Integrand size = 29, antiderivative size = 88 \[ \int \cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} (3 A+4 C) x+\frac {B \sin (c+d x)}{d}+\frac {(3 A+4 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {A \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {B \sin ^3(c+d x)}{3 d} \]
1/8*(3*A+4*C)*x+B*sin(d*x+c)/d+1/8*(3*A+4*C)*cos(d*x+c)*sin(d*x+c)/d+1/4*A *cos(d*x+c)^3*sin(d*x+c)/d-1/3*B*sin(d*x+c)^3/d
Time = 0.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80 \[ \int \cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {36 A c+48 c C+36 A d x+48 C d x+96 B \sin (c+d x)-32 B \sin ^3(c+d x)+24 (A+C) \sin (2 (c+d x))+3 A \sin (4 (c+d x))}{96 d} \]
(36*A*c + 48*c*C + 36*A*d*x + 48*C*d*x + 96*B*Sin[c + d*x] - 32*B*Sin[c + d*x]^3 + 24*(A + C)*Sin[2*(c + d*x)] + 3*A*Sin[4*(c + d*x)])/(96*d)
Time = 0.44 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.98, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 4535, 3042, 3113, 2009, 4533, 3042, 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.
\(\displaystyle \int \cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
\(\Big \downarrow \) 4535 |
\(\displaystyle \int \cos ^4(c+d x) \left (C \sec ^2(c+d x)+A\right )dx+B \int \cos ^3(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx+B \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx\) |
\(\Big \downarrow \) 3113 |
\(\displaystyle \int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx-\frac {B \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx-\frac {B \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 4533 |
\(\displaystyle \frac {1}{4} (3 A+4 C) \int \cos ^2(c+d x)dx+\frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d}-\frac {B \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} (3 A+4 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d}-\frac {B \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {1}{4} (3 A+4 C) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d}-\frac {B \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {1}{4} (3 A+4 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )+\frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d}-\frac {B \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
(A*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + ((3*A + 4*C)*(x/2 + (Cos[c + d*x]* Sin[c + d*x])/(2*d)))/4 - (B*(-Sin[c + d*x] + Sin[c + d*x]^3/3))/d
3.1.62.3.1 Defintions of rubi rules used
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_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Simp[(C*m + A*(m + 1))/(b^2*m) Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
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]
Time = 0.24 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.72
method | result | size |
parallelrisch | \(\frac {24 \left (A +C \right ) \sin \left (2 d x +2 c \right )+8 B \sin \left (3 d x +3 c \right )+3 A \sin \left (4 d x +4 c \right )+72 B \sin \left (d x +c \right )+36 d \left (A +\frac {4 C}{3}\right ) x}{96 d}\) | \(63\) |
risch | \(\frac {3 A x}{8}+\frac {C x}{2}+\frac {3 B \sin \left (d x +c \right )}{4 d}+\frac {A \sin \left (4 d x +4 c \right )}{32 d}+\frac {B \sin \left (3 d x +3 c \right )}{12 d}+\frac {A \sin \left (2 d x +2 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C}{4 d}\) | \(82\) |
derivativedivides | \(\frac {A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(84\) |
default | \(\frac {A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(84\) |
norman | \(\frac {\left (-\frac {3 A}{8}-\frac {C}{2}\right ) x +\left (-\frac {9 A}{8}-\frac {3 C}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-\frac {3 A}{4}-C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (\frac {3 A}{4}+C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (\frac {3 A}{8}+\frac {C}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (\frac {9 A}{8}+\frac {3 C}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\frac {2 \left (3 A -2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {2 \left (3 A +2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {\left (3 A -4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 d}-\frac {\left (5 A -8 B +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {\left (5 A +8 B +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\) | \(259\) |
1/96*(24*(A+C)*sin(2*d*x+2*c)+8*B*sin(3*d*x+3*c)+3*A*sin(4*d*x+4*c)+72*B*s in(d*x+c)+36*d*(A+4/3*C)*x)/d
Time = 0.25 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.74 \[ \int \cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (3 \, A + 4 \, C\right )} d x + {\left (6 \, A \cos \left (d x + c\right )^{3} + 8 \, B \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right ) + 16 \, B\right )} \sin \left (d x + c\right )}{24 \, d} \]
1/24*(3*(3*A + 4*C)*d*x + (6*A*cos(d*x + c)^3 + 8*B*cos(d*x + c)^2 + 3*(3* A + 4*C)*cos(d*x + c) + 16*B)*sin(d*x + c))/d
\[ \int \cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{4}{\left (c + d x \right )}\, dx \]
Time = 0.21 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.88 \[ \int \cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C}{96 \, d} \]
1/96*(3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A - 32*(si n(d*x + c)^3 - 3*sin(d*x + c))*B + 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*C)/ d
Leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (80) = 160\).
Time = 0.29 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.27 \[ \int \cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (d x + c\right )} {\left (3 \, A + 4 \, C\right )} - \frac {2 \, {\left (15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 9 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]
1/24*(3*(d*x + c)*(3*A + 4*C) - 2*(15*A*tan(1/2*d*x + 1/2*c)^7 - 24*B*tan( 1/2*d*x + 1/2*c)^7 + 12*C*tan(1/2*d*x + 1/2*c)^7 - 9*A*tan(1/2*d*x + 1/2*c )^5 - 40*B*tan(1/2*d*x + 1/2*c)^5 + 12*C*tan(1/2*d*x + 1/2*c)^5 + 9*A*tan( 1/2*d*x + 1/2*c)^3 - 40*B*tan(1/2*d*x + 1/2*c)^3 - 12*C*tan(1/2*d*x + 1/2* c)^3 - 15*A*tan(1/2*d*x + 1/2*c) - 24*B*tan(1/2*d*x + 1/2*c) - 12*C*tan(1/ 2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d
Time = 15.06 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.92 \[ \int \cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3\,A\,x}{8}+\frac {C\,x}{2}+\frac {A\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {B\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {3\,B\,\sin \left (c+d\,x\right )}{4\,d} \]