 4274, 3042, 4258, 3042, 3119, 3120, 4336, 3042, 3284}
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.
| \(\displaystyle \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4586 |
| \(\displaystyle -\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (-\left (\left (5 C a^2-3 b B a+3 A b^2-2 b^2 C\right ) \sec ^2(c+d x)\right )+2 b (b B-a (A+C)) \sec (c+d x)+3 \left (A b^2-a (b B-a C)\right )\right )}{2 (a+b \sec (c+d x))}dx}{b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (-\left (\left (5 C a^2-3 b B a+3 A b^2-2 b^2 C\right ) \sec ^2(c+d x)\right )+2 b (b B-a (A+C)) \sec (c+d x)+3 \left (A b^2-a (b B-a C)\right )\right )}{a+b \sec (c+d x)}dx}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\left (-5 C a^2+3 b B a-3 A b^2+2 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 b (b B-a (A+C)) \csc \left (c+d x+\frac {\pi }{2}\right )+3 \left (A b^2-a (b B-a C)\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4590 |
| \(\displaystyle -\frac {\frac {2 \int -\frac {\sqrt {\sec (c+d x)} \left (3 \left (-5 C a^3+3 b B a^2-b^2 (A-4 C) a-2 b^3 B\right ) \sec ^2(c+d x)-2 b \left (2 C a^2-3 b B a+3 A b^2+b^2 C\right ) \sec (c+d x)+a \left (5 C a^2-3 b B a+3 A b^2-2 b^2 C\right )\right )}{2 (a+b \sec (c+d x))}dx}{3 b}-\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {\int \frac {\sqrt {\sec (c+d x)} \left (3 \left (-5 C a^3+3 b B a^2-b^2 (A-4 C) a-2 b^3 B\right ) \sec ^2(c+d x)-2 b \left (2 C a^2-3 b B a+3 A b^2+b^2 C\right ) \sec (c+d x)+a \left (5 C a^2-3 b B a+3 A b^2-2 b^2 C\right )\right )}{a+b \sec (c+d x)}dx}{3 b}-\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (3 \left (-5 C a^3+3 b B a^2-b^2 (A-4 C) a-2 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-2 b \left (2 C a^2-3 b B a+3 A b^2+b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+a \left (5 C a^2-3 b B a+3 A b^2-2 b^2 C\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{3 b}-\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4590 |
| \(\displaystyle -\frac {-\frac {\frac {2 \int -\frac {\left (-15 C a^4+9 b B a^3-b^2 (3 A-16 C) a^2-12 b^3 B a+2 b^4 (3 A+C)\right ) \sec ^2(c+d x)+2 b \left (-10 C a^3+6 b B a^2-b^2 (3 A-7 C) a-3 b^3 B\right ) \sec (c+d x)+3 a \left (-5 C a^3+3 b B a^2-b^2 (A-4 C) a-2 b^3 B\right )}{2 \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}dx}{b}+\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)} \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b d}}{3 b}-\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)} \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b d}-\frac {\int \frac {\left (-15 C a^4+9 b B a^3-b^2 (3 A-16 C) a^2-12 b^3 B a+2 b^4 (3 A+C)\right ) \sec ^2(c+d x)+2 b \left (-10 C a^3+6 b B a^2-b^2 (3 A-7 C) a-3 b^3 B\right ) \sec (c+d x)+3 a \left (-5 C a^3+3 b B a^2-b^2 (A-4 C) a-2 b^3 B\right )}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))}dx}{b}}{3 b}-\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)} \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b d}-\frac {\int \frac {\left (-15 C a^4+9 b B a^3-b^2 (3 A-16 C) a^2-12 b^3 B a+2 b^4 (3 A+C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 b \left (-10 C a^3+6 b B a^2-b^2 (3 A-7 C) a-3 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+3 a \left (-5 C a^3+3 b B a^2-b^2 (A-4 C) a-2 b^3 B\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{3 b}-\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4594 |
| \(\displaystyle -\frac {-\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)} \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b d}-\frac {\frac {\int \frac {3 a^2 \left (-5 C a^3+3 b B a^2-b^2 (A-4 C) a-2 b^3 B\right )-a^2 b \left (5 C a^2-3 b B a+3 A b^2-2 b^2 C\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx}{a^2}+3 \left (-5 a^4 C+3 a^3 b B-a^2 b^2 (A-7 C)-5 a b^3 B+3 A b^4\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)}dx}{b}}{3 b}-\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)} \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b d}-\frac {\frac {\int \frac {3 a^2 \left (-5 C a^3+3 b B a^2-b^2 (A-4 C) a-2 b^3 B\right )-a^2 b \left (5 C a^2-3 b B a+3 A b^2-2 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}+3 \left (-5 a^4 C+3 a^3 b B-a^2 b^2 (A-7 C)-5 a b^3 B+3 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{3 b}-\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle -\frac {-\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)} \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b d}-\frac {\frac {3 a^2 \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}}dx-a^2 b \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right ) \int \sqrt {\sec (c+d x)}dx}{a^2}+3 \left (-5 a^4 C+3 a^3 b B-a^2 b^2 (A-7 C)-5 a b^3 B+3 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{3 b}-\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)} \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b d}-\frac {\frac {3 a^2 \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right ) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-a^2 b \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+3 \left (-5 a^4 C+3 a^3 b B-a^2 b^2 (A-7 C)-5 a b^3 B+3 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{3 b}-\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle -\frac {-\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)} \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b d}-\frac {\frac {3 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right ) \int \sqrt {\cos (c+d x)}dx-a^2 b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a^2}+3 \left (-5 a^4 C+3 a^3 b B-a^2 b^2 (A-7 C)-5 a b^3 B+3 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{3 b}-\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)} \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b d}-\frac {\frac {3 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-a^2 b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}+3 \left (-5 a^4 C+3 a^3 b B-a^2 b^2 (A-7 C)-5 a b^3 B+3 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{3 b}-\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle -\frac {-\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)} \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b d}-\frac {\frac {\frac {6 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{d}-a^2 b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}+3 \left (-5 a^4 C+3 a^3 b B-a^2 b^2 (A-7 C)-5 a b^3 B+3 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{3 b}-\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle -\frac {-\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)} \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b d}-\frac {3 \left (-5 a^4 C+3 a^3 b B-a^2 b^2 (A-7 C)-5 a b^3 B+3 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {\frac {6 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{d}-\frac {2 a^2 b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{d}}{a^2}}{b}}{3 b}-\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4336 |
| \(\displaystyle -\frac {-\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)} \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b d}-\frac {3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-5 a^4 C+3 a^3 b B-a^2 b^2 (A-7 C)-5 a b^3 B+3 A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx+\frac {\frac {6 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{d}-\frac {2 a^2 b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{d}}{a^2}}{b}}{3 b}-\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)} \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b d}-\frac {3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-5 a^4 C+3 a^3 b B-a^2 b^2 (A-7 C)-5 a b^3 B+3 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx+\frac {\frac {6 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{d}-\frac {2 a^2 b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{d}}{a^2}}{b}}{3 b}-\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle -\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {-\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b d}-\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)} \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b d}-\frac {\frac {\frac {6 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{d}-\frac {2 a^2 b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{d}}{a^2}+\frac {6 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-5 a^4 C+3 a^3 b B-a^2 b^2 (A-7 C)-5 a b^3 B+3 A b^4\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{d (a+b)}}{b}}{3 b}}{2 b \left (a^2-b^2\right )}\) |
-(((A*b^2 - a*(b*B - a*C))*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(b*(a^2 - b^2) *d*(a + b*Sec[c + d*x]))) - ((-2*(3*A*b^2 - 3*a*b*B + 5*a^2*C - 2*b^2*C)*S ec[c + d*x]^(3/2)*Sin[c + d*x])/(3*b*d) - (-((((6*a^2*(3*a^2*b*B - 2*b^3*B - a*b^2*(A - 4*C) - 5*a^3*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] *Sqrt[Sec[c + d*x]])/d - (2*a^2*b*(3*A*b^2 - 3*a*b*B + 5*a^2*C - 2*b^2*C)* Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d)/a^2 + (6*(3*A*b^4 + 3*a^3*b*B - 5*a*b^3*B - a^2*b^2*(A - 7*C) - 5*a^4*C)*Sqrt[Co s[c + d*x]]*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/ ((a + b)*d))/b) + (6*(3*a^2*b*B - 2*b^3*B - a*b^2*(A - 4*C) - 5*a^3*C)*Sqr t[Sec[c + d*x]]*Sin[c + d*x])/(b*d))/(3*b))/(2*b*(a^2 - b^2))
3.11.19.3.1 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[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c , d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[d*Sqrt[d*Sin[e + f*x]]*Sqrt[d*Csc[e + f*x]] Int[ 1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[(-d)*(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 1)/(b*f*(a^2 - b^2)*(m + 1)) ), x] + Simp[d/(b*(a^2 - b^2)*(m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*( d*Csc[e + f*x])^(n - 1)*Simp[A*b^2*(n - 1) - a*(b*B - a*C)*(n - 1) + b*(a*A - b*B + a*C)*(m + 1)*Csc[e + f*x] - (b*(A*b - a*B)*(m + n + 1) + C*(a^2*n + b^2*(m + 1)))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C }, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[(-C)*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1 )*((d*Csc[e + f*x])^(n - 1)/(b*f*(m + n + 1))), x] + Simp[d/(b*(m + n + 1)) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1)*Simp[a*C*(n - 1) + ( A*b*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) - a*C*n)*Csc [e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2) Int[(d*Csc[e + f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Simp[1/a^2 Int[(a*A - (A*b - a *B)*Csc[e + f*x])/Sqrt[d*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]
Time = 32.81 (sec) , antiderivative size = 1004, normalized size of antiderivative = 2.25
int(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x, method=_RETURNVERBOSE)
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*C/b^2*(-1/6* cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(c os(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d* x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*E llipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2*(B*b-2*C*a)/b^3/sin(1/2*d*x+1/2*c) ^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^ 2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-EllipticE(cos(1/2*d*x+ 1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1 /2))+2*a^2*(B*b-2*C*a)/b^3/(a^2-a*b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos( 1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^( 1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+2*(A*b^2-B*a*b+C*a^2 )/b^2*(a^2/b/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2 *d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*a-a+b)-1/2/(a+b)/b*(sin(1/2*d *x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c )^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2* a/b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/ 2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2* d*x+1/2*c),2^(1/2))-1/2*a/b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos (1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^ (1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/b/(a^2-b^2)/(a^2-a*b)*a...
Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
integrate(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c) )^2,x, algorithm="fricas")
Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
integrate(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c) )^2,x, algorithm="maxima")
\[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
integrate(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c) )^2,x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sec(d*x + c)^(5/2)/(b*se c(d*x + c) + a)^2, x)
Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]
int(((1/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/cos(c + d*x))^2,x)