Integrand size = 29, antiderivative size = 84 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {1}{2} (A b+a B) x+\frac {(2 a A+3 b B) \sin (c+d x)}{3 d}+\frac {(A b+a B) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a A \cos ^2(c+d x) \sin (c+d x)}{3 d} \] Output:
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)*si n(d*x+c)/d+1/3*a*A*cos(d*x+c)^2*sin(d*x+c)/d
Time = 0.19 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.89 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {6 A b c+6 a B c+6 A b d x+6 a B d x+3 (3 a A+4 b B) \sin (c+d x)+3 (A b+a B) \sin (2 (c+d x))+a A \sin (3 (c+d x))}{12 d} \] Input:
Integrate[Cos[c + d*x]^3*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
Output:
(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)
Time = 0.48 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 4484, 25, 3042, 4274, 3042, 3115, 24, 3117}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \cos ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
\(\Big \downarrow \) 4484 |
\(\displaystyle \frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d}-\frac {1}{3} \int -\cos ^2(c+d x) (3 (A b+a B)+(2 a A+3 b B) \sec (c+d x))dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{3} \int \cos ^2(c+d x) (3 (A b+a B)+(2 a A+3 b B) \sec (c+d x))dx+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \int \frac {3 (A b+a B)+(2 a A+3 b B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle \frac {1}{3} \left (3 (a B+A b) \int \cos ^2(c+d x)dx+(2 a A+3 b B) \int \cos (c+d x)dx\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \left ((2 a A+3 b B) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx+3 (a B+A b) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {1}{3} \left ((2 a A+3 b B) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx+3 (a B+A b) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {1}{3} \left ((2 a A+3 b B) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx+3 (a B+A b) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3117 |
\(\displaystyle \frac {1}{3} \left (\frac {(2 a A+3 b B) \sin (c+d x)}{d}+3 (a B+A b) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d}\) |
Input:
Int[Cos[c + d*x]^3*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
Output:
(a*A*Cos[c + d*x]^2*Sin[c + d*x])/(3*d) + (((2*a*A + 3*b*B)*Sin[c + d*x])/ d + 3*(A*b + a*B)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/3
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] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
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] + Simp[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]
Time = 0.51 (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 | \(\frac {\frac {A a \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+A b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B b \sin \left (d x +c \right )}{d}\) | \(85\) |
default | \(\frac {\frac {A a \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+A b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B b \sin \left (d x +c \right )}{d}\) | \(85\) |
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 | \(\frac {\left (-\frac {A b}{2}-\frac {B a}{2}\right ) x +\left (\frac {A b}{2}+\frac {B a}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-A b -B a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (A b +B a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {\left (2 A a -A b -B a +2 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {\left (2 A a -3 A b -3 B a -6 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}-\frac {\left (2 A a +A b +B a +2 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (2 A a +3 A b +3 B a -6 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\) | \(235\) |
Input:
int(cos(d*x+c)^3*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE )
Output:
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
Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.71 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (B a + A b\right )} d x + {\left (2 \, A a \cos \left (d x + c\right )^{2} + 4 \, A a + 6 \, B b + 3 \, {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="fri cas")
Output:
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
\[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\int \left (A + B \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right ) \cos ^{3}{\left (c + d x \right )}\, dx \] Input:
integrate(cos(d*x+c)**3*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x)
Output:
Integral((A + B*sec(c + d*x))*(a + b*sec(c + d*x))*cos(c + d*x)**3, x)
Time = 0.03 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.94 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b - 12 \, B b \sin \left (d x + c\right )}{12 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="max ima")
Output:
-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
Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (76) = 152\).
Time = 0.12 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.14 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (B a + A b\right )} {\left (d x + c\right )} + \frac {2 \, {\left (6 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="gia c")
Output:
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
Time = 11.66 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {A\,b\,x}{2}+\frac {B\,a\,x}{2}+\frac {3\,A\,a\,\sin \left (c+d\,x\right )}{4\,d}+\frac {B\,b\,\sin \left (c+d\,x\right )}{d}+\frac {A\,a\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {A\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \] Input:
int(cos(c + d*x)^3*(A + B/cos(c + d*x))*(a + b/cos(c + d*x)),x)
Output:
(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)
Time = 0.16 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.75 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b -\sin \left (d x +c \right )^{3} a^{2}+3 \sin \left (d x +c \right ) a^{2}+3 \sin \left (d x +c \right ) b^{2}+3 a b d x}{3 d} \] Input:
int(cos(d*x+c)^3*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x)
Output:
(3*cos(c + d*x)*sin(c + d*x)*a*b - sin(c + d*x)**3*a**2 + 3*sin(c + d*x)*a **2 + 3*sin(c + d*x)*b**2 + 3*a*b*d*x)/(3*d)