Integrand size = 17, antiderivative size = 84 \[ \int \cos ^2(a+b x) \csc ^4(c+b x) \, dx=-\frac {\cos ^2(a-c) \cot (c+b x)}{b}+\frac {\cos (2 (a-c)) \cot (c+b x)}{b}-\frac {\cos ^2(a-c) \cot ^3(c+b x)}{3 b}+\frac {\csc ^2(c+b x) \sin (2 (a-c))}{2 b} \] Output:
-cos(a-c)^2*cot(b*x+c)/b+cos(2*a-2*c)*cot(b*x+c)/b-1/3*cos(a-c)^2*cot(b*x+ c)^3/b+1/2*csc(b*x+c)^2*sin(2*a-2*c)/b
Time = 0.23 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.98 \[ \int \cos ^2(a+b x) \csc ^4(c+b x) \, dx=\frac {\csc (c) \csc ^3(c+b x) (3 \sin (b x)-\sin (2 a-4 c-3 b x)+3 \sin (2 a-2 c-b x)-3 \sin (2 a+b x)+\sin (2 a+3 b x)-\sin (2 c+3 b x))}{12 b} \] Input:
Integrate[Cos[a + b*x]^2*Csc[c + b*x]^4,x]
Output:
(Csc[c]*Csc[c + b*x]^3*(3*Sin[b*x] - Sin[2*a - 4*c - 3*b*x] + 3*Sin[2*a - 2*c - b*x] - 3*Sin[2*a + b*x] + Sin[2*a + 3*b*x] - Sin[2*c + 3*b*x]))/(12* b)
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.
\(\displaystyle \int \cos ^2(a+b x) \csc ^4(b x+c) \, dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \cos ^2(a+b x) \csc ^4(b x+c)dx\) |
Input:
Int[Cos[a + b*x]^2*Csc[c + b*x]^4,x]
Output:
$Aborted
Time = 3.82 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.65
method | result | size |
default | \(-\frac {1}{3 b \left (\cos \left (a \right ) \cos \left (c \right )+\sin \left (a \right ) \sin \left (c \right )\right ) \left (\tan \left (b x +a \right ) \cos \left (a \right ) \cos \left (c \right )+\tan \left (b x +a \right ) \sin \left (a \right ) \sin \left (c \right )-\sin \left (a \right ) \cos \left (c \right )+\cos \left (a \right ) \sin \left (c \right )\right )^{3}}\) | \(55\) |
parallelrisch | \(\frac {\sec \left (\frac {b x}{2}+\frac {c}{2}\right )^{3} \csc \left (\frac {b x}{2}+\frac {c}{2}\right )^{3} \left (-3 \cos \left (b x +c \right )-2 \cos \left (3 b x +2 a +c \right )+\cos \left (3 b x +3 c \right )\right )}{96 b}\) | \(58\) |
risch | \(-\frac {2 i \left (3 \,{\mathrm e}^{2 i \left (2 b x +4 a +c \right )}-3 \,{\mathrm e}^{2 i \left (b x +4 a \right )}+3 \,{\mathrm e}^{2 i \left (b x +3 a +c \right )}+{\mathrm e}^{2 i \left (4 a -c \right )}-{\mathrm e}^{6 i a}+{\mathrm e}^{2 i \left (2 a +c \right )}\right )}{3 \left (-{\mathrm e}^{2 i \left (b x +a +c \right )}+{\mathrm e}^{2 i a}\right )^{3} b}\) | \(97\) |
Input:
int(cos(b*x+a)^2*csc(b*x+c)^4,x,method=_RETURNVERBOSE)
Output:
-1/3/b/(cos(a)*cos(c)+sin(a)*sin(c))/(tan(b*x+a)*cos(a)*cos(c)+tan(b*x+a)* sin(a)*sin(c)-sin(a)*cos(c)+cos(a)*sin(c))^3
Time = 0.07 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.02 \[ \int \cos ^2(a+b x) \csc ^4(c+b x) \, dx=\frac {{\left (4 \, \cos \left (-a + c\right )^{2} - 3\right )} \cos \left (b x + c\right )^{3} + 3 \, \cos \left (-a + c\right ) \sin \left (b x + c\right ) \sin \left (-a + c\right ) - 3 \, {\left (\cos \left (-a + c\right )^{2} - 1\right )} \cos \left (b x + c\right )}{3 \, {\left (b \cos \left (b x + c\right )^{2} - b\right )} \sin \left (b x + c\right )} \] Input:
integrate(cos(b*x+a)^2*csc(b*x+c)^4,x, algorithm="fricas")
Output:
1/3*((4*cos(-a + c)^2 - 3)*cos(b*x + c)^3 + 3*cos(-a + c)*sin(b*x + c)*sin (-a + c) - 3*(cos(-a + c)^2 - 1)*cos(b*x + c))/((b*cos(b*x + c)^2 - b)*sin (b*x + c))
Timed out. \[ \int \cos ^2(a+b x) \csc ^4(c+b x) \, dx=\text {Timed out} \] Input:
integrate(cos(b*x+a)**2*csc(b*x+c)**4,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 908 vs. \(2 (80) = 160\).
Time = 0.05 (sec) , antiderivative size = 908, normalized size of antiderivative = 10.81 \[ \int \cos ^2(a+b x) \csc ^4(c+b x) \, dx=\text {Too large to display} \] Input:
integrate(cos(b*x+a)^2*csc(b*x+c)^4,x, algorithm="maxima")
Output:
-2/3*((3*sin(4*b*x + 4*a + 4*c) - 3*sin(2*b*x + 4*a + 2*c) + 3*sin(2*b*x + 2*a + 4*c) + sin(4*a) - sin(2*a + 2*c) + sin(4*c))*cos(6*b*x + 2*a + 8*c) - 3*(3*sin(2*b*x + 2*a + 4*c) - sin(2*a + 2*c))*cos(4*b*x + 4*a + 4*c) - 3*(3*sin(4*b*x + 4*a + 4*c) - 3*sin(2*b*x + 4*a + 2*c) + 3*sin(2*b*x + 2*a + 4*c) + sin(4*a) - sin(2*a + 2*c) + sin(4*c))*cos(4*b*x + 2*a + 6*c) - 3 *(3*sin(2*b*x + 4*a + 2*c) - sin(4*a) - sin(4*c))*cos(2*b*x + 2*a + 4*c) - (sin(4*a) + sin(4*c))*cos(2*a + 2*c) - (3*cos(4*b*x + 4*a + 4*c) - 3*cos( 2*b*x + 4*a + 2*c) + 3*cos(2*b*x + 2*a + 4*c) + cos(4*a) - cos(2*a + 2*c) + cos(4*c))*sin(6*b*x + 2*a + 8*c) + 3*(3*cos(2*b*x + 2*a + 4*c) - cos(2*a + 2*c))*sin(4*b*x + 4*a + 4*c) + 3*(3*cos(4*b*x + 4*a + 4*c) - 3*cos(2*b* x + 4*a + 2*c) + 3*cos(2*b*x + 2*a + 4*c) + cos(4*a) - cos(2*a + 2*c) + co s(4*c))*sin(4*b*x + 2*a + 6*c) + 3*cos(2*a + 2*c)*sin(2*b*x + 4*a + 2*c) + 3*(3*cos(2*b*x + 4*a + 2*c) - cos(4*a) - cos(4*c))*sin(2*b*x + 2*a + 4*c) + (cos(4*a) + cos(4*c))*sin(2*a + 2*c) - 3*cos(2*b*x + 4*a + 2*c)*sin(2*a + 2*c))/(b*cos(6*b*x + 2*a + 8*c)^2 + 9*b*cos(4*b*x + 2*a + 6*c)^2 + 9*b* cos(2*b*x + 2*a + 4*c)^2 - 6*b*cos(2*b*x + 2*a + 4*c)*cos(2*a + 2*c) + b*c os(2*a + 2*c)^2 + b*sin(6*b*x + 2*a + 8*c)^2 + 9*b*sin(4*b*x + 2*a + 6*c)^ 2 + 9*b*sin(2*b*x + 2*a + 4*c)^2 - 6*b*sin(2*b*x + 2*a + 4*c)*sin(2*a + 2* c) + b*sin(2*a + 2*c)^2 - 2*(3*b*cos(4*b*x + 2*a + 6*c) - 3*b*cos(2*b*x + 2*a + 4*c) + b*cos(2*a + 2*c))*cos(6*b*x + 2*a + 8*c) - 6*(3*b*cos(2*b*...
Leaf count of result is larger than twice the leaf count of optimal. 441 vs. \(2 (80) = 160\).
Time = 0.18 (sec) , antiderivative size = 441, normalized size of antiderivative = 5.25 \[ \int \cos ^2(a+b x) \csc ^4(c+b x) \, dx =\text {Too large to display} \] Input:
integrate(cos(b*x+a)^2*csc(b*x+c)^4,x, algorithm="giac")
Output:
-1/3*(tan(1/2*a)^8*tan(1/2*c)^8 + 4*tan(1/2*a)^8*tan(1/2*c)^6 + 4*tan(1/2* a)^6*tan(1/2*c)^8 + 6*tan(1/2*a)^8*tan(1/2*c)^4 + 16*tan(1/2*a)^6*tan(1/2* c)^6 + 6*tan(1/2*a)^4*tan(1/2*c)^8 + 4*tan(1/2*a)^8*tan(1/2*c)^2 + 24*tan( 1/2*a)^6*tan(1/2*c)^4 + 24*tan(1/2*a)^4*tan(1/2*c)^6 + 4*tan(1/2*a)^2*tan( 1/2*c)^8 + tan(1/2*a)^8 + 16*tan(1/2*a)^6*tan(1/2*c)^2 + 36*tan(1/2*a)^4*t an(1/2*c)^4 + 16*tan(1/2*a)^2*tan(1/2*c)^6 + tan(1/2*c)^8 + 4*tan(1/2*a)^6 + 24*tan(1/2*a)^4*tan(1/2*c)^2 + 24*tan(1/2*a)^2*tan(1/2*c)^4 + 4*tan(1/2 *c)^6 + 6*tan(1/2*a)^4 + 16*tan(1/2*a)^2*tan(1/2*c)^2 + 6*tan(1/2*c)^4 + 4 *tan(1/2*a)^2 + 4*tan(1/2*c)^2 + 1)/((tan(b*x + a)*tan(1/2*a)^2*tan(1/2*c) ^2 - tan(b*x + a)*tan(1/2*a)^2 + 4*tan(b*x + a)*tan(1/2*a)*tan(1/2*c) - 2* tan(1/2*a)^2*tan(1/2*c) - tan(b*x + a)*tan(1/2*c)^2 + 2*tan(1/2*a)*tan(1/2 *c)^2 + tan(b*x + a) - 2*tan(1/2*a) + 2*tan(1/2*c))^3*(tan(1/2*a)^2*tan(1/ 2*c)^2 - tan(1/2*a)^2 + 4*tan(1/2*a)*tan(1/2*c) - tan(1/2*c)^2 + 1)*b)
Timed out. \[ \int \cos ^2(a+b x) \csc ^4(c+b x) \, dx=\text {Hanged} \] Input:
int(cos(a + b*x)^2/sin(c + b*x)^4,x)
Output:
\text{Hanged}
Time = 0.19 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.86 \[ \int \cos ^2(a+b x) \csc ^4(c+b x) \, dx=\frac {-\cos \left (b x +c \right ) \sin \left (b x +c \right )^{2}+\cos \left (b x +c \right ) \sin \left (b x +a \right )^{2}-\cos \left (b x +c \right )+\cos \left (b x +a \right ) \sin \left (b x +c \right ) \sin \left (b x +a \right )}{3 \sin \left (b x +c \right )^{3} b} \] Input:
int(cos(b*x+a)^2*csc(b*x+c)^4,x)
Output:
( - cos(b*x + c)*sin(b*x + c)**2 + cos(b*x + c)*sin(a + b*x)**2 - cos(b*x + c) + cos(a + b*x)*sin(b*x + c)*sin(a + b*x))/(3*sin(b*x + c)**3*b)