Integrand size = 17, antiderivative size = 291 \[ \int \cos ^3(a+b x) \sec ^3(c+d x) \, dx=\frac {i e^{-3 i a-3 i b x+3 i (c+d x)} \operatorname {Hypergeometric2F1}\left (3,\frac {3}{2} \left (1-\frac {b}{d}\right ),\frac {1}{2} \left (5-\frac {3 b}{d}\right ),-e^{2 i (c+d x)}\right )}{3 (b-d)}+\frac {3 i e^{-i a-i b x+3 i (c+d x)} \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2} \left (3-\frac {b}{d}\right ),\frac {1}{2} \left (5-\frac {b}{d}\right ),-e^{2 i (c+d x)}\right )}{b-3 d}-\frac {3 i e^{i a+i b x+3 i (c+d x)} \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2} \left (3+\frac {b}{d}\right ),\frac {1}{2} \left (5+\frac {b}{d}\right ),-e^{2 i (c+d x)}\right )}{b+3 d}-\frac {i e^{3 i a+3 i b x+3 i (c+d x)} \operatorname {Hypergeometric2F1}\left (3,\frac {3 (b+d)}{2 d},\frac {1}{2} \left (5+\frac {3 b}{d}\right ),-e^{2 i (c+d x)}\right )}{3 (b+d)} \] Output:
1/3*I*exp(-3*I*a-3*I*b*x+3*I*(d*x+c))*hypergeom([3, 3/2-3/2*b/d],[5/2-3/2* b/d],-exp(2*I*(d*x+c)))/(b-d)+3*I*exp(-I*a-I*b*x+3*I*(d*x+c))*hypergeom([3 , 3/2-1/2*b/d],[5/2-1/2*b/d],-exp(2*I*(d*x+c)))/(b-3*d)-3*I*exp(I*a+I*b*x+ 3*I*(d*x+c))*hypergeom([3, 3/2+1/2*b/d],[5/2+1/2*b/d],-exp(2*I*(d*x+c)))/( b+3*d)-1/3*I*exp(3*I*a+3*I*b*x+3*I*(d*x+c))*hypergeom([3, 3/2*(b+d)/d],[5/ 2+3/2*b/d],-exp(2*I*(d*x+c)))/(b+d)
Time = 1.77 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.03 \[ \int \cos ^3(a+b x) \sec ^3(c+d x) \, dx=\frac {-2 i e^{-i (3 a-c+3 b x-d x)} \left ((3 b+d) \operatorname {Hypergeometric2F1}\left (1,\frac {-3 b+d}{2 d},\frac {3}{2}-\frac {3 b}{2 d},-e^{2 i (c+d x)}\right )+3 (b+d) e^{2 i (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {-b+d}{2 d},\frac {3}{2}-\frac {b}{2 d},-e^{2 i (c+d x)}\right )-3 (b-d) e^{4 i (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {b+d}{2 d},\frac {1}{2} \left (3+\frac {b}{d}\right ),-e^{2 i (c+d x)}\right )-(3 b-d) e^{6 i (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {3 b+d}{2 d},\frac {3 (b+d)}{2 d},-e^{2 i (c+d x)}\right )\right )+4 \cos ^2(a+b x) \sec ^2(c+d x) ((3 b-d) \sin (a-c+b x-d x)+(3 b+d) \sin (a+c+(b+d) x))}{16 d^2} \] Input:
Integrate[Cos[a + b*x]^3*Sec[c + d*x]^3,x]
Output:
(((-2*I)*((3*b + d)*Hypergeometric2F1[1, (-3*b + d)/(2*d), 3/2 - (3*b)/(2* d), -E^((2*I)*(c + d*x))] + 3*(b + d)*E^((2*I)*(a + b*x))*Hypergeometric2F 1[1, (-b + d)/(2*d), 3/2 - b/(2*d), -E^((2*I)*(c + d*x))] - 3*(b - d)*E^(( 4*I)*(a + b*x))*Hypergeometric2F1[1, (b + d)/(2*d), (3 + b/d)/2, -E^((2*I) *(c + d*x))] - (3*b - d)*E^((6*I)*(a + b*x))*Hypergeometric2F1[1, (3*b + d )/(2*d), (3*(b + d))/(2*d), -E^((2*I)*(c + d*x))]))/E^(I*(3*a - c + 3*b*x - d*x)) + 4*Cos[a + b*x]^2*Sec[c + d*x]^2*((3*b - d)*Sin[a - c + b*x - d*x ] + (3*b + d)*Sin[a + c + (b + d)*x]))/(16*d^2)
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.
\(\displaystyle \int \cos ^3(a+b x) \sec ^3(c+d x) \, dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \cos ^3(a+b x) \sec ^3(c+d x)dx\) |
Input:
Int[Cos[a + b*x]^3*Sec[c + d*x]^3,x]
Output:
$Aborted
\[\int \cos \left (b x +a \right )^{3} \sec \left (d x +c \right )^{3}d x\]
Input:
int(cos(b*x+a)^3*sec(d*x+c)^3,x)
Output:
int(cos(b*x+a)^3*sec(d*x+c)^3,x)
\[ \int \cos ^3(a+b x) \sec ^3(c+d x) \, dx=\int { \cos \left (b x + a\right )^{3} \sec \left (d x + c\right )^{3} \,d x } \] Input:
integrate(cos(b*x+a)^3*sec(d*x+c)^3,x, algorithm="fricas")
Output:
integral(cos(b*x + a)^3*sec(d*x + c)^3, x)
Timed out. \[ \int \cos ^3(a+b x) \sec ^3(c+d x) \, dx=\text {Timed out} \] Input:
integrate(cos(b*x+a)**3*sec(d*x+c)**3,x)
Output:
Timed out
\[ \int \cos ^3(a+b x) \sec ^3(c+d x) \, dx=\int { \cos \left (b x + a\right )^{3} \sec \left (d x + c\right )^{3} \,d x } \] Input:
integrate(cos(b*x+a)^3*sec(d*x+c)^3,x, algorithm="maxima")
Output:
1/8*((3*b - d)*cos(3*b*x + 3*a)*sin((6*b + d)*x + 6*a + c) + 3*(b - d)*cos (3*b*x + 3*a)*sin((4*b + d)*x + 4*a + c) - 3*(b + d)*cos(3*b*x + 3*a)*sin( (2*b + d)*x + 2*a + c) - (3*b - d)*cos((6*b + d)*x + 6*a + c)*sin(3*b*x + 3*a) - 3*(b - d)*cos((4*b + d)*x + 4*a + c)*sin(3*b*x + 3*a) + 3*(b + d)*c os((2*b + d)*x + 2*a + c)*sin(3*b*x + 3*a) + (3*b - d)*cos(3*d*x + 3*c)*si n(3*b*x + 3*a) + (3*b + d)*cos(d*x + c)*sin(3*b*x + 3*a) - (3*b - d)*cos(3 *b*x + 3*a)*sin(3*d*x + 3*c) - (3*b + d)*cos(3*b*x + 3*a)*sin(d*x + c) - 3 *(2*(b + d)*sin((3*b + 2*d)*x + 3*a + 2*c) + (b + d)*sin(3*b*x + 3*a))*cos ((4*b + 3*d)*x + 4*a + 3*c) + ((3*b - d)*sin((6*b + d)*x + 6*a + c) + 3*(b + d)*sin((4*b + 3*d)*x + 4*a + 3*c) + 3*(b - d)*sin((4*b + d)*x + 4*a + c ) - 3*(b - d)*sin((2*b + 3*d)*x + 2*a + 3*c) + (3*b + d)*sin(3*(2*b + d)*x + 6*a + 3*c) - 3*(b + d)*sin((2*b + d)*x + 2*a + c) - (3*b - d)*sin(3*d*x + 3*c) - (3*b + d)*sin(d*x + c))*cos((3*b + 4*d)*x + 3*a + 4*c) + 2*((3*b - d)*sin((6*b + d)*x + 6*a + c) + 3*(b - d)*sin((4*b + d)*x + 4*a + c) - 3*(b + d)*sin((2*b + d)*x + 2*a + c) - (3*b - d)*sin(3*d*x + 3*c) - (3*b + d)*sin(d*x + c))*cos((3*b + 2*d)*x + 3*a + 2*c) + 3*(2*(b - d)*sin((3*b + 2*d)*x + 3*a + 2*c) + (b - d)*sin(3*b*x + 3*a))*cos((2*b + 3*d)*x + 2*a + 3*c) - (2*(3*b + d)*sin((3*b + 2*d)*x + 3*a + 2*c) + (3*b + d)*sin(3*b*x + 3*a))*cos(3*(2*b + d)*x + 6*a + 3*c) + 8*(d^2*cos((3*b + 4*d)*x + 3*a + 4*c)^2 + 4*d^2*cos((3*b + 2*d)*x + 3*a + 2*c)^2 + 4*d^2*cos((3*b + 2*d)...
\[ \int \cos ^3(a+b x) \sec ^3(c+d x) \, dx=\int { \cos \left (b x + a\right )^{3} \sec \left (d x + c\right )^{3} \,d x } \] Input:
integrate(cos(b*x+a)^3*sec(d*x+c)^3,x, algorithm="giac")
Output:
integrate(cos(b*x + a)^3*sec(d*x + c)^3, x)
Timed out. \[ \int \cos ^3(a+b x) \sec ^3(c+d x) \, dx=\int \frac {{\cos \left (a+b\,x\right )}^3}{{\cos \left (c+d\,x\right )}^3} \,d x \] Input:
int(cos(a + b*x)^3/cos(c + d*x)^3,x)
Output:
int(cos(a + b*x)^3/cos(c + d*x)^3, x)
\[ \int \cos ^3(a+b x) \sec ^3(c+d x) \, dx=\int \cos \left (b x +a \right )^{3} \sec \left (d x +c \right )^{3}d x \] Input:
int(cos(b*x+a)^3*sec(d*x+c)^3,x)
Output:
int(cos(a + b*x)**3*sec(c + d*x)**3,x)