\(\int \sec ^2(c+d x) \sin ^2(a+b x) \, dx\) [195]

Optimal result

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

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(295\) vs. \(2(144)=288\).

Time = 2.74 (sec) , antiderivative size = 295, normalized size of antiderivative = 2.05 \[ \int \sec ^2(c+d x) \sin ^2(a+b x) \, dx=\frac {-\frac {i e^{-2 i (a-c)} \left (\frac {b e^{-2 i (b-d) x} \operatorname {Hypergeometric2F1}\left (1,1-\frac {b}{d},2-\frac {b}{d},-e^{2 i (c+d x)}\right )}{b-d}-e^{-2 i b x} \operatorname {Hypergeometric2F1}\left (1,-\frac {b}{d},1-\frac {b}{d},-e^{2 i (c+d x)}\right )\right )}{1+e^{2 i c}}+\frac {i e^{2 i (a+c)} \left ((b+d) e^{2 i b x} \operatorname {Hypergeometric2F1}\left (1,\frac {b}{d},\frac {b+d}{d},-e^{2 i (c+d x)}\right )-b e^{2 i (b+d) x} \operatorname {Hypergeometric2F1}\left (1,\frac {b+d}{d},2+\frac {b}{d},-e^{2 i (c+d x)}\right )\right )}{(b+d) \left (1+e^{2 i c}\right )}+\sec (c) \sec (c+d x) \sin (d x)-\cos (2 a) \cos (2 b x) \sec (c) \sec (c+d x) \sin (d x)+\sec (c) \sec (c+d x) \sin (2 a) \sin (2 b x) \sin (d x)}{2 d} \] Input:

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

Output:

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

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

\(\Big \downarrow \) 7299

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

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 7299

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(-(cos(b*x + a)^2 - 1)*sec(d*x + c)^2, x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate(sec(d*x+c)^2*sin(b*x+a)^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)*c 
os(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) - (co 
s(4*b*x + 4*a) - 2*cos(2*b*x + 2*a) + 1)*sin(2*(b + d)*x + 2*a + 2*c) + co 
s(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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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