Integrand size = 29, antiderivative size = 97 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {1}{8} a (3 A+4 B) x+\frac {a (A+B) \sin (c+d x)}{d}+\frac {a (3 A+4 B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a A \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {a (A+B) \sin ^3(c+d x)}{3 d} \] Output:
1/8*a*(3*A+4*B)*x+a*(A+B)*sin(d*x+c)/d+1/8*a*(3*A+4*B)*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*(A+B)*sin(d*x+c)^3/d
Time = 0.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.77 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a \left (36 A c+48 B c+36 A d x+48 B d x+96 (A+B) \sin (c+d x)-32 (A+B) \sin ^3(c+d x)+24 (A+B) \sin (2 (c+d x))+3 A \sin (4 (c+d x))\right )}{96 d} \] Input:
Integrate[Cos[c + d*x]^4*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
Output:
(a*(36*A*c + 48*B*c + 36*A*d*x + 48*B*d*x + 96*(A + B)*Sin[c + d*x] - 32*( A + B)*Sin[c + d*x]^3 + 24*(A + B)*Sin[2*(c + d*x)] + 3*A*Sin[4*(c + d*x)] ))/(96*d)
Time = 0.50 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.96, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 4484, 25, 3042, 4274, 3042, 3113, 2009, 3115, 24}
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.
| \(\displaystyle \int \cos ^4(c+d x) (a \sec (c+d x)+a) (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 4484 |
| \(\displaystyle \frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}-\frac {1}{4} \int -\cos ^3(c+d x) (4 a (A+B)+a (3 A+4 B) \sec (c+d x))dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{4} \int \cos ^3(c+d x) (4 a (A+B)+a (3 A+4 B) \sec (c+d x))dx+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {4 a (A+B)+a (3 A+4 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {1}{4} \left (4 a (A+B) \int \cos ^3(c+d x)dx+a (3 A+4 B) \int \cos ^2(c+d x)dx\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (a (3 A+4 B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+4 a (A+B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {1}{4} \left (a (3 A+4 B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {4 a (A+B) \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} \left (a (3 A+4 B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {4 a (A+B) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{4} \left (a (3 A+4 B) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {4 a (A+B) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{4} \left (a (3 A+4 B) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {4 a (A+B) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
Input:
Int[Cos[c + d*x]^4*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
Output:
(a*A*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (a*(3*A + 4*B)*(x/2 + (Cos[c + d *x]*Sin[c + d*x])/(2*d)) - (4*a*(A + B)*(-Sin[c + d*x] + Sin[c + d*x]^3/3) )/d)/4
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
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[(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 = 0.59 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.69
| method | result | size |
| parallelrisch | \(\frac {a \left (8 \left (A +B \right ) \sin \left (2 d x +2 c \right )+\frac {8 \left (A +B \right ) \sin \left (3 d x +3 c \right )}{3}+A \sin \left (4 d x +4 c \right )+24 \sin \left (d x +c \right ) \left (A +B \right )+12 d \left (A +\frac {4 B}{3}\right ) x \right )}{32 d}\) | \(67\) |
| derivativedivides | \(\frac {a A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {B a \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(107\) |
| default | \(\frac {a A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {B a \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(107\) |
| risch | \(\frac {3 a A x}{8}+\frac {a x B}{2}+\frac {3 a A \sin \left (d x +c \right )}{4 d}+\frac {3 \sin \left (d x +c \right ) B a}{4 d}+\frac {a A \sin \left (4 d x +4 c \right )}{32 d}+\frac {a A \sin \left (3 d x +3 c \right )}{12 d}+\frac {\sin \left (3 d x +3 c \right ) B a}{12 d}+\frac {a A \sin \left (2 d x +2 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B a}{4 d}\) | \(118\) |
| norman | \(\frac {-\frac {a \left (3 A +4 B \right ) x}{8}+\frac {2 a \left (A -2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {a \left (3 A -4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 d}+\frac {a \left (3 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {3 a \left (3 A +4 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8}-\frac {a \left (3 A +4 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4}+\frac {a \left (3 A +4 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{4}+\frac {3 a \left (3 A +4 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}+\frac {a \left (3 A +4 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8}+\frac {2 a \left (5 A +2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {a \left (13 A +12 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\) | \(270\) |
Input:
int(cos(d*x+c)^4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE )
Output:
1/32*a*(8*(A+B)*sin(2*d*x+2*c)+8/3*(A+B)*sin(3*d*x+3*c)+A*sin(4*d*x+4*c)+2 4*sin(d*x+c)*(A+B)+12*d*(A+4/3*B)*x)/d
Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.76 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (3 \, A + 4 \, B\right )} a d x + {\left (6 \, A a \cos \left (d x + c\right )^{3} + 8 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + 4 \, B\right )} a \cos \left (d x + c\right ) + 16 \, {\left (A + B\right )} a\right )} \sin \left (d x + c\right )}{24 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="fri cas")
Output:
1/24*(3*(3*A + 4*B)*a*d*x + (6*A*a*cos(d*x + c)^3 + 8*(A + B)*a*cos(d*x + c)^2 + 3*(3*A + 4*B)*a*cos(d*x + c) + 16*(A + B)*a)*sin(d*x + c))/d
\[ \int \cos ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=a \left (\int A \cos ^{4}{\left (c + d x \right )}\, dx + \int A \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cos(d*x+c)**4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x)
Output:
a*(Integral(A*cos(c + d*x)**4, x) + Integral(A*cos(c + d*x)**4*sec(c + d*x ), x) + Integral(B*cos(c + d*x)**4*sec(c + d*x), x) + Integral(B*cos(c + d *x)**4*sec(c + d*x)**2, x))
Time = 0.04 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.04 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=-\frac {32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a + 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a}{96 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="max ima")
Output:
-1/96*(32*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a - 3*(12*d*x + 12*c + sin(4 *d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a + 32*(sin(d*x + c)^3 - 3*sin(d*x + c ))*B*a - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a)/d
Time = 0.13 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.61 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (3 \, A a + 4 \, B a\right )} {\left (d x + c\right )} + \frac {2 \, {\left (9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 49 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 28 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 31 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 52 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 39 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 36 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="gia c")
Output:
1/24*(3*(3*A*a + 4*B*a)*(d*x + c) + 2*(9*A*a*tan(1/2*d*x + 1/2*c)^7 + 12*B *a*tan(1/2*d*x + 1/2*c)^7 + 49*A*a*tan(1/2*d*x + 1/2*c)^5 + 28*B*a*tan(1/2 *d*x + 1/2*c)^5 + 31*A*a*tan(1/2*d*x + 1/2*c)^3 + 52*B*a*tan(1/2*d*x + 1/2 *c)^3 + 39*A*a*tan(1/2*d*x + 1/2*c) + 36*B*a*tan(1/2*d*x + 1/2*c))/(tan(1/ 2*d*x + 1/2*c)^2 + 1)^4)/d
Time = 14.24 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.90 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {\left (\frac {3\,A\,a}{4}+B\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {49\,A\,a}{12}+\frac {7\,B\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {31\,A\,a}{12}+\frac {13\,B\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,A\,a}{4}+3\,B\,a\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,A+4\,B\right )}{4\,\left (\frac {3\,A\,a}{4}+B\,a\right )}\right )\,\left (3\,A+4\,B\right )}{4\,d} \] Input:
int(cos(c + d*x)^4*(A + B/cos(c + d*x))*(a + a/cos(c + d*x)),x)
Output:
(tan(c/2 + (d*x)/2)*((13*A*a)/4 + 3*B*a) + tan(c/2 + (d*x)/2)^7*((3*A*a)/4 + B*a) + tan(c/2 + (d*x)/2)^3*((31*A*a)/12 + (13*B*a)/3) + tan(c/2 + (d*x )/2)^5*((49*A*a)/12 + (7*B*a)/3))/(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*ata n((a*tan(c/2 + (d*x)/2)*(3*A + 4*B))/(4*((3*A*a)/4 + B*a)))*(3*A + 4*B))/( 4*d)
Time = 0.15 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.07 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a \left (-6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a +15 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a +12 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -8 \sin \left (d x +c \right )^{3} a -8 \sin \left (d x +c \right )^{3} b +24 \sin \left (d x +c \right ) a +24 \sin \left (d x +c \right ) b +9 a d x +12 b d x \right )}{24 d} \] Input:
int(cos(d*x+c)^4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x)
Output:
(a*( - 6*cos(c + d*x)*sin(c + d*x)**3*a + 15*cos(c + d*x)*sin(c + d*x)*a + 12*cos(c + d*x)*sin(c + d*x)*b - 8*sin(c + d*x)**3*a - 8*sin(c + d*x)**3* b + 24*sin(c + d*x)*a + 24*sin(c + d*x)*b + 9*a*d*x + 12*b*d*x))/(24*d)