\(\int \sec ^5(c+b x) \sin (a+b x) \, dx\) [153]

Optimal result

Integrand size = 15, antiderivative size = 59 \[ \int \sec ^5(c+b x) \sin (a+b x) \, dx=\frac {\cos (a-c) \sec ^4(c+b x)}{4 b}+\frac {\sin (a-c) \tan (c+b x)}{b}+\frac {\sin (a-c) \tan ^3(c+b x)}{3 b} \] Output:

1/4*cos(a-c)*sec(b*x+c)^4/b+sin(a-c)*tan(b*x+c)/b+1/3*sin(a-c)*tan(b*x+c)^ 
3/b
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.81 \[ \int \sec ^5(c+b x) \sin (a+b x) \, dx=\frac {\sec (c) \sec ^4(c+b x) (3 \cos (a)+\sin (a-c) (4 \sin (c+2 b x)+\sin (3 c+4 b x)))}{12 b} \] Input:

Integrate[Sec[c + b*x]^5*Sin[a + b*x],x]
 

Output:

(Sec[c]*Sec[c + b*x]^4*(3*Cos[a] + Sin[a - c]*(4*Sin[c + 2*b*x] + Sin[3*c 
+ 4*b*x])))/(12*b)
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.92, 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, 2009}

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.

\(\displaystyle \int \sin (a+b x) \sec ^5(b x+c) \, dx\)

\(\Big \downarrow \) 5091

\(\displaystyle \sin (a-c) \int \sec ^4(c+b x)dx+\cos (a-c) \int \sec ^4(c+b x) \tan (c+b x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sin (a-c) \int \csc \left (c+b x+\frac {\pi }{2}\right )^4dx+\cos (a-c) \int \sec (c+b x)^4 \tan (c+b x)dx\)

\(\Big \downarrow \) 3086

\(\displaystyle \frac {\cos (a-c) \int \sec ^3(c+b x)d\sec (c+b x)}{b}+\sin (a-c) \int \csc \left (c+b x+\frac {\pi }{2}\right )^4dx\)

\(\Big \downarrow \) 15

\(\displaystyle \sin (a-c) \int \csc \left (c+b x+\frac {\pi }{2}\right )^4dx+\frac {\cos (a-c) \sec ^4(b x+c)}{4 b}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {\cos (a-c) \sec ^4(b x+c)}{4 b}-\frac {\sin (a-c) \int \left (\tan ^2(c+b x)+1\right )d(-\tan (c+b x))}{b}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\cos (a-c) \sec ^4(b x+c)}{4 b}-\frac {\sin (a-c) \left (-\frac {1}{3} \tan ^3(b x+c)-\tan (b x+c)\right )}{b}\)

Input:

Int[Sec[c + b*x]^5*Sin[a + b*x],x]
 

Output:

(Cos[a - c]*Sec[c + b*x]^4)/(4*b) - (Sin[a - c]*(-Tan[c + b*x] - Tan[c + b 
*x]^3/3))/b
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3086

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])
 

rule 4254

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]
 

rule 5091

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]
 

Maple [A] (verified)

Time = 7.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.29

method	result	size
parallelrisch	\(\frac {-5 \cos \left (4 b x +4 c \right )-20 \cos \left (2 b x +2 c \right )-15-32 \cos \left (2 b x +a +c \right )-8 \cos \left (4 b x +a +3 c \right )}{12 b \left (\cos \left (4 b x +4 c \right )+3+4 \cos \left (2 b x +2 c \right )\right )}\)	\(76\)
risch	\(\frac {4 \,{\mathrm e}^{i \left (4 b x +9 a +3 c \right )}+\frac {8 \,{\mathrm e}^{i \left (2 b x +9 a +c \right )}}{3}-\frac {8 \,{\mathrm e}^{i \left (2 b x +7 a +3 c \right )}}{3}+\frac {2 \,{\mathrm e}^{i \left (9 a -c \right )}}{3}-\frac {2 \,{\mathrm e}^{i \left (7 a +c \right )}}{3}}{\left ({\mathrm e}^{2 i \left (b x +a +c \right )}+{\mathrm e}^{2 i a}\right )^{4} b}\)	\(96\)
default	\(\frac {-\frac {1}{\left (\sin \left (a \right ) \cos \left (c \right )-\cos \left (a \right ) \sin \left (c \right )\right )^{4} \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 {3 \cos \left (a \right ) \cos \left (c \right )+3 \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 )^{4} \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}}+\frac {\left (\cos \left (a \right ) \cos \left (c \right )+\sin \left (a \right ) \sin \left (c \right )\right ) \left (\cos \left (a \right )^{2} \cos \left (c \right )^{2}+\sin \left (c \right )^{2} \cos \left (a \right )^{2}+\cos \left (c \right )^{2} \sin \left (a \right )^{2}+\sin \left (a \right )^{2} \sin \left (c \right )^{2}\right )}{4 \left (\sin \left (a \right ) \cos \left (c \right )-\cos \left (a \right ) \sin \left (c \right )\right )^{4} \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 )^{4}}-\frac {\cos \left (c \right )^{2} \sin \left (a \right )^{2}+3 \cos \left (a \right )^{2} \cos \left (c \right )^{2}+4 \cos \left (a \right ) \cos \left (c \right ) \sin \left (a \right ) \sin \left (c \right )+3 \sin \left (a \right )^{2} \sin \left (c \right )^{2}+\sin \left (c \right )^{2} \cos \left (a \right )^{2}}{3 \left (\sin \left (a \right ) \cos \left (c \right )-\cos \left (a \right ) \sin \left (c \right )\right )^{4} \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 )^{3}}}{b}\)	\(324\)

Input:

int(sec(b*x+c)^5*sin(b*x+a),x,method=_RETURNVERBOSE)
 

Output:

1/12*(-5*cos(4*b*x+4*c)-20*cos(2*b*x+2*c)-15-32*cos(2*b*x+a+c)-8*cos(4*b*x 
+a+3*c))/b/(cos(4*b*x+4*c)+3+4*cos(2*b*x+2*c))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.90 \[ \int \sec ^5(c+b x) \sin (a+b x) \, dx=-\frac {4 \, {\left (2 \, \cos \left (b x + c\right )^{3} + \cos \left (b x + c\right )\right )} \sin \left (b x + c\right ) \sin \left (-a + c\right ) - 3 \, \cos \left (-a + c\right )}{12 \, b \cos \left (b x + c\right )^{4}} \] Input:

integrate(sec(b*x+c)^5*sin(b*x+a),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \sec ^5(c+b x) \sin (a+b x) \, dx=\text {Timed out} \] Input:

integrate(sec(b*x+c)**5*sin(b*x+a),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1074 vs. \(2 (55) = 110\).

Time = 0.05 (sec) , antiderivative size = 1074, normalized size of antiderivative = 18.20 \[ \int \sec ^5(c+b x) \sin (a+b x) \, dx=\text {Too large to display} \] Input:

integrate(sec(b*x+c)^5*sin(b*x+a),x, algorithm="maxima")
 

Output:

2/3*((6*cos(4*b*x + 2*a + 4*c) + 4*cos(2*b*x + 2*a + 2*c) - 4*cos(2*b*x + 
4*c) + cos(2*a) - cos(2*c))*cos(8*b*x + a + 9*c) + 4*(6*cos(4*b*x + 2*a + 
4*c) + 4*cos(2*b*x + 2*a + 2*c) - 4*cos(2*b*x + 4*c) + cos(2*a) - cos(2*c) 
)*cos(6*b*x + a + 7*c) + 6*(4*cos(2*b*x + a + 3*c) + cos(a + c))*cos(4*b*x 
 + 2*a + 4*c) + 6*(6*cos(4*b*x + 2*a + 4*c) + 4*cos(2*b*x + 2*a + 2*c) - 4 
*cos(2*b*x + 4*c) + cos(2*a) - cos(2*c))*cos(4*b*x + a + 5*c) + 4*(4*cos(2 
*b*x + 2*a + 2*c) + cos(2*a) - cos(2*c))*cos(2*b*x + a + 3*c) - 4*(4*cos(2 
*b*x + a + 3*c) + cos(a + c))*cos(2*b*x + 4*c) + (cos(2*a) - cos(2*c))*cos 
(a + c) + 4*cos(2*b*x + 2*a + 2*c)*cos(a + c) + (6*sin(4*b*x + 2*a + 4*c) 
+ 4*sin(2*b*x + 2*a + 2*c) - 4*sin(2*b*x + 4*c) + sin(2*a) - sin(2*c))*sin 
(8*b*x + a + 9*c) + 4*(6*sin(4*b*x + 2*a + 4*c) + 4*sin(2*b*x + 2*a + 2*c) 
 - 4*sin(2*b*x + 4*c) + sin(2*a) - sin(2*c))*sin(6*b*x + a + 7*c) + 6*(4*s 
in(2*b*x + a + 3*c) + sin(a + c))*sin(4*b*x + 2*a + 4*c) + 6*(6*sin(4*b*x 
+ 2*a + 4*c) + 4*sin(2*b*x + 2*a + 2*c) - 4*sin(2*b*x + 4*c) + sin(2*a) - 
sin(2*c))*sin(4*b*x + a + 5*c) + 4*(4*sin(2*b*x + 2*a + 2*c) + sin(2*a) - 
sin(2*c))*sin(2*b*x + a + 3*c) - 4*(4*sin(2*b*x + a + 3*c) + sin(a + c))*s 
in(2*b*x + 4*c) + (sin(2*a) - sin(2*c))*sin(a + c) + 4*sin(2*b*x + 2*a + 2 
*c)*sin(a + c))/(b*cos(8*b*x + a + 9*c)^2 + 16*b*cos(6*b*x + a + 7*c)^2 + 
36*b*cos(4*b*x + a + 5*c)^2 + 16*b*cos(2*b*x + a + 3*c)^2 + 8*b*cos(2*b*x 
+ a + 3*c)*cos(a + c) + b*cos(a + c)^2 + b*sin(8*b*x + a + 9*c)^2 + 16*...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 327 vs. \(2 (55) = 110\).

Time = 0.15 (sec) , antiderivative size = 327, normalized size of antiderivative = 5.54 \[ \int \sec ^5(c+b x) \sin (a+b x) \, dx=\frac {3 \, \tan \left (b x + c\right )^{4} \tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - 3 \, \tan \left (b x + c\right )^{4} \tan \left (\frac {1}{2} \, a\right )^{2} + 12 \, \tan \left (b x + c\right )^{4} \tan \left (\frac {1}{2} \, a\right ) \tan \left (\frac {1}{2} \, c\right ) + 8 \, \tan \left (b x + c\right )^{3} \tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - 3 \, \tan \left (b x + c\right )^{4} \tan \left (\frac {1}{2} \, c\right )^{2} - 8 \, \tan \left (b x + c\right )^{3} \tan \left (\frac {1}{2} \, a\right ) \tan \left (\frac {1}{2} \, c\right )^{2} + 6 \, \tan \left (b x + c\right )^{2} \tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + 3 \, \tan \left (b x + c\right )^{4} + 8 \, \tan \left (b x + c\right )^{3} \tan \left (\frac {1}{2} \, a\right ) - 6 \, \tan \left (b x + c\right )^{2} \tan \left (\frac {1}{2} \, a\right )^{2} - 8 \, \tan \left (b x + c\right )^{3} \tan \left (\frac {1}{2} \, c\right ) + 24 \, \tan \left (b x + c\right )^{2} \tan \left (\frac {1}{2} \, a\right ) \tan \left (\frac {1}{2} \, c\right ) + 24 \, \tan \left (b x + c\right ) \tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - 6 \, \tan \left (b x + c\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - 24 \, \tan \left (b x + c\right ) \tan \left (\frac {1}{2} \, a\right ) \tan \left (\frac {1}{2} \, c\right )^{2} + 6 \, \tan \left (b x + c\right )^{2} + 24 \, \tan \left (b x + c\right ) \tan \left (\frac {1}{2} \, a\right ) - 24 \, \tan \left (b x + c\right ) \tan \left (\frac {1}{2} \, c\right )}{12 \, {\left (\tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, a\right )^{2} + \tan \left (\frac {1}{2} \, c\right )^{2} + 1\right )} b} \] Input:

integrate(sec(b*x+c)^5*sin(b*x+a),x, algorithm="giac")
 

Output:

1/12*(3*tan(b*x + c)^4*tan(1/2*a)^2*tan(1/2*c)^2 - 3*tan(b*x + c)^4*tan(1/ 
2*a)^2 + 12*tan(b*x + c)^4*tan(1/2*a)*tan(1/2*c) + 8*tan(b*x + c)^3*tan(1/ 
2*a)^2*tan(1/2*c) - 3*tan(b*x + c)^4*tan(1/2*c)^2 - 8*tan(b*x + c)^3*tan(1 
/2*a)*tan(1/2*c)^2 + 6*tan(b*x + c)^2*tan(1/2*a)^2*tan(1/2*c)^2 + 3*tan(b* 
x + c)^4 + 8*tan(b*x + c)^3*tan(1/2*a) - 6*tan(b*x + c)^2*tan(1/2*a)^2 - 8 
*tan(b*x + c)^3*tan(1/2*c) + 24*tan(b*x + c)^2*tan(1/2*a)*tan(1/2*c) + 24* 
tan(b*x + c)*tan(1/2*a)^2*tan(1/2*c) - 6*tan(b*x + c)^2*tan(1/2*c)^2 - 24* 
tan(b*x + c)*tan(1/2*a)*tan(1/2*c)^2 + 6*tan(b*x + c)^2 + 24*tan(b*x + c)* 
tan(1/2*a) - 24*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)
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^5(c+b x) \sin (a+b x) \, dx=\text {Hanged} \] Input:

int(sin(a + b*x)/cos(c + b*x)^5,x)
 

Output:

\text{Hanged}
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.95 \[ \int \sec ^5(c+b x) \sin (a+b x) \, dx=\frac {8 \cos \left (b x +c \right ) \cos \left (b x +a \right ) \sin \left (b x +c \right )^{2}-10 \cos \left (b x +c \right ) \cos \left (b x +a \right )+3 \sin \left (b x +c \right )^{4}-8 \sin \left (b x +c \right )^{3} \sin \left (b x +a \right )-6 \sin \left (b x +c \right )^{2}+14 \sin \left (b x +c \right ) \sin \left (b x +a \right )+3}{24 b \left (\sin \left (b x +c \right )^{4}-2 \sin \left (b x +c \right )^{2}+1\right )} \] Input:

int(sec(b*x+c)^5*sin(b*x+a),x)
 

Output:

(8*cos(b*x + c)*cos(a + b*x)*sin(b*x + c)**2 - 10*cos(b*x + c)*cos(a + b*x 
) + 3*sin(b*x + c)**4 - 8*sin(b*x + c)**3*sin(a + b*x) - 6*sin(b*x + c)**2 
 + 14*sin(b*x + c)*sin(a + b*x) + 3)/(24*b*(sin(b*x + c)**4 - 2*sin(b*x + 
c)**2 + 1))