\(\int \cos ^4(c+d x) (a+b \sec (c+d x)) (A+C \sec ^2(c+d x)) \, dx\) [644]

Optimal result

Integrand size = 31, antiderivative size = 95 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} a (3 A+4 C) x+\frac {b (A+C) \sin (c+d x)}{d}+\frac {a (3 A+4 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a A \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {A b \sin ^3(c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4160, 4132, 2715, 8, 4129, 3092} \[ \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a (3 A+4 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {1}{8} a x (3 A+4 C)+\frac {b (A+C) \sin (c+d x)}{d}-\frac {A b \sin ^3(c+d x)}{3 d} \]

[In]

Int[Cos[c + d*x]^4*(a + b*Sec[c + d*x])*(A + C*Sec[c + d*x]^2),x]

[Out]

(a*(3*A + 4*C)*x)/8 + (b*(A + C)*Sin[c + d*x])/d + (a*(3*A + 4*C)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*A*Cos[
c + d*x]^3*Sin[c + d*x])/(4*d) - (A*b*Sin[c + d*x]^3)/(3*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 3092

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[-f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

Rule 4129

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Int[(C + A*Sin[e + f*
x]^2)/Sin[e + f*x]^(m + 2), x] /; FreeQ[{e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && ILtQ[(m + 1)/2, 0]

Rule 4132

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 4160

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(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[A*b*n + a*(C*n + A*(n + 1))*Csc[e + f*x] + b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b,
d, e, f, A, C}, x] && LtQ[n, -1]

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.88 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {36 a A c+48 a c C+36 a A d x+48 a C d x+24 b (3 A+4 C) \sin (c+d x)+24 a (A+C) \sin (2 (c+d x))+8 A b \sin (3 (c+d x))+3 a A \sin (4 (c+d x))}{96 d} \]

[In]

Integrate[Cos[c + d*x]^4*(a + b*Sec[c + d*x])*(A + C*Sec[c + d*x]^2),x]

[Out]

(36*a*A*c + 48*a*c*C + 36*a*A*d*x + 48*a*C*d*x + 24*b*(3*A + 4*C)*Sin[c + d*x] + 24*a*(A + C)*Sin[2*(c + d*x)]
+ 8*A*b*Sin[3*(c + d*x)] + 3*a*A*Sin[4*(c + d*x)])/(96*d)

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.73

method	result	size
parallelrisch	\(\frac {24 a \left (A +C \right ) \sin \left (2 d x +2 c \right )+8 A b \sin \left (3 d x +3 c \right )+3 a A \sin \left (4 d x +4 c \right )+36 \left (A +\frac {4 C}{3}\right ) \left (a x d +2 b \sin \left (d x +c \right )\right )}{96 d}\)	\(69\)
derivativedivides	\(\frac {a A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C a \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \sin \left (d x +c \right ) b}{d}\)	\(96\)
default	\(\frac {a A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C a \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \sin \left (d x +c \right ) b}{d}\)	\(96\)
risch	\(\frac {3 a A x}{8}+\frac {a x C}{2}+\frac {3 A b \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (d x +c \right ) C b}{d}+\frac {a A \sin \left (4 d x +4 c \right )}{32 d}+\frac {A b \sin \left (3 d x +3 c \right )}{12 d}+\frac {a A \sin \left (2 d x +2 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C a}{4 d}\)	\(101\)
norman	\(\frac {\left (\frac {3}{8} a A +\frac {1}{2} C a \right ) x +\left (-\frac {3}{2} a A -2 C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-\frac {3}{8} a A -\frac {1}{2} C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-\frac {3}{8} a A -\frac {1}{2} C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {3}{4} a A +C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {3}{4} a A +C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (\frac {3}{8} a A +\frac {1}{2} C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}-\frac {\left (5 a A -8 A b +4 C a -8 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}+\frac {\left (5 a A +8 A b +4 C a +8 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (21 a A -8 A b -12 C a -24 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{6 d}-\frac {\left (21 a A +8 A b -12 C a +24 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}+\frac {\left (39 a A -8 A b +12 C a +24 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}-\frac {\left (39 a A +8 A b +12 C a -24 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}\)	\(370\)

[In]

int(cos(d*x+c)^4*(a+b*sec(d*x+c))*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.80 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (3 \, A + 4 \, C\right )} a d x + {\left (6 \, A a \cos \left (d x + c\right )^{3} + 8 \, A b \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + 4 \, C\right )} a \cos \left (d x + c\right ) + 8 \, {\left (2 \, A + 3 \, C\right )} b\right )} \sin \left (d x + c\right )}{24 \, d} \]

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right ) \cos ^{4}{\left (c + d x \right )}\, dx \]

[In]

integrate(cos(d*x+c)**4*(a+b*sec(d*x+c))*(A+C*sec(d*x+c)**2),x)

[Out]

Integral((A + C*sec(c + d*x)**2)*(a + b*sec(c + d*x))*cos(c + d*x)**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.95 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b + 96 \, C b \sin \left (d x + c\right )}{96 \, d} \]

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))*(A+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))*A*a + 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a
- 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*b + 96*C*b*sin(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 272 vs. \(2 (87) = 174\).

Time = 0.32 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.86 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (3 \, A a + 4 \, C a\right )} {\left (d x + c\right )} - \frac {2 \, {\left (15 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 72 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/24*(3*(3*A*a + 4*C*a)*(d*x + c) - 2*(15*A*a*tan(1/2*d*x + 1/2*c)^7 + 12*C*a*tan(1/2*d*x + 1/2*c)^7 - 24*A*b*
tan(1/2*d*x + 1/2*c)^7 - 24*C*b*tan(1/2*d*x + 1/2*c)^7 - 9*A*a*tan(1/2*d*x + 1/2*c)^5 + 12*C*a*tan(1/2*d*x + 1
/2*c)^5 - 40*A*b*tan(1/2*d*x + 1/2*c)^5 - 72*C*b*tan(1/2*d*x + 1/2*c)^5 + 9*A*a*tan(1/2*d*x + 1/2*c)^3 - 12*C*
a*tan(1/2*d*x + 1/2*c)^3 - 40*A*b*tan(1/2*d*x + 1/2*c)^3 - 72*C*b*tan(1/2*d*x + 1/2*c)^3 - 15*A*a*tan(1/2*d*x
+ 1/2*c) - 12*C*a*tan(1/2*d*x + 1/2*c) - 24*A*b*tan(1/2*d*x + 1/2*c) - 24*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 = 18.54 (sec) , antiderivative size = 215, normalized size of antiderivative = 2.26 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (2\,A\,b-\frac {5\,A\,a}{4}-C\,a+2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {3\,A\,a}{4}+\frac {10\,A\,b}{3}-C\,a+6\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {10\,A\,b}{3}-\frac {3\,A\,a}{4}+C\,a+6\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,A\,a}{4}+2\,A\,b+C\,a+2\,C\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,A+4\,C\right )}{4\,\left (\frac {3\,A\,a}{4}+C\,a\right )}\right )\,\left (3\,A+4\,C\right )}{4\,d} \]

[In]

int(cos(c + d*x)^4*(A + C/cos(c + d*x)^2)*(a + b/cos(c + d*x)),x)

[Out]

(tan(c/2 + (d*x)/2)*((5*A*a)/4 + 2*A*b + C*a + 2*C*b) - tan(c/2 + (d*x)/2)^7*((5*A*a)/4 - 2*A*b + C*a - 2*C*b)
+ tan(c/2 + (d*x)/2)^3*((10*A*b)/3 - (3*A*a)/4 + C*a + 6*C*b) + tan(c/2 + (d*x)/2)^5*((3*A*a)/4 + (10*A*b)/3
- C*a + 6*C*b))/(d*(4*tan(c/2 + (d*x)/2)^2 + 6*tan(c/2 + (d*x)/2)^4 + 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)
/2)^8 + 1)) + (a*atan((a*tan(c/2 + (d*x)/2)*(3*A + 4*C))/(4*((3*A*a)/4 + C*a)))*(3*A + 4*C))/(4*d)