Integrand size = 16, antiderivative size = 37 \[ \int \sec ^3(c-b x) \sin (a+b x) \, dx=\frac {\cos (a+c) \sec ^2(c-b x)}{2 b}-\frac {\sin (a+c) \tan (c-b x)}{b} \] Output:
1/2*cos(a+c)*sec(b*x-c)^2/b+sin(a+c)*tan(b*x-c)/b
Time = 0.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92 \[ \int \sec ^3(c-b x) \sin (a+b x) \, dx=\frac {\sec (c) \sec ^2(c-b x) (\cos (a)-\sin (a+c) \sin (c-2 b x))}{2 b} \] Input:
Integrate[Sec[c - b*x]^3*Sin[a + b*x],x]
Output:
(Sec[c]*Sec[c - b*x]^2*(Cos[a] - Sin[a + c]*Sin[c - 2*b*x]))/(2*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.
\(\displaystyle \int \sin (a+b x) \sec ^3(c-b x) \, dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \sin (a+b x) \sec ^3(c-b x)dx\) |
Input:
Int[Sec[c - b*x]^3*Sin[a + b*x],x]
Output:
$Aborted
Time = 2.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.14
method | result | size |
parallelrisch | \(\frac {1+\cos \left (2 b x -2 c \right )-2 \cos \left (2 b x +a -c \right )}{2 b \left (1+\cos \left (2 b x -2 c \right )\right )}\) | \(42\) |
risch | \(\frac {{\mathrm e}^{5 i \left (a +c \right )}+2 \,{\mathrm e}^{i \left (2 b x +5 a +3 c \right )}-{\mathrm e}^{3 i \left (a +c \right )}}{\left ({\mathrm e}^{2 i \left (a +c \right )}+{\mathrm e}^{2 i \left (b x +a \right )}\right )^{2} b}\) | \(58\) |
default | \(\frac {-\frac {1}{\left (\sin \left (a \right ) \cos \left (c \right )+\cos \left (a \right ) \sin \left (c \right )\right )^{2} \left (\tan \left (b x +a \right ) \sin \left (a \right ) \cos \left (c \right )+\tan \left (b x +a \right ) \cos \left (a \right ) \sin \left (c \right )+\cos \left (a \right ) \cos \left (c \right )-\sin \left (a \right ) \sin \left (c \right )\right )}-\frac {-\cos \left (a \right ) \cos \left (c \right )+\sin \left (a \right ) \sin \left (c \right )}{2 \left (\sin \left (a \right ) \cos \left (c \right )+\cos \left (a \right ) \sin \left (c \right )\right )^{2} \left (\tan \left (b x +a \right ) \sin \left (a \right ) \cos \left (c \right )+\tan \left (b x +a \right ) \cos \left (a \right ) \sin \left (c \right )+\cos \left (a \right ) \cos \left (c \right )-\sin \left (a \right ) \sin \left (c \right )\right )^{2}}}{b}\) | \(120\) |
Input:
int(sec(b*x-c)^3*sin(b*x+a),x,method=_RETURNVERBOSE)
Output:
1/2*(1+cos(2*b*x-2*c)-2*cos(2*b*x+a-c))/b/(1+cos(2*b*x-2*c))
Leaf count of result is larger than twice the leaf count of optimal. 126 vs. \(2 (36) = 72\).
Time = 0.08 (sec) , antiderivative size = 126, normalized size of antiderivative = 3.41 \[ \int \sec ^3(c-b x) \sin (a+b x) \, dx=\frac {2 \, {\left (2 \, \cos \left (a + c\right )^{2} - 1\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (a + c\right ) + 4 \, {\left (\cos \left (a + c\right )^{3} - \cos \left (a + c\right )\right )} \cos \left (b x + a\right )^{2} - 2 \, \cos \left (a + c\right )^{3} + 3 \, \cos \left (a + c\right )}{2 \, {\left (2 \, b \cos \left (b x + a\right ) \cos \left (a + c\right ) \sin \left (b x + a\right ) \sin \left (a + c\right ) + {\left (2 \, b \cos \left (a + c\right )^{2} - b\right )} \cos \left (b x + a\right )^{2} - b \cos \left (a + c\right )^{2} + b\right )}} \] Input:
integrate(sec(b*x-c)^3*sin(b*x+a),x, algorithm="fricas")
Output:
1/2*(2*(2*cos(a + c)^2 - 1)*cos(b*x + a)*sin(b*x + a)*sin(a + c) + 4*(cos( a + c)^3 - cos(a + c))*cos(b*x + a)^2 - 2*cos(a + c)^3 + 3*cos(a + c))/(2* b*cos(b*x + a)*cos(a + c)*sin(b*x + a)*sin(a + c) + (2*b*cos(a + c)^2 - b) *cos(b*x + a)^2 - b*cos(a + c)^2 + b)
Exception generated. \[ \int \sec ^3(c-b x) \sin (a+b x) \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:
integrate(sec(b*x-c)**3*sin(b*x+a),x)
Output:
Exception raised: HeuristicGCDFailed >> no luck
Leaf count of result is larger than twice the leaf count of optimal. 425 vs. \(2 (36) = 72\).
Time = 0.04 (sec) , antiderivative size = 425, normalized size of antiderivative = 11.49 \[ \int \sec ^3(c-b x) \sin (a+b x) \, dx=\frac {{\left (2 \, \cos \left (2 \, b x + 2 \, a + 3 \, c\right ) + \cos \left (2 \, a + 5 \, c\right ) - \cos \left (3 \, c\right )\right )} \cos \left (4 \, b x + a\right ) + 2 \, {\left (2 \, \cos \left (2 \, b x + a + 2 \, c\right ) + \cos \left (a + 4 \, c\right )\right )} \cos \left (2 \, b x + 2 \, a + 3 \, c\right ) + 2 \, {\left (\cos \left (2 \, a + 5 \, c\right ) - \cos \left (3 \, c\right )\right )} \cos \left (2 \, b x + a + 2 \, c\right ) + \cos \left (2 \, a + 5 \, c\right ) \cos \left (a + 4 \, c\right ) - \cos \left (a + 4 \, c\right ) \cos \left (3 \, c\right ) + {\left (2 \, \sin \left (2 \, b x + 2 \, a + 3 \, c\right ) + \sin \left (2 \, a + 5 \, c\right ) - \sin \left (3 \, c\right )\right )} \sin \left (4 \, b x + a\right ) + 2 \, {\left (2 \, \sin \left (2 \, b x + a + 2 \, c\right ) + \sin \left (a + 4 \, c\right )\right )} \sin \left (2 \, b x + 2 \, a + 3 \, c\right ) + 2 \, {\left (\sin \left (2 \, a + 5 \, c\right ) - \sin \left (3 \, c\right )\right )} \sin \left (2 \, b x + a + 2 \, c\right ) + \sin \left (2 \, a + 5 \, c\right ) \sin \left (a + 4 \, c\right ) - \sin \left (a + 4 \, c\right ) \sin \left (3 \, c\right )}{b \cos \left (4 \, b x + a\right )^{2} + 4 \, b \cos \left (2 \, b x + a + 2 \, c\right )^{2} + 4 \, b \cos \left (2 \, b x + a + 2 \, c\right ) \cos \left (a + 4 \, c\right ) + b \cos \left (a + 4 \, c\right )^{2} + b \sin \left (4 \, b x + a\right )^{2} + 4 \, b \sin \left (2 \, b x + a + 2 \, c\right )^{2} + 4 \, b \sin \left (2 \, b x + a + 2 \, c\right ) \sin \left (a + 4 \, c\right ) + b \sin \left (a + 4 \, c\right )^{2} + 2 \, {\left (2 \, b \cos \left (2 \, b x + a + 2 \, c\right ) + b \cos \left (a + 4 \, c\right )\right )} \cos \left (4 \, b x + a\right ) + 2 \, {\left (2 \, b \sin \left (2 \, b x + a + 2 \, c\right ) + b \sin \left (a + 4 \, c\right )\right )} \sin \left (4 \, b x + a\right )} \] Input:
integrate(sec(b*x-c)^3*sin(b*x+a),x, algorithm="maxima")
Output:
((2*cos(2*b*x + 2*a + 3*c) + cos(2*a + 5*c) - cos(3*c))*cos(4*b*x + a) + 2 *(2*cos(2*b*x + a + 2*c) + cos(a + 4*c))*cos(2*b*x + 2*a + 3*c) + 2*(cos(2 *a + 5*c) - cos(3*c))*cos(2*b*x + a + 2*c) + cos(2*a + 5*c)*cos(a + 4*c) - cos(a + 4*c)*cos(3*c) + (2*sin(2*b*x + 2*a + 3*c) + sin(2*a + 5*c) - sin( 3*c))*sin(4*b*x + a) + 2*(2*sin(2*b*x + a + 2*c) + sin(a + 4*c))*sin(2*b*x + 2*a + 3*c) + 2*(sin(2*a + 5*c) - sin(3*c))*sin(2*b*x + a + 2*c) + sin(2 *a + 5*c)*sin(a + 4*c) - sin(a + 4*c)*sin(3*c))/(b*cos(4*b*x + a)^2 + 4*b* cos(2*b*x + a + 2*c)^2 + 4*b*cos(2*b*x + a + 2*c)*cos(a + 4*c) + b*cos(a + 4*c)^2 + b*sin(4*b*x + a)^2 + 4*b*sin(2*b*x + a + 2*c)^2 + 4*b*sin(2*b*x + a + 2*c)*sin(a + 4*c) + b*sin(a + 4*c)^2 + 2*(2*b*cos(2*b*x + a + 2*c) + b*cos(a + 4*c))*cos(4*b*x + a) + 2*(2*b*sin(2*b*x + a + 2*c) + b*sin(a + 4*c))*sin(4*b*x + a))
Leaf count of result is larger than twice the leaf count of optimal. 192 vs. \(2 (36) = 72\).
Time = 0.13 (sec) , antiderivative size = 192, normalized size of antiderivative = 5.19 \[ \int \sec ^3(c-b x) \sin (a+b x) \, dx=\frac {\tan \left (b x - c\right )^{2} \tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - \tan \left (b x - c\right )^{2} \tan \left (\frac {1}{2} \, a\right )^{2} - 4 \, \tan \left (b x - c\right )^{2} \tan \left (\frac {1}{2} \, a\right ) \tan \left (\frac {1}{2} \, c\right ) - 4 \, \tan \left (b x - c\right ) \tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - \tan \left (b x - c\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - 4 \, \tan \left (b x - c\right ) \tan \left (\frac {1}{2} \, a\right ) \tan \left (\frac {1}{2} \, c\right )^{2} + \tan \left (b x - c\right )^{2} + 4 \, \tan \left (b x - c\right ) \tan \left (\frac {1}{2} \, a\right ) + 4 \, \tan \left (b x - c\right ) \tan \left (\frac {1}{2} \, c\right )}{2 \, {\left (\tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, a\right )^{2} + \tan \left (\frac {1}{2} \, c\right )^{2} + 1\right )} b} \] Input:
integrate(sec(b*x-c)^3*sin(b*x+a),x, algorithm="giac")
Output:
1/2*(tan(b*x - c)^2*tan(1/2*a)^2*tan(1/2*c)^2 - tan(b*x - c)^2*tan(1/2*a)^ 2 - 4*tan(b*x - c)^2*tan(1/2*a)*tan(1/2*c) - 4*tan(b*x - c)*tan(1/2*a)^2*t an(1/2*c) - tan(b*x - c)^2*tan(1/2*c)^2 - 4*tan(b*x - c)*tan(1/2*a)*tan(1/ 2*c)^2 + tan(b*x - c)^2 + 4*tan(b*x - c)*tan(1/2*a) + 4*tan(b*x - c)*tan(1 /2*c))/((tan(1/2*a)^2*tan(1/2*c)^2 + tan(1/2*a)^2 + tan(1/2*c)^2 + 1)*b)
Timed out. \[ \int \sec ^3(c-b x) \sin (a+b x) \, dx=\text {Hanged} \] Input:
int(sin(a + b*x)/cos(c - b*x)^3,x)
Output:
\text{Hanged}
Time = 0.16 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.38 \[ \int \sec ^3(c-b x) \sin (a+b x) \, dx=\frac {\cos \left (b x -c \right ) \cos \left (b x +a \right )-\sin \left (b x -c \right ) \sin \left (b x +a \right )}{2 b \left (\sin \left (b x -c \right )^{2}-1\right )} \] Input:
int(sec(b*x-c)^3*sin(b*x+a),x)
Output:
(cos(b*x - c)*cos(a + b*x) - sin(b*x - c)*sin(a + b*x))/(2*b*(sin(b*x - c) **2 - 1))