\(\int \sin ^2(a+b x) \tan ^3(c+b x) \, dx\) [237]

Optimal result

Integrand size = 17, antiderivative size = 1 \[ \int \sin ^2(a+b x) \tan ^3(c+b x) \, dx=0 \] Output:

0
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 3 vs. order 1 in optimal.

Time = 1.62 (sec) , antiderivative size = 239, normalized size of antiderivative = 239.00 \[ \int \sin ^2(a+b x) \tan ^3(c+b x) \, dx=\frac {-\cos (2 a) \cos (2 b x)+2 \log (\cos (c+b x))+2 \cos (2 a-2 c-b x) \sec (c) \sec (c+b x)-2 \cos (2 a-2 c+b x) \sec (c) \sec (c+b x)+2 \cos ^2(a-c) \sec ^2(c+b x)+6 b x \cos (a) \sin (a)-6 b x \cos (a) \cos ^2(c) \sin (a)-6 b x \sin (2 (a-c))+9 b x \cos ^2(a) \cos (c) \sin (c)-9 b x \cos (c) \sin ^2(a) \sin (c)+9 b x \sin (2 a) \sin ^2(c)+\sin (2 a) \sin (2 b x)-3 b x \cos ^2(a) \tan (c)+3 b x \sin ^2(a) \tan (c)-3 b x \cos ^2(a) \sin ^2(c) \tan (c)+3 b x \sin ^2(a) \sin ^2(c) \tan (c)+6 \cos (2 (a-c)) (\log (\cos (c+b x))-b x \tan (c))}{4 b} \] Input:

Integrate[Sin[a + b*x]^2*Tan[c + b*x]^3,x]
 

Output:

(-(Cos[2*a]*Cos[2*b*x]) + 2*Log[Cos[c + b*x]] + 2*Cos[2*a - 2*c - b*x]*Sec 
[c]*Sec[c + b*x] - 2*Cos[2*a - 2*c + b*x]*Sec[c]*Sec[c + b*x] + 2*Cos[a - 
c]^2*Sec[c + b*x]^2 + 6*b*x*Cos[a]*Sin[a] - 6*b*x*Cos[a]*Cos[c]^2*Sin[a] - 
 6*b*x*Sin[2*(a - c)] + 9*b*x*Cos[a]^2*Cos[c]*Sin[c] - 9*b*x*Cos[c]*Sin[a] 
^2*Sin[c] + 9*b*x*Sin[2*a]*Sin[c]^2 + Sin[2*a]*Sin[2*b*x] - 3*b*x*Cos[a]^2 
*Tan[c] + 3*b*x*Sin[a]^2*Tan[c] - 3*b*x*Cos[a]^2*Sin[c]^2*Tan[c] + 3*b*x*S 
in[a]^2*Sin[c]^2*Tan[c] + 6*Cos[2*(a - c)]*(Log[Cos[c + b*x]] - b*x*Tan[c] 
))/(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.

\(\displaystyle \int \sin ^2(a+b x) \tan ^3(b x+c) \, dx\)

\(\Big \downarrow \) 7299

\(\displaystyle \int \sin ^2(a+b x) \tan ^3(b x+c)dx\)

Input:

Int[Sin[a + b*x]^2*Tan[c + b*x]^3,x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 7299

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 3 vs. order 1.

Time = 0.40 (sec) , antiderivative size = 226, normalized size of antiderivative = 226.00

method	result	size
risch	\(\frac {3 i x \,{\mathrm e}^{2 i \left (a -c \right )}}{2}-\frac {i x}{2}-\frac {{\mathrm e}^{2 i \left (b x +a \right )}}{8 b}-\frac {{\mathrm e}^{-2 i \left (b x +a \right )}}{8 b}-3 i \cos \left (2 a -2 c \right ) x -\frac {3 i \cos \left (2 a -2 c \right ) a}{b}-\frac {i a}{b}+\frac {3 \,{\mathrm e}^{2 i \left (b x +3 a \right )}+2 \,{\mathrm e}^{2 i \left (b x +2 a +c \right )}-{\mathrm e}^{2 i \left (b x +a +2 c \right )}+2 \,{\mathrm e}^{2 i \left (3 a -c \right )}-2 \,{\mathrm e}^{2 i \left (a +c \right )}}{2 b \left ({\mathrm e}^{2 i \left (b x +a +c \right )}+{\mathrm e}^{2 i a}\right )^{2}}+\frac {\ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+{\mathrm e}^{2 i \left (a -c \right )}\right )}{2 b}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+{\mathrm e}^{2 i \left (a -c \right )}\right ) \cos \left (2 a -2 c \right )}{2 b}\)	\(226\)

Input:

int(sin(b*x+a)^2*tan(b*x+c)^3,x,method=_RETURNVERBOSE)
 

Output:

3/2*I*x*exp(2*I*(a-c))-1/2*I*x-1/8/b*exp(2*I*(b*x+a))-1/8/b*exp(-2*I*(b*x+ 
a))-3*I*cos(2*a-2*c)*x-3*I/b*cos(2*a-2*c)*a-I/b*a+1/2/b/(exp(2*I*(b*x+a+c) 
)+exp(2*I*a))^2*(3*exp(2*I*(b*x+3*a))+2*exp(2*I*(b*x+2*a+c))-exp(2*I*(b*x+ 
a+2*c))+2*exp(2*I*(3*a-c))-2*exp(2*I*(a+c)))+1/2*ln(exp(2*I*(b*x+a))+exp(2 
*I*(a-c)))/b+3/2*ln(exp(2*I*(b*x+a))+exp(2*I*(a-c)))/b*cos(2*a-2*c)
 

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 3 vs. order 1.

Time = 0.10 (sec) , antiderivative size = 412, normalized size of antiderivative = 412.00 \[ \int \sin ^2(a+b x) \tan ^3(c+b x) \, dx=-\frac {4 \, \cos \left (b x + a\right )^{4} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, {\left (4 \, \cos \left (-2 \, a + 2 \, c\right )^{2} + 2 \, \cos \left (-2 \, a + 2 \, c\right ) - 5\right )} \cos \left (b x + a\right )^{2} + 4 \, \cos \left (-2 \, a + 2 \, c\right )^{2} + {\left (2 \, {\left (3 \, \cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - 2 \, {\left (3 \, \cos \left (-2 \, a + 2 \, c\right )^{2} + \cos \left (-2 \, a + 2 \, c\right )\right )} \cos \left (b x + a\right )^{2} + 3 \, \cos \left (-2 \, a + 2 \, c\right )^{2} - 2 \, \cos \left (-2 \, a + 2 \, c\right ) - 1\right )} \log \left (\frac {2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) + 1}{\cos \left (-2 \, a + 2 \, c\right ) + 1}\right ) - 2 \, {\left (6 \, {\left (b x \cos \left (-2 \, a + 2 \, c\right )^{2} - b x\right )} \cos \left (b x + a\right ) + {\left (2 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \cos \left (b x + a\right )\right )} \sin \left (-2 \, a + 2 \, c\right )\right )} \sin \left (b x + a\right ) - 6 \, {\left (2 \, b x \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - b x \cos \left (-2 \, a + 2 \, c\right ) + b x\right )} \sin \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) - 7}{4 \, {\left (2 \, b \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, b \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - b \cos \left (-2 \, a + 2 \, c\right ) + b\right )}} \] Input:

integrate(sin(b*x+a)^2*tan(b*x+c)^3,x, algorithm="fricas")
 

Output:

-1/4*(4*cos(b*x + a)^4*cos(-2*a + 2*c) - 2*(4*cos(-2*a + 2*c)^2 + 2*cos(-2 
*a + 2*c) - 5)*cos(b*x + a)^2 + 4*cos(-2*a + 2*c)^2 + (2*(3*cos(-2*a + 2*c 
) + 1)*cos(b*x + a)*sin(b*x + a)*sin(-2*a + 2*c) - 2*(3*cos(-2*a + 2*c)^2 
+ cos(-2*a + 2*c))*cos(b*x + a)^2 + 3*cos(-2*a + 2*c)^2 - 2*cos(-2*a + 2*c 
) - 1)*log((2*cos(b*x + a)^2*cos(-2*a + 2*c) - 2*cos(b*x + a)*sin(b*x + a) 
*sin(-2*a + 2*c) - cos(-2*a + 2*c) + 1)/(cos(-2*a + 2*c) + 1)) - 2*(6*(b*x 
*cos(-2*a + 2*c)^2 - b*x)*cos(b*x + a) + (2*cos(b*x + a)^3 - (4*cos(-2*a + 
 2*c) + 1)*cos(b*x + a))*sin(-2*a + 2*c))*sin(b*x + a) - 6*(2*b*x*cos(b*x 
+ a)^2*cos(-2*a + 2*c) - b*x*cos(-2*a + 2*c) + b*x)*sin(-2*a + 2*c) - cos( 
-2*a + 2*c) - 7)/(2*b*cos(b*x + a)^2*cos(-2*a + 2*c) - 2*b*cos(b*x + a)*si 
n(b*x + a)*sin(-2*a + 2*c) - b*cos(-2*a + 2*c) + b)
 

Sympy [F]

\[ \int \sin ^2(a+b x) \tan ^3(c+b x) \, dx=\int \sin ^{2}{\left (a + b x \right )} \tan ^{3}{\left (b x + c \right )}\, dx \] Input:

integrate(sin(b*x+a)**2*tan(b*x+c)**3,x)
 

Output:

Integral(sin(a + b*x)**2*tan(b*x + c)**3, x)
 

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 3 vs. order 1.

Time = 0.12 (sec) , antiderivative size = 1755, normalized size of antiderivative = 1755.00 \[ \int \sin ^2(a+b x) \tan ^3(c+b x) \, dx=\text {Too large to display} \] Input:

integrate(sin(b*x+a)^2*tan(b*x+c)^3,x, algorithm="maxima")
 

Output:

-1/8*((cos(6*b*x + 2*a + 6*c) + 2*cos(4*b*x + 2*a + 4*c) + cos(2*b*x + 2*a 
 + 2*c))*cos(8*b*x + 4*a + 6*c) + 2*(2*cos(4*b*x + 2*a + 4*c) + cos(2*b*x 
+ 2*a + 2*c))*cos(6*b*x + 4*a + 4*c) - (24*b*x*sin(4*b*x + 6*c) + 12*b*x*s 
in(2*b*x + 4*c) - 2*cos(6*b*x + 4*a + 4*c) + 11*cos(4*b*x + 4*a + 2*c) + 8 
*cos(4*b*x + 2*a + 4*c) - 5*cos(4*b*x + 6*c) + 8*cos(2*b*x + 4*a) - 10*cos 
(2*b*x + 4*c) - cos(2*c))*cos(6*b*x + 2*a + 6*c) + 12*(b*x*sin(6*b*x + 2*a 
 + 6*c) + 2*b*x*sin(4*b*x + 2*a + 4*c) + b*x*sin(2*b*x + 2*a + 2*c))*cos(6 
*b*x + 8*c) - 2*(12*b*x*sin(2*b*x + 4*c) + 11*cos(4*b*x + 4*a + 2*c) + 8*c 
os(2*b*x + 4*a) + 4*cos(2*b*x + 2*a + 2*c) - 10*cos(2*b*x + 4*c) - cos(2*c 
))*cos(4*b*x + 2*a + 4*c) - 16*cos(4*b*x + 2*a + 4*c)^2 + (48*b*x*sin(4*b* 
x + 2*a + 4*c) + 24*b*x*sin(2*b*x + 2*a + 2*c) + 10*cos(4*b*x + 2*a + 4*c) 
 + 5*cos(2*b*x + 2*a + 2*c))*cos(4*b*x + 6*c) - (8*cos(2*b*x + 4*a) - cos( 
2*c))*cos(2*b*x + 2*a + 2*c) - 11*cos(4*b*x + 4*a + 2*c)*cos(2*b*x + 2*a + 
 2*c) + 2*(6*b*x*sin(2*b*x + 2*a + 2*c) + 5*cos(2*b*x + 2*a + 2*c))*cos(2* 
b*x + 4*c) - 2*((3*cos(-2*a + 2*c) + 1)*cos(6*b*x + 2*a + 6*c)^2 + 4*(3*co 
s(-2*a + 2*c) + 1)*cos(4*b*x + 2*a + 4*c)^2 + 4*(3*cos(-2*a + 2*c) + 1)*co 
s(4*b*x + 2*a + 4*c)*cos(2*b*x + 2*a + 2*c) + (3*cos(-2*a + 2*c) + 1)*cos( 
2*b*x + 2*a + 2*c)^2 + (3*cos(-2*a + 2*c) + 1)*sin(6*b*x + 2*a + 6*c)^2 + 
4*(3*cos(-2*a + 2*c) + 1)*sin(4*b*x + 2*a + 4*c)^2 + 4*(3*cos(-2*a + 2*c) 
+ 1)*sin(4*b*x + 2*a + 4*c)*sin(2*b*x + 2*a + 2*c) + (3*cos(-2*a + 2*c)...
 

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 3 vs. order 1.

Time = 88.12 (sec) , antiderivative size = 1957357, normalized size of antiderivative = 1957357.00 \[ \int \sin ^2(a+b x) \tan ^3(c+b x) \, dx=\text {Too large to display} \] Input:

integrate(sin(b*x+a)^2*tan(b*x+c)^3,x, algorithm="giac")
 

Output:

-1/32*(48*b*x*tan(b*x)^4*tan(a + 3*c)^2*tan(a + 2*c)^2*tan(a + c)^2*tan(a 
- c)^2*tan(a - 2*c)^2*tan(a - 3*c)^2*tan(a)^2*tan(c)^9 - 48*b*x*tan(b*x)^4 
*tan(a + 3*c)^2*tan(a + 2*c)^2*tan(a + c)^2*tan(a - c)^2*tan(a - 2*c)^2*ta 
n(a - 3*c)^2*tan(a)*tan(c)^10 + 48*b*x*tan(b*x)^4*tan(a + 3*c)^2*tan(a + 2 
*c)^2*tan(a + c)^2*tan(a - c)*tan(a - 2*c)^2*tan(a - 3*c)^2*tan(a)^2*tan(c 
)^10 - 8*log(4*(tan(b*x)^2*tan(c)^2 - 2*tan(b*x)*tan(c) + 1)/(tan(b*x)^2*t 
an(c)^2 + tan(b*x)^2 + tan(c)^2 + 1))*tan(b*x)^4*tan(a + 3*c)^2*tan(a + 2* 
c)^2*tan(a + c)^2*tan(a - c)^2*tan(a - 2*c)^2*tan(a - 3*c)^2*tan(a)^2*tan( 
c)^10 + 48*arctan((tan(b*x) + tan(c))/(tan(b*x)*tan(c) - 1))*tan(b*x)^4*ta 
n(a + 3*c)^2*tan(a + 2*c)^2*tan(a + c)^2*tan(a - c)^2*tan(a - 2*c)^2*tan(a 
 - 3*c)^2*tan(a)^2*tan(c)^9 - 24*arctan((tan(b*x) + tan(c))/(tan(b*x)*tan( 
c) - 1))*tan(b*x)^4*tan(a + 3*c)^2*tan(a + 2*c)^2*tan(a + c)^2*tan(a - c)^ 
2*tan(a - 2*c)^2*tan(a - 3*c)^2*tan(a)*tan(c)^10 + 24*arctan((tan(b*x) + t 
an(c))/(tan(b*x)*tan(c) - 1))*tan(b*x)^4*tan(a + 3*c)^2*tan(a + 2*c)^2*tan 
(a + c)^2*tan(a - c)*tan(a - 2*c)^2*tan(a - 3*c)^2*tan(a)^2*tan(c)^10 + 24 
*arctan((tan(b*x) + tan(c))/(tan(b*x)*tan(c) - 1))*tan(b*x)^4*tan(a + 3*c) 
^2*tan(a + 2*c)*tan(a + c)^2*tan(a - c)^2*tan(a - 2*c)^2*tan(a - 3*c)^2*ta 
n(a)^2*tan(c)^10 - 24*arctan((tan(b*x) + tan(c))/(tan(b*x)*tan(c) - 1))*ta 
n(b*x)^4*tan(a + 3*c)*tan(a + 2*c)^2*tan(a + c)^2*tan(a - c)^2*tan(a - 2*c 
)^2*tan(a - 3*c)^2*tan(a)^2*tan(c)^10 - 8*tan(b*x)^4*tan(a + 3*c)^2*tan...
 

Mupad [F(-1)]

Timed out. \[ \int \sin ^2(a+b x) \tan ^3(c+b x) \, dx=\text {Hanged} \] Input:

int(sin(a + b*x)^2*tan(c + b*x)^3,x)
 

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \sin ^2(a+b x) \tan ^3(c+b x) \, dx=\int \sin \left (b x +a \right )^{2} \tan \left (b x +c \right )^{3}d x \] Input:

int(sin(b*x+a)^2*tan(b*x+c)^3,x)
 

Output:

int(sin(a + b*x)**2*tan(b*x + c)**3,x)