\(\int \frac {\sec ^3(c+d x) (A+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^3} \, dx\) [692]

Optimal result

Integrand size = 33, antiderivative size = 271 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=-\frac {3 a C \text {arctanh}(\sin (c+d x))}{b^4 d}+\frac {\left (2 A b^6+6 a^6 C-15 a^4 b^2 C+a^2 b^4 (A+12 C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} b^4 (a+b)^{5/2} d}+\frac {\left (A b^2+3 a^2 C-2 b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac {\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {a \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \] Output:

-3*a*C*arctanh(sin(d*x+c))/b^4/d+(2*A*b^6+6*a^6*C-15*a^4*b^2*C+a^2*b^4*(A+ 
12*C))*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(5/2)/b^4 
/(a+b)^(5/2)/d+1/2*(A*b^2+3*C*a^2-2*C*b^2)*tan(d*x+c)/b^3/(a^2-b^2)/d-1/2* 
(A*b^2+C*a^2)*sec(d*x+c)^2*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^2-1/2 
*a*(2*A*b^4-3*a^4*C+a^2*b^2*(A+6*C))*tan(d*x+c)/b^3/(a^2-b^2)^2/d/(a+b*sec 
(d*x+c))
 

Mathematica [A] (warning: unable to verify)

Time = 2.99 (sec) , antiderivative size = 421, normalized size of antiderivative = 1.55 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\frac {(b+a \cos (c+d x)) \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (-\frac {2 \left (2 A b^6+6 a^6 C-15 a^4 b^2 C+a^2 b^4 (A+12 C)\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right ) (b+a \cos (c+d x))^2}{\left (a^2-b^2\right )^{5/2}}+6 a C (b+a \cos (c+d x))^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-6 a C (b+a \cos (c+d x))^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {2 b C (b+a \cos (c+d x))^2 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {2 b C (b+a \cos (c+d x))^2 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {a b^2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{(a-b) (a+b)}+\frac {a b \left (-3 A b^4+4 a^4 C-7 a^2 b^2 C\right ) (b+a \cos (c+d x)) \sin (c+d x)}{(a-b)^2 (a+b)^2}\right )}{b^4 d (A+2 C+A \cos (2 (c+d x))) (a+b \sec (c+d x))^3} \] Input:

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

Output:

((b + a*Cos[c + d*x])*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*((-2*(2*A*b^6 + 
6*a^6*C - 15*a^4*b^2*C + a^2*b^4*(A + 12*C))*ArcTanh[((-a + b)*Tan[(c + d* 
x)/2])/Sqrt[a^2 - b^2]]*(b + a*Cos[c + d*x])^2)/(a^2 - b^2)^(5/2) + 6*a*C* 
(b + a*Cos[c + d*x])^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 6*a*C*(b 
 + a*Cos[c + d*x])^2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (2*b*C*(b 
+ a*Cos[c + d*x])^2*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2] 
) + (2*b*C*(b + a*Cos[c + d*x])^2*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Si 
n[(c + d*x)/2]) + (a*b^2*(A*b^2 + a^2*C)*Sin[c + d*x])/((a - b)*(a + b)) + 
 (a*b*(-3*A*b^4 + 4*a^4*C - 7*a^2*b^2*C)*(b + a*Cos[c + d*x])*Sin[c + d*x] 
)/((a - b)^2*(a + b)^2)))/(b^4*d*(A + 2*C + A*Cos[2*(c + d*x)])*(a + b*Sec 
[c + d*x])^3)
 

Rubi [A] (verified)

Time = 2.03 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.13, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4587, 3042, 4578, 25, 3042, 4570, 3042, 4486, 3042, 4257, 4318, 3042, 3138, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4587

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4578

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {a \left (-3 a^4 C+a^2 b^2 (A+6 C)+2 A b^4\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (a^2-b^2\right ) \left (3 C a^2+A b^2-2 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a \left (a^2-b^2\right ) \left (-3 C a^2+A b^2+4 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+b \left (-3 C a^4+b^2 (A+6 C) a^2+2 A b^4\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2 \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 4570

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

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {a \left (-3 a^4 C+a^2 b^2 (A+6 C)+2 A b^4\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b^2 \left (-3 C a^4+b^2 (A+6 C) a^2+2 A b^4\right )-6 a b \left (a^2-b^2\right )^2 C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {\left (a^2-b^2\right ) \left (3 a^2 C+A b^2-2 b^2 C\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 4486

\(\displaystyle -\frac {\frac {a \left (-3 a^4 C+a^2 b^2 (A+6 C)+2 A b^4\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\left (6 a^6 C-15 a^4 b^2 C+a^2 b^4 (A+12 C)+2 A b^6\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx-6 a C \left (a^2-b^2\right )^2 \int \sec (c+d x)dx}{b}+\frac {\left (a^2-b^2\right ) \left (3 a^2 C+A b^2-2 b^2 C\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {a \left (-3 a^4 C+a^2 b^2 (A+6 C)+2 A b^4\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\left (6 a^6 C-15 a^4 b^2 C+a^2 b^4 (A+12 C)+2 A b^6\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx-6 a C \left (a^2-b^2\right )^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{b}+\frac {\left (a^2-b^2\right ) \left (3 a^2 C+A b^2-2 b^2 C\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 4257

\(\displaystyle -\frac {\frac {a \left (-3 a^4 C+a^2 b^2 (A+6 C)+2 A b^4\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\left (6 a^6 C-15 a^4 b^2 C+a^2 b^4 (A+12 C)+2 A b^6\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {6 a C \left (a^2-b^2\right )^2 \text {arctanh}(\sin (c+d x))}{d}}{b}+\frac {\left (a^2-b^2\right ) \left (3 a^2 C+A b^2-2 b^2 C\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 4318

\(\displaystyle -\frac {\frac {a \left (-3 a^4 C+a^2 b^2 (A+6 C)+2 A b^4\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\frac {\left (6 a^6 C-15 a^4 b^2 C+a^2 b^4 (A+12 C)+2 A b^6\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b}-\frac {6 a C \left (a^2-b^2\right )^2 \text {arctanh}(\sin (c+d x))}{d}}{b}+\frac {\left (a^2-b^2\right ) \left (3 a^2 C+A b^2-2 b^2 C\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {a \left (-3 a^4 C+a^2 b^2 (A+6 C)+2 A b^4\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\frac {\left (6 a^6 C-15 a^4 b^2 C+a^2 b^4 (A+12 C)+2 A b^6\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b}-\frac {6 a C \left (a^2-b^2\right )^2 \text {arctanh}(\sin (c+d x))}{d}}{b}+\frac {\left (a^2-b^2\right ) \left (3 a^2 C+A b^2-2 b^2 C\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 3138

\(\displaystyle -\frac {\frac {a \left (-3 a^4 C+a^2 b^2 (A+6 C)+2 A b^4\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\frac {2 \left (6 a^6 C-15 a^4 b^2 C+a^2 b^4 (A+12 C)+2 A b^6\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}-\frac {6 a C \left (a^2-b^2\right )^2 \text {arctanh}(\sin (c+d x))}{d}}{b}+\frac {\left (a^2-b^2\right ) \left (3 a^2 C+A b^2-2 b^2 C\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {a \left (-3 a^4 C+a^2 b^2 (A+6 C)+2 A b^4\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\left (a^2-b^2\right ) \left (3 a^2 C+A b^2-2 b^2 C\right ) \tan (c+d x)}{d}+\frac {\frac {2 \left (6 a^6 C-15 a^4 b^2 C+a^2 b^4 (A+12 C)+2 A b^6\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b}}-\frac {6 a C \left (a^2-b^2\right )^2 \text {arctanh}(\sin (c+d x))}{d}}{b}}{b^2 \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}\)

Input:

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

Output:

-1/2*((A*b^2 + a^2*C)*Sec[c + d*x]^2*Tan[c + d*x])/(b*(a^2 - b^2)*d*(a + b 
*Sec[c + d*x])^2) - ((a*(2*A*b^4 - 3*a^4*C + a^2*b^2*(A + 6*C))*Tan[c + d* 
x])/(b^2*(a^2 - b^2)*d*(a + b*Sec[c + d*x])) - (((-6*a*(a^2 - b^2)^2*C*Arc 
Tanh[Sin[c + d*x]])/d + (2*(2*A*b^6 + 6*a^6*C - 15*a^4*b^2*C + a^2*b^4*(A 
+ 12*C))*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b] 
*Sqrt[a + b]*d))/b + ((a^2 - b^2)*(A*b^2 + 3*a^2*C - 2*b^2*C)*Tan[c + d*x] 
)/d)/(b^2*(a^2 - b^2)))/(2*b*(a^2 - b^2))
 

Defintions of rubi rules used

rule 25

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

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[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 4257

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

rule 4318

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo 
l] :> Simp[1/b   Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, 
f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4486

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[( 
e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[B/b   Int[Csc[e + f*x], 
 x], x] + Simp[(A*b - a*B)/b   Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x 
] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]
 

rule 4570

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e 
_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S 
ymbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2) 
)), x] + Simp[1/(b*(m + 2))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[ 
b*A*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; 
 FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]
 

rule 4578

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[ 
(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x 
_Symbol] :> Simp[a*(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*((a + b*Csc[e + f*x 
])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - Simp[1/(b^2*(m + 1)*(a^2 - b^ 
2))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[b*(m + 1)*((-a)*(b 
*B - a*C) + A*b^2) + (b*B*(a^2 + b^2*(m + 1)) - a*(A*b^2*(m + 2) + C*(a^2 + 
 b^2*(m + 1))))*Csc[e + f*x] - b*C*(m + 1)*(a^2 - b^2)*Csc[e + f*x]^2, x], 
x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1 
]
 

rule 4587

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. 
))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-d) 
*(A*b^2 + a^2*C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x] 
)^(n - 1)/(b*f*(a^2 - b^2)*(m + 1))), x] + Simp[d/(b*(a^2 - b^2)*(m + 1)) 
 Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1)*Simp[A*b^2*(n - 
1) + a^2*C*(n - 1) + a*b*(A + C)*(m + 1)*Csc[e + f*x] - (A*b^2*(m + n + 1) 
+ C*(a^2*n + b^2*(m + 1)))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, 
 f, A, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 0]
 

Maple [A] (verified)

Time = 1.20 (sec) , antiderivative size = 356, normalized size of antiderivative = 1.31

method	result	size
derivativedivides	\(\frac {-\frac {C}{b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 C a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{4}}-\frac {2 \left (\frac {-\frac {\left (a A \,b^{3}+4 A \,b^{4}-4 a^{4} C +a^{3} b C +8 C \,a^{2} b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (a A \,b^{3}-4 A \,b^{4}+4 a^{4} C +a^{3} b C -8 C \,a^{2} b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}-\frac {\left (a^{2} A \,b^{4}+2 A \,b^{6}+6 a^{6} C -15 a^{4} b^{2} C +12 C \,a^{2} b^{4}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{4}}-\frac {C}{b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 C a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{4}}}{d}\)	\(356\)
default	\(\frac {-\frac {C}{b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 C a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{4}}-\frac {2 \left (\frac {-\frac {\left (a A \,b^{3}+4 A \,b^{4}-4 a^{4} C +a^{3} b C +8 C \,a^{2} b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (a A \,b^{3}-4 A \,b^{4}+4 a^{4} C +a^{3} b C -8 C \,a^{2} b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}-\frac {\left (a^{2} A \,b^{4}+2 A \,b^{6}+6 a^{6} C -15 a^{4} b^{2} C +12 C \,a^{2} b^{4}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{4}}-\frac {C}{b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 C a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{4}}}{d}\)	\(356\)
risch	\(\text {Expression too large to display}\)	\(1403\)

Input:

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

Output:

1/d*(-C/b^3/(tan(1/2*d*x+1/2*c)+1)-3*C*a/b^4*ln(tan(1/2*d*x+1/2*c)+1)-2/b^ 
4*((-1/2*(A*a*b^3+4*A*b^4-4*C*a^4+C*a^3*b+8*C*a^2*b^2)*a*b/(a-b)/(a^2+2*a* 
b+b^2)*tan(1/2*d*x+1/2*c)^3-1/2*(A*a*b^3-4*A*b^4+4*C*a^4+C*a^3*b-8*C*a^2*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*(A*a^2*b^4+2*A*b^6+6*C*a^6-15*C*a^4*b^2+ 
12*C*a^2*b^4)/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/ 
2*d*x+1/2*c)/((a+b)*(a-b))^(1/2)))-C/b^3/(tan(1/2*d*x+1/2*c)-1)+3*C*a/b^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. 747 vs. \(2 (257) = 514\).

Time = 6.27 (sec) , antiderivative size = 1552, normalized size of antiderivative = 5.73 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/4*(((6*C*a^8 - 15*C*a^6*b^2 + (A + 12*C)*a^4*b^4 + 2*A*a^2*b^6)*cos(d*x 
 + c)^3 + 2*(6*C*a^7*b - 15*C*a^5*b^3 + (A + 12*C)*a^3*b^5 + 2*A*a*b^7)*co 
s(d*x + c)^2 + (6*C*a^6*b^2 - 15*C*a^4*b^4 + (A + 12*C)*a^2*b^6 + 2*A*b^8) 
*cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos 
(d*x + c)^2 + 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 
- b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) - 6*((C*a^9 - 3*C* 
a^7*b^2 + 3*C*a^5*b^4 - C*a^3*b^6)*cos(d*x + c)^3 + 2*(C*a^8*b - 3*C*a^6*b 
^3 + 3*C*a^4*b^5 - C*a^2*b^7)*cos(d*x + c)^2 + (C*a^7*b^2 - 3*C*a^5*b^4 + 
3*C*a^3*b^6 - C*a*b^8)*cos(d*x + c))*log(sin(d*x + c) + 1) + 6*((C*a^9 - 3 
*C*a^7*b^2 + 3*C*a^5*b^4 - C*a^3*b^6)*cos(d*x + c)^3 + 2*(C*a^8*b - 3*C*a^ 
6*b^3 + 3*C*a^4*b^5 - C*a^2*b^7)*cos(d*x + c)^2 + (C*a^7*b^2 - 3*C*a^5*b^4 
 + 3*C*a^3*b^6 - C*a*b^8)*cos(d*x + c))*log(-sin(d*x + c) + 1) + 2*(2*C*a^ 
6*b^3 - 6*C*a^4*b^5 + 6*C*a^2*b^7 - 2*C*b^9 + (6*C*a^8*b - 17*C*a^6*b^3 - 
(3*A - 13*C)*a^4*b^5 + (3*A - 2*C)*a^2*b^7)*cos(d*x + c)^2 + (9*C*a^7*b^2 
+ (A - 25*C)*a^5*b^4 - 5*(A - 4*C)*a^3*b^6 + 4*(A - C)*a*b^8)*cos(d*x + c) 
)*sin(d*x + c))/((a^8*b^4 - 3*a^6*b^6 + 3*a^4*b^8 - a^2*b^10)*d*cos(d*x + 
c)^3 + 2*(a^7*b^5 - 3*a^5*b^7 + 3*a^3*b^9 - a*b^11)*d*cos(d*x + c)^2 + (a^ 
6*b^6 - 3*a^4*b^8 + 3*a^2*b^10 - b^12)*d*cos(d*x + c)), 1/2*(((6*C*a^8 - 1 
5*C*a^6*b^2 + (A + 12*C)*a^4*b^4 + 2*A*a^2*b^6)*cos(d*x + c)^3 + 2*(6*C*a^ 
7*b - 15*C*a^5*b^3 + (A + 12*C)*a^3*b^5 + 2*A*a*b^7)*cos(d*x + c)^2 + (...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(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*a^2-4*b^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. 521 vs. \(2 (257) = 514\).

Time = 0.40 (sec) , antiderivative size = 521, normalized size of antiderivative = 1.92 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

((6*C*a^6 - 15*C*a^4*b^2 + A*a^2*b^4 + 12*C*a^2*b^4 + 2*A*b^6)*(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^4*b^4 - 2*a^2*b^6 + b^8)*s 
qrt(-a^2 + b^2)) - 3*C*a*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^4 + 3*C*a*lo 
g(abs(tan(1/2*d*x + 1/2*c) - 1))/b^4 - (4*C*a^6*tan(1/2*d*x + 1/2*c)^3 - 5 
*C*a^5*b*tan(1/2*d*x + 1/2*c)^3 - 7*C*a^4*b^2*tan(1/2*d*x + 1/2*c)^3 - A*a 
^3*b^3*tan(1/2*d*x + 1/2*c)^3 + 8*C*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 - 3*A*a 
^2*b^4*tan(1/2*d*x + 1/2*c)^3 + 4*A*a*b^5*tan(1/2*d*x + 1/2*c)^3 - 4*C*a^6 
*tan(1/2*d*x + 1/2*c) - 5*C*a^5*b*tan(1/2*d*x + 1/2*c) + 7*C*a^4*b^2*tan(1 
/2*d*x + 1/2*c) - A*a^3*b^3*tan(1/2*d*x + 1/2*c) + 8*C*a^3*b^3*tan(1/2*d*x 
 + 1/2*c) + 3*A*a^2*b^4*tan(1/2*d*x + 1/2*c) + 4*A*a*b^5*tan(1/2*d*x + 1/2 
*c))/((a^4*b^3 - 2*a^2*b^5 + b^7)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d* 
x + 1/2*c)^2 - a - b)^2) - 2*C*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c) 
^2 - 1)*b^3))/d
 

Mupad [B] (verification not implemented)

Time = 22.17 (sec) , antiderivative size = 7197, normalized size of antiderivative = 26.56 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

((tan(c/2 + (d*x)/2)*(6*C*a^5 + 2*C*b^5 + A*a^2*b^3 - 4*C*a^2*b^3 - 12*C*a 
^3*b^2 - 4*A*a*b^4 + 2*C*a*b^4 + 3*C*a^4*b))/((a + b)*(b^5 - 2*a*b^4 + a^2 
*b^3)) - (tan(c/2 + (d*x)/2)^5*(2*C*b^5 - 6*C*a^5 + A*a^2*b^3 - 4*C*a^2*b^ 
3 + 12*C*a^3*b^2 + 4*A*a*b^4 - 2*C*a*b^4 + 3*C*a^4*b))/((a*b^3 - b^4)*(a + 
 b)^2) + (2*tan(c/2 + (d*x)/2)^3*(2*C*b^6 - 6*C*a^6 + 3*A*a^2*b^4 - 6*C*a^ 
2*b^4 + 13*C*a^4*b^2))/(b*(a*b^2 - b^3)*(a + b)^2*(a - b)))/(d*(2*a*b - ta 
n(c/2 + (d*x)/2)^2*(2*a*b + 3*a^2 - b^2) - tan(c/2 + (d*x)/2)^6*(a^2 - 2*a 
*b + b^2) + a^2 + b^2 - tan(c/2 + (d*x)/2)^4*(2*a*b - 3*a^2 + b^2))) + (C* 
a*atan(((C*a*((8*tan(c/2 + (d*x)/2)*(4*A^2*b^12 + 72*C^2*a^12 - 72*C^2*a^1 
1*b + 4*A^2*a^2*b^10 + A^2*a^4*b^8 + 36*C^2*a^2*b^10 - 72*C^2*a^3*b^9 + 36 
*C^2*a^4*b^8 + 288*C^2*a^5*b^7 - 288*C^2*a^6*b^6 - 432*C^2*a^7*b^5 + 441*C 
^2*a^8*b^4 + 288*C^2*a^9*b^3 - 288*C^2*a^10*b^2 + 48*A*C*a^2*b^10 - 36*A*C 
*a^4*b^8 - 6*A*C*a^6*b^6 + 12*A*C*a^8*b^4))/(a*b^12 + b^13 - 3*a^2*b^11 - 
3*a^3*b^10 + 3*a^4*b^9 + 3*a^5*b^8 - a^6*b^7 - a^7*b^6) - (3*C*a*((8*(4*A* 
b^18 - 6*A*a^2*b^16 + 6*A*a^3*b^15 + 2*A*a^6*b^12 - 2*A*a^7*b^11 + 24*C*a^ 
2*b^16 + 36*C*a^3*b^15 - 78*C*a^4*b^14 - 42*C*a^5*b^13 + 96*C*a^6*b^12 + 2 
4*C*a^7*b^11 - 54*C*a^8*b^10 - 6*C*a^9*b^9 + 12*C*a^10*b^8 - 4*A*a*b^17 - 
12*C*a*b^17))/(a*b^15 + b^16 - 3*a^2*b^14 - 3*a^3*b^13 + 3*a^4*b^12 + 3*a^ 
5*b^11 - a^6*b^10 - a^7*b^9) - (24*C*a*tan(c/2 + (d*x)/2)*(8*a*b^17 - 8*a^ 
2*b^16 - 32*a^3*b^15 + 32*a^4*b^14 + 48*a^5*b^13 - 48*a^6*b^12 - 32*a^7...
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 2828, normalized size of antiderivative = 10.44 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(12*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**8*c - 30*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**6*b**2*c + 2*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 + 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**4*b**4*c + 4*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 - 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)*a**8*c + 18*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**2*c - 2*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 + 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)*a**4*b**4*c - 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)*a**3*b**6 - 
24*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**2*b**6*c - 4*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 +...