\(\int \cos ^2(a+b x) \sec ^3(c+d x) \, dx\) [338]

Optimal result

Integrand size = 17, antiderivative size = 174 \[ \int \cos ^2(a+b x) \sec ^3(c+d x) \, dx=\frac {\text {arctanh}(\sin (c+d x))}{4 d}+\frac {2 i e^{-2 i a-2 i b x+3 i (c+d x)} \operatorname {Hypergeometric2F1}\left (3,\frac {3}{2}-\frac {b}{d},\frac {5}{2}-\frac {b}{d},-e^{2 i (c+d x)}\right )}{2 b-3 d}-\frac {2 i e^{2 i a+2 i b x+3 i (c+d x)} \operatorname {Hypergeometric2F1}\left (3,\frac {3}{2}+\frac {b}{d},\frac {5}{2}+\frac {b}{d},-e^{2 i (c+d x)}\right )}{2 b+3 d}+\frac {\sec (c+d x) \tan (c+d x)}{4 d} \] Output:

1/4*arctanh(sin(d*x+c))/d+2*I*exp(-2*I*a-2*I*b*x+3*I*(d*x+c))*hypergeom([3 
, 3/2-b/d],[5/2-b/d],-exp(2*I*(d*x+c)))/(2*b-3*d)-2*I*exp(2*I*a+2*I*b*x+3* 
I*(d*x+c))*hypergeom([3, 3/2+b/d],[5/2+b/d],-exp(2*I*(d*x+c)))/(2*b+3*d)+1 
/4*sec(d*x+c)*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 6.37 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.18 \[ \int \cos ^2(a+b x) \sec ^3(c+d x) \, dx=\frac {-4 i d \arctan \left (e^{i (c+d x)}\right )-2 i (2 b+d) e^{-i (2 a-c+2 b x-d x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}-\frac {b}{d},\frac {3}{2}-\frac {b}{d},-e^{2 i (c+d x)}\right )+2 i (2 b-d) e^{i (2 a+c+(2 b+d) x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+\frac {b}{d},\frac {3}{2}+\frac {b}{d},-e^{2 i (c+d x)}\right )+2 \cos (a+b x) \sec ^2(c+d x) ((2 b-d) \sin (a-c+b x-d x)+(2 b+d) \sin (a+c+(b+d) x))}{8 d^2} \] Input:

Integrate[Cos[a + b*x]^2*Sec[c + d*x]^3,x]
 

Output:

((-4*I)*d*ArcTan[E^(I*(c + d*x))] - ((2*I)*(2*b + d)*Hypergeometric2F1[1, 
1/2 - b/d, 3/2 - b/d, -E^((2*I)*(c + d*x))])/E^(I*(2*a - c + 2*b*x - d*x)) 
 + (2*I)*(2*b - d)*E^(I*(2*a + c + (2*b + d)*x))*Hypergeometric2F1[1, 1/2 
+ b/d, 3/2 + b/d, -E^((2*I)*(c + d*x))] + 2*Cos[a + b*x]*Sec[c + d*x]^2*(( 
2*b - d)*Sin[a - c + b*x - d*x] + (2*b + d)*Sin[a + c + (b + d)*x]))/(8*d^ 
2)
 

Rubi [F]

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.

\(\displaystyle \int \cos ^2(a+b x) \sec ^3(c+d x) \, dx\)

\(\Big \downarrow \) 7299

\(\displaystyle \int \cos ^2(a+b x) \sec ^3(c+d x)dx\)

Input:

Int[Cos[a + b*x]^2*Sec[c + d*x]^3,x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 7299

Int[u_, x_] :> CannotIntegrate[u, x]
 

Maple [F]

\[\int \cos \left (b x +a \right )^{2} \sec \left (d x +c \right )^{3}d x\]

Input:

int(cos(b*x+a)^2*sec(d*x+c)^3,x)
 

Output:

int(cos(b*x+a)^2*sec(d*x+c)^3,x)
 

Fricas [F]

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

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

Output:

integral(cos(b*x + a)^2*sec(d*x + c)^3, x)
 

Sympy [F(-1)]

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

integrate(cos(b*x+a)**2*sec(d*x+c)**3,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(cos(b*x+a)^2*sec(d*x+c)^3,x, algorithm="maxima")
 

Output:

1/4*((2*b - d)*cos(2*b*x + 2*a)*sin((4*b + d)*x + 4*a + c) - 2*d*cos(2*b*x 
 + 2*a)*sin((2*b + d)*x + 2*a + c) - (2*b - d)*cos((4*b + d)*x + 4*a + c)* 
sin(2*b*x + 2*a) + 2*d*cos((2*b + d)*x + 2*a + c)*sin(2*b*x + 2*a) + (2*b 
- d)*cos(3*d*x + 3*c)*sin(2*b*x + 2*a) + (2*b + d)*cos(d*x + c)*sin(2*b*x 
+ 2*a) - (2*b - d)*cos(2*b*x + 2*a)*sin(3*d*x + 3*c) - (2*b + d)*cos(2*b*x 
 + 2*a)*sin(d*x + c) - (2*(2*b + d)*sin(2*(b + d)*x + 2*a + 2*c) + (2*b + 
d)*sin(2*b*x + 2*a))*cos((4*b + 3*d)*x + 4*a + 3*c) - 2*(2*d*sin(2*(b + d) 
*x + 2*a + 2*c) + d*sin(2*b*x + 2*a))*cos((2*b + 3*d)*x + 2*a + 3*c) + ((2 
*b + d)*sin((4*b + 3*d)*x + 4*a + 3*c) + (2*b - d)*sin((4*b + d)*x + 4*a + 
 c) + 2*d*sin((2*b + 3*d)*x + 2*a + 3*c) - 2*d*sin((2*b + d)*x + 2*a + c) 
- (2*b - d)*sin(3*d*x + 3*c) - (2*b + d)*sin(d*x + c))*cos(2*(b + 2*d)*x + 
 2*a + 4*c) + 2*((2*b - d)*sin((4*b + d)*x + 4*a + c) - 2*d*sin((2*b + d)* 
x + 2*a + c) - (2*b - d)*sin(3*d*x + 3*c) - (2*b + d)*sin(d*x + c))*cos(2* 
(b + d)*x + 2*a + 2*c) + 4*(d^2*cos(2*(b + 2*d)*x + 2*a + 4*c)^2 + 4*d^2*c 
os(2*(b + d)*x + 2*a + 2*c)^2 + 4*d^2*cos(2*(b + d)*x + 2*a + 2*c)*cos(2*b 
*x + 2*a) + d^2*cos(2*b*x + 2*a)^2 + d^2*sin(2*(b + 2*d)*x + 2*a + 4*c)^2 
+ 4*d^2*sin(2*(b + d)*x + 2*a + 2*c)^2 + 4*d^2*sin(2*(b + d)*x + 2*a + 2*c 
)*sin(2*b*x + 2*a) + d^2*sin(2*b*x + 2*a)^2 + 2*(2*d^2*cos(2*(b + d)*x + 2 
*a + 2*c) + d^2*cos(2*b*x + 2*a))*cos(2*(b + 2*d)*x + 2*a + 4*c) + 2*(2*d^ 
2*sin(2*(b + d)*x + 2*a + 2*c) + d^2*sin(2*b*x + 2*a))*sin(2*(b + 2*d)*...
 

Giac [F]

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

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

Output:

integrate(cos(b*x + a)^2*sec(d*x + c)^3, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(cos(b*x+a)^2*sec(d*x+c)^3,x)
 

Output:

int(cos(a + b*x)**2*sec(c + d*x)**3,x)