\(\int \csc ^4(c+b x) \sin ^3(a+b x) \, dx\) [100]

Optimal result

Integrand size = 17, antiderivative size = 627 \[ \int \csc ^4(c+b x) \sin ^3(a+b x) \, dx=\frac {e^{-i (a-2 c)+i b x}}{b \left (1-e^{2 i (c+b x)}\right )^3}-\frac {e^{-3 i a+4 i c+i b x}}{3 b \left (1-e^{2 i (c+b x)}\right )^3}-\frac {e^{i (a+2 c)+3 i b x}}{b \left (1-e^{2 i (c+b x)}\right )^3}+\frac {e^{i (3 a+2 c)+5 i b x}}{3 b \left (1-e^{2 i (c+b x)}\right )^3}+\frac {3 e^{i a+i b x}}{4 b \left (1-e^{2 i (c+b x)}\right )^2}-\frac {e^{-i (a-2 c)+i b x}}{4 b \left (1-e^{2 i (c+b x)}\right )^2}-\frac {5 e^{-3 i a+4 i c+i b x}}{12 b \left (1-e^{2 i (c+b x)}\right )^2}-\frac {5 e^{3 i a+3 i b x}}{12 b \left (1-e^{2 i (c+b x)}\right )^2}-\frac {3 e^{i a+i b x}}{8 b \left (1-e^{2 i (c+b x)}\right )}-\frac {3 e^{-i (a-2 c)+i b x}}{8 b \left (1-e^{2 i (c+b x)}\right )}+\frac {5 e^{i (3 a-2 c)+i b x}}{8 b \left (1-e^{2 i (c+b x)}\right )}-\frac {5 e^{-3 i a+4 i c+i b x}}{8 b \left (1-e^{2 i (c+b x)}\right )}-\frac {3 e^{-i (a-c)} \text {arctanh}\left (e^{i c+i b x}\right )}{8 b}-\frac {3 e^{i (a-c)} \text {arctanh}\left (e^{i c+i b x}\right )}{8 b}-\frac {5 e^{-3 i (a-c)} \text {arctanh}\left (e^{i c+i b x}\right )}{8 b}-\frac {5 e^{3 i (a-c)} \text {arctanh}\left (e^{i c+i b x}\right )}{8 b} \] Output:

exp(-I*(a-2*c)+I*b*x)/b/(1-exp(2*I*(b*x+c)))^3-1/3*exp(-3*I*a+4*I*c+I*b*x) 
/b/(1-exp(2*I*(b*x+c)))^3-exp(I*(a+2*c)+3*I*b*x)/b/(1-exp(2*I*(b*x+c)))^3+ 
1/3*exp(I*(3*a+2*c)+5*I*b*x)/b/(1-exp(2*I*(b*x+c)))^3+3/4*exp(I*a+I*b*x)/b 
/(1-exp(2*I*(b*x+c)))^2-1/4*exp(-I*(a-2*c)+I*b*x)/b/(1-exp(2*I*(b*x+c)))^2 
-5/12*exp(-3*I*a+4*I*c+I*b*x)/b/(1-exp(2*I*(b*x+c)))^2-5/12*exp(3*I*a+3*I* 
b*x)/b/(1-exp(2*I*(b*x+c)))^2-3/8*exp(I*a+I*b*x)/b/(1-exp(2*I*(b*x+c)))-3/ 
8*exp(-I*(a-2*c)+I*b*x)/b/(1-exp(2*I*(b*x+c)))+5/8*exp(I*(3*a-2*c)+I*b*x)/ 
b/(1-exp(2*I*(b*x+c)))-5/8*exp(-3*I*a+4*I*c+I*b*x)/b/(1-exp(2*I*(b*x+c)))- 
3/8*arctanh(exp(I*c+I*b*x))/b/exp(I*(a-c))-3/8*exp(I*(a-c))*arctanh(exp(I* 
c+I*b*x))/b-5/8*arctanh(exp(I*c+I*b*x))/b/exp(3*I*(a-c))-5/8*exp(3*I*(a-c) 
)*arctanh(exp(I*c+I*b*x))/b
 

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.17 \[ \int \csc ^4(c+b x) \sin ^3(a+b x) \, dx=\frac {-12 \text {arctanh}\left (\cos (c)-\sin (c) \tan \left (\frac {b x}{2}\right )\right ) (3 \cos (a-c)+5 \cos (3 (a-c)))+(-32-40 \cos (2 (a-c))+15 \cos (2 (a-2 c-b x))+33 \cos (2 (a+b x))+24 \cos (2 (c+b x))) \csc ^3(c+b x) \sin (a-c)}{48 b} \] Input:

Integrate[Csc[c + b*x]^4*Sin[a + b*x]^3,x]
 

Output:

(-12*ArcTanh[Cos[c] - Sin[c]*Tan[(b*x)/2]]*(3*Cos[a - c] + 5*Cos[3*(a - c) 
]) + (-32 - 40*Cos[2*(a - c)] + 15*Cos[2*(a - 2*c - b*x)] + 33*Cos[2*(a + 
b*x)] + 24*Cos[2*(c + b*x)])*Csc[c + b*x]^3*Sin[a - c])/(48*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.

\(\displaystyle \int \sin ^3(a+b x) \csc ^4(b x+c) \, dx\)

\(\Big \downarrow \) 7299

\(\displaystyle \int \sin ^3(a+b x) \csc ^4(b x+c)dx\)

Input:

Int[Csc[c + b*x]^4*Sin[a + b*x]^3,x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 7299

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

Maple [A] (verified)

Time = 6.58 (sec) , antiderivative size = 350, normalized size of antiderivative = 0.56

method	result	size
risch	\(\frac {33 \,{\mathrm e}^{i \left (5 b x +9 a +2 c \right )}-9 \,{\mathrm e}^{i \left (5 b x +7 a +4 c \right )}-9 \,{\mathrm e}^{i \left (5 b x +5 a +6 c \right )}-15 \,{\mathrm e}^{i \left (5 b x +3 a +8 c \right )}-40 \,{\mathrm e}^{3 i \left (b x +3 a \right )}-24 \,{\mathrm e}^{i \left (3 b x +7 a +2 c \right )}+24 \,{\mathrm e}^{i \left (3 b x +5 a +4 c \right )}+40 \,{\mathrm e}^{3 i \left (b x +a +2 c \right )}+15 \,{\mathrm e}^{i \left (b x +9 a -2 c \right )}+9 \,{\mathrm e}^{i \left (b x +7 a \right )}+9 \,{\mathrm e}^{i \left (b x +5 a +2 c \right )}-33 \,{\mathrm e}^{i \left (b x +3 a +4 c \right )}}{24 \left (-{\mathrm e}^{2 i \left (b x +a +c \right )}+{\mathrm e}^{2 i a}\right )^{3} b}-\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+{\mathrm e}^{i \left (a -c \right )}\right ) \cos \left (a -c \right )}{8 b}-\frac {5 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+{\mathrm e}^{i \left (a -c \right )}\right ) \cos \left (3 a -3 c \right )}{8 b}+\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-{\mathrm e}^{i \left (a -c \right )}\right ) \cos \left (a -c \right )}{8 b}+\frac {5 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-{\mathrm e}^{i \left (a -c \right )}\right ) \cos \left (3 a -3 c \right )}{8 b}\)	\(350\)
default	\(\text {Expression too large to display}\)	\(2823\)

Input:

int(csc(b*x+c)^4*sin(b*x+a)^3,x,method=_RETURNVERBOSE)
 

Output:

1/24/(-exp(2*I*(b*x+a+c))+exp(2*I*a))^3/b*(33*exp(I*(5*b*x+9*a+2*c))-9*exp 
(I*(5*b*x+7*a+4*c))-9*exp(I*(5*b*x+5*a+6*c))-15*exp(I*(5*b*x+3*a+8*c))-40* 
exp(3*I*(b*x+3*a))-24*exp(I*(3*b*x+7*a+2*c))+24*exp(I*(3*b*x+5*a+4*c))+40* 
exp(3*I*(b*x+a+2*c))+15*exp(I*(b*x+9*a-2*c))+9*exp(I*(b*x+7*a))+9*exp(I*(b 
*x+5*a+2*c))-33*exp(I*(b*x+3*a+4*c)))-3/8*ln(exp(I*(b*x+a))+exp(I*(a-c)))/ 
b*cos(a-c)-5/8*ln(exp(I*(b*x+a))+exp(I*(a-c)))/b*cos(3*a-3*c)+3/8*ln(exp(I 
*(b*x+a))-exp(I*(a-c)))/b*cos(a-c)+5/8*ln(exp(I*(b*x+a))-exp(I*(a-c)))/b*c 
os(3*a-3*c)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.37 \[ \int \csc ^4(c+b x) \sin ^3(a+b x) \, dx=-\frac {18 \, {\left (\cos \left (-a + c\right )^{3} - \cos \left (-a + c\right )\right )} \cos \left (b x + c\right ) \sin \left (b x + c\right ) + 3 \, {\left ({\left (5 \, \cos \left (-a + c\right )^{3} - 3 \, \cos \left (-a + c\right )\right )} \cos \left (b x + c\right )^{2} - 5 \, \cos \left (-a + c\right )^{3} + 3 \, \cos \left (-a + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (b x + c\right ) + \frac {1}{2}\right ) \sin \left (b x + c\right ) - 3 \, {\left ({\left (5 \, \cos \left (-a + c\right )^{3} - 3 \, \cos \left (-a + c\right )\right )} \cos \left (b x + c\right )^{2} - 5 \, \cos \left (-a + c\right )^{3} + 3 \, \cos \left (-a + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + c\right ) + \frac {1}{2}\right ) \sin \left (b x + c\right ) - 4 \, {\left (3 \, {\left (4 \, \cos \left (-a + c\right )^{2} - 1\right )} \cos \left (b x + c\right )^{2} - 11 \, \cos \left (-a + c\right )^{2} + 2\right )} \sin \left (-a + c\right )}{12 \, {\left (b \cos \left (b x + c\right )^{2} - b\right )} \sin \left (b x + c\right )} \] Input:

integrate(csc(b*x+c)^4*sin(b*x+a)^3,x, algorithm="fricas")
 

Output:

-1/12*(18*(cos(-a + c)^3 - cos(-a + c))*cos(b*x + c)*sin(b*x + c) + 3*((5* 
cos(-a + c)^3 - 3*cos(-a + c))*cos(b*x + c)^2 - 5*cos(-a + c)^3 + 3*cos(-a 
 + c))*log(1/2*cos(b*x + c) + 1/2)*sin(b*x + c) - 3*((5*cos(-a + c)^3 - 3* 
cos(-a + c))*cos(b*x + c)^2 - 5*cos(-a + c)^3 + 3*cos(-a + c))*log(-1/2*co 
s(b*x + c) + 1/2)*sin(b*x + c) - 4*(3*(4*cos(-a + c)^2 - 1)*cos(b*x + c)^2 
 - 11*cos(-a + c)^2 + 2)*sin(-a + c))/((b*cos(b*x + c)^2 - b)*sin(b*x + c) 
)
 

Sympy [F(-1)]

Timed out. \[ \int \csc ^4(c+b x) \sin ^3(a+b x) \, dx=\text {Timed out} \] Input:

integrate(csc(b*x+c)**4*sin(b*x+a)**3,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3532 vs. \(2 (443) = 886\).

Time = 0.19 (sec) , antiderivative size = 3532, normalized size of antiderivative = 5.63 \[ \int \csc ^4(c+b x) \sin ^3(a+b x) \, dx=\text {Too large to display} \] Input:

integrate(csc(b*x+c)^4*sin(b*x+a)^3,x, algorithm="maxima")
 

Output:

-1/48*(2*(33*cos(5*b*x + 6*a + 4*c) - 9*cos(5*b*x + 4*a + 6*c) - 9*cos(5*b 
*x + 2*a + 8*c) - 15*cos(5*b*x + 10*c) - 40*cos(3*b*x + 6*a + 2*c) - 24*co 
s(3*b*x + 4*a + 4*c) + 24*cos(3*b*x + 2*a + 6*c) + 40*cos(3*b*x + 8*c) + 1 
5*cos(b*x + 6*a) + 9*cos(b*x + 4*a + 2*c) + 9*cos(b*x + 2*a + 4*c) - 33*co 
s(b*x + 6*c))*cos(6*b*x + 3*a + 8*c) - 66*(3*cos(4*b*x + 3*a + 6*c) - 3*co 
s(2*b*x + 3*a + 4*c) + cos(3*a + 2*c))*cos(5*b*x + 6*a + 4*c) + 18*(3*cos( 
4*b*x + 3*a + 6*c) - 3*cos(2*b*x + 3*a + 4*c) + cos(3*a + 2*c))*cos(5*b*x 
+ 4*a + 6*c) + 18*(3*cos(4*b*x + 3*a + 6*c) - 3*cos(2*b*x + 3*a + 4*c) + c 
os(3*a + 2*c))*cos(5*b*x + 2*a + 8*c) + 30*(3*cos(4*b*x + 3*a + 6*c) - 3*c 
os(2*b*x + 3*a + 4*c) + cos(3*a + 2*c))*cos(5*b*x + 10*c) + 6*(40*cos(3*b* 
x + 6*a + 2*c) + 24*cos(3*b*x + 4*a + 4*c) - 24*cos(3*b*x + 2*a + 6*c) - 4 
0*cos(3*b*x + 8*c) - 15*cos(b*x + 6*a) - 9*cos(b*x + 4*a + 2*c) - 9*cos(b* 
x + 2*a + 4*c) + 33*cos(b*x + 6*c))*cos(4*b*x + 3*a + 6*c) - 80*(3*cos(2*b 
*x + 3*a + 4*c) - cos(3*a + 2*c))*cos(3*b*x + 6*a + 2*c) - 48*(3*cos(2*b*x 
 + 3*a + 4*c) - cos(3*a + 2*c))*cos(3*b*x + 4*a + 4*c) + 48*(3*cos(2*b*x + 
 3*a + 4*c) - cos(3*a + 2*c))*cos(3*b*x + 2*a + 6*c) + 80*(3*cos(2*b*x + 3 
*a + 4*c) - cos(3*a + 2*c))*cos(3*b*x + 8*c) + 18*(5*cos(b*x + 6*a) + 3*co 
s(b*x + 4*a + 2*c) + 3*cos(b*x + 2*a + 4*c) - 11*cos(b*x + 6*c))*cos(2*b*x 
 + 3*a + 4*c) - 30*cos(b*x + 6*a)*cos(3*a + 2*c) - 18*cos(b*x + 4*a + 2*c) 
*cos(3*a + 2*c) - 18*cos(b*x + 2*a + 4*c)*cos(3*a + 2*c) + 66*cos(b*x +...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 16081 vs. \(2 (443) = 886\).

Time = 0.31 (sec) , antiderivative size = 16081, normalized size of antiderivative = 25.65 \[ \int \csc ^4(c+b x) \sin ^3(a+b x) \, dx=\text {Too large to display} \] Input:

integrate(csc(b*x+c)^4*sin(b*x+a)^3,x, algorithm="giac")
 

Output:

1/6*(6*(tan(1/2*a)^6*tan(1/2*c)^6 - 9*tan(1/2*a)^6*tan(1/2*c)^4 + 24*tan(1 
/2*a)^5*tan(1/2*c)^5 - 9*tan(1/2*a)^4*tan(1/2*c)^6 + 9*tan(1/2*a)^6*tan(1/ 
2*c)^2 - 72*tan(1/2*a)^5*tan(1/2*c)^3 + 141*tan(1/2*a)^4*tan(1/2*c)^4 - 72 
*tan(1/2*a)^3*tan(1/2*c)^5 + 9*tan(1/2*a)^2*tan(1/2*c)^6 - tan(1/2*a)^6 + 
24*tan(1/2*a)^5*tan(1/2*c) - 141*tan(1/2*a)^4*tan(1/2*c)^2 + 256*tan(1/2*a 
)^3*tan(1/2*c)^3 - 141*tan(1/2*a)^2*tan(1/2*c)^4 + 24*tan(1/2*a)*tan(1/2*c 
)^5 - tan(1/2*c)^6 + 9*tan(1/2*a)^4 - 72*tan(1/2*a)^3*tan(1/2*c) + 141*tan 
(1/2*a)^2*tan(1/2*c)^2 - 72*tan(1/2*a)*tan(1/2*c)^3 + 9*tan(1/2*c)^4 - 9*t 
an(1/2*a)^2 + 24*tan(1/2*a)*tan(1/2*c) - 9*tan(1/2*c)^2 + 1)*log(abs(tan(1 
/2*b*x + 1/2*c)))/(tan(1/2*a)^6*tan(1/2*c)^6 + 3*tan(1/2*a)^6*tan(1/2*c)^4 
 + 3*tan(1/2*a)^4*tan(1/2*c)^6 + 3*tan(1/2*a)^6*tan(1/2*c)^2 + 9*tan(1/2*a 
)^4*tan(1/2*c)^4 + 3*tan(1/2*a)^2*tan(1/2*c)^6 + tan(1/2*a)^6 + 9*tan(1/2* 
a)^4*tan(1/2*c)^2 + 9*tan(1/2*a)^2*tan(1/2*c)^4 + tan(1/2*c)^6 + 3*tan(1/2 
*a)^4 + 9*tan(1/2*a)^2*tan(1/2*c)^2 + 3*tan(1/2*c)^4 + 3*tan(1/2*a)^2 + 3* 
tan(1/2*c)^2 + 1) - (2*tan(1/2*b*x + 1/2*c)^3*tan(1/2*a)^18*tan(1/2*c)^15 
- 6*tan(1/2*b*x + 1/2*c)^3*tan(1/2*a)^17*tan(1/2*c)^16 - 9*tan(1/2*b*x + 1 
/2*c)^2*tan(1/2*a)^18*tan(1/2*c)^16 + 6*tan(1/2*b*x + 1/2*c)^3*tan(1/2*a)^ 
16*tan(1/2*c)^17 + 18*tan(1/2*b*x + 1/2*c)^2*tan(1/2*a)^17*tan(1/2*c)^17 + 
 18*tan(1/2*b*x + 1/2*c)*tan(1/2*a)^18*tan(1/2*c)^17 - 2*tan(1/2*b*x + 1/2 
*c)^3*tan(1/2*a)^15*tan(1/2*c)^18 - 9*tan(1/2*b*x + 1/2*c)^2*tan(1/2*a)...
 

Mupad [F(-1)]

Timed out. \[ \int \csc ^4(c+b x) \sin ^3(a+b x) \, dx=\text {Hanged} \] Input:

int(sin(a + b*x)^3/sin(c + b*x)^4,x)
 

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \csc ^4(c+b x) \sin ^3(a+b x) \, dx=\int \csc \left (b x +c \right )^{4} \sin \left (b x +a \right )^{3}d x \] Input:

int(csc(b*x+c)^4*sin(b*x+a)^3,x)
 

Output:

int(csc(b*x + c)**4*sin(a + b*x)**3,x)