Integrand size = 17, antiderivative size = 174 \[ \int \sec ^3(c+d x) \sin ^2(a+b x) \, dx=\frac {\text {arctanh}(\sin (c+d x))}{4 d}-\frac {2 i e^{-2 i a-2 i b x+3 i (c+d x)} \operatorname {Hypergeometric2F1}\left (3,\frac {3}{2}-\frac {b}{d},\frac {5}{2}-\frac {b}{d},-e^{2 i (c+d x)}\right )}{2 b-3 d}+\frac {2 i e^{2 i a+2 i b x+3 i (c+d x)} \operatorname {Hypergeometric2F1}\left (3,\frac {3}{2}+\frac {b}{d},\frac {5}{2}+\frac {b}{d},-e^{2 i (c+d x)}\right )}{2 b+3 d}+\frac {\sec (c+d x) \tan (c+d x)}{4 d} \] 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
Time = 4.64 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.36 \[ \int \sec ^3(c+d x) \sin ^2(a+b x) \, dx=\frac {i (2 b+d) e^{-i (2 a-c+2 b x-d x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}-\frac {b}{d},\frac {3}{2}-\frac {b}{d},-e^{2 i (c+d x)}\right )+i (-2 b+d) e^{i (2 a+c+(2 b+d) x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+\frac {b}{d},\frac {3}{2}+\frac {b}{d},-e^{2 i (c+d x)}\right )-d \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+d \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-((2 b-d) \cos (a-c+b x-d x)+(2 b+d) \cos (a+c+(b+d) x)) \sec ^2(c+d x) \sin (a+b x)}{4 d^2} \] Input:
Integrate[Sec[c + d*x]^3*Sin[a + b*x]^2,x]
Output:
((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)) + 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))] - d*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + d*Log[Cos[(c + d*x)/2] + Sin[ (c + d*x)/2]] - ((2*b - d)*Cos[a - c + b*x - d*x] + (2*b + d)*Cos[a + c + (b + d)*x])*Sec[c + d*x]^2*Sin[a + b*x])/(4*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.
\(\displaystyle \int \sin ^2(a+b x) \sec ^3(c+d x) \, dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \sin ^2(a+b x) \sec ^3(c+d x)dx\) |
Input:
Int[Sec[c + d*x]^3*Sin[a + b*x]^2,x]
Output:
$Aborted
\[\int \sec \left (d x +c \right )^{3} \sin \left (b x +a \right )^{2}d x\]
Input:
int(sec(d*x+c)^3*sin(b*x+a)^2,x)
Output:
int(sec(d*x+c)^3*sin(b*x+a)^2,x)
\[ \int \sec ^3(c+d x) \sin ^2(a+b x) \, dx=\int { \sec \left (d x + c\right )^{3} \sin \left (b x + a\right )^{2} \,d x } \] Input:
integrate(sec(d*x+c)^3*sin(b*x+a)^2,x, algorithm="fricas")
Output:
integral(-(cos(b*x + a)^2 - 1)*sec(d*x + c)^3, x)
Timed out. \[ \int \sec ^3(c+d x) \sin ^2(a+b x) \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**3*sin(b*x+a)**2,x)
Output:
Timed out
\[ \int \sec ^3(c+d x) \sin ^2(a+b x) \, dx=\int { \sec \left (d x + c\right )^{3} \sin \left (b x + a\right )^{2} \,d x } \] Input:
integrate(sec(d*x+c)^3*sin(b*x+a)^2,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* cos(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)...
\[ \int \sec ^3(c+d x) \sin ^2(a+b x) \, dx=\int { \sec \left (d x + c\right )^{3} \sin \left (b x + a\right )^{2} \,d x } \] Input:
integrate(sec(d*x+c)^3*sin(b*x+a)^2,x, algorithm="giac")
Output:
integrate(sec(d*x + c)^3*sin(b*x + a)^2, x)
Timed out. \[ \int \sec ^3(c+d x) \sin ^2(a+b x) \, dx=\int \frac {{\sin \left (a+b\,x\right )}^2}{{\cos \left (c+d\,x\right )}^3} \,d x \] Input:
int(sin(a + b*x)^2/cos(c + d*x)^3,x)
Output:
int(sin(a + b*x)^2/cos(c + d*x)^3, x)
\[ \int \sec ^3(c+d x) \sin ^2(a+b x) \, dx=\int \sec \left (d x +c \right )^{3} \sin \left (b x +a \right )^{2}d x \] Input:
int(sec(d*x+c)^3*sin(b*x+a)^2,x)
Output:
int(sec(c + d*x)**3*sin(a + b*x)**2,x)