\(\int \cos ^5(c+d x) (a+a \sec (c+d x))^2 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [321]

Optimal result

Integrand size = 40, antiderivative size = 135 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} a^2 (7 B+8 C) x+\frac {a^2 (4 B+5 C) \sin (c+d x)}{3 d}+\frac {a^2 (7 B+8 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 (5 B+4 C) \cos ^2(c+d x) \sin (c+d x)}{12 d}+\frac {B \cos ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{4 d} \]

[Out]

1/8*a^2*(7*B+8*C)*x+1/3*a^2*(4*B+5*C)*sin(d*x+c)/d+1/8*a^2*(7*B+8*C)*cos(d*x+c)*sin(d*x+c)/d+1/12*a^2*(5*B+4*C
)*cos(d*x+c)^2*sin(d*x+c)/d+1/4*B*cos(d*x+c)^3*(a^2+a^2*sec(d*x+c))*sin(d*x+c)/d

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {4157, 4102, 4081, 3872, 2715, 8, 2717} \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 (4 B+5 C) \sin (c+d x)}{3 d}+\frac {a^2 (5 B+4 C) \sin (c+d x) \cos ^2(c+d x)}{12 d}+\frac {a^2 (7 B+8 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {B \sin (c+d x) \cos ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{4 d}+\frac {1}{8} a^2 x (7 B+8 C) \]

[In]

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

[Out]

(a^2*(7*B + 8*C)*x)/8 + (a^2*(4*B + 5*C)*Sin[c + d*x])/(3*d) + (a^2*(7*B + 8*C)*Cos[c + d*x]*Sin[c + d*x])/(8*
d) + (a^2*(5*B + 4*C)*Cos[c + d*x]^2*Sin[c + d*x])/(12*d) + (B*Cos[c + d*x]^3*(a^2 + a^2*Sec[c + d*x])*Sin[c +
d*x])/(4*d)

Rule 8

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

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 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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 4102

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x]
- Dist[b/(a*d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*(m - n - 1) - b*B*n - (a*
B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2
- b^2, 0] && GtQ[m, 1/2] && LtQ[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+a \sec (c+d x))^2 (B+C \sec (c+d x)) \, dx \\ & = \frac {B \cos ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{4 d}+\frac {1}{4} \int \cos ^3(c+d x) (a+a \sec (c+d x)) (a (5 B+4 C)+2 a (B+2 C) \sec (c+d x)) \, dx \\ & = \frac {a^2 (5 B+4 C) \cos ^2(c+d x) \sin (c+d x)}{12 d}+\frac {B \cos ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{4 d}-\frac {1}{12} \int \cos ^2(c+d x) \left (-3 a^2 (7 B+8 C)-4 a^2 (4 B+5 C) \sec (c+d x)\right ) \, dx \\ & = \frac {a^2 (5 B+4 C) \cos ^2(c+d x) \sin (c+d x)}{12 d}+\frac {B \cos ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{4 d}+\frac {1}{3} \left (a^2 (4 B+5 C)\right ) \int \cos (c+d x) \, dx+\frac {1}{4} \left (a^2 (7 B+8 C)\right ) \int \cos ^2(c+d x) \, dx \\ & = \frac {a^2 (4 B+5 C) \sin (c+d x)}{3 d}+\frac {a^2 (7 B+8 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 (5 B+4 C) \cos ^2(c+d x) \sin (c+d x)}{12 d}+\frac {B \cos ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{4 d}+\frac {1}{8} \left (a^2 (7 B+8 C)\right ) \int 1 \, dx \\ & = \frac {1}{8} a^2 (7 B+8 C) x+\frac {a^2 (4 B+5 C) \sin (c+d x)}{3 d}+\frac {a^2 (7 B+8 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 (5 B+4 C) \cos ^2(c+d x) \sin (c+d x)}{12 d}+\frac {B \cos ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.64 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 (84 B c+84 B d x+96 C d x+24 (6 B+7 C) \sin (c+d x)+48 (B+C) \sin (2 (c+d x))+16 B \sin (3 (c+d x))+8 C \sin (3 (c+d x))+3 B \sin (4 (c+d x)))}{96 d} \]

[In]

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

[Out]

(a^2*(84*B*c + 84*B*d*x + 96*C*d*x + 24*(6*B + 7*C)*Sin[c + d*x] + 48*(B + C)*Sin[2*(c + d*x)] + 16*B*Sin[3*(c
+ d*x)] + 8*C*Sin[3*(c + d*x)] + 3*B*Sin[4*(c + d*x)]))/(96*d)

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.56

method	result	size
parallelrisch	\(\frac {a^{2} \left (16 \left (B +C \right ) \sin \left (2 d x +2 c \right )+\frac {8 \left (2 B +C \right ) \sin \left (3 d x +3 c \right )}{3}+B \sin \left (4 d x +4 c \right )+8 \left (6 B +7 C \right ) \sin \left (d x +c \right )+28 x d \left (B +\frac {8 C}{7}\right )\right )}{32 d}\)	\(75\)
risch	\(\frac {7 a^{2} B x}{8}+a^{2} x C +\frac {3 a^{2} B \sin \left (d x +c \right )}{2 d}+\frac {7 \sin \left (d x +c \right ) C \,a^{2}}{4 d}+\frac {B \,a^{2} \sin \left (4 d x +4 c \right )}{32 d}+\frac {B \,a^{2} \sin \left (3 d x +3 c \right )}{6 d}+\frac {\sin \left (3 d x +3 c \right ) C \,a^{2}}{12 d}+\frac {B \,a^{2} \sin \left (2 d x +2 c \right )}{2 d}+\frac {\sin \left (2 d x +2 c \right ) C \,a^{2}}{2 d}\)	\(135\)
derivativedivides	\(\frac {B \,a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \,a^{2} \sin \left (d x +c \right )+\frac {2 B \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+2 C \,a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {C \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(154\)
default	\(\frac {B \,a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \,a^{2} \sin \left (d x +c \right )+\frac {2 B \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+2 C \,a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {C \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(154\)
norman	\(\frac {-\frac {a^{2} \left (7 B +8 C \right ) x}{8}+\frac {5 a^{2} \left (7 B +8 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{12 d}+\frac {a^{2} \left (7 B +8 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{4 d}-\frac {a^{2} \left (7 B +8 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{4}+\frac {a^{2} \left (7 B +8 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4}+\frac {3 a^{2} \left (7 B +8 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{4}-\frac {3 a^{2} \left (7 B +8 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{4}-\frac {a^{2} \left (7 B +8 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{4}+\frac {a^{2} \left (7 B +8 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{4}+\frac {a^{2} \left (7 B +8 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}}{8}-\frac {a^{2} \left (17 B -8 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{12 d}-\frac {a^{2} \left (25 B +24 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {a^{2} \left (49 B +152 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}+\frac {a^{2} \left (67 B +8 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}-\frac {a^{2} \left (71 B +40 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{12 d}+\frac {a^{2} \left (89 B +184 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}\)	\(420\)

[In]

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

[Out]

1/32*a^2*(16*(B+C)*sin(2*d*x+2*c)+8/3*(2*B+C)*sin(3*d*x+3*c)+B*sin(4*d*x+4*c)+8*(6*B+7*C)*sin(d*x+c)+28*x*d*(B
+8/7*C))/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.67 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (7 \, B + 8 \, C\right )} a^{2} d x + {\left (6 \, B a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (2 \, B + C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right ) + 8 \, {\left (4 \, B + 5 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{24 \, d} \]

[In]

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

[Out]

1/24*(3*(7*B + 8*C)*a^2*d*x + (6*B*a^2*cos(d*x + c)^3 + 8*(2*B + C)*a^2*cos(d*x + c)^2 + 3*(7*B + 8*C)*a^2*cos
(d*x + c) + 8*(4*B + 5*C)*a^2)*sin(d*x + c))/d

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.07 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {64 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} + 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} - 48 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} - 96 \, C a^{2} \sin \left (d x + c\right )}{96 \, d} \]

[In]

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

[Out]

-1/96*(64*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^2 - 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*
B*a^2 - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^2 + 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^2 - 48*(2*d*x + 2
*c + sin(2*d*x + 2*c))*C*a^2 - 96*C*a^2*sin(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.30 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (7 \, B a^{2} + 8 \, C a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (21 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 77 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 88 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 83 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 136 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 75 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 72 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]

[In]

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

[Out]

1/24*(3*(7*B*a^2 + 8*C*a^2)*(d*x + c) + 2*(21*B*a^2*tan(1/2*d*x + 1/2*c)^7 + 24*C*a^2*tan(1/2*d*x + 1/2*c)^7 +
77*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 88*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 83*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 136*C*a
^2*tan(1/2*d*x + 1/2*c)^3 + 75*B*a^2*tan(1/2*d*x + 1/2*c) + 72*C*a^2*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 = 16.19 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.99 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {7\,B\,a^2\,x}{8}+C\,a^2\,x+\frac {3\,B\,a^2\,\sin \left (c+d\,x\right )}{2\,d}+\frac {7\,C\,a^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {B\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{6\,d}+\frac {B\,a^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {C\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {C\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d} \]

[In]

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

[Out]

(7*B*a^2*x)/8 + C*a^2*x + (3*B*a^2*sin(c + d*x))/(2*d) + (7*C*a^2*sin(c + d*x))/(4*d) + (B*a^2*sin(2*c + 2*d*x
))/(2*d) + (B*a^2*sin(3*c + 3*d*x))/(6*d) + (B*a^2*sin(4*c + 4*d*x))/(32*d) + (C*a^2*sin(2*c + 2*d*x))/(2*d) +
(C*a^2*sin(3*c + 3*d*x))/(12*d)