\(\int \cos ^2(a+b x) \sec ^2(c+d x) \, dx\) [337]

Optimal result

Integrand size = 17, antiderivative size = 144 \[ \int \cos ^2(a+b x) \sec ^2(c+d x) \, dx=\frac {i e^{-2 i a-2 i b x+2 i (c+d x)} \operatorname {Hypergeometric2F1}\left (2,1-\frac {b}{d},2-\frac {b}{d},-e^{2 i (c+d x)}\right )}{2 (b-d)}-\frac {i e^{2 i a+2 i b x+2 i (c+d x)} \operatorname {Hypergeometric2F1}\left (2,\frac {b+d}{d},2+\frac {b}{d},-e^{2 i (c+d x)}\right )}{2 (b+d)}+\frac {\tan (c+d x)}{2 d} \] Output:

1/2*I*exp(-2*I*a-2*I*b*x+2*I*(d*x+c))*hypergeom([2, 1-b/d],[2-b/d],-exp(2* 
I*(d*x+c)))/(b-d)-1/2*I*exp(2*I*a+2*I*b*x+2*I*(d*x+c))*hypergeom([2, (b+d) 
/d],[2+b/d],-exp(2*I*(d*x+c)))/(b+d)+1/2*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 1.46 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.10 \[ \int \cos ^2(a+b x) \sec ^2(c+d x) \, dx=-\frac {i e^{-2 i (a+b x)} \left (-1-e^{4 i (a+b x)}+\left (1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (1,-\frac {b}{d},1-\frac {b}{d},-e^{2 i (c+d x)}\right )+e^{4 i (a+b x)} \left (1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {b}{d},\frac {b+d}{d},-e^{2 i (c+d x)}\right )\right )}{2 d \left (1+e^{2 i c}\right )}+\frac {\cos ^2(a+b x) \sec (c) \sec (c+d x) \sin (d x)}{d} \] Input:

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

Output:

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

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.

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

\(\Big \downarrow \) 7299

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

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 7299

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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