\(\int \frac {(A+C \cos ^2(c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^4} \, dx\) [590]

Optimal result

Integrand size = 31, antiderivative size = 301 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^4} \, dx=-\frac {b \left (7 a^2 A b^4-2 A b^6-a^4 b^2 (8 A-C)+4 a^6 (2 A+C)\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 (a-b)^{7/2} (a+b)^{7/2} d}+\frac {A \text {arctanh}(\sin (c+d x))}{a^4 d}+\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac {\left (3 A b^4-2 a^4 C-a^2 b^2 (8 A+3 C)\right ) \sin (c+d x)}{6 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}-\frac {\left (17 a^2 A b^4-6 A b^6-2 a^6 C-13 a^4 b^2 (2 A+C)\right ) \sin (c+d x)}{6 a^3 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))} \] Output:

-b*(7*a^2*A*b^4-2*A*b^6-a^4*b^2*(8*A-C)+4*a^6*(2*A+C))*arctan((a-b)^(1/2)* 
tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^4/(a-b)^(7/2)/(a+b)^(7/2)/d+A*arctanh(si 
n(d*x+c))/a^4/d+1/3*(A*b^2+C*a^2)*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c) 
)^3-1/6*(3*A*b^4-2*a^4*C-a^2*b^2*(8*A+3*C))*sin(d*x+c)/a^2/(a^2-b^2)^2/d/( 
a+b*cos(d*x+c))^2-1/6*(17*a^2*A*b^4-6*A*b^6-2*a^6*C-13*a^4*b^2*(2*A+C))*si 
n(d*x+c)/a^3/(a^2-b^2)^3/d/(a+b*cos(d*x+c))
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 7.58 (sec) , antiderivative size = 498, normalized size of antiderivative = 1.65 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {\cos (c+d x) (C \cos (c+d x)+A \sec (c+d x)) \left (-\frac {6 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^4}+\frac {6 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^4}+\frac {6 b \left (7 a^2 A b^4-2 A b^6+a^4 b^2 (-8 A+C)+4 a^6 (2 A+C)\right ) \arctan \left (\frac {(i \cos (c)+\sin (c)) \left (b \sin (c)+(-a+b \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {-\left (\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2\right )}}\right ) (i \cos (c)+\sin (c))}{a^4 \left (a^2-b^2\right )^3 \sqrt {-\left (\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2\right )}}+\frac {2 \left (A b^2+a^2 C\right ) \sec (c) (-a \sin (c)+b \sin (d x))}{a b \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}+\frac {\left (-17 a^2 A b^4+6 A b^6+2 a^6 C+13 a^4 b^2 (2 A+C)\right ) \sec (c) \sin (d x)-3 a b \left (A b^4+a^2 b^2 (-2 A+C)+a^4 (6 A+4 C)\right ) \tan (c)}{\left (a^3-a b^2\right )^3 (a+b \cos (c+d x))}+\frac {\left (-3 A b^4+2 a^4 C+a^2 b^2 (8 A+3 C)\right ) \sec (c) \sin (d x)+a b \left (A b^2-a^2 (6 A+5 C)\right ) \tan (c)}{a^2 \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}\right )}{3 d (2 A+C+C \cos (2 (c+d x)))} \] Input:

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x])^4,x]
 

Output:

(Cos[c + d*x]*(C*Cos[c + d*x] + A*Sec[c + d*x])*((-6*A*Log[Cos[(c + d*x)/2 
] - Sin[(c + d*x)/2]])/a^4 + (6*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] 
)/a^4 + (6*b*(7*a^2*A*b^4 - 2*A*b^6 + a^4*b^2*(-8*A + C) + 4*a^6*(2*A + C) 
)*ArcTan[((I*Cos[c] + Sin[c])*(b*Sin[c] + (-a + b*Cos[c])*Tan[(d*x)/2]))/S 
qrt[-((a^2 - b^2)*(Cos[c] - I*Sin[c])^2)]]*(I*Cos[c] + Sin[c]))/(a^4*(a^2 
- b^2)^3*Sqrt[-((a^2 - b^2)*(Cos[c] - I*Sin[c])^2)]) + (2*(A*b^2 + a^2*C)* 
Sec[c]*(-(a*Sin[c]) + b*Sin[d*x]))/(a*b*(a^2 - b^2)*(a + b*Cos[c + d*x])^3 
) + ((-17*a^2*A*b^4 + 6*A*b^6 + 2*a^6*C + 13*a^4*b^2*(2*A + C))*Sec[c]*Sin 
[d*x] - 3*a*b*(A*b^4 + a^2*b^2*(-2*A + C) + a^4*(6*A + 4*C))*Tan[c])/((a^3 
 - a*b^2)^3*(a + b*Cos[c + d*x])) + ((-3*A*b^4 + 2*a^4*C + a^2*b^2*(8*A + 
3*C))*Sec[c]*Sin[d*x] + a*b*(A*b^2 - a^2*(6*A + 5*C))*Tan[c])/(a^2*(a^2 - 
b^2)^2*(a + b*Cos[c + d*x])^2)))/(3*d*(2*A + C + C*Cos[2*(c + d*x)]))
 

Rubi [A] (verified)

Time = 1.92 (sec) , antiderivative size = 366, normalized size of antiderivative = 1.22, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 3535, 3042, 3534, 3042, 3534, 27, 3042, 3480, 3042, 3138, 218, 4257}

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.

\(\displaystyle \int \frac {\sec (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^4}dx\)

\(\Big \downarrow \) 3535

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

\(\displaystyle \frac {\frac {\int \frac {\left (6 A \left (a^2-b^2\right )^2-\left (-2 C a^4-b^2 (8 A+3 C) a^2+3 A b^4\right ) \cos ^2(c+d x)+2 a b \left (A b^2-a^2 (6 A+5 C)\right ) \cos (c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^2}dx}{2 a \left (a^2-b^2\right )}-\frac {\left (-2 a^4 C-a^2 b^2 (8 A+3 C)+3 A b^4\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {6 A \left (a^2-b^2\right )^2+\left (2 C a^4+b^2 (8 A+3 C) a^2-3 A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 a b \left (A b^2-a^2 (6 A+5 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 a \left (a^2-b^2\right )}-\frac {\left (-2 a^4 C-a^2 b^2 (8 A+3 C)+3 A b^4\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3534

\(\displaystyle \frac {\frac {\frac {\int \frac {3 \left (2 A \left (a^2-b^2\right )^3-a b \left ((6 A+4 C) a^4-b^2 (2 A-C) a^2+A b^4\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}-\frac {\left (-2 a^6 C-13 a^4 b^2 (2 A+C)+17 a^2 A b^4-6 A b^6\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}-\frac {\left (-2 a^4 C-a^2 b^2 (8 A+3 C)+3 A b^4\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {3 \int \frac {\left (2 A \left (a^2-b^2\right )^3-a b \left ((6 A+4 C) a^4-b^2 (2 A-C) a^2+A b^4\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}-\frac {\left (-2 a^6 C-13 a^4 b^2 (2 A+C)+17 a^2 A b^4-6 A b^6\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}-\frac {\left (-2 a^4 C-a^2 b^2 (8 A+3 C)+3 A b^4\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 \int \frac {2 A \left (a^2-b^2\right )^3-a b \left ((6 A+4 C) a^4-b^2 (2 A-C) a^2+A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a \left (a^2-b^2\right )}-\frac {\left (-2 a^6 C-13 a^4 b^2 (2 A+C)+17 a^2 A b^4-6 A b^6\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}-\frac {\left (-2 a^4 C-a^2 b^2 (8 A+3 C)+3 A b^4\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3480

\(\displaystyle \frac {\frac {\frac {3 \left (\frac {2 A \left (a^2-b^2\right )^3 \int \sec (c+d x)dx}{a}-\frac {b \left (4 a^6 (2 A+C)-a^4 b^2 (8 A-C)+7 a^2 A b^4-2 A b^6\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a}\right )}{a \left (a^2-b^2\right )}-\frac {\left (-2 a^6 C-13 a^4 b^2 (2 A+C)+17 a^2 A b^4-6 A b^6\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}-\frac {\left (-2 a^4 C-a^2 b^2 (8 A+3 C)+3 A b^4\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 \left (\frac {2 A \left (a^2-b^2\right )^3 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {b \left (4 a^6 (2 A+C)-a^4 b^2 (8 A-C)+7 a^2 A b^4-2 A b^6\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}\right )}{a \left (a^2-b^2\right )}-\frac {\left (-2 a^6 C-13 a^4 b^2 (2 A+C)+17 a^2 A b^4-6 A b^6\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}-\frac {\left (-2 a^4 C-a^2 b^2 (8 A+3 C)+3 A b^4\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {\frac {\frac {3 \left (\frac {2 A \left (a^2-b^2\right )^3 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b \left (4 a^6 (2 A+C)-a^4 b^2 (8 A-C)+7 a^2 A b^4-2 A b^6\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}\right )}{a \left (a^2-b^2\right )}-\frac {\left (-2 a^6 C-13 a^4 b^2 (2 A+C)+17 a^2 A b^4-6 A b^6\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}-\frac {\left (-2 a^4 C-a^2 b^2 (8 A+3 C)+3 A b^4\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {3 \left (\frac {2 A \left (a^2-b^2\right )^3 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b \left (4 a^6 (2 A+C)-a^4 b^2 (8 A-C)+7 a^2 A b^4-2 A b^6\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a \left (a^2-b^2\right )}-\frac {\left (-2 a^6 C-13 a^4 b^2 (2 A+C)+17 a^2 A b^4-6 A b^6\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}-\frac {\left (-2 a^4 C-a^2 b^2 (8 A+3 C)+3 A b^4\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}+\frac {\frac {\frac {3 \left (\frac {2 A \left (a^2-b^2\right )^3 \text {arctanh}(\sin (c+d x))}{a d}-\frac {2 b \left (4 a^6 (2 A+C)-a^4 b^2 (8 A-C)+7 a^2 A b^4-2 A b^6\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a \left (a^2-b^2\right )}-\frac {\left (-2 a^6 C-13 a^4 b^2 (2 A+C)+17 a^2 A b^4-6 A b^6\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}-\frac {\left (-2 a^4 C-a^2 b^2 (8 A+3 C)+3 A b^4\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 a \left (a^2-b^2\right )}\)

Input:

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x])^4,x]
 

Output:

((A*b^2 + a^2*C)*Sin[c + d*x])/(3*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^3) 
+ (-1/2*((3*A*b^4 - 2*a^4*C - a^2*b^2*(8*A + 3*C))*Sin[c + d*x])/(a*(a^2 - 
 b^2)*d*(a + b*Cos[c + d*x])^2) + ((3*((-2*b*(7*a^2*A*b^4 - 2*A*b^6 - a^4* 
b^2*(8*A - C) + 4*a^6*(2*A + C))*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqr 
t[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + (2*A*(a^2 - b^2)^3*ArcTanh[Sin[ 
c + d*x]])/(a*d)))/(a*(a^2 - b^2)) - ((17*a^2*A*b^4 - 6*A*b^6 - 2*a^6*C - 
13*a^4*b^2*(2*A + C))*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])) 
)/(2*a*(a^2 - b^2)))/(3*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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042

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

rule 3138

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ 
e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + b + 
(a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] 
 && NeQ[a^2 - b^2, 0]
 

rule 3480

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ 
.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A*b 
- a*B)/(b*c - a*d)   Int[1/(a + b*Sin[e + f*x]), x], x] + Simp[(B*c - A*d)/ 
(b*c - a*d)   Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f 
, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3534

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* 
c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2))   Int 
[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* 
d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - 
a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A 
*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && 
NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ 
[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) | 
| EqQ[a, 0])))
 

rule 3535

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S 
in[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m 
+ 1)*(b*c - a*d)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin 
[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n 
+ 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d 
*(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, 
d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 
- d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) || 
 !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 
 0])))
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 1.47 (sec) , antiderivative size = 498, normalized size of antiderivative = 1.65

method	result	size
derivativedivides	\(\frac {-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4}}-\frac {2 \left (\frac {-\frac {\left (12 A \,a^{4} b^{2}+4 A \,a^{3} b^{3}-6 a^{2} A \,b^{4}-A a \,b^{5}+2 A \,b^{6}+2 a^{6} C +2 C \,a^{5} b +6 a^{4} b^{2} C +C \,a^{3} b^{3}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {2 \left (18 A \,a^{4} b^{2}-11 a^{2} A \,b^{4}+3 A \,b^{6}+3 a^{6} C +7 a^{4} b^{2} C \right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (12 A \,a^{4} b^{2}-4 A \,a^{3} b^{3}-6 a^{2} A \,b^{4}+A a \,b^{5}+2 A \,b^{6}+2 a^{6} C -2 C \,a^{5} b +6 a^{4} b^{2} C -C \,a^{3} b^{3}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{3}}+\frac {b \left (8 A \,a^{6}-8 A \,a^{4} b^{2}+7 a^{2} A \,b^{4}-2 A \,b^{6}+4 a^{6} C +a^{4} b^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4}}}{d}\)	\(498\)
default	\(\frac {-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4}}-\frac {2 \left (\frac {-\frac {\left (12 A \,a^{4} b^{2}+4 A \,a^{3} b^{3}-6 a^{2} A \,b^{4}-A a \,b^{5}+2 A \,b^{6}+2 a^{6} C +2 C \,a^{5} b +6 a^{4} b^{2} C +C \,a^{3} b^{3}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {2 \left (18 A \,a^{4} b^{2}-11 a^{2} A \,b^{4}+3 A \,b^{6}+3 a^{6} C +7 a^{4} b^{2} C \right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (12 A \,a^{4} b^{2}-4 A \,a^{3} b^{3}-6 a^{2} A \,b^{4}+A a \,b^{5}+2 A \,b^{6}+2 a^{6} C -2 C \,a^{5} b +6 a^{4} b^{2} C -C \,a^{3} b^{3}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{3}}+\frac {b \left (8 A \,a^{6}-8 A \,a^{4} b^{2}+7 a^{2} A \,b^{4}-2 A \,b^{6}+4 a^{6} C +a^{4} b^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4}}}{d}\)	\(498\)
risch	\(\text {Expression too large to display}\)	\(1714\)

Input:

int((A+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^4,x,method=_RETURNVERBO 
SE)
 

Output:

1/d*(-A/a^4*ln(tan(1/2*d*x+1/2*c)-1)-2/a^4*((-1/2*(12*A*a^4*b^2+4*A*a^3*b^ 
3-6*A*a^2*b^4-A*a*b^5+2*A*b^6+2*C*a^6+2*C*a^5*b+6*C*a^4*b^2+C*a^3*b^3)*a/( 
a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5-2/3*(18*A*a^4*b^2-11*A 
*a^2*b^4+3*A*b^6+3*C*a^6+7*C*a^4*b^2)*a/(a^2+2*a*b+b^2)/(a^2-2*a*b+b^2)*ta 
n(1/2*d*x+1/2*c)^3-1/2*(12*A*a^4*b^2-4*A*a^3*b^3-6*A*a^2*b^4+A*a*b^5+2*A*b 
^6+2*C*a^6-2*C*a^5*b+6*C*a^4*b^2-C*a^3*b^3)*a/(a+b)/(a^3-3*a^2*b+3*a*b^2-b 
^3)*tan(1/2*d*x+1/2*c))/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b 
)^3+1/2*b*(8*A*a^6-8*A*a^4*b^2+7*A*a^2*b^4-2*A*b^6+4*C*a^6+C*a^4*b^2)/(a^6 
-3*a^4*b^2+3*a^2*b^4-b^6)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2 
*c)/((a+b)*(a-b))^(1/2)))+A/a^4*ln(tan(1/2*d*x+1/2*c)+1))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1045 vs. \(2 (285) = 570\).

Time = 13.59 (sec) , antiderivative size = 2159, normalized size of antiderivative = 7.17 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Too large to display} \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^4,x, algorithm="f 
ricas")
 

Output:

[-1/12*(3*(4*(2*A + C)*a^9*b - (8*A - C)*a^7*b^3 + 7*A*a^5*b^5 - 2*A*a^3*b 
^7 + (4*(2*A + C)*a^6*b^4 - (8*A - C)*a^4*b^6 + 7*A*a^2*b^8 - 2*A*b^10)*co 
s(d*x + c)^3 + 3*(4*(2*A + C)*a^7*b^3 - (8*A - C)*a^5*b^5 + 7*A*a^3*b^7 - 
2*A*a*b^9)*cos(d*x + c)^2 + 3*(4*(2*A + C)*a^8*b^2 - (8*A - C)*a^6*b^4 + 7 
*A*a^4*b^6 - 2*A*a^2*b^8)*cos(d*x + c))*sqrt(-a^2 + b^2)*log((2*a*b*cos(d* 
x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 - 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) 
 + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) 
 + a^2)) - 6*(A*a^11 - 4*A*a^9*b^2 + 6*A*a^7*b^4 - 4*A*a^5*b^6 + A*a^3*b^8 
 + (A*a^8*b^3 - 4*A*a^6*b^5 + 6*A*a^4*b^7 - 4*A*a^2*b^9 + A*b^11)*cos(d*x 
+ c)^3 + 3*(A*a^9*b^2 - 4*A*a^7*b^4 + 6*A*a^5*b^6 - 4*A*a^3*b^8 + A*a*b^10 
)*cos(d*x + c)^2 + 3*(A*a^10*b - 4*A*a^8*b^3 + 6*A*a^6*b^5 - 4*A*a^4*b^7 + 
 A*a^2*b^9)*cos(d*x + c))*log(sin(d*x + c) + 1) + 6*(A*a^11 - 4*A*a^9*b^2 
+ 6*A*a^7*b^4 - 4*A*a^5*b^6 + A*a^3*b^8 + (A*a^8*b^3 - 4*A*a^6*b^5 + 6*A*a 
^4*b^7 - 4*A*a^2*b^9 + A*b^11)*cos(d*x + c)^3 + 3*(A*a^9*b^2 - 4*A*a^7*b^4 
 + 6*A*a^5*b^6 - 4*A*a^3*b^8 + A*a*b^10)*cos(d*x + c)^2 + 3*(A*a^10*b - 4* 
A*a^8*b^3 + 6*A*a^6*b^5 - 4*A*a^4*b^7 + A*a^2*b^9)*cos(d*x + c))*log(-sin( 
d*x + c) + 1) - 2*(6*C*a^11 + 4*(9*A + C)*a^9*b^2 - (68*A + 11*C)*a^7*b^4 
+ (43*A + C)*a^5*b^6 - 11*A*a^3*b^8 + (2*C*a^9*b^2 + (26*A + 11*C)*a^7*b^4 
 - (43*A + 13*C)*a^5*b^6 + 23*A*a^3*b^8 - 6*A*a*b^10)*cos(d*x + c)^2 + 3*( 
2*C*a^10*b + (20*A + 7*C)*a^8*b^3 - 5*(7*A + 2*C)*a^6*b^5 + (20*A + C)*...
 

Sympy [F]

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

integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)/(a+b*cos(d*x+c))**4,x)
 

Output:

Integral((A + C*cos(c + d*x)**2)*sec(c + d*x)/(a + b*cos(c + d*x))**4, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^4,x, algorithm="m 
axima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 868 vs. \(2 (285) = 570\).

Time = 0.21 (sec) , antiderivative size = 868, normalized size of antiderivative = 2.88 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Too large to display} \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^4,x, algorithm="g 
iac")
 

Output:

-1/3*(3*(8*A*a^6*b + 4*C*a^6*b - 8*A*a^4*b^3 + C*a^4*b^3 + 7*A*a^2*b^5 - 2 
*A*b^7)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1 
/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/((a^10 - 3*a^8 
*b^2 + 3*a^6*b^4 - a^4*b^6)*sqrt(a^2 - b^2)) - 3*A*log(abs(tan(1/2*d*x + 1 
/2*c) + 1))/a^4 + 3*A*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 - (6*C*a^8*ta 
n(1/2*d*x + 1/2*c)^5 - 6*C*a^7*b*tan(1/2*d*x + 1/2*c)^5 + 36*A*a^6*b^2*tan 
(1/2*d*x + 1/2*c)^5 + 12*C*a^6*b^2*tan(1/2*d*x + 1/2*c)^5 - 60*A*a^5*b^3*t 
an(1/2*d*x + 1/2*c)^5 - 27*C*a^5*b^3*tan(1/2*d*x + 1/2*c)^5 - 6*A*a^4*b^4* 
tan(1/2*d*x + 1/2*c)^5 + 12*C*a^4*b^4*tan(1/2*d*x + 1/2*c)^5 + 45*A*a^3*b^ 
5*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^3*b^5*tan(1/2*d*x + 1/2*c)^5 - 6*A*a^2*b^ 
6*tan(1/2*d*x + 1/2*c)^5 - 15*A*a*b^7*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^8*tan 
(1/2*d*x + 1/2*c)^5 + 12*C*a^8*tan(1/2*d*x + 1/2*c)^3 + 72*A*a^6*b^2*tan(1 
/2*d*x + 1/2*c)^3 + 16*C*a^6*b^2*tan(1/2*d*x + 1/2*c)^3 - 116*A*a^4*b^4*ta 
n(1/2*d*x + 1/2*c)^3 - 28*C*a^4*b^4*tan(1/2*d*x + 1/2*c)^3 + 56*A*a^2*b^6* 
tan(1/2*d*x + 1/2*c)^3 - 12*A*b^8*tan(1/2*d*x + 1/2*c)^3 + 6*C*a^8*tan(1/2 
*d*x + 1/2*c) + 6*C*a^7*b*tan(1/2*d*x + 1/2*c) + 36*A*a^6*b^2*tan(1/2*d*x 
+ 1/2*c) + 12*C*a^6*b^2*tan(1/2*d*x + 1/2*c) + 60*A*a^5*b^3*tan(1/2*d*x + 
1/2*c) + 27*C*a^5*b^3*tan(1/2*d*x + 1/2*c) - 6*A*a^4*b^4*tan(1/2*d*x + 1/2 
*c) + 12*C*a^4*b^4*tan(1/2*d*x + 1/2*c) - 45*A*a^3*b^5*tan(1/2*d*x + 1/2*c 
) - 3*C*a^3*b^5*tan(1/2*d*x + 1/2*c) - 6*A*a^2*b^6*tan(1/2*d*x + 1/2*c)...
 

Mupad [B] (verification not implemented)

Time = 11.33 (sec) , antiderivative size = 9766, normalized size of antiderivative = 32.45 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Too large to display} \] Input:

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)*(a + b*cos(c + d*x))^4),x)
 

Output:

- ((tan(c/2 + (d*x)/2)*(2*A*b^6 + 2*C*a^6 - 6*A*a^2*b^4 - 4*A*a^3*b^3 + 12 
*A*a^4*b^2 - C*a^3*b^3 + 6*C*a^4*b^2 + A*a*b^5 - 2*C*a^5*b))/((a + b)*(3*a 
^5*b - a^6 + a^3*b^3 - 3*a^4*b^2)) - (4*tan(c/2 + (d*x)/2)^3*(3*A*b^6 + 3* 
C*a^6 - 11*A*a^2*b^4 + 18*A*a^4*b^2 + 7*C*a^4*b^2))/(3*(a + b)^2*(a^5 - 2* 
a^4*b + a^3*b^2)) + (tan(c/2 + (d*x)/2)^5*(2*A*b^6 + 2*C*a^6 - 6*A*a^2*b^4 
 + 4*A*a^3*b^3 + 12*A*a^4*b^2 + C*a^3*b^3 + 6*C*a^4*b^2 - A*a*b^5 + 2*C*a^ 
5*b))/((a^3*b - a^4)*(a + b)^3))/(d*(3*a*b^2 - tan(c/2 + (d*x)/2)^4*(3*a*b 
^2 + 3*a^2*b - 3*a^3 - 3*b^3) - tan(c/2 + (d*x)/2)^2*(3*a*b^2 - 3*a^2*b - 
3*a^3 + 3*b^3) + 3*a^2*b + a^3 + b^3 + tan(c/2 + (d*x)/2)^6*(3*a*b^2 - 3*a 
^2*b + a^3 - b^3))) - (A*atan(((A*((8*tan(c/2 + (d*x)/2)*(4*A^2*a^14 + 8*A 
^2*b^14 - 8*A^2*a*b^13 - 8*A^2*a^13*b - 48*A^2*a^2*b^12 + 48*A^2*a^3*b^11 
+ 117*A^2*a^4*b^10 - 120*A^2*a^5*b^9 - 164*A^2*a^6*b^8 + 160*A^2*a^7*b^7 + 
 156*A^2*a^8*b^6 - 120*A^2*a^9*b^5 - 92*A^2*a^10*b^4 + 48*A^2*a^11*b^3 + 4 
4*A^2*a^12*b^2 + C^2*a^8*b^6 + 8*C^2*a^10*b^4 + 16*C^2*a^12*b^2 - 4*A*C*a^ 
4*b^10 - 2*A*C*a^6*b^8 + 40*A*C*a^8*b^6 - 48*A*C*a^10*b^4 + 64*A*C*a^12*b^ 
2))/(a^16*b + a^17 - a^6*b^11 - a^7*b^10 + 5*a^8*b^9 + 5*a^9*b^8 - 10*a^10 
*b^7 - 10*a^11*b^6 + 10*a^12*b^5 + 10*a^13*b^4 - 5*a^14*b^3 - 5*a^15*b^2) 
+ (A*((8*(4*A*a^21 - 4*A*a^8*b^13 + 2*A*a^9*b^12 + 26*A*a^10*b^11 - 14*A*a 
^11*b^10 - 70*A*a^12*b^9 + 30*A*a^13*b^8 + 110*A*a^14*b^7 - 30*A*a^15*b^6 
- 110*A*a^16*b^5 + 20*A*a^17*b^4 + 64*A*a^18*b^3 - 12*A*a^19*b^2 - 2*C*...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 3572, normalized size of antiderivative = 11.87 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^4} \, dx =\text {Too large to display} \] Input:

int((A+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^4,x)
 

Output:

( - 48*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sq 
rt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*a**6*b**4 - 24*sqrt(a**2 - b 
**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos 
(c + d*x)*sin(c + d*x)**2*a**5*b**4*c + 48*sqrt(a**2 - b**2)*atan((tan((c 
+ d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + 
d*x)**2*a**4*b**6 - 6*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c 
+ d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*a**3*b**6*c - 
 42*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( 
a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*a**2*b**8 + 12*sqrt(a**2 - b**2 
)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c 
+ d*x)*sin(c + d*x)**2*b**10 + 144*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2 
)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a**8*b**2 + 72*s 
qrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 
- b**2))*cos(c + d*x)*a**7*b**2*c - 96*sqrt(a**2 - b**2)*atan((tan((c + d* 
x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a**6*b**4 + 
42*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a 
**2 - b**2))*cos(c + d*x)*a**5*b**4*c + 78*sqrt(a**2 - b**2)*atan((tan((c 
+ d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a**4*b** 
6 + 6*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr 
t(a**2 - b**2))*cos(c + d*x)*a**3*b**6*c + 6*sqrt(a**2 - b**2)*atan((ta...