Integrand size = 21, antiderivative size = 111 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {3 a b x}{4}+\frac {\left (a^2+b^2\right ) \sin (c+d x)}{d}+\frac {3 a b \cos (c+d x) \sin (c+d x)}{4 d}+\frac {a b \cos ^3(c+d x) \sin (c+d x)}{2 d}-\frac {\left (2 a^2+b^2\right ) \sin ^3(c+d x)}{3 d}+\frac {a^2 \sin ^5(c+d x)}{5 d} \]
3/4*a*b*x+(a^2+b^2)*sin(d*x+c)/d+3/4*a*b*cos(d*x+c)*sin(d*x+c)/d+1/2*a*b*c os(d*x+c)^3*sin(d*x+c)/d-1/3*(2*a^2+b^2)*sin(d*x+c)^3/d+1/5*a^2*sin(d*x+c) ^5/d
Time = 0.14 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.77 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {240 \left (a^2+b^2\right ) \sin (c+d x)-80 \left (2 a^2+b^2\right ) \sin ^3(c+d x)+48 a^2 \sin ^5(c+d x)+15 a b (12 (c+d x)+8 \sin (2 (c+d x))+\sin (4 (c+d x)))}{240 d} \]
(240*(a^2 + b^2)*Sin[c + d*x] - 80*(2*a^2 + b^2)*Sin[c + d*x]^3 + 48*a^2*S in[c + d*x]^5 + 15*a*b*(12*(c + d*x) + 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x )]))/(240*d)
Time = 0.60 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.02, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4275, 3042, 3115, 3042, 3115, 24, 4532, 3042, 3492, 290, 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.
| \(\displaystyle \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 4275 |
| \(\displaystyle \int \cos ^5(c+d x) \left (a^2+b^2 \sec ^2(c+d x)\right )dx+2 a b \int \cos ^4(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a^2+b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx+2 a b \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \int \frac {a^2+b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx+2 a b \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a^2+b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx+2 a b \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \int \frac {a^2+b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx+2 a b \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \int \frac {a^2+b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx+2 a b \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )\) |
| \(\Big \downarrow \) 4532 |
| \(\displaystyle \int \cos ^3(c+d x) \left (b^2+a^2 \cos ^2(c+d x)\right )dx+2 a b \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (b^2+a^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )\) |
| \(\Big \downarrow \) 3492 |
| \(\displaystyle 2 a b \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {\int \left (1-\sin ^2(c+d x)\right ) \left (-\sin ^2(c+d x) a^2+a^2+b^2\right )d(-\sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 290 |
| \(\displaystyle 2 a b \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {\int \left (a^2 \sin ^4(c+d x)-\left (2 a^2+b^2\right ) \sin ^2(c+d x)+a^2 \left (\frac {b^2}{a^2}+1\right )\right )d(-\sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 a b \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {\frac {1}{3} \left (2 a^2+b^2\right ) \sin ^3(c+d x)-\left (a^2+b^2\right ) \sin (c+d x)-\frac {1}{5} a^2 \sin ^5(c+d x)}{d}\) |
-((-((a^2 + b^2)*Sin[c + d*x]) + ((2*a^2 + b^2)*Sin[c + d*x]^3)/3 - (a^2*S in[c + d*x]^5)/5)/d) + 2*a*b*((Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (3*(x/ 2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4)
3.5.65.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> I nt[ExpandIntegrand[(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d }, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 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[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_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Simp[2*a*(b/d) Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, n}, x]
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]
Time = 1.17 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.86
| method | result | size |
| derivativedivides | \(\frac {\frac {a^{2} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+2 a b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(95\) |
| default | \(\frac {\frac {a^{2} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+2 a b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(95\) |
| parallelrisch | \(\frac {180 a b x d +3 a^{2} \sin \left (5 d x +5 c \right )+15 a b \sin \left (4 d x +4 c \right )+25 a^{2} \sin \left (3 d x +3 c \right )+20 b^{2} \sin \left (3 d x +3 c \right )+120 a b \sin \left (2 d x +2 c \right )+150 a^{2} \sin \left (d x +c \right )+180 b^{2} \sin \left (d x +c \right )}{240 d}\) | \(103\) |
| risch | \(\frac {3 a b x}{4}+\frac {5 a^{2} \sin \left (d x +c \right )}{8 d}+\frac {3 \sin \left (d x +c \right ) b^{2}}{4 d}+\frac {a^{2} \sin \left (5 d x +5 c \right )}{80 d}+\frac {a b \sin \left (4 d x +4 c \right )}{16 d}+\frac {5 a^{2} \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) b^{2}}{12 d}+\frac {a b \sin \left (2 d x +2 c \right )}{2 d}\) | \(118\) |
| norman | \(\frac {-\frac {3 a b x}{4}-\frac {\left (4 a^{2}-9 a b +20 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}+\frac {\left (4 a^{2}-5 a b +4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{2 d}-\frac {\left (4 a^{2}+5 a b +4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {\left (4 a^{2}+9 a b +20 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{6 d}-\frac {\left (76 a^{2}-15 a b +20 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {\left (76 a^{2}+15 a b +20 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{15 d}-3 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {15 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4}+\frac {15 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{4}+3 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\frac {3 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{4}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\) | \(302\) |
1/d*(1/5*a^2*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+2*a*b*(1/4*(co s(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+1/3*b^2*(2+cos(d*x+c) ^2)*sin(d*x+c))
Time = 0.27 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.77 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {45 \, a b d x + {\left (12 \, a^{2} \cos \left (d x + c\right )^{4} + 30 \, a b \cos \left (d x + c\right )^{3} + 45 \, a b \cos \left (d x + c\right ) + 4 \, {\left (4 \, a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 32 \, a^{2} + 40 \, b^{2}\right )} \sin \left (d x + c\right )}{60 \, d} \]
1/60*(45*a*b*d*x + (12*a^2*cos(d*x + c)^4 + 30*a*b*cos(d*x + c)^3 + 45*a*b *cos(d*x + c) + 4*(4*a^2 + 5*b^2)*cos(d*x + c)^2 + 32*a^2 + 40*b^2)*sin(d* x + c))/d
\[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \cos ^{5}{\left (c + d x \right )}\, dx \]
Time = 0.21 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.85 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {16 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{2} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a b - 80 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} b^{2}}{240 \, d} \]
1/240*(16*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^2 + 1 5*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a*b - 80*(sin(d* x + c)^3 - 3*sin(d*x + c))*b^2)/d
Leaf count of result is larger than twice the leaf count of optimal. 247 vs. \(2 (101) = 202\).
Time = 0.31 (sec) , antiderivative size = 247, normalized size of antiderivative = 2.23 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {45 \, {\left (d x + c\right )} a b + \frac {2 \, {\left (60 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 80 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 160 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 232 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 200 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 80 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 160 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 60 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{60 \, d} \]
1/60*(45*(d*x + c)*a*b + 2*(60*a^2*tan(1/2*d*x + 1/2*c)^9 - 75*a*b*tan(1/2 *d*x + 1/2*c)^9 + 60*b^2*tan(1/2*d*x + 1/2*c)^9 + 80*a^2*tan(1/2*d*x + 1/2 *c)^7 - 30*a*b*tan(1/2*d*x + 1/2*c)^7 + 160*b^2*tan(1/2*d*x + 1/2*c)^7 + 2 32*a^2*tan(1/2*d*x + 1/2*c)^5 + 200*b^2*tan(1/2*d*x + 1/2*c)^5 + 80*a^2*ta n(1/2*d*x + 1/2*c)^3 + 30*a*b*tan(1/2*d*x + 1/2*c)^3 + 160*b^2*tan(1/2*d*x + 1/2*c)^3 + 60*a^2*tan(1/2*d*x + 1/2*c) + 75*a*b*tan(1/2*d*x + 1/2*c) + 60*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d
Time = 13.46 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.05 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {5\,a^2\,\sin \left (c+d\,x\right )}{8\,d}+\frac {3\,b^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {3\,a\,b\,x}{4}+\frac {5\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {a^2\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}+\frac {b^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {a\,b\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {a\,b\,\sin \left (4\,c+4\,d\,x\right )}{16\,d} \]