\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx\) [1023]

Optimal result

Integrand size = 43, antiderivative size = 406 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\frac {\left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^3 \left (a^2-b^2\right ) d}-\frac {\left (15 A b^4+12 a^3 b B-9 a b^3 B-a^2 b^2 (16 A-3 C)-2 a^4 (A+3 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^4 \left (a^2-b^2\right ) d}+\frac {b \left (5 A b^4+5 a^3 b B-3 a b^3 B-a^2 b^2 (7 A-C)-3 a^4 C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^4 (a-b) (a+b)^2 d}-\frac {\left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \] Output:

(5*A*b^3+2*B*a^3-3*B*a*b^2-a^2*b*(4*A-C))*cos(d*x+c)^(1/2)*EllipticE(sin(1 
/2*d*x+1/2*c),2^(1/2))*sec(d*x+c)^(1/2)/a^3/(a^2-b^2)/d-1/3*(15*A*b^4+12*B 
*a^3*b-9*B*a*b^3-a^2*b^2*(16*A-3*C)-2*a^4*(A+3*C))*cos(d*x+c)^(1/2)*Invers 
eJacobiAM(1/2*d*x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/a^4/(a^2-b^2)/d+b*(5*A*b 
^4+5*B*a^3*b-3*B*a*b^3-a^2*b^2*(7*A-C)-3*a^4*C)*cos(d*x+c)^(1/2)*EllipticP 
i(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))*sec(d*x+c)^(1/2)/a^4/(a-b)/(a+b)^2 
/d-1/3*(5*A*b^2-3*B*a*b-a^2*(2*A-3*C))*sin(d*x+c)/a^2/(a^2-b^2)/d/sec(d*x+ 
c)^(1/2)+(A*b^2-a*(B*b-C*a))*sin(d*x+c)/a/(a^2-b^2)/d/sec(d*x+c)^(1/2)/(a+ 
b*sec(d*x+c))
                                                                                    
                                                                                    
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(882\) vs. \(2(406)=812\).

Time = 8.14 (sec) , antiderivative size = 882, normalized size of antiderivative = 2.17 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:

Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(3/2)*(a + 
 b*Sec[c + d*x])^2),x]
 

Output:

((b + a*Cos[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((2*(-8*a^ 
2*A*b + 5*A*b^3 + 6*a^3*B - 3*a*b^2*B - 3*a^2*b*C)*Cos[c + d*x]^2*(Ellipti 
cF[ArcSin[Sqrt[Sec[c + d*x]]], -1] - 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]) 
/(b*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + (2*(4*a^3*A + 8*a*A*b^2 - 
 12*a^2*b*B + 12*a^3*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)) + ((-12*a^2*A*b + 15*A*b^3 
 + 6*a^3*B - 9*a*b^2*B + 3*a^2*b*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)*Ellipti 
cF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x 
]^2] + 2*a^2*EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c 
 + d*x]]*Sqrt[1 - Sec[c + d*x]^2] - 4*b^2*EllipticPi[-(b/a), ArcSin[Sqrt[S 
ec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2])*Sin[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))))/(6*a^2*(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) + ((b + a*Cos[c + d*x])^2*S 
qrt[Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((2*b*(A*b^2 - a 
*b*B + a^2*C)*Sin[c + d*x])/(a^3*(-a^2 + b^2)) + (2*(A*b^4*Sin[c + d*x]...
 

Rubi [A] (verified)

Time = 2.65 (sec) , antiderivative size = 393, normalized size of antiderivative = 0.97, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 4588, 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.

\(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\)

\(\Big \downarrow \) 4588

\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}-\frac {\int \frac {-\left ((2 A-3 C) a^2\right )-3 b B a+2 (A b+C b-a B) \sec (c+d x) a+5 A b^2-3 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x)}{2 \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}dx}{a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}-\frac {\int \frac {-\left ((2 A-3 C) a^2\right )-3 b B a+2 (A b+C b-a B) \sec (c+d x) a+5 A b^2-3 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}dx}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}-\frac {\int \frac {-\left ((2 A-3 C) a^2\right )-3 b B a+2 (A b+C b-a B) \csc \left (c+d x+\frac {\pi }{2}\right ) a+5 A b^2-3 \left (A b^2-a (b B-a C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 4592

\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {2 \int \frac {-b \left (-\left ((2 A-3 C) a^2\right )-3 b B a+5 A b^2\right ) \sec ^2(c+d x)+2 a \left ((A+3 C) a^2-3 b B a+2 A b^2\right ) \sec (c+d x)+3 \left (2 B a^3-b (4 A-C) a^2-3 b^2 B a+5 A b^3\right )}{2 \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}dx}{3 a}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\int \frac {-b \left (-\left ((2 A-3 C) a^2\right )-3 b B a+5 A b^2\right ) \sec ^2(c+d x)+2 a \left ((A+3 C) a^2-3 b B a+2 A b^2\right ) \sec (c+d x)+3 \left (2 B a^3-b (4 A-C) a^2-3 b^2 B a+5 A b^3\right )}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))}dx}{3 a}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\int \frac {-b \left (-\left ((2 A-3 C) a^2\right )-3 b B a+5 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 a \left ((A+3 C) a^2-3 b B a+2 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+3 \left (2 B a^3-b (4 A-C) a^2-3 b^2 B a+5 A b^3\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 4594

\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {3 b \left (-3 a^4 C+5 a^3 b B-a^2 b^2 (7 A-C)-3 a b^3 B+5 A b^4\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)}dx}{a^2}+\frac {\int \frac {3 a \left (2 B a^3-b (4 A-C) a^2-3 b^2 B a+5 A b^3\right )-\left (-2 (A+3 C) a^4+12 b B a^3-b^2 (16 A-3 C) a^2-9 b^3 B a+15 A b^4\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx}{a^2}}{3 a}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {3 b \left (-3 a^4 C+5 a^3 b B-a^2 b^2 (7 A-C)-3 a b^3 B+5 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {\int \frac {3 a \left (2 B a^3-b (4 A-C) a^2-3 b^2 B a+5 A b^3\right )+\left (2 (A+3 C) a^4-12 b B a^3+b^2 (16 A-3 C) a^2+9 b^3 B a-15 A b^4\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}}{3 a}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {3 b \left (-3 a^4 C+5 a^3 b B-a^2 b^2 (7 A-C)-3 a b^3 B+5 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {3 a \left (2 a^3 B-a^2 b (4 A-C)-3 a b^2 B+5 A b^3\right ) \int \frac {1}{\sqrt {\sec (c+d x)}}dx-\left (-2 a^4 (A+3 C)+12 a^3 b B-a^2 b^2 (16 A-3 C)-9 a b^3 B+15 A b^4\right ) \int \sqrt {\sec (c+d x)}dx}{a^2}}{3 a}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {3 a \left (2 a^3 B-a^2 b (4 A-C)-3 a b^2 B+5 A b^3\right ) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (-2 a^4 (A+3 C)+12 a^3 b B-a^2 b^2 (16 A-3 C)-9 a b^3 B+15 A b^4\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {3 b \left (-3 a^4 C+5 a^3 b B-a^2 b^2 (7 A-C)-3 a b^3 B+5 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{3 a}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {3 b \left (-3 a^4 C+5 a^3 b B-a^2 b^2 (7 A-C)-3 a b^3 B+5 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {3 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (2 a^3 B-a^2 b (4 A-C)-3 a b^2 B+5 A b^3\right ) \int \sqrt {\cos (c+d x)}dx-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-2 a^4 (A+3 C)+12 a^3 b B-a^2 b^2 (16 A-3 C)-9 a b^3 B+15 A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a^2}}{3 a}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {3 b \left (-3 a^4 C+5 a^3 b B-a^2 b^2 (7 A-C)-3 a b^3 B+5 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {3 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (2 a^3 B-a^2 b (4 A-C)-3 a b^2 B+5 A b^3\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-2 a^4 (A+3 C)+12 a^3 b B-a^2 b^2 (16 A-3 C)-9 a b^3 B+15 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}}{3 a}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {3 b \left (-3 a^4 C+5 a^3 b B-a^2 b^2 (7 A-C)-3 a b^3 B+5 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {\frac {6 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (2 a^3 B-a^2 b (4 A-C)-3 a b^2 B+5 A b^3\right )}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-2 a^4 (A+3 C)+12 a^3 b B-a^2 b^2 (16 A-3 C)-9 a b^3 B+15 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}}{3 a}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {3 b \left (-3 a^4 C+5 a^3 b B-a^2 b^2 (7 A-C)-3 a b^3 B+5 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {\frac {6 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (2 a^3 B-a^2 b (4 A-C)-3 a b^2 B+5 A b^3\right )}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-2 a^4 (A+3 C)+12 a^3 b B-a^2 b^2 (16 A-3 C)-9 a b^3 B+15 A b^4\right )}{d}}{a^2}}{3 a}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 4336

\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {3 b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-3 a^4 C+5 a^3 b B-a^2 b^2 (7 A-C)-3 a b^3 B+5 A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a^2}+\frac {\frac {6 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (2 a^3 B-a^2 b (4 A-C)-3 a b^2 B+5 A b^3\right )}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-2 a^4 (A+3 C)+12 a^3 b B-a^2 b^2 (16 A-3 C)-9 a b^3 B+15 A b^4\right )}{d}}{a^2}}{3 a}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {3 b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-3 a^4 C+5 a^3 b B-a^2 b^2 (7 A-C)-3 a b^3 B+5 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a^2}+\frac {\frac {6 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (2 a^3 B-a^2 b (4 A-C)-3 a b^2 B+5 A b^3\right )}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-2 a^4 (A+3 C)+12 a^3 b B-a^2 b^2 (16 A-3 C)-9 a b^3 B+15 A b^4\right )}{d}}{a^2}}{3 a}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {6 b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-3 a^4 C+5 a^3 b B-a^2 b^2 (7 A-C)-3 a b^3 B+5 A b^4\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a^2 d (a+b)}+\frac {\frac {6 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (2 a^3 B-a^2 b (4 A-C)-3 a b^2 B+5 A b^3\right )}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-2 a^4 (A+3 C)+12 a^3 b B-a^2 b^2 (16 A-3 C)-9 a b^3 B+15 A b^4\right )}{d}}{a^2}}{3 a}}{2 a \left (a^2-b^2\right )}\)

Input:

Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(3/2)*(a + b*Sec 
[c + d*x])^2),x]
 

Output:

((A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]] 
*(a + b*Sec[c + d*x])) - (-1/3*(((6*a*(5*A*b^3 + 2*a^3*B - 3*a*b^2*B - a^2 
*b*(4*A - C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d* 
x]])/d - (2*(15*A*b^4 + 12*a^3*b*B - 9*a*b^3*B - a^2*b^2*(16*A - 3*C) - 2* 
a^4*(A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d 
*x]])/d)/a^2 + (6*b*(5*A*b^4 + 5*a^3*b*B - 3*a*b^3*B - a^2*b^2*(7*A - C) - 
 3*a^4*C)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]*Sqr 
t[Sec[c + d*x]])/(a^2*(a + b)*d))/a + (2*(5*A*b^2 - 3*a*b*B - a^2*(2*A - 3 
*C))*Sin[c + d*x])/(3*a*d*Sqrt[Sec[c + d*x]]))/(2*a*(a^2 - b^2))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3119

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]
 

rule 3120

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]
 

rule 3284

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]
 

rule 4258

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]
 

rule 4274

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]
 

rule 4336

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]
 

rule 4588

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])
 

rule 4592

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]
 

rule 4594

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1122\) vs. \(2(391)=782\).

Time = 11.36 (sec) , antiderivative size = 1123, normalized size of antiderivative = 2.77

method	result	size
default	\(\text {Expression too large to display}\)	\(1123\)

Input:

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+b*sec(d*x+c))^2,x, 
method=_RETURNVERBOSE)
 

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/3/a^4*(4*A*c 
os(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a^2-2*A*cos(1/2*d*x+1/2*c)*sin(1/2* 
d*x+1/2*c)^2*a^2+A*EllipticF(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)*a^2+9*A*EllipticF(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)*b^2+6*A*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)*a*b-6*B*EllipticF(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)*a 
*b-3*B*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)*a^2+3*C*EllipticF(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)*a^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*b^2-B*a*b+C*a^ 
2)/a^4*(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*a*cos(1/2*d*x+1/2*c)^2-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*co 
s(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)...
 

Fricas [F]

\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+b*sec(d*x+c) 
)^2,x, algorithm="fricas")
 

Output:

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)^4 + 2*a*b*sec(d*x + c)^3 + a^2*sec(d*x + c)^2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(3/2)/(a+b*sec(d*x+ 
c))**2,x)
                                                                                    
                                                                                    
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+b*sec(d*x+c) 
)^2,x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+b*sec(d*x+c) 
)^2,x, algorithm="giac")
 

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)^2* 
sec(d*x + c)^(3/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:

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))^(3/2)),x)
 

Output:

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))^(3/2)), x)
 

Reduce [F]

\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{4} b^{2}+2 \sec \left (d x +c \right )^{3} a b +\sec \left (d x +c \right )^{2} a^{2}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{3} b^{2}+2 \sec \left (d x +c \right )^{2} a b +\sec \left (d x +c \right ) a^{2}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{2} b^{2}+2 \sec \left (d x +c \right ) a b +a^{2}}d x \right ) c \] Input:

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+b*sec(d*x+c))^2,x)
 

Output:

int(sqrt(sec(c + d*x))/(sec(c + d*x)**4*b**2 + 2*sec(c + d*x)**3*a*b + sec 
(c + d*x)**2*a**2),x)*a + int(sqrt(sec(c + d*x))/(sec(c + d*x)**3*b**2 + 2 
*sec(c + d*x)**2*a*b + sec(c + d*x)*a**2),x)*b + int(sqrt(sec(c + d*x))/(s 
ec(c + d*x)**2*b**2 + 2*sec(c + d*x)*a*b + a**2),x)*c