\(\int \cos ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx\) [283]

Optimal result

Integrand size = 29, antiderivative size = 84 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {1}{2} (A b+a B) x+\frac {(2 a A+3 b B) \sin (c+d x)}{3 d}+\frac {(A b+a B) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a A \cos ^2(c+d x) \sin (c+d x)}{3 d} \]

[Out]

1/2*(A*b+B*a)*x+1/3*(2*A*a+3*B*b)*sin(d*x+c)/d+1/2*(A*b+B*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 = 0.13 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {4081, 3872, 2715, 8, 2717} \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {(2 a A+3 b B) \sin (c+d x)}{3 d}+\frac {(a B+A b) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {1}{2} x (a B+A b)+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d} \]

[In]

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

[Out]

((A*b + a*B)*x)/2 + ((2*a*A + 3*b*B)*Sin[c + d*x])/(3*d) + ((A*b + a*B)*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 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]

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) (-3 (A b+a B)-(2 a A+3 b B) \sec (c+d x)) \, dx \\ & = \frac {a A \cos ^2(c+d x) \sin (c+d x)}{3 d}-(-A b-a B) \int \cos ^2(c+d x) \, dx-\frac {1}{3} (-2 a A-3 b B) \int \cos (c+d x) \, dx \\ & = \frac {(2 a A+3 b B) \sin (c+d x)}{3 d}+\frac {(A b+a B) \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 b-a B) \int 1 \, dx \\ & = \frac {1}{2} (A b+a B) x+\frac {(2 a A+3 b B) \sin (c+d x)}{3 d}+\frac {(A b+a B) \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 = 0.11 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.89 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {6 A b c+6 a B c+6 A b d x+6 a B d x+3 (3 a A+4 b B) \sin (c+d x)+3 (A b+a B) \sin (2 (c+d x))+a A \sin (3 (c+d x))}{12 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.74 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.77

method	result	size
parallelrisch	\(\frac {3 \left (A b +B a \right ) \sin \left (2 d x +2 c \right )+a A \sin \left (3 d x +3 c \right )+3 \left (3 a A +4 B b \right ) \sin \left (d x +c \right )+6 \left (A b +B a \right ) x d}{12 d}\)	\(65\)
derivativedivides	\(\frac {\frac {a A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+A b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \sin \left (d x +c \right ) b}{d}\)	\(85\)
default	\(\frac {\frac {a A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+A b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \sin \left (d x +c \right ) b}{d}\)	\(85\)
risch	\(\frac {A b x}{2}+\frac {B a x}{2}+\frac {3 a A \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (d x +c \right ) B b}{d}+\frac {a A \sin \left (3 d x +3 c \right )}{12 d}+\frac {\sin \left (2 d x +2 c \right ) A b}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B a}{4 d}\)	\(85\)
norman	\(\frac {\left (-\frac {A b}{2}-\frac {B a}{2}\right ) x +\left (\frac {A b}{2}+\frac {B a}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-A b -B a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (A b +B a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {\left (2 a A -A b -B a +2 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {\left (2 a A -3 A b -3 B a -6 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}-\frac {\left (2 a A +A b +B a +2 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (2 a A +3 A b +3 B a -6 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\)	\(235\)

[In]

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

[Out]

1/12*(3*(A*b+B*a)*sin(2*d*x+2*c)+a*A*sin(3*d*x+3*c)+3*(3*A*a+4*B*b)*sin(d*x+c)+6*(A*b+B*a)*x*d)/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.71 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (B a + A b\right )} d x + {\left (2 \, A a \cos \left (d x + c\right )^{2} + 4 \, A a + 6 \, B b + 3 \, {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate(cos(d*x+c)**3*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x)

[Out]

Integral((A + B*sec(c + d*x))*(a + b*sec(c + d*x))*cos(c + d*x)**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.94 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b - 12 \, B b \sin \left (d x + c\right )}{12 \, d} \]

[In]

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),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))*B*a - 3*(2*d*x + 2*c + sin
(2*d*x + 2*c))*A*b - 12*B*b*sin(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (76) = 152\).

Time = 0.29 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.14 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (B a + A b\right )} {\left (d x + c\right )} + \frac {2 \, {\left (6 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

[In]

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

[Out]

1/6*(3*(B*a + A*b)*(d*x + c) + 2*(6*A*a*tan(1/2*d*x + 1/2*c)^5 - 3*B*a*tan(1/2*d*x + 1/2*c)^5 - 3*A*b*tan(1/2*
d*x + 1/2*c)^5 + 6*B*b*tan(1/2*d*x + 1/2*c)^5 + 4*A*a*tan(1/2*d*x + 1/2*c)^3 + 12*B*b*tan(1/2*d*x + 1/2*c)^3 +
6*A*a*tan(1/2*d*x + 1/2*c) + 3*B*a*tan(1/2*d*x + 1/2*c) + 3*A*b*tan(1/2*d*x + 1/2*c) + 6*B*b*tan(1/2*d*x + 1/
2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d

Mupad [B] (verification not implemented)

Time = 14.81 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {A\,b\,x}{2}+\frac {B\,a\,x}{2}+\frac {3\,A\,a\,\sin \left (c+d\,x\right )}{4\,d}+\frac {B\,b\,\sin \left (c+d\,x\right )}{d}+\frac {A\,a\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {A\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]

[In]

int(cos(c + d*x)^3*(A + B/cos(c + d*x))*(a + b/cos(c + d*x)),x)

[Out]

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