Integrand size = 31, antiderivative size = 95 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} a (3 A+4 C) x+\frac {b (A+C) \sin (c+d x)}{d}+\frac {a (3 A+4 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a A \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {A b \sin ^3(c+d x)}{3 d} \]
1/8*a*(3*A+4*C)*x+b*(A+C)*sin(d*x+c)/d+1/8*a*(3*A+4*C)*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*b*sin(d*x+c)^3/d
Time = 0.15 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.88 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {36 a A c+48 a c C+36 a A d x+48 a C d x+24 b (3 A+4 C) \sin (c+d x)+24 a (A+C) \sin (2 (c+d x))+8 A b \sin (3 (c+d x))+3 a A \sin (4 (c+d x))}{96 d} \]
(36*a*A*c + 48*a*c*C + 36*a*A*d*x + 48*a*C*d*x + 24*b*(3*A + 4*C)*Sin[c + d*x] + 24*a*(A + C)*Sin[2*(c + d*x)] + 8*A*b*Sin[3*(c + d*x)] + 3*a*A*Sin[ 4*(c + d*x)])/(96*d)
Time = 0.64 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 4563, 25, 3042, 4535, 3042, 3115, 24, 4532, 3042, 3492, 2009}
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.
| \(\displaystyle \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 4563 |
| \(\displaystyle \frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}-\frac {1}{4} \int -\cos ^3(c+d x) \left (4 b C \sec ^2(c+d x)+a (3 A+4 C) \sec (c+d x)+4 A b\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{4} \int \cos ^3(c+d x) \left (4 b C \sec ^2(c+d x)+a (3 A+4 C) \sec (c+d x)+4 A b\right )dx+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {4 b C \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (3 A+4 C) \csc \left (c+d x+\frac {\pi }{2}\right )+4 A b}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{4} \left (a (3 A+4 C) \int \cos ^2(c+d x)dx+\int \cos ^3(c+d x) \left (4 b C \sec ^2(c+d x)+4 A b\right )dx\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (a (3 A+4 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\int \frac {4 b C \csc \left (c+d x+\frac {\pi }{2}\right )^2+4 A b}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{4} \left (a (3 A+4 C) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\int \frac {4 b C \csc \left (c+d x+\frac {\pi }{2}\right )^2+4 A b}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{4} \left (\int \frac {4 b C \csc \left (c+d x+\frac {\pi }{2}\right )^2+4 A b}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+a (3 A+4 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4532 |
| \(\displaystyle \frac {1}{4} \left (\int \cos (c+d x) \left (4 A b \cos ^2(c+d x)+4 b C\right )dx+a (3 A+4 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (4 A b \sin \left (c+d x+\frac {\pi }{2}\right )^2+4 b C\right )dx+a (3 A+4 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3492 |
| \(\displaystyle \frac {1}{4} \left (a (3 A+4 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {\int \left (4 b (A+C)-4 A b \sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} \left (a (3 A+4 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {\frac {4}{3} A b \sin ^3(c+d x)-4 b (A+C) \sin (c+d x)}{d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
(a*A*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (a*(3*A + 4*C)*(x/2 + (Cos[c + d *x]*Sin[c + d*x])/(2*d)) - (-4*b*(A + C)*Sin[c + d*x] + (4*A*b*Sin[c + d*x ]^3)/3)/d)/4
3.7.44.3.1 Defintions of rubi rules used
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[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[-f^(-1) Subst[Int[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2 ), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2, 0]
Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Int[(C + A*Sin[e + f*x]^2)/Sin[e + f*x]^(m + 2), x] /; FreeQ[ {e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && ILtQ[(m + 1)/2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[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]
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] + Simp[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]
Time = 0.52 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.73
| method | result | size |
| parallelrisch | \(\frac {24 a \left (A +C \right ) \sin \left (2 d x +2 c \right )+8 A b \sin \left (3 d x +3 c \right )+3 a A \sin \left (4 d x +4 c \right )+36 \left (A +\frac {4 C}{3}\right ) \left (a x d +2 b \sin \left (d x +c \right )\right )}{96 d}\) | \(69\) |
| derivativedivides | \(\frac {a A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C a \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \sin \left (d x +c \right ) b}{d}\) | \(96\) |
| default | \(\frac {a A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C a \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \sin \left (d x +c \right ) b}{d}\) | \(96\) |
| risch | \(\frac {3 a A x}{8}+\frac {a x C}{2}+\frac {3 A b \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (d x +c \right ) C b}{d}+\frac {a A \sin \left (4 d x +4 c \right )}{32 d}+\frac {A b \sin \left (3 d x +3 c \right )}{12 d}+\frac {a A \sin \left (2 d x +2 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C a}{4 d}\) | \(101\) |
| norman | \(\frac {\left (\frac {3}{8} a A +\frac {1}{2} C a \right ) x +\left (-\frac {3}{2} a A -2 C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-\frac {3}{8} a A -\frac {1}{2} C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-\frac {3}{8} a A -\frac {1}{2} C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {3}{4} a A +C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {3}{4} a A +C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (\frac {3}{8} a A +\frac {1}{2} C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}-\frac {\left (5 a A -8 A b +4 C a -8 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}+\frac {\left (5 a A +8 A b +4 C a +8 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (21 a A -8 A b -12 C a -24 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{6 d}-\frac {\left (21 a A +8 A b -12 C a +24 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}+\frac {\left (39 a A -8 A b +12 C a +24 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}-\frac {\left (39 a A +8 A b +12 C a -24 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}\) | \(370\) |
1/96*(24*a*(A+C)*sin(2*d*x+2*c)+8*A*b*sin(3*d*x+3*c)+3*a*A*sin(4*d*x+4*c)+ 36*(A+4/3*C)*(a*x*d+2*b*sin(d*x+c)))/d
Time = 0.27 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.80 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (3 \, A + 4 \, C\right )} a d x + {\left (6 \, A a \cos \left (d x + c\right )^{3} + 8 \, A b \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + 4 \, C\right )} a \cos \left (d x + c\right ) + 8 \, {\left (2 \, A + 3 \, C\right )} b\right )} \sin \left (d x + c\right )}{24 \, d} \]
1/24*(3*(3*A + 4*C)*a*d*x + (6*A*a*cos(d*x + c)^3 + 8*A*b*cos(d*x + c)^2 + 3*(3*A + 4*C)*a*cos(d*x + c) + 8*(2*A + 3*C)*b)*sin(d*x + c))/d
\[ \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right ) \cos ^{4}{\left (c + d x \right )}\, dx \]
Time = 0.22 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.95 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b + 96 \, C b \sin \left (d x + c\right )}{96 \, d} \]
1/96*(3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a + 24*( 2*d*x + 2*c + sin(2*d*x + 2*c))*C*a - 32*(sin(d*x + c)^3 - 3*sin(d*x + c)) *A*b + 96*C*b*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 272 vs. \(2 (87) = 174\).
Time = 0.32 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.86 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (3 \, A a + 4 \, C a\right )} {\left (d x + c\right )} - \frac {2 \, {\left (15 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 72 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]
1/24*(3*(3*A*a + 4*C*a)*(d*x + c) - 2*(15*A*a*tan(1/2*d*x + 1/2*c)^7 + 12* C*a*tan(1/2*d*x + 1/2*c)^7 - 24*A*b*tan(1/2*d*x + 1/2*c)^7 - 24*C*b*tan(1/ 2*d*x + 1/2*c)^7 - 9*A*a*tan(1/2*d*x + 1/2*c)^5 + 12*C*a*tan(1/2*d*x + 1/2 *c)^5 - 40*A*b*tan(1/2*d*x + 1/2*c)^5 - 72*C*b*tan(1/2*d*x + 1/2*c)^5 + 9* A*a*tan(1/2*d*x + 1/2*c)^3 - 12*C*a*tan(1/2*d*x + 1/2*c)^3 - 40*A*b*tan(1/ 2*d*x + 1/2*c)^3 - 72*C*b*tan(1/2*d*x + 1/2*c)^3 - 15*A*a*tan(1/2*d*x + 1/ 2*c) - 12*C*a*tan(1/2*d*x + 1/2*c) - 24*A*b*tan(1/2*d*x + 1/2*c) - 24*C*b* tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d
Time = 18.54 (sec) , antiderivative size = 215, normalized size of antiderivative = 2.26 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (2\,A\,b-\frac {5\,A\,a}{4}-C\,a+2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {3\,A\,a}{4}+\frac {10\,A\,b}{3}-C\,a+6\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {10\,A\,b}{3}-\frac {3\,A\,a}{4}+C\,a+6\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,A\,a}{4}+2\,A\,b+C\,a+2\,C\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,A+4\,C\right )}{4\,\left (\frac {3\,A\,a}{4}+C\,a\right )}\right )\,\left (3\,A+4\,C\right )}{4\,d} \]
(tan(c/2 + (d*x)/2)*((5*A*a)/4 + 2*A*b + C*a + 2*C*b) - tan(c/2 + (d*x)/2) ^7*((5*A*a)/4 - 2*A*b + C*a - 2*C*b) + tan(c/2 + (d*x)/2)^3*((10*A*b)/3 - (3*A*a)/4 + C*a + 6*C*b) + tan(c/2 + (d*x)/2)^5*((3*A*a)/4 + (10*A*b)/3 - C*a + 6*C*b))/(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*atan((a*tan(c/2 + (d*x) /2)*(3*A + 4*C))/(4*((3*A*a)/4 + C*a)))*(3*A + 4*C))/(4*d)