\(\int \cos (a+b x) \sec (c+d x) \, dx\) [330]

Optimal result

Integrand size = 13, antiderivative size = 142 \[ \int \cos (a+b x) \sec (c+d x) \, dx=\frac {i e^{-i a-i b x+i (c+d x)} \operatorname {Hypergeometric2F1}\left (1,-\frac {b-d}{2 d},\frac {1}{2} \left (3-\frac {b}{d}\right ),-e^{2 i (c+d x)}\right )}{b-d}-\frac {i e^{i a+i b x+i (c+d x)} \operatorname {Hypergeometric2F1}\left (1,\frac {b+d}{2 d},\frac {1}{2} \left (3+\frac {b}{d}\right ),-e^{2 i (c+d x)}\right )}{b+d} \] Output:

I*exp(-I*a-I*b*x+I*(d*x+c))*hypergeom([1, -1/2*(b-d)/d],[3/2-1/2*b/d],-exp 
(2*I*(d*x+c)))/(b-d)-I*exp(I*a+I*b*x+I*(d*x+c))*hypergeom([1, 1/2*(b+d)/d] 
,[3/2+1/2*b/d],-exp(2*I*(d*x+c)))/(b+d)
 

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.93 \[ \int \cos (a+b x) \sec (c+d x) \, dx=\frac {i e^{-i (a-c+(b-d) x)} \left ((b+d) \operatorname {Hypergeometric2F1}\left (1,\frac {-b+d}{2 d},\frac {3}{2}-\frac {b}{2 d},-e^{2 i (c+d x)}\right )-(b-d) e^{2 i (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {b+d}{2 d},\frac {1}{2} \left (3+\frac {b}{d}\right ),-e^{2 i (c+d x)}\right )\right )}{(b-d) (b+d)} \] Input:

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

Output:

(I*((b + d)*Hypergeometric2F1[1, (-b + d)/(2*d), 3/2 - b/(2*d), -E^((2*I)* 
(c + d*x))] - (b - d)*E^((2*I)*(a + b*x))*Hypergeometric2F1[1, (b + d)/(2* 
d), (3 + b/d)/2, -E^((2*I)*(c + d*x))]))/((b - d)*(b + d)*E^(I*(a - c + (b 
 - d)*x)))
 

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.

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

\(\Big \downarrow \) 7299

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

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 7299

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \cos (a+b x) \sec (c+d x) \, dx=\frac {-4 \left (\int \frac {1}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}d x \right ) b d +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -\sin \left (b x +a \right ) d -b d x}{b d} \] Input:

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

Output:

( - 4*int(1/(tan((a + b*x)/2)**2*tan((c + d*x)/2)**2 - tan((a + b*x)/2)**2 
 + tan((c + d*x)/2)**2 - 1),x)*b*d + log(tan((c + d*x)/2) - 1)*b - log(tan 
((c + d*x)/2) + 1)*b - sin(a + b*x)*d - b*d*x)/(b*d)