\(\int \frac {\sec (c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^4} \, dx\) [926]

Optimal result

Integrand size = 39, antiderivative size = 299 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=-\frac {\left (4 a^2 b B+b^3 B-a^3 (2 A+C)-a b^2 (3 A+4 C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (2 a^2 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (2 a^3 b B+13 a b^3 B+a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))} \] Output:

-(4*B*a^2*b+B*b^3-a^3*(2*A+C)-a*b^2*(3*A+4*C))*arctanh((a-b)^(1/2)*tan(1/2 
*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(7/2)/(a+b)^(7/2)/d-1/3*(A*b^2-a*(B*b-C*a)) 
*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^3+1/6*(2*B*a^2*b+3*B*b^3+a^3*C- 
a*b^2*(5*A+6*C))*tan(d*x+c)/b/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^2+1/6*(2*B*a^ 
3*b+13*B*a*b^3+a^4*C-2*b^4*(2*A+3*C)-a^2*b^2*(11*A+10*C))*tan(d*x+c)/b/(a^ 
2-b^2)^3/d/(a+b*sec(d*x+c))
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 7.69 (sec) , antiderivative size = 538, normalized size of antiderivative = 1.80 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\frac {(b+a \cos (c+d x)) \sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-\frac {6 i \left (-4 a^2 b B-b^3 B+a^3 (2 A+C)+a b^2 (3 A+4 C)\right ) \arctan \left (\frac {(i \cos (c)+\sin (c)) \left (a \sin (c)+(-b+a \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right ) (b+a \cos (c+d x))^3 (\cos (c)-i \sin (c))}{\left (a^2-b^2\right )^{7/2} \sqrt {(\cos (c)-i \sin (c))^2}}+\frac {2 b \left (A b^2+a (-b B+a C)\right ) \sec (c) (b \sin (c)-a \sin (d x))}{a^5-a^3 b^2}+\frac {(b+a \cos (c+d x)) \sec (c) \left (b \left (-11 a^2 A b^2+6 A b^4+8 a^3 b B-3 a b^3 B-5 a^4 C\right ) \sin (c)+a \left (-4 A b^4-6 a^3 b B+a b^3 B+3 a^4 C+a^2 b^2 (9 A+2 C)\right ) \sin (d x)\right )}{a^3 \left (a^2-b^2\right )^2}+\frac {(b+a \cos (c+d x))^2 \sec (c) \left (3 \left (-6 a^2 A b^4+2 A b^6-4 a^5 b B-a^3 b^3 B+a^6 C+a^4 b^2 (9 A+4 C)\right ) \sin (c)+a \left (-2 A b^5+6 a^5 B+10 a^3 b^2 B-a b^4 B+a^2 b^3 (5 A-2 C)-a^4 b (18 A+13 C)\right ) \sin (d x)\right )}{\left (a^3-a b^2\right )^3}\right )}{3 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (a+b \sec (c+d x))^4} \] Input:

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

Output:

((b + a*Cos[c + d*x])*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^ 
2)*(((-6*I)*(-4*a^2*b*B - b^3*B + a^3*(2*A + C) + a*b^2*(3*A + 4*C))*ArcTa 
n[((I*Cos[c] + Sin[c])*(a*Sin[c] + (-b + a*Cos[c])*Tan[(d*x)/2]))/(Sqrt[a^ 
2 - b^2]*Sqrt[(Cos[c] - I*Sin[c])^2])]*(b + a*Cos[c + d*x])^3*(Cos[c] - I* 
Sin[c]))/((a^2 - b^2)^(7/2)*Sqrt[(Cos[c] - I*Sin[c])^2]) + (2*b*(A*b^2 + a 
*(-(b*B) + a*C))*Sec[c]*(b*Sin[c] - a*Sin[d*x]))/(a^5 - a^3*b^2) + ((b + a 
*Cos[c + d*x])*Sec[c]*(b*(-11*a^2*A*b^2 + 6*A*b^4 + 8*a^3*b*B - 3*a*b^3*B 
- 5*a^4*C)*Sin[c] + a*(-4*A*b^4 - 6*a^3*b*B + a*b^3*B + 3*a^4*C + a^2*b^2* 
(9*A + 2*C))*Sin[d*x]))/(a^3*(a^2 - b^2)^2) + ((b + a*Cos[c + d*x])^2*Sec[ 
c]*(3*(-6*a^2*A*b^4 + 2*A*b^6 - 4*a^5*b*B - a^3*b^3*B + a^6*C + a^4*b^2*(9 
*A + 4*C))*Sin[c] + a*(-2*A*b^5 + 6*a^5*B + 10*a^3*b^2*B - a*b^4*B + a^2*b 
^3*(5*A - 2*C) - a^4*b*(18*A + 13*C))*Sin[d*x]))/(a^3 - a*b^2)^3))/(3*d*(A 
 + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x])^4)
 

Rubi [A] (verified)

Time = 1.40 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.13, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4568, 3042, 4491, 25, 3042, 4491, 27, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4568

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4491

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (2 b \left (-\left ((3 A+2 C) a^2\right )+5 b B a-b^2 (2 A+3 C)\right )+\left (-C a^3-2 b B a^2+b^2 (5 A+6 C) a-3 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (a^3 C+2 a^2 b B-a b^2 (5 A+6 C)+3 b^3 B\right )}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 4491

\(\displaystyle -\frac {\frac {-\frac {\int -\frac {3 b \left (-\left ((2 A+C) a^3\right )+4 b B a^2-b^2 (3 A+4 C) a+b^3 B\right ) \sec (c+d x)}{a+b \sec (c+d x)}dx}{a^2-b^2}-\frac {\tan (c+d x) \left (a^4 C+2 a^3 b B-a^2 b^2 (11 A+10 C)+13 a b^3 B-2 b^4 (2 A+3 C)\right )}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (a^3 C+2 a^2 b B-a b^2 (5 A+6 C)+3 b^3 B\right )}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\frac {3 b \left (-\left (a^3 (2 A+C)\right )+4 a^2 b B-a b^2 (3 A+4 C)+b^3 B\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{a^2-b^2}-\frac {\tan (c+d x) \left (a^4 C+2 a^3 b B-a^2 b^2 (11 A+10 C)+13 a b^3 B-2 b^4 (2 A+3 C)\right )}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (a^3 C+2 a^2 b B-a b^2 (5 A+6 C)+3 b^3 B\right )}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\frac {3 b \left (-\left (a^3 (2 A+C)\right )+4 a^2 b B-a b^2 (3 A+4 C)+b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2-b^2}-\frac {\tan (c+d x) \left (a^4 C+2 a^3 b B-a^2 b^2 (11 A+10 C)+13 a b^3 B-2 b^4 (2 A+3 C)\right )}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (a^3 C+2 a^2 b B-a b^2 (5 A+6 C)+3 b^3 B\right )}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 4318

\(\displaystyle -\frac {\frac {\frac {3 \left (-\left (a^3 (2 A+C)\right )+4 a^2 b B-a b^2 (3 A+4 C)+b^3 B\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{a^2-b^2}-\frac {\tan (c+d x) \left (a^4 C+2 a^3 b B-a^2 b^2 (11 A+10 C)+13 a b^3 B-2 b^4 (2 A+3 C)\right )}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (a^3 C+2 a^2 b B-a b^2 (5 A+6 C)+3 b^3 B\right )}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\frac {3 \left (-\left (a^3 (2 A+C)\right )+4 a^2 b B-a b^2 (3 A+4 C)+b^3 B\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{a^2-b^2}-\frac {\tan (c+d x) \left (a^4 C+2 a^3 b B-a^2 b^2 (11 A+10 C)+13 a b^3 B-2 b^4 (2 A+3 C)\right )}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (a^3 C+2 a^2 b B-a b^2 (5 A+6 C)+3 b^3 B\right )}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 3138

\(\displaystyle -\frac {\frac {\frac {6 \left (-\left (a^3 (2 A+C)\right )+4 a^2 b B-a b^2 (3 A+4 C)+b^3 B\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{d \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (a^4 C+2 a^3 b B-a^2 b^2 (11 A+10 C)+13 a b^3 B-2 b^4 (2 A+3 C)\right )}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (a^3 C+2 a^2 b B-a b^2 (5 A+6 C)+3 b^3 B\right )}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}-\frac {\frac {\frac {6 b \left (-\left (a^3 (2 A+C)\right )+4 a^2 b B-a b^2 (3 A+4 C)+b^3 B\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (a^4 C+2 a^3 b B-a^2 b^2 (11 A+10 C)+13 a b^3 B-2 b^4 (2 A+3 C)\right )}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (a^3 C+2 a^2 b B-a b^2 (5 A+6 C)+3 b^3 B\right )}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 b \left (a^2-b^2\right )}\)

Input:

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

Output:

-1/3*((A*b^2 - a*(b*B - a*C))*Tan[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Sec[c 
+ d*x])^3) - (-1/2*((2*a^2*b*B + 3*b^3*B + a^3*C - a*b^2*(5*A + 6*C))*Tan[ 
c + d*x])/((a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) + ((6*b*(4*a^2*b*B + b^3* 
B - a^3*(2*A + C) - a*b^2*(3*A + 4*C))*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/ 
2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*(a^2 - b^2)*d) - ((2*a^3*b*B + 
13*a*b^3*B + a^4*C - 2*b^4*(2*A + 3*C) - a^2*b^2*(11*A + 10*C))*Tan[c + d* 
x])/((a^2 - b^2)*d*(a + b*Sec[c + d*x])))/(2*(a^2 - b^2)))/(3*b*(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 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 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 4491

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

rule 4568

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[(-(A*b^2 - a*b*B + a^2*C))*Cot[e + f*x]*((a + b*Csc[e + f*x] 
)^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) 
 Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m 
+ 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Csc[e + f*x], 
x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^ 
2, 0]
 

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 452, normalized size of antiderivative = 1.51

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-\frac {\left (6 A \,a^{2} b +3 a A \,b^{2}+2 A \,b^{3}-2 B \,a^{3}-2 B \,a^{2} b -6 B a \,b^{2}-B \,b^{3}+a^{3} C +6 a^{2} b C +2 C a \,b^{2}+2 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 \left (9 A \,a^{2} b +A \,b^{3}-3 B \,a^{3}-7 B a \,b^{2}+7 a^{2} b C +3 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (6 A \,a^{2} b -3 a A \,b^{2}+2 A \,b^{3}-2 B \,a^{3}+2 B \,a^{2} b -6 B a \,b^{2}+B \,b^{3}-a^{3} C +6 a^{2} b C -2 C a \,b^{2}+2 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{3}}+\frac {\left (2 a^{3} A +3 a A \,b^{2}-4 B \,a^{2} b -B \,b^{3}+a^{3} C +4 C a \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(452\)
default	\(\frac {-\frac {2 \left (-\frac {\left (6 A \,a^{2} b +3 a A \,b^{2}+2 A \,b^{3}-2 B \,a^{3}-2 B \,a^{2} b -6 B a \,b^{2}-B \,b^{3}+a^{3} C +6 a^{2} b C +2 C a \,b^{2}+2 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 \left (9 A \,a^{2} b +A \,b^{3}-3 B \,a^{3}-7 B a \,b^{2}+7 a^{2} b C +3 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (6 A \,a^{2} b -3 a A \,b^{2}+2 A \,b^{3}-2 B \,a^{3}+2 B \,a^{2} b -6 B a \,b^{2}+B \,b^{3}-a^{3} C +6 a^{2} b C -2 C a \,b^{2}+2 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{3}}+\frac {\left (2 a^{3} A +3 a A \,b^{2}-4 B \,a^{2} b -B \,b^{3}+a^{3} C +4 C a \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(452\)
risch	\(\text {Expression too large to display}\)	\(1968\)

Input:

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

Output:

1/d*(-2*(-1/2*(6*A*a^2*b+3*A*a*b^2+2*A*b^3-2*B*a^3-2*B*a^2*b-6*B*a*b^2-B*b 
^3+C*a^3+6*C*a^2*b+2*C*a*b^2+2*C*b^3)/(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*(9*A*a^2*b+A*b^3-3*B*a^3-7*B*a*b^2+7*C*a^2*b+3*C*b^3) 
/(a^2+2*a*b+b^2)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3-1/2*(6*A*a^2*b-3*A*a 
*b^2+2*A*b^3-2*B*a^3+2*B*a^2*b-6*B*a*b^2+B*b^3-C*a^3+6*C*a^2*b-2*C*a*b^2+2 
*C*b^3)/(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+(2*A*a^3+3*A*a*b^2-4*B*a^2*b-B*b^3 
+C*a^3+4*C*a*b^2)/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a+b)*(a-b))^(1/2)*arctan 
h((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. 678 vs. \(2 (281) = 562\).

Time = 0.18 (sec) , antiderivative size = 1414, normalized size of antiderivative = 4.73 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/12*(3*((2*A + C)*a^3*b^3 - 4*B*a^2*b^4 + (3*A + 4*C)*a*b^5 - B*b^6 + (( 
2*A + C)*a^6 - 4*B*a^5*b + (3*A + 4*C)*a^4*b^2 - B*a^3*b^3)*cos(d*x + c)^3 
 + 3*((2*A + C)*a^5*b - 4*B*a^4*b^2 + (3*A + 4*C)*a^3*b^3 - B*a^2*b^4)*cos 
(d*x + c)^2 + 3*((2*A + C)*a^4*b^2 - 4*B*a^3*b^3 + (3*A + 4*C)*a^2*b^4 - B 
*a*b^5)*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)) + 2*(C*a^6* 
b + 2*B*a^5*b^2 - 11*(A + C)*a^4*b^3 + 11*B*a^3*b^4 + (7*A + 4*C)*a^2*b^5 
- 13*B*a*b^6 + 2*(2*A + 3*C)*b^7 + (6*B*a^7 - (18*A + 13*C)*a^6*b + 4*B*a^ 
5*b^2 + (23*A + 11*C)*a^4*b^3 - 11*B*a^3*b^4 - (7*A - 2*C)*a^2*b^5 + B*a*b 
^6 + 2*A*b^7)*cos(d*x + c)^2 + 3*(C*a^7 + 2*B*a^6*b - (9*A + 10*C)*a^5*b^2 
 + 7*B*a^4*b^3 + (8*A + 7*C)*a^3*b^4 - 10*B*a^2*b^5 + (A + 2*C)*a*b^6 + B* 
b^7)*cos(d*x + c))*sin(d*x + c))/((a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^ 
6 + a^3*b^8)*d*cos(d*x + c)^3 + 3*(a^10*b - 4*a^8*b^3 + 6*a^6*b^5 - 4*a^4* 
b^7 + a^2*b^9)*d*cos(d*x + c)^2 + 3*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a 
^3*b^8 + a*b^10)*d*cos(d*x + c) + (a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2 
*b^9 + b^11)*d), 1/6*(3*((2*A + C)*a^3*b^3 - 4*B*a^2*b^4 + (3*A + 4*C)*a*b 
^5 - B*b^6 + ((2*A + C)*a^6 - 4*B*a^5*b + (3*A + 4*C)*a^4*b^2 - B*a^3*b^3) 
*cos(d*x + c)^3 + 3*((2*A + C)*a^5*b - 4*B*a^4*b^2 + (3*A + 4*C)*a^3*b^3 - 
 B*a^2*b^4)*cos(d*x + c)^2 + 3*((2*A + C)*a^4*b^2 - 4*B*a^3*b^3 + (3*A ...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^4,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. 968 vs. \(2 (281) = 562\).

Time = 0.31 (sec) , antiderivative size = 968, normalized size of antiderivative = 3.24 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/3*(3*(2*A*a^3 + C*a^3 - 4*B*a^2*b + 3*A*a*b^2 + 4*C*a*b^2 - B*b^3)*(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^6 - 3*a^4*b^2 + 3*a^2 
*b^4 - b^6)*sqrt(-a^2 + b^2)) + (6*B*a^5*tan(1/2*d*x + 1/2*c)^5 - 3*C*a^5* 
tan(1/2*d*x + 1/2*c)^5 - 18*A*a^4*b*tan(1/2*d*x + 1/2*c)^5 - 6*B*a^4*b*tan 
(1/2*d*x + 1/2*c)^5 - 12*C*a^4*b*tan(1/2*d*x + 1/2*c)^5 + 27*A*a^3*b^2*tan 
(1/2*d*x + 1/2*c)^5 + 12*B*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 + 27*C*a^3*b^2*t 
an(1/2*d*x + 1/2*c)^5 - 6*A*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 - 27*B*a^2*b^3* 
tan(1/2*d*x + 1/2*c)^5 - 12*C*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*A*a*b^4*t 
an(1/2*d*x + 1/2*c)^5 + 12*B*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 6*C*a*b^4*tan( 
1/2*d*x + 1/2*c)^5 - 6*A*b^5*tan(1/2*d*x + 1/2*c)^5 + 3*B*b^5*tan(1/2*d*x 
+ 1/2*c)^5 - 6*C*b^5*tan(1/2*d*x + 1/2*c)^5 - 12*B*a^5*tan(1/2*d*x + 1/2*c 
)^3 + 36*A*a^4*b*tan(1/2*d*x + 1/2*c)^3 + 28*C*a^4*b*tan(1/2*d*x + 1/2*c)^ 
3 - 16*B*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 32*A*a^2*b^3*tan(1/2*d*x + 1/2*c 
)^3 - 16*C*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + 28*B*a*b^4*tan(1/2*d*x + 1/2*c 
)^3 - 4*A*b^5*tan(1/2*d*x + 1/2*c)^3 - 12*C*b^5*tan(1/2*d*x + 1/2*c)^3 + 6 
*B*a^5*tan(1/2*d*x + 1/2*c) + 3*C*a^5*tan(1/2*d*x + 1/2*c) - 18*A*a^4*b*ta 
n(1/2*d*x + 1/2*c) + 6*B*a^4*b*tan(1/2*d*x + 1/2*c) - 12*C*a^4*b*tan(1/2*d 
*x + 1/2*c) - 27*A*a^3*b^2*tan(1/2*d*x + 1/2*c) + 12*B*a^3*b^2*tan(1/2*d*x 
 + 1/2*c) - 27*C*a^3*b^2*tan(1/2*d*x + 1/2*c) - 6*A*a^2*b^3*tan(1/2*d*x...
 

Mupad [B] (verification not implemented)

Time = 15.97 (sec) , antiderivative size = 516, normalized size of antiderivative = 1.73 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\frac {\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}}\right )\,\left (2\,A\,a^3-B\,b^3+C\,a^3+3\,A\,a\,b^2-4\,B\,a^2\,b+4\,C\,a\,b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}-\frac {\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,b^3-2\,B\,a^3+B\,b^3-C\,a^3+2\,C\,b^3-3\,A\,a\,b^2+6\,A\,a^2\,b-6\,B\,a\,b^2+2\,B\,a^2\,b-2\,C\,a\,b^2+6\,C\,a^2\,b\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A\,b^3-3\,B\,a^3+3\,C\,b^3+9\,A\,a^2\,b-7\,B\,a\,b^2+7\,C\,a^2\,b\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,A\,b^3-2\,B\,a^3-B\,b^3+C\,a^3+2\,C\,b^3+3\,A\,a\,b^2+6\,A\,a^2\,b-6\,B\,a\,b^2-2\,B\,a^2\,b+2\,C\,a\,b^2+6\,C\,a^2\,b\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )+3\,a\,b^2+3\,a^2\,b+a^3+b^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )} \] Input:

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

Output:

(atanh((tan(c/2 + (d*x)/2)*(2*a - 2*b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3))/(2 
*(a + b)^(1/2)*(a - b)^(7/2)))*(2*A*a^3 - B*b^3 + C*a^3 + 3*A*a*b^2 - 4*B* 
a^2*b + 4*C*a*b^2))/(d*(a + b)^(7/2)*(a - b)^(7/2)) - ((tan(c/2 + (d*x)/2) 
*(2*A*b^3 - 2*B*a^3 + B*b^3 - C*a^3 + 2*C*b^3 - 3*A*a*b^2 + 6*A*a^2*b - 6* 
B*a*b^2 + 2*B*a^2*b - 2*C*a*b^2 + 6*C*a^2*b))/((a + b)*(3*a*b^2 - 3*a^2*b 
+ a^3 - b^3)) - (4*tan(c/2 + (d*x)/2)^3*(A*b^3 - 3*B*a^3 + 3*C*b^3 + 9*A*a 
^2*b - 7*B*a*b^2 + 7*C*a^2*b))/(3*(a + b)^2*(a^2 - 2*a*b + b^2)) + (tan(c/ 
2 + (d*x)/2)^5*(2*A*b^3 - 2*B*a^3 - B*b^3 + C*a^3 + 2*C*b^3 + 3*A*a*b^2 + 
6*A*a^2*b - 6*B*a*b^2 - 2*B*a^2*b + 2*C*a*b^2 + 6*C*a^2*b))/((a + b)^3*(a 
- b)))/(d*(tan(c/2 + (d*x)/2)^2*(3*a*b^2 - 3*a^2*b - 3*a^3 + 3*b^3) - tan( 
c/2 + (d*x)/2)^4*(3*a*b^2 + 3*a^2*b - 3*a^3 - 3*b^3) + 3*a*b^2 + 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)))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 2134, normalized size of antiderivative = 7.14 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx =\text {Too large to display} \] Input:

int(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^4,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**7 + 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**6*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)*sin(c 
 + d*x)**2*a**5*b**2 + 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**2*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)*sin(c + d*x)**2*a**3*b**4 - 12*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 - 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**6*c - 
 30*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)*a**5*b**2 - 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**4*b**2*c + 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)*a**3*b**4 - 72*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**2*b**4*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*b**6...