Integrand size = 17, antiderivative size = 283 \[ \int \sec ^3(c+d x) \sin ^3(a+b x) \, dx=\frac {e^{-3 i a-3 i b x+3 i (c+d x)} \operatorname {Hypergeometric2F1}\left (3,\frac {3}{2} \left (1-\frac {b}{d}\right ),\frac {1}{2} \left (5-\frac {3 b}{d}\right ),-e^{2 i (c+d x)}\right )}{3 (b-d)}-\frac {3 e^{-i a-i b x+3 i (c+d x)} \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2} \left (3-\frac {b}{d}\right ),\frac {1}{2} \left (5-\frac {b}{d}\right ),-e^{2 i (c+d x)}\right )}{b-3 d}-\frac {3 e^{i a+i b x+3 i (c+d x)} \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2} \left (3+\frac {b}{d}\right ),\frac {1}{2} \left (5+\frac {b}{d}\right ),-e^{2 i (c+d x)}\right )}{b+3 d}+\frac {e^{3 i a+3 i b x+3 i (c+d x)} \operatorname {Hypergeometric2F1}\left (3,\frac {3 (b+d)}{2 d},\frac {1}{2} \left (5+\frac {3 b}{d}\right ),-e^{2 i (c+d x)}\right )}{3 (b+d)} \] Output:
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)))/(3*b-3*d)-3*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*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) +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)))/(3*b+3*d)
Time = 1.78 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.11 \[ \int \sec ^3(c+d x) \sin ^3(a+b x) \, dx=\frac {-4 (3 b+d) e^{-i (3 a-c+3 b x-d x)} \operatorname {Hypergeometric2F1}\left (1,\frac {-3 b+d}{2 d},\frac {3}{2}-\frac {3 b}{2 d},-e^{2 i (c+d x)}\right )+12 (b+d) e^{-i (a-c+(b-d) x)} \operatorname {Hypergeometric2F1}\left (1,\frac {-b+d}{2 d},\frac {3}{2}-\frac {b}{2 d},-e^{2 i (c+d x)}\right )+12 (b-d) e^{i (a+c+(b+d) x)} \operatorname {Hypergeometric2F1}\left (1,\frac {b+d}{2 d},\frac {1}{2} \left (3+\frac {b}{d}\right ),-e^{2 i (c+d x)}\right )-4 (3 b-d) e^{i (3 a+c+(3 b+d) x)} \operatorname {Hypergeometric2F1}\left (1,\frac {3 b+d}{2 d},\frac {3 (b+d)}{2 d},-e^{2 i (c+d x)}\right )-8 ((3 b-d) \cos (a-c+b x-d x)+(3 b+d) \cos (a+c+(b+d) x)) \sec ^2(c+d x) \sin ^2(a+b x)}{32 d^2} \] Input:
Integrate[Sec[c + d*x]^3*Sin[a + b*x]^3,x]
Output:
((-4*(3*b + d)*Hypergeometric2F1[1, (-3*b + d)/(2*d), 3/2 - (3*b)/(2*d), - E^((2*I)*(c + d*x))])/E^(I*(3*a - c + 3*b*x - d*x)) + (12*(b + d)*Hypergeo metric2F1[1, (-b + d)/(2*d), 3/2 - b/(2*d), -E^((2*I)*(c + d*x))])/E^(I*(a - c + (b - d)*x)) + 12*(b - d)*E^(I*(a + c + (b + d)*x))*Hypergeometric2F 1[1, (b + d)/(2*d), (3 + b/d)/2, -E^((2*I)*(c + d*x))] - 4*(3*b - d)*E^(I* (3*a + c + (3*b + d)*x))*Hypergeometric2F1[1, (3*b + d)/(2*d), (3*(b + d)) /(2*d), -E^((2*I)*(c + d*x))] - 8*((3*b - d)*Cos[a - c + b*x - d*x] + (3*b + d)*Cos[a + c + (b + d)*x])*Sec[c + d*x]^2*Sin[a + b*x]^2)/(32*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.
\(\displaystyle \int \sin ^3(a+b x) \sec ^3(c+d x) \, dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \sin ^3(a+b x) \sec ^3(c+d x)dx\) |
Input:
Int[Sec[c + d*x]^3*Sin[a + b*x]^3,x]
Output:
$Aborted
\[\int \sec \left (d x +c \right )^{3} \sin \left (b x +a \right )^{3}d x\]
Input:
int(sec(d*x+c)^3*sin(b*x+a)^3,x)
Output:
int(sec(d*x+c)^3*sin(b*x+a)^3,x)
\[ \int \sec ^3(c+d x) \sin ^3(a+b x) \, dx=\int { \sec \left (d x + c\right )^{3} \sin \left (b x + a\right )^{3} \,d x } \] Input:
integrate(sec(d*x+c)^3*sin(b*x+a)^3,x, algorithm="fricas")
Output:
integral(-(cos(b*x + a)^2 - 1)*sec(d*x + c)^3*sin(b*x + a), x)
Timed out. \[ \int \sec ^3(c+d x) \sin ^3(a+b x) \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**3*sin(b*x+a)**3,x)
Output:
Timed out
\[ \int \sec ^3(c+d x) \sin ^3(a+b x) \, dx=\int { \sec \left (d x + c\right )^{3} \sin \left (b x + a\right )^{3} \,d x } \] Input:
integrate(sec(d*x+c)^3*sin(b*x+a)^3,x, algorithm="maxima")
Output:
1/8*((3*b - d)*cos((6*b + d)*x + 6*a + c)*cos(3*b*x + 3*a) - 3*(b - d)*cos ((4*b + d)*x + 4*a + c)*cos(3*b*x + 3*a) - 3*(b + d)*cos((2*b + d)*x + 2*a + c)*cos(3*b*x + 3*a) + (3*b - d)*cos(3*b*x + 3*a)*cos(3*d*x + 3*c) + (3* b + d)*cos(3*b*x + 3*a)*cos(d*x + c) + (3*b - d)*sin((6*b + d)*x + 6*a + c )*sin(3*b*x + 3*a) - 3*(b - d)*sin((4*b + d)*x + 4*a + c)*sin(3*b*x + 3*a) - 3*(b + d)*sin((2*b + d)*x + 2*a + c)*sin(3*b*x + 3*a) + (3*b - d)*sin(3 *b*x + 3*a)*sin(3*d*x + 3*c) + (3*b + d)*sin(3*b*x + 3*a)*sin(d*x + c) - 3 *(2*(b + d)*cos((3*b + 2*d)*x + 3*a + 2*c) + (b + d)*cos(3*b*x + 3*a))*cos ((4*b + 3*d)*x + 4*a + 3*c) + ((3*b - d)*cos((6*b + d)*x + 6*a + c) - 3*(b + d)*cos((4*b + 3*d)*x + 4*a + 3*c) - 3*(b - d)*cos((4*b + d)*x + 4*a + c ) - 3*(b - d)*cos((2*b + 3*d)*x + 2*a + 3*c) + (3*b + d)*cos(3*(2*b + d)*x + 6*a + 3*c) - 3*(b + d)*cos((2*b + d)*x + 2*a + c) + (3*b - d)*cos(3*d*x + 3*c) + (3*b + d)*cos(d*x + c))*cos((3*b + 4*d)*x + 3*a + 4*c) + 2*((3*b - d)*cos((6*b + d)*x + 6*a + c) - 3*(b - d)*cos((4*b + d)*x + 4*a + c) - 3*(b + d)*cos((2*b + d)*x + 2*a + c) + (3*b - d)*cos(3*d*x + 3*c) + (3*b + d)*cos(d*x + c))*cos((3*b + 2*d)*x + 3*a + 2*c) - 3*(2*(b - d)*cos((3*b + 2*d)*x + 3*a + 2*c) + (b - d)*cos(3*b*x + 3*a))*cos((2*b + 3*d)*x + 2*a + 3*c) + (2*(3*b + d)*cos((3*b + 2*d)*x + 3*a + 2*c) + (3*b + d)*cos(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)...
\[ \int \sec ^3(c+d x) \sin ^3(a+b x) \, dx=\int { \sec \left (d x + c\right )^{3} \sin \left (b x + a\right )^{3} \,d x } \] Input:
integrate(sec(d*x+c)^3*sin(b*x+a)^3,x, algorithm="giac")
Output:
integrate(sec(d*x + c)^3*sin(b*x + a)^3, x)
Timed out. \[ \int \sec ^3(c+d x) \sin ^3(a+b x) \, dx=\int \frac {{\sin \left (a+b\,x\right )}^3}{{\cos \left (c+d\,x\right )}^3} \,d x \] Input:
int(sin(a + b*x)^3/cos(c + d*x)^3,x)
Output:
int(sin(a + b*x)^3/cos(c + d*x)^3, x)
\[ \int \sec ^3(c+d x) \sin ^3(a+b x) \, dx=\int \sec \left (d x +c \right )^{3} \sin \left (b x +a \right )^{3}d x \] Input:
int(sec(d*x+c)^3*sin(b*x+a)^3,x)
Output:
int(sec(c + d*x)**3*sin(a + b*x)**3,x)