\(\int \cos ^3(c+d x) (a+a \sec (c+d x)) (A+C \sec ^2(c+d x)) \, dx\) [90]

Optimal result

Integrand size = 31, antiderivative size = 77 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} a (A+2 C) x+\frac {a (2 A+3 C) \sin (c+d x)}{3 d}+\frac {a A \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a A \cos ^2(c+d x) \sin (c+d x)}{3 d} \]

[Out]

1/2*a*(A+2*C)*x+1/3*a*(2*A+3*C)*sin(d*x+c)/d+1/2*a*A*cos(d*x+c)*sin(d*x+c)/d+1/3*a*A*cos(d*x+c)^2*sin(d*x+c)/d

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {4160, 4132, 2717, 4130, 8} \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a (2 A+3 C) \sin (c+d x)}{3 d}+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d}+\frac {a A \sin (c+d x) \cos (c+d x)}{2 d}+\frac {1}{2} a x (A+2 C) \]

[In]

Int[Cos[c + d*x]^3*(a + a*Sec[c + d*x])*(A + C*Sec[c + d*x]^2),x]

[Out]

(a*(A + 2*C)*x)/2 + (a*(2*A + 3*C)*Sin[c + d*x])/(3*d) + (a*A*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (a*A*Cos[c +
d*x]^2*Sin[c + d*x])/(3*d)

Rule 8

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

Rule 2717

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

Rule 4130

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e
+ f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

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 ^2(c+d x) \sin (c+d x)}{3 d}-\frac {1}{3} \int \cos ^2(c+d x) \left (-3 a A-a (2 A+3 C) \sec (c+d x)-3 a C \sec ^2(c+d x)\right ) \, dx \\ & = \frac {a A \cos ^2(c+d x) \sin (c+d x)}{3 d}-\frac {1}{3} \int \cos ^2(c+d x) \left (-3 a A-3 a C \sec ^2(c+d x)\right ) \, dx+\frac {1}{3} (a (2 A+3 C)) \int \cos (c+d x) \, dx \\ & = \frac {a (2 A+3 C) \sin (c+d x)}{3 d}+\frac {a A \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a A \cos ^2(c+d x) \sin (c+d x)}{3 d}+\frac {1}{2} (a (A+2 C)) \int 1 \, dx \\ & = \frac {1}{2} a (A+2 C) x+\frac {a (2 A+3 C) \sin (c+d x)}{3 d}+\frac {a A \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a A \cos ^2(c+d x) \sin (c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.77 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a (6 A c+6 A d x+12 C d x+3 (3 A+4 C) \sin (c+d x)+3 A \sin (2 (c+d x))+A \sin (3 (c+d x)))}{12 d} \]

[In]

Integrate[Cos[c + d*x]^3*(a + a*Sec[c + d*x])*(A + C*Sec[c + d*x]^2),x]

[Out]

(a*(6*A*c + 6*A*d*x + 12*C*d*x + 3*(3*A + 4*C)*Sin[c + d*x] + 3*A*Sin[2*(c + d*x)] + A*Sin[3*(c + d*x)]))/(12*
d)

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(\frac {a \left (\frac {A \sin \left (2 d x +2 c \right )}{2}+\frac {A \sin \left (3 d x +3 c \right )}{6}+\left (\frac {3 A}{2}+2 C \right ) \sin \left (d x +c \right )+\left (A +2 C \right ) x d \right )}{2 d}\)	\(54\)
derivativedivides	\(\frac {\frac {a A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a A \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C a \sin \left (d x +c \right )+C a \left (d x +c \right )}{d}\)	\(68\)
default	\(\frac {\frac {a A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a A \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C a \sin \left (d x +c \right )+C a \left (d x +c \right )}{d}\)	\(68\)
risch	\(\frac {a A x}{2}+a x C +\frac {3 a A \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (d x +c \right ) C a}{d}+\frac {a A \sin \left (3 d x +3 c \right )}{12 d}+\frac {a A \sin \left (2 d x +2 c \right )}{4 d}\)	\(68\)
norman	\(\frac {\frac {a \left (A +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {a \left (3 A +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a \left (A +2 C \right ) x}{2}-\frac {14 a A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {2 a A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}+\frac {4 a \left (A -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}+\frac {a \left (A +2 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}-a \left (A +2 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-a \left (A +2 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {a \left (A +2 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2}+\frac {a \left (A +2 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{2}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}\)	\(242\)

[In]

int(cos(d*x+c)^3*(a+a*sec(d*x+c))*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/2*a*(1/2*A*sin(2*d*x+2*c)+1/6*A*sin(3*d*x+3*c)+(3/2*A+2*C)*sin(d*x+c)+(A+2*C)*x*d)/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.73 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (A + 2 \, C\right )} a d x + {\left (2 \, A a \cos \left (d x + c\right )^{2} + 3 \, A a \cos \left (d x + c\right ) + 2 \, {\left (2 \, A + 3 \, C\right )} a\right )} \sin \left (d x + c\right )}{6 \, d} \]

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=a \left (\int A \cos ^{3}{\left (c + d x \right )}\, dx + \int A \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int C \cos ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{3}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx\right ) \]

[In]

integrate(cos(d*x+c)**3*(a+a*sec(d*x+c))*(A+C*sec(d*x+c)**2),x)

[Out]

a*(Integral(A*cos(c + d*x)**3, x) + Integral(A*cos(c + d*x)**3*sec(c + d*x), x) + Integral(C*cos(c + d*x)**3*s
ec(c + d*x)**2, x) + Integral(C*cos(c + d*x)**3*sec(c + d*x)**3, x))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.87 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 12 \, {\left (d x + c\right )} C a - 12 \, C a \sin \left (d x + c\right )}{12 \, d} \]

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a - 12*(d*x + c)*C*a - 1
2*C*a*sin(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.62 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (A a + 2 \, C a\right )} {\left (d x + c\right )} + \frac {2 \, {\left (3 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(3*(A*a + 2*C*a)*(d*x + c) + 2*(3*A*a*tan(1/2*d*x + 1/2*c)^5 + 6*C*a*tan(1/2*d*x + 1/2*c)^5 + 4*A*a*tan(1/
2*d*x + 1/2*c)^3 + 12*C*a*tan(1/2*d*x + 1/2*c)^3 + 9*A*a*tan(1/2*d*x + 1/2*c) + 6*C*a*tan(1/2*d*x + 1/2*c))/(t
an(1/2*d*x + 1/2*c)^2 + 1)^3)/d

Mupad [B] (verification not implemented)

Time = 15.97 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.87 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {A\,a\,x}{2}+C\,a\,x+\frac {3\,A\,a\,\sin \left (c+d\,x\right )}{4\,d}+\frac {C\,a\,\sin \left (c+d\,x\right )}{d}+\frac {A\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,a\,\sin \left (3\,c+3\,d\,x\right )}{12\,d} \]

[In]

int(cos(c + d*x)^3*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x)),x)

[Out]

(A*a*x)/2 + C*a*x + (3*A*a*sin(c + d*x))/(4*d) + (C*a*sin(c + d*x))/d + (A*a*sin(2*c + 2*d*x))/(4*d) + (A*a*si
n(3*c + 3*d*x))/(12*d)