\(\int \cos ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx\) [50]

Optimal result

Integrand size = 29, antiderivative size = 77 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {1}{2} a (A+B) x+\frac {a (2 A+3 B) \sin (c+d x)}{3 d}+\frac {a (A+B) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a A \cos ^2(c+d x) \sin (c+d x)}{3 d} \]

[Out]

1/2*a*(A+B)*x+1/3*a*(2*A+3*B)*sin(d*x+c)/d+1/2*a*(A+B)*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 = 0.17 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {4081, 3872, 2715, 8, 2717} \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a (2 A+3 B) \sin (c+d x)}{3 d}+\frac {a (A+B) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {1}{2} a x (A+B)+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d} \]

[In]

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

[Out]

(a*(A + B)*x)/2 + (a*(2*A + 3*B)*Sin[c + d*x])/(3*d) + (a*(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 (A+B)-a (2 A+3 B) \sec (c+d x)) \, dx \\ & = \frac {a A \cos ^2(c+d x) \sin (c+d x)}{3 d}+(a (A+B)) \int \cos ^2(c+d x) \, dx+\frac {1}{3} (a (2 A+3 B)) \int \cos (c+d x) \, dx \\ & = \frac {a (2 A+3 B) \sin (c+d x)}{3 d}+\frac {a (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 (A+B)) \int 1 \, dx \\ & = \frac {1}{2} a (A+B) x+\frac {a (2 A+3 B) \sin (c+d x)}{3 d}+\frac {a (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 = 0.13 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.84 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a (6 A c+6 B c+6 A d x+6 B d x+3 (3 A+4 B) \sin (c+d x)+3 (A+B) \sin (2 (c+d x))+A \sin (3 (c+d x)))}{12 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.52 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(\frac {\left (\frac {\left (A +B \right ) \sin \left (2 d x +2 c \right )}{2}+\frac {A \sin \left (3 d x +3 c \right )}{6}+\left (\frac {3 A}{2}+2 B \right ) \sin \left (d x +c \right )+\left (A +B \right ) x d \right ) a}{2 d}\)	\(54\)
derivativedivides	\(\frac {\frac {a A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B a \sin \left (d x +c \right )}{d}\)	\(85\)
default	\(\frac {\frac {a A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B a \sin \left (d x +c \right )}{d}\)	\(85\)
risch	\(\frac {a A x}{2}+\frac {a x B}{2}+\frac {3 a A \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (d x +c \right ) B a}{d}+\frac {a A \sin \left (3 d x +3 c \right )}{12 d}+\frac {a A \sin \left (2 d x +2 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B a}{4 d}\)	\(85\)
norman	\(\frac {\frac {a \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+a \left (A +B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-\frac {a \left (A +B \right ) x}{2}-\frac {3 a \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-a \left (A +B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\frac {a \left (A +B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2}+\frac {a \left (A +9 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}+\frac {a \left (5 A -3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\)	\(176\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.73 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (A + B\right )} a d x + {\left (2 \, A a \cos \left (d x + c\right )^{2} + 3 \, {\left (A + B\right )} a \cos \left (d x + c\right ) + 2 \, {\left (2 \, A + 3 \, B\right )} a\right )} \sin \left (d x + c\right )}{6 \, d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.03 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a - 12 \, B a \sin \left (d x + c\right )}{12 \, d} \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.61 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (A a + B a\right )} {\left (d x + c\right )} + \frac {2 \, {\left (3 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

[In]

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

[Out]

1/6*(3*(A*a + B*a)*(d*x + c) + 2*(3*A*a*tan(1/2*d*x + 1/2*c)^5 + 3*B*a*tan(1/2*d*x + 1/2*c)^5 + 4*A*a*tan(1/2*
d*x + 1/2*c)^3 + 12*B*a*tan(1/2*d*x + 1/2*c)^3 + 9*A*a*tan(1/2*d*x + 1/2*c) + 9*B*a*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 = 13.62 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.09 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {A\,a\,x}{2}+\frac {B\,a\,x}{2}+\frac {3\,A\,a\,\sin \left (c+d\,x\right )}{4\,d}+\frac {B\,a\,\sin \left (c+d\,x\right )}{d}+\frac {A\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,a\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {B\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]

[In]

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

[Out]

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