\(\int \frac {\sec ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx\) [477]

Optimal result

Integrand size = 21, antiderivative size = 305 \[ \int \frac {\sec ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {b^3 \left (20 a^4-29 a^2 b^2+12 b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^5 (a-b)^{5/2} (a+b)^{5/2} d}+\frac {\left (a^2+12 b^2\right ) \text {arctanh}(\sin (c+d x))}{2 a^5 d}-\frac {3 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) \tan (c+d x)}{2 a^4 \left (a^2-b^2\right )^2 d}+\frac {\left (a^4-10 a^2 b^2+6 b^4\right ) \sec (c+d x) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac {b^2 \sec (c+d x) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {b^2 \left (7 a^2-4 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \] Output:

-b^3*(20*a^4-29*a^2*b^2+12*b^4)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b 
)^(1/2))/a^5/(a-b)^(5/2)/(a+b)^(5/2)/d+1/2*(a^2+12*b^2)*arctanh(sin(d*x+c) 
)/a^5/d-3/2*b*(2*a^4-7*a^2*b^2+4*b^4)*tan(d*x+c)/a^4/(a^2-b^2)^2/d+1/2*(a^ 
4-10*a^2*b^2+6*b^4)*sec(d*x+c)*tan(d*x+c)/a^3/(a^2-b^2)^2/d+1/2*b^2*sec(d* 
x+c)*tan(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))^2+1/2*b^2*(7*a^2-4*b^2)*sec 
(d*x+c)*tan(d*x+c)/a^2/(a^2-b^2)^2/d/(a+b*cos(d*x+c))
 

Mathematica [A] (verified)

Time = 6.82 (sec) , antiderivative size = 427, normalized size of antiderivative = 1.40 \[ \int \frac {\sec ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {b^3 \left (20 a^4-29 a^2 b^2+12 b^4\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{a^5 \left (a^2-b^2\right )^2 \sqrt {-a^2+b^2} d}+\frac {\left (-a^2-12 b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 a^5 d}+\frac {\left (a^2+12 b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 a^5 d}+\frac {1}{4 a^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {3 b \sin \left (\frac {1}{2} (c+d x)\right )}{a^4 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {1}{4 a^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {3 b \sin \left (\frac {1}{2} (c+d x)\right )}{a^4 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {b^4 \sin (c+d x)}{2 a^3 (a-b) (a+b) d (a+b \cos (c+d x))^2}+\frac {3 \left (3 a^2 b^4 \sin (c+d x)-2 b^6 \sin (c+d x)\right )}{2 a^4 (a-b)^2 (a+b)^2 d (a+b \cos (c+d x))} \] Input:

Integrate[Sec[c + d*x]^3/(a + b*Cos[c + d*x])^3,x]
 

Output:

(b^3*(20*a^4 - 29*a^2*b^2 + 12*b^4)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqr 
t[-a^2 + b^2]])/(a^5*(a^2 - b^2)^2*Sqrt[-a^2 + b^2]*d) + ((-a^2 - 12*b^2)* 
Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/(2*a^5*d) + ((a^2 + 12*b^2)*Log[ 
Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/(2*a^5*d) + 1/(4*a^3*d*(Cos[(c + d*x 
)/2] - Sin[(c + d*x)/2])^2) - (3*b*Sin[(c + d*x)/2])/(a^4*d*(Cos[(c + d*x) 
/2] - Sin[(c + d*x)/2])) - 1/(4*a^3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2] 
)^2) - (3*b*Sin[(c + d*x)/2])/(a^4*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) 
) + (b^4*Sin[c + d*x])/(2*a^3*(a - b)*(a + b)*d*(a + b*Cos[c + d*x])^2) + 
(3*(3*a^2*b^4*Sin[c + d*x] - 2*b^6*Sin[c + d*x]))/(2*a^4*(a - b)^2*(a + b) 
^2*d*(a + b*Cos[c + d*x]))
 

Rubi [A] (verified)

Time = 2.14 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.08, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.762, Rules used = {3042, 3281, 3042, 3534, 3042, 3534, 27, 3042, 3534, 25, 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.

\(\displaystyle \int \frac {\sec ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\)

\(\Big \downarrow \) 3281

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

\(\displaystyle \frac {\frac {\int \frac {\left (2 b^2 \left (7 a^2-4 b^2\right ) \cos ^2(c+d x)-a b \left (4 a^2-b^2\right ) \cos (c+d x)+2 \left (a^4-10 b^2 a^2+6 b^4\right )\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {b^2 \left (7 a^2-4 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {2 b^2 \left (7 a^2-4 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-a b \left (4 a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+2 \left (a^4-10 b^2 a^2+6 b^4\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a \left (a^2-b^2\right )}+\frac {b^2 \left (7 a^2-4 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3534

\(\displaystyle \frac {\frac {\frac {\int -\frac {2 \left (-b \left (a^4-10 b^2 a^2+6 b^4\right ) \cos ^2(c+d x)-a \left (a^4+4 b^2 a^2-2 b^4\right ) \cos (c+d x)+3 b \left (2 a^4-7 b^2 a^2+4 b^4\right )\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}+\frac {\left (a^4-10 a^2 b^2+6 b^4\right ) \tan (c+d x) \sec (c+d x)}{a d}}{a \left (a^2-b^2\right )}+\frac {b^2 \left (7 a^2-4 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\left (a^4-10 a^2 b^2+6 b^4\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\int \frac {\left (-b \left (a^4-10 b^2 a^2+6 b^4\right ) \cos ^2(c+d x)-a \left (a^4+4 b^2 a^2-2 b^4\right ) \cos (c+d x)+3 b \left (2 a^4-7 b^2 a^2+4 b^4\right )\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{a}}{a \left (a^2-b^2\right )}+\frac {b^2 \left (7 a^2-4 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\left (a^4-10 a^2 b^2+6 b^4\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\int \frac {-b \left (a^4-10 b^2 a^2+6 b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-a \left (a^4+4 b^2 a^2-2 b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 b \left (2 a^4-7 b^2 a^2+4 b^4\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{a \left (a^2-b^2\right )}+\frac {b^2 \left (7 a^2-4 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3534

\(\displaystyle \frac {\frac {\frac {\left (a^4-10 a^2 b^2+6 b^4\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {\int -\frac {\left (\left (a^2+12 b^2\right ) \left (a^2-b^2\right )^2+a b \left (a^4-10 b^2 a^2+6 b^4\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {3 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) \tan (c+d x)}{a d}}{a}}{a \left (a^2-b^2\right )}+\frac {b^2 \left (7 a^2-4 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {\left (a^4-10 a^2 b^2+6 b^4\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {3 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) \tan (c+d x)}{a d}-\frac {\int \frac {\left (\left (a^2+12 b^2\right ) \left (a^2-b^2\right )^2+a b \left (a^4-10 b^2 a^2+6 b^4\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}}{a}}{a \left (a^2-b^2\right )}+\frac {b^2 \left (7 a^2-4 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\left (a^4-10 a^2 b^2+6 b^4\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {3 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) \tan (c+d x)}{a d}-\frac {\int \frac {\left (a^2+12 b^2\right ) \left (a^2-b^2\right )^2+a b \left (a^4-10 b^2 a^2+6 b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{a}}{a \left (a^2-b^2\right )}+\frac {b^2 \left (7 a^2-4 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3480

\(\displaystyle \frac {\frac {\frac {\left (a^4-10 a^2 b^2+6 b^4\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {3 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2-b^2\right )^2 \left (a^2+12 b^2\right ) \int \sec (c+d x)dx}{a}-\frac {b^3 \left (20 a^4-29 a^2 b^2+12 b^4\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a}}{a}}{a}}{a \left (a^2-b^2\right )}+\frac {b^2 \left (7 a^2-4 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\left (a^4-10 a^2 b^2+6 b^4\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {3 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2-b^2\right )^2 \left (a^2+12 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {b^3 \left (20 a^4-29 a^2 b^2+12 b^4\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a}}{a}}{a \left (a^2-b^2\right )}+\frac {b^2 \left (7 a^2-4 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {\frac {\frac {\left (a^4-10 a^2 b^2+6 b^4\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {3 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2-b^2\right )^2 \left (a^2+12 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b^3 \left (20 a^4-29 a^2 b^2+12 b^4\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}}{a}}{a}}{a \left (a^2-b^2\right )}+\frac {b^2 \left (7 a^2-4 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {\left (a^4-10 a^2 b^2+6 b^4\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {3 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2-b^2\right )^2 \left (a^2+12 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b^3 \left (20 a^4-29 a^2 b^2+12 b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{a}}{a \left (a^2-b^2\right )}+\frac {b^2 \left (7 a^2-4 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {b^2 \tan (c+d x) \sec (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac {\frac {b^2 \left (7 a^2-4 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\frac {\left (a^4-10 a^2 b^2+6 b^4\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {3 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2-b^2\right )^2 \left (a^2+12 b^2\right ) \text {arctanh}(\sin (c+d x))}{a d}-\frac {2 b^3 \left (20 a^4-29 a^2 b^2+12 b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{a}}{a \left (a^2-b^2\right )}}{2 a \left (a^2-b^2\right )}\)

Input:

Int[Sec[c + d*x]^3/(a + b*Cos[c + d*x])^3,x]
 

Output:

(b^2*Sec[c + d*x]*Tan[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) 
 + ((b^2*(7*a^2 - 4*b^2)*Sec[c + d*x]*Tan[c + d*x])/(a*(a^2 - b^2)*d*(a + 
b*Cos[c + d*x])) + (((a^4 - 10*a^2*b^2 + 6*b^4)*Sec[c + d*x]*Tan[c + d*x]) 
/(a*d) - (-(((-2*b^3*(20*a^4 - 29*a^2*b^2 + 12*b^4)*ArcTan[(Sqrt[a - b]*Ta 
n[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + ((a^2 - b^2) 
^2*(a^2 + 12*b^2)*ArcTanh[Sin[c + d*x]])/(a*d))/a) + (3*b*(2*a^4 - 7*a^2*b 
^2 + 4*b^4)*Tan[c + d*x])/(a*d))/a)/(a*(a^2 - b^2)))/(2*a*(a^2 - b^2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 3281

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*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[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n 
 + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* 
d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2* 
n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 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 4257

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

Maple [A] (verified)

Time = 2.60 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.18

method	result	size
derivativedivides	\(\frac {\frac {1}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-a -6 b}{2 a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-a^{2}-12 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{5}}-\frac {1}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-a -6 b}{2 a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (a^{2}+12 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{5}}-\frac {2 b^{3} \left (\frac {-\frac {\left (10 a^{2}+a b -6 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (10 a^{2}-a b -6 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (20 a^{4}-29 a^{2} b^{2}+12 b^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{5}}}{d}\)	\(360\)
default	\(\frac {\frac {1}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-a -6 b}{2 a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-a^{2}-12 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{5}}-\frac {1}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-a -6 b}{2 a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (a^{2}+12 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{5}}-\frac {2 b^{3} \left (\frac {-\frac {\left (10 a^{2}+a b -6 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (10 a^{2}-a b -6 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (20 a^{4}-29 a^{2} b^{2}+12 b^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{5}}}{d}\)	\(360\)
risch	\(\text {Expression too large to display}\)	\(1108\)

Input:

int(sec(d*x+c)^3/(a+cos(d*x+c)*b)^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*(1/2/a^3/(tan(1/2*d*x+1/2*c)-1)^2-1/2*(-a-6*b)/a^4/(tan(1/2*d*x+1/2*c) 
-1)+1/2/a^5*(-a^2-12*b^2)*ln(tan(1/2*d*x+1/2*c)-1)-1/2/a^3/(tan(1/2*d*x+1/ 
2*c)+1)^2-1/2*(-a-6*b)/a^4/(tan(1/2*d*x+1/2*c)+1)+1/2*(a^2+12*b^2)/a^5*ln( 
tan(1/2*d*x+1/2*c)+1)-2*b^3/a^5*((-1/2*(10*a^2+a*b-6*b^2)*a*b/(a-b)/(a^2+2 
*a*b+b^2)*tan(1/2*d*x+1/2*c)^3-1/2*(10*a^2-a*b-6*b^2)*a*b/(a+b)/(a^2-2*a*b 
+b^2)*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)^2+1/2*(20*a^4-29*a^2*b^2+12*b^4)/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))^(1/ 
2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 728 vs. \(2 (286) = 572\).

Time = 1.31 (sec) , antiderivative size = 1524, normalized size of antiderivative = 5.00 \[ \int \frac {\sec ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \] Input:

integrate(sec(d*x+c)^3/(a+b*cos(d*x+c))^3,x, algorithm="fricas")
 

Output:

[-1/4*(((20*a^4*b^5 - 29*a^2*b^7 + 12*b^9)*cos(d*x + c)^4 + 2*(20*a^5*b^4 
- 29*a^3*b^6 + 12*a*b^8)*cos(d*x + c)^3 + (20*a^6*b^3 - 29*a^4*b^5 + 12*a^ 
2*b^7)*cos(d*x + c)^2)*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)) - ((a^8* 
b^2 + 9*a^6*b^4 - 33*a^4*b^6 + 35*a^2*b^8 - 12*b^10)*cos(d*x + c)^4 + 2*(a 
^9*b + 9*a^7*b^3 - 33*a^5*b^5 + 35*a^3*b^7 - 12*a*b^9)*cos(d*x + c)^3 + (a 
^10 + 9*a^8*b^2 - 33*a^6*b^4 + 35*a^4*b^6 - 12*a^2*b^8)*cos(d*x + c)^2)*lo 
g(sin(d*x + c) + 1) + ((a^8*b^2 + 9*a^6*b^4 - 33*a^4*b^6 + 35*a^2*b^8 - 12 
*b^10)*cos(d*x + c)^4 + 2*(a^9*b + 9*a^7*b^3 - 33*a^5*b^5 + 35*a^3*b^7 - 1 
2*a*b^9)*cos(d*x + c)^3 + (a^10 + 9*a^8*b^2 - 33*a^6*b^4 + 35*a^4*b^6 - 12 
*a^2*b^8)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(a^10 - 3*a^8*b^2 + 3 
*a^6*b^4 - a^4*b^6 - 3*(2*a^7*b^3 - 9*a^5*b^5 + 11*a^3*b^7 - 4*a*b^9)*cos( 
d*x + c)^3 - (11*a^8*b^2 - 43*a^6*b^4 + 50*a^4*b^6 - 18*a^2*b^8)*cos(d*x + 
 c)^2 - 4*(a^9*b - 3*a^7*b^3 + 3*a^5*b^5 - a^3*b^7)*cos(d*x + c))*sin(d*x 
+ c))/((a^11*b^2 - 3*a^9*b^4 + 3*a^7*b^6 - a^5*b^8)*d*cos(d*x + c)^4 + 2*( 
a^12*b - 3*a^10*b^3 + 3*a^8*b^5 - a^6*b^7)*d*cos(d*x + c)^3 + (a^13 - 3*a^ 
11*b^2 + 3*a^9*b^4 - a^7*b^6)*d*cos(d*x + c)^2), -1/4*(2*((20*a^4*b^5 - 29 
*a^2*b^7 + 12*b^9)*cos(d*x + c)^4 + 2*(20*a^5*b^4 - 29*a^3*b^6 + 12*a*b^8) 
*cos(d*x + c)^3 + (20*a^6*b^3 - 29*a^4*b^5 + 12*a^2*b^7)*cos(d*x + c)^2...
 

Sympy [F]

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

integrate(sec(d*x+c)**3/(a+b*cos(d*x+c))**3,x)
 

Output:

Integral(sec(c + d*x)**3/(a + b*cos(c + d*x))**3, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sec ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(sec(d*x+c)^3/(a+b*cos(d*x+c))^3,x, algorithm="maxima")
 

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. 801 vs. \(2 (286) = 572\).

Time = 0.57 (sec) , antiderivative size = 801, normalized size of antiderivative = 2.63 \[ \int \frac {\sec ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \] Input:

integrate(sec(d*x+c)^3/(a+b*cos(d*x+c))^3,x, algorithm="giac")
 

Output:

1/2*(2*(20*a^4*b^3 - 29*a^2*b^5 + 12*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^9 - 2*a^7*b^2 + a^5*b^4)*sqrt(a^2 - b^2)) + 2*(a 
^7*tan(1/2*d*x + 1/2*c)^7 + 4*a^6*b*tan(1/2*d*x + 1/2*c)^7 - 13*a^5*b^2*ta 
n(1/2*d*x + 1/2*c)^7 - 2*a^4*b^3*tan(1/2*d*x + 1/2*c)^7 + 33*a^3*b^4*tan(1 
/2*d*x + 1/2*c)^7 - 17*a^2*b^5*tan(1/2*d*x + 1/2*c)^7 - 18*a*b^6*tan(1/2*d 
*x + 1/2*c)^7 + 12*b^7*tan(1/2*d*x + 1/2*c)^7 + 3*a^7*tan(1/2*d*x + 1/2*c) 
^5 + 4*a^6*b*tan(1/2*d*x + 1/2*c)^5 + 5*a^5*b^2*tan(1/2*d*x + 1/2*c)^5 - 2 
6*a^4*b^3*tan(1/2*d*x + 1/2*c)^5 - 29*a^3*b^4*tan(1/2*d*x + 1/2*c)^5 + 67* 
a^2*b^5*tan(1/2*d*x + 1/2*c)^5 + 18*a*b^6*tan(1/2*d*x + 1/2*c)^5 - 36*b^7* 
tan(1/2*d*x + 1/2*c)^5 + 3*a^7*tan(1/2*d*x + 1/2*c)^3 - 4*a^6*b*tan(1/2*d* 
x + 1/2*c)^3 + 5*a^5*b^2*tan(1/2*d*x + 1/2*c)^3 + 26*a^4*b^3*tan(1/2*d*x + 
 1/2*c)^3 - 29*a^3*b^4*tan(1/2*d*x + 1/2*c)^3 - 67*a^2*b^5*tan(1/2*d*x + 1 
/2*c)^3 + 18*a*b^6*tan(1/2*d*x + 1/2*c)^3 + 36*b^7*tan(1/2*d*x + 1/2*c)^3 
+ a^7*tan(1/2*d*x + 1/2*c) - 4*a^6*b*tan(1/2*d*x + 1/2*c) - 13*a^5*b^2*tan 
(1/2*d*x + 1/2*c) + 2*a^4*b^3*tan(1/2*d*x + 1/2*c) + 33*a^3*b^4*tan(1/2*d* 
x + 1/2*c) + 17*a^2*b^5*tan(1/2*d*x + 1/2*c) - 18*a*b^6*tan(1/2*d*x + 1/2* 
c) - 12*b^7*tan(1/2*d*x + 1/2*c))/((a^8 - 2*a^6*b^2 + a^4*b^4)*(a*tan(1/2* 
d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 + 2*b*tan(1/2*d*x + 1/2*c)^2 - a 
 - b)^2) + (a^2 + 12*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^5 - (a^2...
 

Mupad [B] (verification not implemented)

Time = 49.09 (sec) , antiderivative size = 5910, normalized size of antiderivative = 19.38 \[ \int \frac {\sec ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

((tan(c/2 + (d*x)/2)^3*(18*a*b^6 - 4*a^6*b + 3*a^7 + 36*b^7 - 67*a^2*b^5 - 
 29*a^3*b^4 + 26*a^4*b^3 + 5*a^5*b^2))/((a + b)^2*(a^6 - 2*a^5*b + a^4*b^2 
)) + (tan(c/2 + (d*x)/2)^5*(18*a*b^6 + 4*a^6*b + 3*a^7 - 36*b^7 + 67*a^2*b 
^5 - 29*a^3*b^4 - 26*a^4*b^3 + 5*a^5*b^2))/((a + b)^2*(a^6 - 2*a^5*b + a^4 
*b^2)) - (tan(c/2 + (d*x)/2)^7*(6*a*b^5 + 5*a^5*b + a^6 - 12*b^6 + 23*a^2* 
b^4 - 10*a^3*b^3 - 8*a^4*b^2))/((a^4*b - a^5)*(a + b)^2) - (tan(c/2 + (d*x 
)/2)*(6*a*b^5 + 5*a^5*b - a^6 + 12*b^6 - 23*a^2*b^4 - 10*a^3*b^3 + 8*a^4*b 
^2))/((a + b)*(a^6 - 2*a^5*b + a^4*b^2)))/(d*(2*a*b - tan(c/2 + (d*x)/2)^4 
*(2*a^2 - 6*b^2) - tan(c/2 + (d*x)/2)^2*(4*a*b + 4*b^2) + tan(c/2 + (d*x)/ 
2)^6*(4*a*b - 4*b^2) + tan(c/2 + (d*x)/2)^8*(a^2 - 2*a*b + b^2) + a^2 + b^ 
2)) - (atan((((a^2 + 12*b^2)*((8*tan(c/2 + (d*x)/2)*(a^14 - 2*a^13*b - 288 
*a*b^13 + 288*b^14 - 1104*a^2*b^12 + 1104*a^3*b^11 + 1538*a^4*b^10 - 1538* 
a^5*b^9 - 827*a^6*b^8 + 872*a^7*b^7 + 18*a^8*b^6 - 108*a^9*b^5 + 74*a^10*b 
^4 - 40*a^11*b^3 + 21*a^12*b^2))/(a^14*b + a^15 - a^8*b^7 - a^9*b^6 + 3*a^ 
10*b^5 + 3*a^11*b^4 - 3*a^12*b^3 - 3*a^13*b^2) - ((a^2 + 12*b^2)*((4*(4*a^ 
21 - 48*a^10*b^11 + 24*a^11*b^10 + 212*a^12*b^9 - 100*a^13*b^8 - 360*a^14* 
b^7 + 164*a^15*b^6 + 276*a^16*b^5 - 120*a^17*b^4 - 80*a^18*b^3 + 28*a^19*b 
^2))/(a^18*b + a^19 - a^12*b^7 - a^13*b^6 + 3*a^14*b^5 + 3*a^15*b^4 - 3*a^ 
16*b^3 - 3*a^17*b^2) - (4*tan(c/2 + (d*x)/2)*(a^2 + 12*b^2)*(8*a^19*b - 8* 
a^10*b^10 + 8*a^11*b^9 + 32*a^12*b^8 - 32*a^13*b^7 - 48*a^14*b^6 + 48*a...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 3080, normalized size of antiderivative = 10.10 \[ \int \frac {\sec ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

int(sec(d*x+c)^3/(a+b*cos(d*x+c))^3,x)
 

Output:

( - 80*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**5*b**4 + 116*sqrt(a**2 - 
b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*co 
s(c + d*x)*sin(c + d*x)**2*a**3*b**6 - 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*b**8 + 80*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 - 116*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**3*b**6 + 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)*a*b**8 + 40*sqrt(a**2 - b**2) 
*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*sin(c + 
 d*x)**4*a**4*b**5 - 58*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan(( 
c + d*x)/2)*b)/sqrt(a**2 - b**2))*sin(c + d*x)**4*a**2*b**7 + 24*sqrt(a**2 
 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2)) 
*sin(c + d*x)**4*b**9 - 40*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - ta 
n((c + d*x)/2)*b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a**6*b**3 - 22*sqrt(a 
**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b** 
2))*sin(c + d*x)**2*a**4*b**5 + 92*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2 
)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a**2*b**7 - 4 
8*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt...