\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3} \, dx\) [1029]

Optimal result

Integrand size = 43, antiderivative size = 486 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3} \, dx=\frac {\left (15 A b^4+9 a^3 b B-3 a b^3 B+a^4 (8 A-5 C)-a^2 b^2 (29 A+C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 a^3 \left (a^2-b^2\right )^2 d}-\frac {\left (15 A b^5-8 a^5 B+5 a^3 b^2 B-3 a b^4 B-a^2 b^3 (33 A+C)+a^4 b (24 A+7 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 a^4 \left (a^2-b^2\right )^2 d}+\frac {\left (15 A b^6-15 a^5 b B+6 a^3 b^3 B-3 a b^5 B+3 a^6 C-a^2 b^4 (38 A+C)+5 a^4 b^2 (7 A+2 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 a^4 (a-b)^2 (a+b)^3 d}+\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {\left (5 A b^4+7 a^3 b B-a b^3 B-3 a^4 C-a^2 b^2 (11 A+3 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \] Output:

1/4*(15*A*b^4+9*B*a^3*b-3*B*a*b^3+a^4*(8*A-5*C)-a^2*b^2*(29*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)^2/d-1/4*(15*A*b^5-8*a^5*B+5*a^3*b^2*B-3*a*b^4*B-a^2*b^3*(33*A+C)+a^4*b 
*(24*A+7*C))*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*sec(d 
*x+c)^(1/2)/a^4/(a^2-b^2)^2/d+1/4*(15*A*b^6-15*a^5*b*B+6*a^3*b^3*B-3*a*b^5 
*B+3*a^6*C-a^2*b^4*(38*A+C)+5*a^4*b^2*(7*A+2*C))*cos(d*x+c)^(1/2)*Elliptic 
Pi(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)^2/(a+b 
)^3/d+1/2*(A*b^2-a*(B*b-C*a))*sec(d*x+c)^(1/2)*sin(d*x+c)/a/(a^2-b^2)/d/(a 
+b*sec(d*x+c))^2-1/4*(5*A*b^4+7*B*a^3*b-B*a*b^3-3*a^4*C-a^2*b^2*(11*A+3*C) 
)*sec(d*x+c)^(1/2)*sin(d*x+c)/a^2/(a^2-b^2)^2/d/(a+b*sec(d*x+c))
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1059\) vs. \(2(486)=972\).

Time = 9.48 (sec) , antiderivative size = 1059, normalized size of antiderivative = 2.18 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

((b + a*Cos[c + d*x])^3*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^ 
2)*((2*(8*a^4*A - 7*a^2*A*b^2 + 5*A*b^4 - 5*a^3*b*B - a*b^3*B + a^4*C + 5* 
a^2*b^2*C)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1] - Ell 
ipticPi[-(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*(-32*a^3*A*b + 8*a*A*b^3 + 16*a^4*B + 8*a^2*b^2*B - 24*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)) + ((8*a^4*A - 29*a^2*A*b^2 + 15*A*b^4 + 9*a^3*b* 
B - 3*a*b^3*B - 5*a^4*C - 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)*Ellipt 
icF[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[ 
Sec[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))))/(8*a^2*(a - b)^2*(a + b)^2*d*(A + 2*C + 2*B*Cos[c + 
 d*x] + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^3) + ((b + a*Cos[c + d*x] 
)^3*Sec[c + d*x]^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(-1/2*((...
 

Rubi [A] (verified)

Time = 3.20 (sec) , antiderivative size = 486, normalized size of antiderivative = 1.00, 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, 4588, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4588

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4588

\(\displaystyle \frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (-3 a^4 C+7 a^3 b B-a^2 b^2 (11 A+3 C)-a b^3 B+5 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\int \frac {(8 A-5 C) a^4+9 b B a^3-b^2 (29 A+C) a^2-3 b^3 B a+4 \left (2 B a^3-b (4 A+3 C) a^2+b^2 B a+A b^3\right ) \sec (c+d x) a+15 A b^4-\left (-3 C a^4+7 b B a^3-b^2 (11 A+3 C) a^2-b^3 B a+5 A b^4\right ) \sec ^2(c+d x)}{2 \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}dx}{a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (-3 a^4 C+7 a^3 b B-a^2 b^2 (11 A+3 C)-a b^3 B+5 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\int \frac {(8 A-5 C) a^4+9 b B a^3-b^2 (29 A+C) a^2-3 b^3 B a+4 \left (2 B a^3-b (4 A+3 C) a^2+b^2 B a+A b^3\right ) \sec (c+d x) a+15 A b^4-\left (-3 C a^4+7 b B a^3-b^2 (11 A+3 C) a^2-b^3 B a+5 A b^4\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))}dx}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (-3 a^4 C+7 a^3 b B-a^2 b^2 (11 A+3 C)-a b^3 B+5 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\int \frac {(8 A-5 C) a^4+9 b B a^3-b^2 (29 A+C) a^2-3 b^3 B a+4 \left (2 B a^3-b (4 A+3 C) a^2+b^2 B a+A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) a+15 A b^4+\left (3 C a^4-7 b B a^3+b^2 (11 A+3 C) a^2+b^3 B a-5 A b^4\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 4594

\(\displaystyle \frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (-3 a^4 C+7 a^3 b B-a^2 b^2 (11 A+3 C)-a b^3 B+5 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\int \frac {a \left ((8 A-5 C) a^4+9 b B a^3-b^2 (29 A+C) a^2-3 b^3 B a+15 A b^4\right )-\left (-8 B a^5+b (24 A+7 C) a^4+5 b^2 B a^3-b^3 (33 A+C) a^2-3 b^4 B a+15 A b^5\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx}{a^2}+\frac {\left (3 a^6 C-15 a^5 b B+5 a^4 b^2 (7 A+2 C)+6 a^3 b^3 B-a^2 b^4 (38 A+C)-3 a b^5 B+15 A b^6\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)}dx}{a^2}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (-3 a^4 C+7 a^3 b B-a^2 b^2 (11 A+3 C)-a b^3 B+5 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\int \frac {a \left ((8 A-5 C) a^4+9 b B a^3-b^2 (29 A+C) a^2-3 b^3 B a+15 A b^4\right )+\left (8 B a^5-b (24 A+7 C) a^4-5 b^2 B a^3+b^3 (33 A+C) a^2+3 b^4 B a-15 A b^5\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}+\frac {\left (3 a^6 C-15 a^5 b B+5 a^4 b^2 (7 A+2 C)+6 a^3 b^3 B-a^2 b^4 (38 A+C)-3 a b^5 B+15 A b^6\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (-3 a^4 C+7 a^3 b B-a^2 b^2 (11 A+3 C)-a b^3 B+5 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {a \left (a^4 (8 A-5 C)+9 a^3 b B-a^2 b^2 (29 A+C)-3 a b^3 B+15 A b^4\right ) \int \frac {1}{\sqrt {\sec (c+d x)}}dx-\left (-8 a^5 B+a^4 b (24 A+7 C)+5 a^3 b^2 B-a^2 b^3 (33 A+C)-3 a b^4 B+15 A b^5\right ) \int \sqrt {\sec (c+d x)}dx}{a^2}+\frac {\left (3 a^6 C-15 a^5 b B+5 a^4 b^2 (7 A+2 C)+6 a^3 b^3 B-a^2 b^4 (38 A+C)-3 a b^5 B+15 A b^6\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (-3 a^4 C+7 a^3 b B-a^2 b^2 (11 A+3 C)-a b^3 B+5 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {a \left (a^4 (8 A-5 C)+9 a^3 b B-a^2 b^2 (29 A+C)-3 a b^3 B+15 A b^4\right ) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (-8 a^5 B+a^4 b (24 A+7 C)+5 a^3 b^2 B-a^2 b^3 (33 A+C)-3 a b^4 B+15 A b^5\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {\left (3 a^6 C-15 a^5 b B+5 a^4 b^2 (7 A+2 C)+6 a^3 b^3 B-a^2 b^4 (38 A+C)-3 a b^5 B+15 A b^6\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (-3 a^4 C+7 a^3 b B-a^2 b^2 (11 A+3 C)-a b^3 B+5 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (a^4 (8 A-5 C)+9 a^3 b B-a^2 b^2 (29 A+C)-3 a b^3 B+15 A b^4\right ) \int \sqrt {\cos (c+d x)}dx-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-8 a^5 B+a^4 b (24 A+7 C)+5 a^3 b^2 B-a^2 b^3 (33 A+C)-3 a b^4 B+15 A b^5\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a^2}+\frac {\left (3 a^6 C-15 a^5 b B+5 a^4 b^2 (7 A+2 C)+6 a^3 b^3 B-a^2 b^4 (38 A+C)-3 a b^5 B+15 A b^6\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (-3 a^4 C+7 a^3 b B-a^2 b^2 (11 A+3 C)-a b^3 B+5 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (a^4 (8 A-5 C)+9 a^3 b B-a^2 b^2 (29 A+C)-3 a b^3 B+15 A b^4\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-8 a^5 B+a^4 b (24 A+7 C)+5 a^3 b^2 B-a^2 b^3 (33 A+C)-3 a b^4 B+15 A b^5\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}+\frac {\left (3 a^6 C-15 a^5 b B+5 a^4 b^2 (7 A+2 C)+6 a^3 b^3 B-a^2 b^4 (38 A+C)-3 a b^5 B+15 A b^6\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (-3 a^4 C+7 a^3 b B-a^2 b^2 (11 A+3 C)-a b^3 B+5 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\frac {2 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^4 (8 A-5 C)+9 a^3 b B-a^2 b^2 (29 A+C)-3 a b^3 B+15 A b^4\right )}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-8 a^5 B+a^4 b (24 A+7 C)+5 a^3 b^2 B-a^2 b^3 (33 A+C)-3 a b^4 B+15 A b^5\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}+\frac {\left (3 a^6 C-15 a^5 b B+5 a^4 b^2 (7 A+2 C)+6 a^3 b^3 B-a^2 b^4 (38 A+C)-3 a b^5 B+15 A b^6\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (-3 a^4 C+7 a^3 b B-a^2 b^2 (11 A+3 C)-a b^3 B+5 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\left (3 a^6 C-15 a^5 b B+5 a^4 b^2 (7 A+2 C)+6 a^3 b^3 B-a^2 b^4 (38 A+C)-3 a b^5 B+15 A b^6\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {\frac {2 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^4 (8 A-5 C)+9 a^3 b B-a^2 b^2 (29 A+C)-3 a b^3 B+15 A b^4\right )}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-8 a^5 B+a^4 b (24 A+7 C)+5 a^3 b^2 B-a^2 b^3 (33 A+C)-3 a b^4 B+15 A b^5\right )}{d}}{a^2}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 4336

\(\displaystyle \frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (-3 a^4 C+7 a^3 b B-a^2 b^2 (11 A+3 C)-a b^3 B+5 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (3 a^6 C-15 a^5 b B+5 a^4 b^2 (7 A+2 C)+6 a^3 b^3 B-a^2 b^4 (38 A+C)-3 a b^5 B+15 A b^6\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a^2}+\frac {\frac {2 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^4 (8 A-5 C)+9 a^3 b B-a^2 b^2 (29 A+C)-3 a b^3 B+15 A b^4\right )}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-8 a^5 B+a^4 b (24 A+7 C)+5 a^3 b^2 B-a^2 b^3 (33 A+C)-3 a b^4 B+15 A b^5\right )}{d}}{a^2}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (-3 a^4 C+7 a^3 b B-a^2 b^2 (11 A+3 C)-a b^3 B+5 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (3 a^6 C-15 a^5 b B+5 a^4 b^2 (7 A+2 C)+6 a^3 b^3 B-a^2 b^4 (38 A+C)-3 a b^5 B+15 A b^6\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a^2}+\frac {\frac {2 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^4 (8 A-5 C)+9 a^3 b B-a^2 b^2 (29 A+C)-3 a b^3 B+15 A b^4\right )}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-8 a^5 B+a^4 b (24 A+7 C)+5 a^3 b^2 B-a^2 b^3 (33 A+C)-3 a b^4 B+15 A b^5\right )}{d}}{a^2}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (-3 a^4 C+7 a^3 b B-a^2 b^2 (11 A+3 C)-a b^3 B+5 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\frac {2 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^4 (8 A-5 C)+9 a^3 b B-a^2 b^2 (29 A+C)-3 a b^3 B+15 A b^4\right )}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-8 a^5 B+a^4 b (24 A+7 C)+5 a^3 b^2 B-a^2 b^3 (33 A+C)-3 a b^4 B+15 A b^5\right )}{d}}{a^2}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (3 a^6 C-15 a^5 b B+5 a^4 b^2 (7 A+2 C)+6 a^3 b^3 B-a^2 b^4 (38 A+C)-3 a b^5 B+15 A b^6\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a^2 d (a+b)}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

Input:

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

Output:

((A*b^2 - a*(b*B - a*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(2*a*(a^2 - b^2) 
*d*(a + b*Sec[c + d*x])^2) - (-1/2*(((2*a*(15*A*b^4 + 9*a^3*b*B - 3*a*b^3* 
B + a^4*(8*A - 5*C) - a^2*b^2*(29*A + C))*Sqrt[Cos[c + d*x]]*EllipticE[(c 
+ d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d - (2*(15*A*b^5 - 8*a^5*B + 5*a^3*b^2*B 
- 3*a*b^4*B - a^2*b^3*(33*A + C) + a^4*b*(24*A + 7*C))*Sqrt[Cos[c + d*x]]* 
EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d)/a^2 + (2*(15*A*b^6 - 15*a 
^5*b*B + 6*a^3*b^3*B - 3*a*b^5*B + 3*a^6*C - a^2*b^4*(38*A + C) + 5*a^4*b^ 
2*(7*A + 2*C))*Sqrt[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*(a^2 - b^2)) + ((5*A*b^4 + 7*a^3 
*b*B - a*b^3*B - 3*a^4*C - a^2*b^2*(11*A + 3*C))*Sqrt[Sec[c + d*x]]*Sin[c 
+ d*x])/(a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])))/(4*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 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. \(2021\) vs. \(2(465)=930\).

Time = 8.71 (sec) , antiderivative size = 2022, normalized size of antiderivative = 4.16

method	result	size
default	\(\text {Expression too large to display}\)	\(2022\)

Input:

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,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/a^4/(-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)^2-1)^( 
1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(3*A*b*EllipticF(cos(1/2*d*x+1/2*c),2^(1 
/2))+a*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-B*EllipticF(cos(1/2*d*x+1/2 
*c),2^(1/2))*a)-2/a^3*(6*A*b^2-3*B*a*b+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^2*(A*b^2-B*a*b+C*a^2)/a^4*(1/2*a^2/b/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*s 
in(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)^2+3/4*a^2*(a^2-3*b^2)/b^2/(a^2-b^2)^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)-3/ 
8/(a+b)/(a^2-b^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))*a^2-1/4/(a+b)/(a^2-b^2)/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))*a+7/8/(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))+3/8*a^3/b^2/(a^2-b^2)^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)^...
 

Fricas [F(-1)]

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

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2)/(a+b*sec(d*x+c) 
)^3,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

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

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(1/2)/(a+b*sec(d*x+ 
c))**3,x)
 

Output:

Timed out
 

Maxima [F(-1)]

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

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2)/(a+b*sec(d*x+c) 
)^3,x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2)/(a+b*sec(d*x+c) 
)^3,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/((a + b/cos(c + d*x))^3*(1/cos 
(c + d*x))^(1/2)),x)
 

Output:

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/((a + b/cos(c + d*x))^3*(1/cos 
(c + d*x))^(1/2)), x)
 

Reduce [F]

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

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,x)
 

Output:

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