\(\int \cos ^5(c+d x) (a+b \sec (c+d x)) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [773]

Optimal result

Integrand size = 38, antiderivative size = 105 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} (3 a B+4 b C) x+\frac {(b B+a C) \sin (c+d x)}{d}+\frac {(3 a B+4 b C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a B \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {(b B+a C) \sin ^3(c+d x)}{3 d} \]

[Out]

1/8*(3*B*a+4*C*b)*x+(B*b+C*a)*sin(d*x+c)/d+1/8*(3*B*a+4*C*b)*cos(d*x+c)*sin(d*x+c)/d+1/4*a*B*cos(d*x+c)^3*sin(
d*x+c)/d-1/3*(B*b+C*a)*sin(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4157, 4081, 3872, 2713, 2715, 8} \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {(a C+b B) \sin ^3(c+d x)}{3 d}+\frac {(a C+b B) \sin (c+d x)}{d}+\frac {(3 a B+4 b C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x (3 a B+4 b C)+\frac {a B \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

[In]

Int[Cos[c + d*x]^5*(a + b*Sec[c + d*x])*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

((3*a*B + 4*b*C)*x)/8 + ((b*B + a*C)*Sin[c + d*x])/d + ((3*a*B + 4*b*C)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*
B*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) - ((b*B + a*C)*Sin[c + d*x]^3)/(3*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4081

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.)
+ (A_)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Dist[1/(d*n), Int[(d*Csc[e + f*x
])^(n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B},
x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]

Rule 4157

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rubi steps \begin{align*} \text {integral}& = \int \cos ^4(c+d x) (a+b \sec (c+d x)) (B+C \sec (c+d x)) \, dx \\ & = \frac {a B \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {1}{4} \int \cos ^3(c+d x) (-4 (b B+a C)-(3 a B+4 b C) \sec (c+d x)) \, dx \\ & = \frac {a B \cos ^3(c+d x) \sin (c+d x)}{4 d}-(-b B-a C) \int \cos ^3(c+d x) \, dx-\frac {1}{4} (-3 a B-4 b C) \int \cos ^2(c+d x) \, dx \\ & = \frac {(3 a B+4 b C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a B \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {1}{8} (-3 a B-4 b C) \int 1 \, dx-\frac {(b B+a C) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d} \\ & = \frac {1}{8} (3 a B+4 b C) x+\frac {(b B+a C) \sin (c+d x)}{d}+\frac {(3 a B+4 b C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a B \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {(b B+a C) \sin ^3(c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.87 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {36 a B c+48 b c C+36 a B d x+48 b C d x+96 (b B+a C) \sin (c+d x)-32 (b B+a C) \sin ^3(c+d x)+24 (a B+b C) \sin (2 (c+d x))+3 a B \sin (4 (c+d x))}{96 d} \]

[In]

Integrate[Cos[c + d*x]^5*(a + b*Sec[c + d*x])*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(36*a*B*c + 48*b*c*C + 36*a*B*d*x + 48*b*C*d*x + 96*(b*B + a*C)*Sin[c + d*x] - 32*(b*B + a*C)*Sin[c + d*x]^3 +
24*(a*B + b*C)*Sin[2*(c + d*x)] + 3*a*B*Sin[4*(c + d*x)])/(96*d)

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.82

method	result	size
parallelrisch	\(\frac {\left (24 a B +24 C b \right ) \sin \left (2 d x +2 c \right )+\left (8 B b +8 C a \right ) \sin \left (3 d x +3 c \right )+3 a B \sin \left (4 d x +4 c \right )+\left (72 B b +72 C a \right ) \sin \left (d x +c \right )+36 x d \left (a B +\frac {4 C b}{3}\right )}{96 d}\)	\(86\)
derivativedivides	\(\frac {a B \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {C a \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C b \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(107\)
default	\(\frac {a B \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {C a \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C b \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(107\)
risch	\(\frac {3 a B x}{8}+\frac {x C b}{2}+\frac {3 \sin \left (d x +c \right ) B b}{4 d}+\frac {3 \sin \left (d x +c \right ) C a}{4 d}+\frac {a B \sin \left (4 d x +4 c \right )}{32 d}+\frac {\sin \left (3 d x +3 c \right ) B b}{12 d}+\frac {\sin \left (3 d x +3 c \right ) C a}{12 d}+\frac {a B \sin \left (2 d x +2 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C b}{4 d}\)	\(118\)
norman	\(\frac {\left (\frac {3 a B}{8}+\frac {C b}{2}\right ) x +\left (-\frac {15 a B}{8}-\frac {5 C b}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-\frac {15 a B}{8}-\frac {5 C b}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {3 a B}{8}+\frac {C b}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (\frac {3 a B}{8}+\frac {C b}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (\frac {3 a B}{8}+\frac {C b}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (\frac {9 a B}{8}+\frac {3 C b}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {9 a B}{8}+\frac {3 C b}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}-\frac {8 \left (B b +C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {2 \left (3 a B -2 B b -2 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {2 \left (3 a B +2 B b +2 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}+\frac {\left (a B -8 B b -8 C a -12 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}-\frac {\left (a B +8 B b +8 C a -12 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {\left (5 a B -8 B b -8 C a +4 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d}+\frac {\left (5 a B +8 B b +8 C a +4 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}\)	\(407\)

[In]

int(cos(d*x+c)^5*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/96*((24*B*a+24*C*b)*sin(2*d*x+2*c)+(8*B*b+8*C*a)*sin(3*d*x+3*c)+3*a*B*sin(4*d*x+4*c)+(72*B*b+72*C*a)*sin(d*x
+c)+36*x*d*(a*B+4/3*C*b))/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.77 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (3 \, B a + 4 \, C b\right )} d x + {\left (6 \, B a \cos \left (d x + c\right )^{3} + 8 \, {\left (C a + B b\right )} \cos \left (d x + c\right )^{2} + 16 \, C a + 16 \, B b + 3 \, {\left (3 \, B a + 4 \, C b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]

[In]

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/24*(3*(3*B*a + 4*C*b)*d*x + (6*B*a*cos(d*x + c)^3 + 8*(C*a + B*b)*cos(d*x + c)^2 + 16*C*a + 16*B*b + 3*(3*B*
a + 4*C*b)*cos(d*x + c))*sin(d*x + c))/d

Sympy [F(-1)]

Timed out. \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.96 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B b + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b}{96 \, d} \]

[In]

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/96*(3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a
- 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*b + 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*b)/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 272 vs. \(2 (97) = 194\).

Time = 0.30 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.59 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (3 \, B a + 4 \, C b\right )} {\left (d x + c\right )} - \frac {2 \, {\left (15 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 9 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]

[In]

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/24*(3*(3*B*a + 4*C*b)*(d*x + c) - 2*(15*B*a*tan(1/2*d*x + 1/2*c)^7 - 24*C*a*tan(1/2*d*x + 1/2*c)^7 - 24*B*b*
tan(1/2*d*x + 1/2*c)^7 + 12*C*b*tan(1/2*d*x + 1/2*c)^7 - 9*B*a*tan(1/2*d*x + 1/2*c)^5 - 40*C*a*tan(1/2*d*x + 1
/2*c)^5 - 40*B*b*tan(1/2*d*x + 1/2*c)^5 + 12*C*b*tan(1/2*d*x + 1/2*c)^5 + 9*B*a*tan(1/2*d*x + 1/2*c)^3 - 40*C*
a*tan(1/2*d*x + 1/2*c)^3 - 40*B*b*tan(1/2*d*x + 1/2*c)^3 - 12*C*b*tan(1/2*d*x + 1/2*c)^3 - 15*B*a*tan(1/2*d*x
+ 1/2*c) - 24*C*a*tan(1/2*d*x + 1/2*c) - 24*B*b*tan(1/2*d*x + 1/2*c) - 12*C*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d
*x + 1/2*c)^2 + 1)^4)/d

Mupad [B] (verification not implemented)

Time = 17.46 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.11 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3\,B\,a\,x}{8}+\frac {C\,b\,x}{2}+\frac {3\,B\,b\,\sin \left (c+d\,x\right )}{4\,d}+\frac {3\,C\,a\,\sin \left (c+d\,x\right )}{4\,d}+\frac {B\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {B\,b\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,a\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]

[In]

int(cos(c + d*x)^5*(B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + b/cos(c + d*x)),x)

[Out]

(3*B*a*x)/8 + (C*b*x)/2 + (3*B*b*sin(c + d*x))/(4*d) + (3*C*a*sin(c + d*x))/(4*d) + (B*a*sin(2*c + 2*d*x))/(4*
d) + (B*a*sin(4*c + 4*d*x))/(32*d) + (B*b*sin(3*c + 3*d*x))/(12*d) + (C*a*sin(3*c + 3*d*x))/(12*d) + (C*b*sin(
2*c + 2*d*x))/(4*d)