\(\int \sin ^2(a+b x) \tan ^2(c+d x) \, dx\) [258]

Optimal result

Integrand size = 17, antiderivative size = 168 \[ \int \sin ^2(a+b x) \tan ^2(c+d x) \, dx=\frac {i e^{-2 i (a+b x)} \operatorname {Hypergeometric2F1}\left (1,-\frac {b}{d},1-\frac {b}{d},-e^{2 i (c+d x)}\right )}{2 d}+\frac {i e^{2 i (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {b}{d},\frac {b+d}{d},-e^{2 i (c+d x)}\right )}{2 d}+\frac {\sec (c) \sec (c+d x) \sin (d x) \sin ^2(a+b x)}{d}+\frac {-2 b x+\sin (2 (a+b x))}{4 b}+\frac {\cos (2 a+2 b x) \sec ^2(c)}{2 i d-2 d \tan (c)} \] Output:

1/2*I*hypergeom([1, -b/d],[1-b/d],-exp(2*I*(d*x+c)))/d/exp(2*I*(b*x+a))+1/ 
2*I*exp(2*I*(b*x+a))*hypergeom([1, b/d],[(b+d)/d],-exp(2*I*(d*x+c)))/d+sec 
(c)*sec(d*x+c)*sin(d*x)*sin(b*x+a)^2/d+1/4*(-2*b*x+sin(2*b*x+2*a))/b+cos(2 
*b*x+2*a)*sec(c)^2/(2*I*d-2*d*tan(c))
 

Mathematica [A] (verified)

Time = 1.66 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.06 \[ \int \sin ^2(a+b x) \tan ^2(c+d x) \, dx=\frac {i e^{-2 i (a+b x)} \left (-1-e^{4 i (a+b x)}+\left (1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (1,-\frac {b}{d},1-\frac {b}{d},-e^{2 i (c+d x)}\right )+e^{4 i (a+b x)} \left (1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {b}{d},\frac {b+d}{d},-e^{2 i (c+d x)}\right )\right )}{2 d \left (1+e^{2 i c}\right )}+\frac {\sec (c) \sec (c+d x) \sin (d x) \sin ^2(a+b x)}{d}+\frac {-2 b x+\sin (2 (a+b x))}{4 b} \] Input:

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

Output:

((I/2)*(-1 - E^((4*I)*(a + b*x)) + (1 + E^((2*I)*c))*Hypergeometric2F1[1, 
-(b/d), 1 - b/d, -E^((2*I)*(c + d*x))] + E^((4*I)*(a + b*x))*(1 + E^((2*I) 
*c))*Hypergeometric2F1[1, b/d, (b + d)/d, -E^((2*I)*(c + d*x))]))/(d*E^((2 
*I)*(a + b*x))*(1 + E^((2*I)*c))) + (Sec[c]*Sec[c + d*x]*Sin[d*x]*Sin[a + 
b*x]^2)/d + (-2*b*x + Sin[2*(a + b*x)])/(4*b)
 

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.

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

\(\Big \downarrow \) 7299

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

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 7299

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral(sin(a + b*x)**2*tan(c + d*x)**2, x)
 

Maxima [F]

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

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

Output:

-1/8*(4*b*d*x*cos(2*(b + d)*x + 2*a + 2*c)^2 + 4*b*d*x*cos(2*b*x + 2*a)^2 
+ 4*b*d*x*sin(2*(b + d)*x + 2*a + 2*c)^2 + 4*b*d*x*sin(2*b*x + 2*a)^2 - (4 
*b + d)*cos(2*b*x + 2*a)*sin(4*b*x + 4*a) + (4*b + d)*cos(4*b*x + 4*a)*sin 
(2*b*x + 2*a) - d*cos(2*d*x + 2*c)*sin(2*b*x + 2*a) + d*cos(2*b*x + 2*a)*s 
in(2*d*x + 2*c) + (d*sin(2*(b + d)*x + 2*a + 2*c) + d*sin(2*b*x + 2*a))*co 
s(2*(2*b + d)*x + 4*a + 2*c) + (8*b*d*x*cos(2*b*x + 2*a) - (4*b + d)*sin(4 
*b*x + 4*a) + 8*b*sin(2*b*x + 2*a) + d*sin(2*d*x + 2*c))*cos(2*(b + d)*x + 
 2*a + 2*c) + 8*(b*d*cos(2*(b + d)*x + 2*a + 2*c)^2 + 2*b*d*cos(2*(b + d)* 
x + 2*a + 2*c)*cos(2*b*x + 2*a) + b*d*cos(2*b*x + 2*a)^2 + b*d*sin(2*(b + 
d)*x + 2*a + 2*c)^2 + 2*b*d*sin(2*(b + d)*x + 2*a + 2*c)*sin(2*b*x + 2*a) 
+ b*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*c 
os(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) - (d*cos(2*(b + d)*x + 2*a + 2*c) + d*cos(2*b*x + 2*a)) 
*sin(2*(2*b + d)*x + 4*a + 2*c) + (8*b*d*x*sin(2*b*x + 2*a) + (4*b + d)*co 
s(4*b*x + 4*a) - 8*b*cos(2*b*x + 2*a) - d*cos(2*d*x + 2*c) + 4*b - d)*sin( 
2*(b + d)*x + 2*a + 2*c) + (4*b - d)*sin(2*b*x + 2*a))/(b*d*cos(2*(b + ...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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