Integrand size = 15, antiderivative size = 38 \[ \int \sec ^3(c+b x) \sin (a+b x) \, dx=\frac {\cos (a-c) \sec ^2(c+b x)}{2 b}+\frac {\sin (a-c) \tan (c+b x)}{b} \] Output:
1/2*cos(a-c)*sec(b*x+c)^2/b+sin(a-c)*tan(b*x+c)/b
Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \sec ^3(c+b x) \sin (a+b x) \, dx=\frac {\sec (c) \sec ^2(c+b x) (\cos (a)+\sin (a-c) \sin (c+2 b x))}{2 b} \] 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)
Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5091, 3042, 3086, 15, 4254, 24}
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.
\(\displaystyle \int \sin (a+b x) \sec ^3(b x+c) \, dx\) |
\(\Big \downarrow \) 5091 |
\(\displaystyle \sin (a-c) \int \sec ^2(c+b x)dx+\cos (a-c) \int \sec ^2(c+b x) \tan (c+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sin (a-c) \int \csc \left (c+b x+\frac {\pi }{2}\right )^2dx+\cos (a-c) \int \sec (c+b x)^2 \tan (c+b x)dx\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle \frac {\cos (a-c) \int \sec (c+b x)d\sec (c+b x)}{b}+\sin (a-c) \int \csc \left (c+b x+\frac {\pi }{2}\right )^2dx\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \sin (a-c) \int \csc \left (c+b x+\frac {\pi }{2}\right )^2dx+\frac {\cos (a-c) \sec ^2(b x+c)}{2 b}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {\cos (a-c) \sec ^2(b x+c)}{2 b}-\frac {\sin (a-c) \int 1d(-\tan (c+b x))}{b}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\sin (a-c) \tan (b x+c)}{b}+\frac {\cos (a-c) \sec ^2(b x+c)}{2 b}\) |
Input:
Int[Sec[c + b*x]^3*Sin[a + b*x],x]
Output:
(Cos[a - c]*Sec[c + b*x]^2)/(2*b) + (Sin[a - c]*Tan[c + b*x])/b
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[a/f Subst[Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2 ), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2 ] && !(IntegerQ[m/2] && LtQ[0, m, n + 1])
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[Sec[w_]^(n_.)*Sin[v_], x_Symbol] :> Simp[Cos[v - w] Int[Tan[w]*Sec[w] ^(n - 1), x], x] + Simp[Sin[v - w] Int[Sec[w]^(n - 1), x], x] /; GtQ[n, 0 ] && FreeQ[v - w, x] && NeQ[w, v]
Time = 1.31 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.05
method | result | size |
parallelrisch | \(\frac {1+\cos \left (2 b x +2 c \right )-2 \cos \left (2 b x +a +c \right )}{2 b \left (1+\cos \left (2 b x +2 c \right )\right )}\) | \(40\) |
risch | \(\frac {2 \,{\mathrm e}^{i \left (2 b x +5 a +c \right )}+{\mathrm e}^{i \left (5 a -c \right )}-{\mathrm e}^{i \left (3 a +c \right )}}{\left ({\mathrm e}^{2 i \left (b x +a +c \right )}+{\mathrm e}^{2 i a}\right )^{2} b}\) | \(61\) |
default | \(\frac {-\frac {1}{\left (\sin \left (a \right ) \cos \left (c \right )-\cos \left (a \right ) \sin \left (c \right )\right )^{2} \left (\tan \left (b x +a \right ) \sin \left (a \right ) \cos \left (c \right )-\tan \left (b x +a \right ) \cos \left (a \right ) \sin \left (c \right )+\cos \left (a \right ) \cos \left (c \right )+\sin \left (a \right ) \sin \left (c \right )\right )}+\frac {\cos \left (a \right ) \cos \left (c \right )+\sin \left (a \right ) \sin \left (c \right )}{2 \left (\sin \left (a \right ) \cos \left (c \right )-\cos \left (a \right ) \sin \left (c \right )\right )^{2} \left (\tan \left (b x +a \right ) \sin \left (a \right ) \cos \left (c \right )-\tan \left (b x +a \right ) \cos \left (a \right ) \sin \left (c \right )+\cos \left (a \right ) \cos \left (c \right )+\sin \left (a \right ) \sin \left (c \right )\right )^{2}}}{b}\) | \(121\) |
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))
Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.11 \[ \int \sec ^3(c+b x) \sin (a+b x) \, dx=-\frac {2 \, \cos \left (b x + c\right ) \sin \left (b x + c\right ) \sin \left (-a + c\right ) - \cos \left (-a + c\right )}{2 \, b \cos \left (b x + c\right )^{2}} \] Input:
integrate(sec(b*x+c)^3*sin(b*x+a),x, algorithm="fricas")
Output:
-1/2*(2*cos(b*x + c)*sin(b*x + c)*sin(-a + c) - cos(-a + c))/(b*cos(b*x + c)^2)
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. 391 vs. \(2 (36) = 72\).
Time = 0.04 (sec) , antiderivative size = 391, normalized size of antiderivative = 10.29 \[ \int \sec ^3(c+b x) \sin (a+b x) \, dx=\frac {{\left (2 \, \cos \left (2 \, b x + 2 \, a + 2 \, c\right ) + \cos \left (2 \, a\right ) - \cos \left (2 \, c\right )\right )} \cos \left (4 \, b x + a + 5 \, c\right ) + 2 \, {\left (2 \, \cos \left (2 \, b x + 2 \, a + 2 \, c\right ) + \cos \left (2 \, a\right ) - \cos \left (2 \, c\right )\right )} \cos \left (2 \, b x + a + 3 \, c\right ) + {\left (\cos \left (2 \, a\right ) - \cos \left (2 \, c\right )\right )} \cos \left (a + c\right ) + 2 \, \cos \left (2 \, b x + 2 \, a + 2 \, c\right ) \cos \left (a + c\right ) + {\left (2 \, \sin \left (2 \, b x + 2 \, a + 2 \, c\right ) + \sin \left (2 \, a\right ) - \sin \left (2 \, c\right )\right )} \sin \left (4 \, b x + a + 5 \, c\right ) + 2 \, {\left (2 \, \sin \left (2 \, b x + 2 \, a + 2 \, c\right ) + \sin \left (2 \, a\right ) - \sin \left (2 \, c\right )\right )} \sin \left (2 \, b x + a + 3 \, c\right ) + {\left (\sin \left (2 \, a\right ) - \sin \left (2 \, c\right )\right )} \sin \left (a + c\right ) + 2 \, \sin \left (2 \, b x + 2 \, a + 2 \, c\right ) \sin \left (a + c\right )}{b \cos \left (4 \, b x + a + 5 \, c\right )^{2} + 4 \, b \cos \left (2 \, b x + a + 3 \, c\right )^{2} + 4 \, b \cos \left (2 \, b x + a + 3 \, c\right ) \cos \left (a + c\right ) + b \cos \left (a + c\right )^{2} + b \sin \left (4 \, b x + a + 5 \, c\right )^{2} + 4 \, b \sin \left (2 \, b x + a + 3 \, c\right )^{2} + 4 \, b \sin \left (2 \, b x + a + 3 \, c\right ) \sin \left (a + c\right ) + b \sin \left (a + c\right )^{2} + 2 \, {\left (2 \, b \cos \left (2 \, b x + a + 3 \, c\right ) + b \cos \left (a + c\right )\right )} \cos \left (4 \, b x + a + 5 \, c\right ) + 2 \, {\left (2 \, b \sin \left (2 \, b x + a + 3 \, c\right ) + b \sin \left (a + c\right )\right )} \sin \left (4 \, b x + a + 5 \, c\right )} \] Input:
integrate(sec(b*x+c)^3*sin(b*x+a),x, algorithm="maxima")
Output:
((2*cos(2*b*x + 2*a + 2*c) + cos(2*a) - cos(2*c))*cos(4*b*x + a + 5*c) + 2 *(2*cos(2*b*x + 2*a + 2*c) + cos(2*a) - cos(2*c))*cos(2*b*x + a + 3*c) + ( cos(2*a) - cos(2*c))*cos(a + c) + 2*cos(2*b*x + 2*a + 2*c)*cos(a + c) + (2 *sin(2*b*x + 2*a + 2*c) + sin(2*a) - sin(2*c))*sin(4*b*x + a + 5*c) + 2*(2 *sin(2*b*x + 2*a + 2*c) + sin(2*a) - sin(2*c))*sin(2*b*x + a + 3*c) + (sin (2*a) - sin(2*c))*sin(a + c) + 2*sin(2*b*x + 2*a + 2*c)*sin(a + c))/(b*cos (4*b*x + a + 5*c)^2 + 4*b*cos(2*b*x + a + 3*c)^2 + 4*b*cos(2*b*x + a + 3*c )*cos(a + c) + b*cos(a + c)^2 + b*sin(4*b*x + a + 5*c)^2 + 4*b*sin(2*b*x + a + 3*c)^2 + 4*b*sin(2*b*x + a + 3*c)*sin(a + c) + b*sin(a + c)^2 + 2*(2* b*cos(2*b*x + a + 3*c) + b*cos(a + c))*cos(4*b*x + a + 5*c) + 2*(2*b*sin(2 *b*x + a + 3*c) + b*sin(a + c))*sin(4*b*x + a + 5*c))
Leaf count of result is larger than twice the leaf count of optimal. 174 vs. \(2 (36) = 72\).
Time = 0.15 (sec) , antiderivative size = 174, normalized size of antiderivative = 4.58 \[ \int \sec ^3(c+b x) \sin (a+b x) \, dx=\frac {\tan \left (b x + c\right )^{2} \tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - \tan \left (b x + c\right )^{2} \tan \left (\frac {1}{2} \, a\right )^{2} + 4 \, \tan \left (b x + c\right )^{2} \tan \left (\frac {1}{2} \, a\right ) \tan \left (\frac {1}{2} \, c\right ) + 4 \, \tan \left (b x + c\right ) \tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - \tan \left (b x + c\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - 4 \, \tan \left (b x + c\right ) \tan \left (\frac {1}{2} \, a\right ) \tan \left (\frac {1}{2} \, c\right )^{2} + \tan \left (b x + c\right )^{2} + 4 \, \tan \left (b x + c\right ) \tan \left (\frac {1}{2} \, a\right ) - 4 \, \tan \left (b x + c\right ) \tan \left (\frac {1}{2} \, c\right )}{2 \, {\left (\tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, a\right )^{2} + \tan \left (\frac {1}{2} \, c\right )^{2} + 1\right )} b} \] 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 = 45, normalized size of antiderivative = 1.18 \[ \int \sec ^3(c+b x) \sin (a+b x) \, dx=\frac {\cos \left (b x +c \right ) \cos \left (b x +a \right )-\sin \left (b x +c \right ) \sin \left (b x +a \right )}{2 b \left (\sin \left (b x +c \right )^{2}-1\right )} \] 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))