Integrand size = 43, antiderivative size = 507 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=-\frac {\left (35 A b^4+20 a^3 b B-25 a b^3 B-3 a^2 b^2 (8 A-5 C)-2 a^4 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a^4 \left (a^2-b^2\right ) d}+\frac {\left (21 A b^5+2 a^5 B+16 a^3 b^2 B-15 a b^4 B-a^2 b^3 (20 A-9 C)-4 a^4 b (A+3 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^5 \left (a^2-b^2\right ) d}-\frac {b^2 \left (7 A b^4+7 a^3 b B-5 a b^3 B-3 a^2 b^2 (3 A-C)-5 a^4 C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^5 (a-b) (a+b)^2 d}-\frac {\left (7 A b^2-5 a b B-a^2 (2 A-5 C)\right ) \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (7 A b^3+2 a^3 B-5 a b^2 B-a^2 (4 A b-3 b C)\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))} \]
-1/5*(7*A*b^2-5*B*a*b-a^2*(2*A-5*C))*sin(d*x+c)/a^2/(a^2-b^2)/d/sec(d*x+c) ^(3/2)+(A*b^2-a*(B*b-C*a))*sin(d*x+c)/a/(a^2-b^2)/d/sec(d*x+c)^(3/2)/(a+b* sec(d*x+c))+1/3*(7*A*b^3+2*B*a^3-5*B*a*b^2-a^2*(4*A*b-3*C*b))*sin(d*x+c)/a ^3/(a^2-b^2)/d/sec(d*x+c)^(1/2)-1/5*(35*A*b^4+20*B*a^3*b-25*B*a*b^3-3*a^2* b^2*(8*A-5*C)-2*a^4*(3*A+5*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)/a^4/(a^2-b^2)/d+1/3*(21*A*b^5+2*a^5*B+16*a^3*b^2*B-15*a*b^4*B-a^2*b^3*( 20*A-9*C)-4*a^4*b*(A+3*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c) *EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a ^5/(a^2-b^2)/d-b^2*(7*A*b^4+7*B*a^3*b-5*B*a*b^3-3*a^2*b^2*(3*A-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^5/(a-b)/(a+b)^ 2/d
Time = 14.38 (sec) , antiderivative size = 971, normalized size of antiderivative = 1.92 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (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 (-18 a^4 A-32 a^2 A b^2+35 A b^4+40 a^3 b B-25 a b^3 B-30 a^4 C+15 a^2 b^2 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 (4 a^3 A b+56 a A b^3-20 a^4 B-40 a^2 b^2 B+60 a^3 b 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 (-18 a^4 A-72 a^2 A b^2+105 A b^4+60 a^3 b B-75 a b^3 B-30 a^4 C+45 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 )}{30 a^3 (-a+b) (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 {\left (a^4 A-a^2 A b^2+10 A b^4-10 a b^3 B+10 a^2 b^2 C\right ) \sin (c+d x)}{5 a^4 \left (a^2-b^2\right )}-\frac {2 \left (A b^5 \sin (c+d x)-a b^4 B \sin (c+d x)+a^2 b^3 C \sin (c+d x)\right )}{a^4 \left (a^2-b^2\right ) (b+a \cos (c+d x))}+\frac {2 (-2 A b+a B) \sin (2 (c+d x))}{3 a^3}+\frac {A \sin (3 (c+d x))}{5 a^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[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(5/2)*(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*(-18*a ^4*A - 32*a^2*A*b^2 + 35*A*b^4 + 40*a^3*b*B - 25*a*b^3*B - 30*a^4*C + 15*a ^2*b^2*C)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1] - Elli pticPi[-(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*(4*a^3*A*b + 56*a*A*b^3 - 20*a^4*B - 40*a^2*b^2*B + 60*a^3*b*C) *Cos[c + d*x]^2*EllipticPi[-(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])/(a*(b + a*Cos[c + d*x ])*(1 - Cos[c + d*x]^2)) + ((-18*a^4*A - 72*a^2*A*b^2 + 105*A*b^4 + 60*a^3 *b*B - 75*a*b^3*B - 30*a^4*C + 45*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 - Se c[c + d*x]^2] + 2*a^2*EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*S qrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] - 4*b^2*EllipticPi[-(b/a), ArcS in[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2])*S in[c + d*x])/(a^2*b*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2)*Sqrt[Sec[c + d*x]]*(2 - Sec[c + d*x]^2))))/(30*a^3*(-a + b)*(a + b)*d*(A + 2*C + 2*B*C os[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)*((...
Time = 3.56 (sec) , antiderivative size = 487, 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, 4588, 27, 3042, 4592, 27, 3042, 4592, 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.
| \(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4588 |
| \(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\int \frac {-\left ((2 A-5 C) a^2\right )-5 b B a+2 (A b+C b-a B) \sec (c+d x) a+7 A b^2-5 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x)}{2 \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))}dx}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\int \frac {-\left ((2 A-5 C) a^2\right )-5 b B a+2 (A b+C b-a B) \sec (c+d x) a+7 A b^2-5 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))}dx}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\int \frac {-\left ((2 A-5 C) a^2\right )-5 b B a+2 (A b+C b-a B) \csc \left (c+d x+\frac {\pi }{2}\right ) a+7 A b^2-5 \left (A b^2-a (b B-a C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 4592 |
| \(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-5 C)\right )-5 a b B+7 A b^2\right )}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 \int \frac {-3 b \left (-\left ((2 A-5 C) a^2\right )-5 b B a+7 A b^2\right ) \sec ^2(c+d x)+2 a \left ((3 A+5 C) a^2-5 b B a+2 A b^2\right ) \sec (c+d x)+5 \left (2 B a^3-(4 A b-3 b C) a^2-5 b^2 B a+7 A b^3\right )}{2 \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}dx}{5 a}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-5 C)\right )-5 a b B+7 A b^2\right )}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-3 b \left (-\left ((2 A-5 C) a^2\right )-5 b B a+7 A b^2\right ) \sec ^2(c+d x)+2 a \left ((3 A+5 C) a^2-5 b B a+2 A b^2\right ) \sec (c+d x)+5 \left (2 B a^3-(4 A b-3 b C) a^2-5 b^2 B a+7 A b^3\right )}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}dx}{5 a}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-5 C)\right )-5 a b B+7 A b^2\right )}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-3 b \left (-\left ((2 A-5 C) a^2\right )-5 b B a+7 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 a \left ((3 A+5 C) a^2-5 b B a+2 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+5 \left (2 B a^3-(4 A b-3 b C) a^2-5 b^2 B a+7 A b^3\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{5 a}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 4592 |
| \(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-5 C)\right )-5 a b B+7 A b^2\right )}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 \sin (c+d x) \left (2 a^3 B-a^2 (4 A b-3 b C)-5 a b^2 B+7 A b^3\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {2 \int \frac {-5 b \left (2 B a^3-(4 A b-3 b C) a^2-5 b^2 B a+7 A b^3\right ) \sec ^2(c+d x)+2 a \left (-5 B a^3+b (A+15 C) a^2-10 b^2 B a+14 A b^3\right ) \sec (c+d x)+3 \left (-2 (3 A+5 C) a^4+20 b B a^3-3 b^2 (8 A-5 C) a^2-25 b^3 B a+35 A b^4\right )}{2 \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}dx}{3 a}}{5 a}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-5 C)\right )-5 a b B+7 A b^2\right )}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 \sin (c+d x) \left (2 a^3 B-a^2 (4 A b-3 b C)-5 a b^2 B+7 A b^3\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\int \frac {-5 b \left (2 B a^3-(4 A b-3 b C) a^2-5 b^2 B a+7 A b^3\right ) \sec ^2(c+d x)+2 a \left (-5 B a^3+b (A+15 C) a^2-10 b^2 B a+14 A b^3\right ) \sec (c+d x)+3 \left (-2 (3 A+5 C) a^4+20 b B a^3-3 b^2 (8 A-5 C) a^2-25 b^3 B a+35 A b^4\right )}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))}dx}{3 a}}{5 a}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-5 C)\right )-5 a b B+7 A b^2\right )}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 \sin (c+d x) \left (2 a^3 B-a^2 (4 A b-3 b C)-5 a b^2 B+7 A b^3\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\int \frac {-5 b \left (2 B a^3-(4 A b-3 b C) a^2-5 b^2 B a+7 A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 a \left (-5 B a^3+b (A+15 C) a^2-10 b^2 B a+14 A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+3 \left (-2 (3 A+5 C) a^4+20 b B a^3-3 b^2 (8 A-5 C) a^2-25 b^3 B a+35 A b^4\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}}{5 a}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 4594 |
| \(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-5 C)\right )-5 a b B+7 A b^2\right )}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 \sin (c+d x) \left (2 a^3 B-a^2 (4 A b-3 b C)-5 a b^2 B+7 A b^3\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^2 \left (-5 a^4 C+7 a^3 b B-3 a^2 b^2 (3 A-C)-5 a b^3 B+7 A b^4\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)}dx}{a^2}+\frac {\int \frac {3 a \left (-2 (3 A+5 C) a^4+20 b B a^3-3 b^2 (8 A-5 C) a^2-25 b^3 B a+35 A b^4\right )-5 \left (2 B a^5-4 b (A+3 C) a^4+16 b^2 B a^3-b^3 (20 A-9 C) a^2-15 b^4 B a+21 A b^5\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx}{a^2}}{3 a}}{5 a}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-5 C)\right )-5 a b B+7 A b^2\right )}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 \sin (c+d x) \left (2 a^3 B-a^2 (4 A b-3 b C)-5 a b^2 B+7 A b^3\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^2 \left (-5 a^4 C+7 a^3 b B-3 a^2 b^2 (3 A-C)-5 a b^3 B+7 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {\int \frac {3 a \left (-2 (3 A+5 C) a^4+20 b B a^3-3 b^2 (8 A-5 C) a^2-25 b^3 B a+35 A b^4\right )-5 \left (2 B a^5-4 b (A+3 C) a^4+16 b^2 B a^3-b^3 (20 A-9 C) a^2-15 b^4 B a+21 A b^5\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}}{3 a}}{5 a}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-5 C)\right )-5 a b B+7 A b^2\right )}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 \sin (c+d x) \left (2 a^3 B-a^2 (4 A b-3 b C)-5 a b^2 B+7 A b^3\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^2 \left (-5 a^4 C+7 a^3 b B-3 a^2 b^2 (3 A-C)-5 a b^3 B+7 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {3 a \left (-2 a^4 (3 A+5 C)+20 a^3 b B-3 a^2 b^2 (8 A-5 C)-25 a b^3 B+35 A b^4\right ) \int \frac {1}{\sqrt {\sec (c+d x)}}dx-5 \left (2 a^5 B-4 a^4 b (A+3 C)+16 a^3 b^2 B-a^2 b^3 (20 A-9 C)-15 a b^4 B+21 A b^5\right ) \int \sqrt {\sec (c+d x)}dx}{a^2}}{3 a}}{5 a}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-5 C)\right )-5 a b B+7 A b^2\right )}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 \sin (c+d x) \left (2 a^3 B-a^2 (4 A b-3 b C)-5 a b^2 B+7 A b^3\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^2 \left (-5 a^4 C+7 a^3 b B-3 a^2 b^2 (3 A-C)-5 a b^3 B+7 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {3 a \left (-2 a^4 (3 A+5 C)+20 a^3 b B-3 a^2 b^2 (8 A-5 C)-25 a b^3 B+35 A b^4\right ) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-5 \left (2 a^5 B-4 a^4 b (A+3 C)+16 a^3 b^2 B-a^2 b^3 (20 A-9 C)-15 a b^4 B+21 A b^5\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{3 a}}{5 a}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-5 C)\right )-5 a b B+7 A b^2\right )}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 \sin (c+d x) \left (2 a^3 B-a^2 (4 A b-3 b C)-5 a b^2 B+7 A b^3\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^2 \left (-5 a^4 C+7 a^3 b B-3 a^2 b^2 (3 A-C)-5 a b^3 B+7 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {3 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-2 a^4 (3 A+5 C)+20 a^3 b B-3 a^2 b^2 (8 A-5 C)-25 a b^3 B+35 A b^4\right ) \int \sqrt {\cos (c+d x)}dx-5 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (2 a^5 B-4 a^4 b (A+3 C)+16 a^3 b^2 B-a^2 b^3 (20 A-9 C)-15 a b^4 B+21 A b^5\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a^2}}{3 a}}{5 a}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-5 C)\right )-5 a b B+7 A b^2\right )}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 \sin (c+d x) \left (2 a^3 B-a^2 (4 A b-3 b C)-5 a b^2 B+7 A b^3\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^2 \left (-5 a^4 C+7 a^3 b B-3 a^2 b^2 (3 A-C)-5 a b^3 B+7 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {3 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-2 a^4 (3 A+5 C)+20 a^3 b B-3 a^2 b^2 (8 A-5 C)-25 a b^3 B+35 A b^4\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-5 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (2 a^5 B-4 a^4 b (A+3 C)+16 a^3 b^2 B-a^2 b^3 (20 A-9 C)-15 a b^4 B+21 A b^5\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}}{3 a}}{5 a}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-5 C)\right )-5 a b B+7 A b^2\right )}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 \sin (c+d x) \left (2 a^3 B-a^2 (4 A b-3 b C)-5 a b^2 B+7 A b^3\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^2 \left (-5 a^4 C+7 a^3 b B-3 a^2 b^2 (3 A-C)-5 a b^3 B+7 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {\frac {6 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-2 a^4 (3 A+5 C)+20 a^3 b B-3 a^2 b^2 (8 A-5 C)-25 a b^3 B+35 A b^4\right )}{d}-5 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (2 a^5 B-4 a^4 b (A+3 C)+16 a^3 b^2 B-a^2 b^3 (20 A-9 C)-15 a b^4 B+21 A b^5\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}}{3 a}}{5 a}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-5 C)\right )-5 a b B+7 A b^2\right )}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 \sin (c+d x) \left (2 a^3 B-a^2 (4 A b-3 b C)-5 a b^2 B+7 A b^3\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^2 \left (-5 a^4 C+7 a^3 b B-3 a^2 b^2 (3 A-C)-5 a b^3 B+7 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {\frac {6 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-2 a^4 (3 A+5 C)+20 a^3 b B-3 a^2 b^2 (8 A-5 C)-25 a b^3 B+35 A b^4\right )}{d}-\frac {10 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (2 a^5 B-4 a^4 b (A+3 C)+16 a^3 b^2 B-a^2 b^3 (20 A-9 C)-15 a b^4 B+21 A b^5\right )}{d}}{a^2}}{3 a}}{5 a}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 4336 |
| \(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-5 C)\right )-5 a b B+7 A b^2\right )}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 \sin (c+d x) \left (2 a^3 B-a^2 (4 A b-3 b C)-5 a b^2 B+7 A b^3\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-5 a^4 C+7 a^3 b B-3 a^2 b^2 (3 A-C)-5 a b^3 B+7 A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a^2}+\frac {\frac {6 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-2 a^4 (3 A+5 C)+20 a^3 b B-3 a^2 b^2 (8 A-5 C)-25 a b^3 B+35 A b^4\right )}{d}-\frac {10 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (2 a^5 B-4 a^4 b (A+3 C)+16 a^3 b^2 B-a^2 b^3 (20 A-9 C)-15 a b^4 B+21 A b^5\right )}{d}}{a^2}}{3 a}}{5 a}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-5 C)\right )-5 a b B+7 A b^2\right )}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 \sin (c+d x) \left (2 a^3 B-a^2 (4 A b-3 b C)-5 a b^2 B+7 A b^3\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-5 a^4 C+7 a^3 b B-3 a^2 b^2 (3 A-C)-5 a b^3 B+7 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a^2}+\frac {\frac {6 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-2 a^4 (3 A+5 C)+20 a^3 b B-3 a^2 b^2 (8 A-5 C)-25 a b^3 B+35 A b^4\right )}{d}-\frac {10 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (2 a^5 B-4 a^4 b (A+3 C)+16 a^3 b^2 B-a^2 b^3 (20 A-9 C)-15 a b^4 B+21 A b^5\right )}{d}}{a^2}}{3 a}}{5 a}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-5 C)\right )-5 a b B+7 A b^2\right )}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 \sin (c+d x) \left (2 a^3 B-a^2 (4 A b-3 b C)-5 a b^2 B+7 A b^3\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {30 b^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-5 a^4 C+7 a^3 b B-3 a^2 b^2 (3 A-C)-5 a b^3 B+7 A b^4\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a^2 d (a+b)}+\frac {\frac {6 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-2 a^4 (3 A+5 C)+20 a^3 b B-3 a^2 b^2 (8 A-5 C)-25 a b^3 B+35 A b^4\right )}{d}-\frac {10 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (2 a^5 B-4 a^4 b (A+3 C)+16 a^3 b^2 B-a^2 b^3 (20 A-9 C)-15 a b^4 B+21 A b^5\right )}{d}}{a^2}}{3 a}}{5 a}}{2 a \left (a^2-b^2\right )}\) |
((A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(a*(a^2 - b^2)*d*Sec[c + d*x]^(3/2) *(a + b*Sec[c + d*x])) - ((2*(7*A*b^2 - 5*a*b*B - a^2*(2*A - 5*C))*Sin[c + d*x])/(5*a*d*Sec[c + d*x]^(3/2)) - (-1/3*(((6*a*(35*A*b^4 + 20*a^3*b*B - 25*a*b^3*B - 3*a^2*b^2*(8*A - 5*C) - 2*a^4*(3*A + 5*C))*Sqrt[Cos[c + d*x]] *EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d - (10*(21*A*b^5 + 2*a^5*B + 16*a^3*b^2*B - 15*a*b^4*B - a^2*b^3*(20*A - 9*C) - 4*a^4*b*(A + 3*C))*S qrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d)/a^2 + ( 30*b^2*(7*A*b^4 + 7*a^3*b*B - 5*a*b^3*B - 3*a^2*b^2*(3*A - C) - 5*a^4*C)*S qrt[Cos[c + d*x]]*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d *x]])/(a^2*(a + b)*d))/a + (10*(7*A*b^3 + 2*a^3*B - 5*a*b^2*B - a^2*(4*A*b - 3*b*C))*Sin[c + d*x])/(3*a*d*Sqrt[Sec[c + d*x]]))/(5*a))/(2*a*(a^2 - b^ 2))
3.11.24.3.1 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[(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/(a*f*(m + 1)*(a^2 - b^2))), x] + Sim p[1/(a*(m + 1)*(a^2 - b^2)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f *x])^n*Simp[a*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C)*(m + n + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x ] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && !(ILtQ[m + 1/2, 0] && ILtQ[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[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d *Csc[e + f*x])^n/(a*f*n)), x] + Simp[1/(a*d*n) Int[(a + b*Csc[e + f*x])^m *(d*Csc[e + f*x])^(n + 1)*Simp[a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)* Csc[e + f*x] + A*b*(m + n + 2)*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] && LeQ[n, -1]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(1376\) vs. \(2(559)=1118\).
Time = 3.53 (sec) , antiderivative size = 1377, normalized size of antiderivative = 2.72
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(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)*(-4/5*A/a^2/(-2 *sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*sin(1/2*d*x+1/2*c)^6* cos(1/2*d*x+1/2*c)-14*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d* x+1/2*c)^2*cos(1/2*d*x+1/2*c)-5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d* x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+9*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))-4/3/a^3*(3*A*a+2*A*b-B*a)*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1 /2*c)^4-sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+2*(sin(1/2*d*x+1/2*c)^2)^( 1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2) )-3*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*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c) ^2)^(1/2)-2*b^2/a^4*(5*A*b^2-4*B*a*b+3*C*a^2)/(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 *b^3*(A*b^2-B*a*b+C*a^2)/a^5*(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...
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c) )^2,x, algorithm="fricas")
integral((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(sec(d*x + c))/(b^2*s ec(d*x + c)^5 + 2*a*b*sec(d*x + c)^4 + a^2*sec(d*x + c)^3), x)
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c) )^2,x, algorithm="maxima")
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+b*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)^2* sec(d*x + c)^(5/2)), x)
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/((a + b/cos(c + d*x))^2*(1/cos (c + d*x))^(5/2)),x)