Integrand size = 21, antiderivative size = 103 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {3 a^2 x}{4}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {3 a^2 \cos (c+d x) \sin (c+d x)}{4 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{2 d}-\frac {a^2 \sin ^3(c+d x)}{d}+\frac {a^2 \sin ^5(c+d x)}{5 d} \]
3/4*a^2*x+2*a^2*sin(d*x+c)/d+3/4*a^2*cos(d*x+c)*sin(d*x+c)/d+1/2*a^2*cos(d *x+c)^3*sin(d*x+c)/d-a^2*sin(d*x+c)^3/d+1/5*a^2*sin(d*x+c)^5/d
Time = 0.14 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.59 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {a^2 (60 d x+110 \sin (c+d x)+40 \sin (2 (c+d x))+15 \sin (3 (c+d x))+5 \sin (4 (c+d x))+\sin (5 (c+d x)))}{80 d} \]
(a^2*(60*d*x + 110*Sin[c + d*x] + 40*Sin[2*(c + d*x)] + 15*Sin[3*(c + d*x) ] + 5*Sin[4*(c + d*x)] + Sin[5*(c + d*x)]))/(80*d)
Time = 0.58 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.91, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 4275, 3042, 3115, 3042, 3115, 24, 4532, 3042, 3492, 27, 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.
| \(\displaystyle \int \cos ^5(c+d x) (a \sec (c+d x)+a)^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 4275 |
| \(\displaystyle 2 a^2 \int \cos ^4(c+d x)dx+\int \cos ^5(c+d x) \left (\sec ^2(c+d x) a^2+a^2\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 a^2 \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx+\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 a^2+a^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 a^2+a^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx+2 a^2 \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 a^2+a^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx+2 a^2 \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 a^2+a^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx+2 a^2 \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 a^2+a^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx+2 a^2 \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )\) |
| \(\Big \downarrow \) 4532 |
| \(\displaystyle \int \cos ^3(c+d x) \left (\cos ^2(c+d x) a^2+a^2\right )dx+2 a^2 \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (\sin \left (c+d x+\frac {\pi }{2}\right )^2 a^2+a^2\right )dx+2 a^2 \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )\) |
| \(\Big \downarrow \) 3492 |
| \(\displaystyle 2 a^2 \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {\int a^2 \left (1-\sin ^2(c+d x)\right ) \left (2-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 a^2 \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {a^2 \int \left (1-\sin ^2(c+d x)\right ) \left (2-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 290 |
| \(\displaystyle 2 a^2 \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {a^2 \int \left (\sin ^4(c+d x)-3 \sin ^2(c+d x)+2\right )d(-\sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 a^2 \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {a^2 \left (-\frac {1}{5} \sin ^5(c+d x)+\sin ^3(c+d x)-2 \sin (c+d x)\right )}{d}\) |
-((a^2*(-2*Sin[c + d*x] + Sin[c + d*x]^3 - Sin[c + d*x]^5/5))/d) + 2*a^2*( (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)
3.1.19.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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 = 0.67 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.62
| method | result | size |
| parallelrisch | \(\frac {a^{2} \left (60 d x +\sin \left (5 d x +5 c \right )+5 \sin \left (4 d x +4 c \right )+15 \sin \left (3 d x +3 c \right )+40 \sin \left (2 d x +2 c \right )+110 \sin \left (d x +c \right )\right )}{80 d}\) | \(64\) |
| risch | \(\frac {3 a^{2} x}{4}+\frac {11 a^{2} \sin \left (d x +c \right )}{8 d}+\frac {a^{2} \sin \left (5 d x +5 c \right )}{80 d}+\frac {a^{2} \sin \left (4 d x +4 c \right )}{16 d}+\frac {3 a^{2} \sin \left (3 d x +3 c \right )}{16 d}+\frac {a^{2} \sin \left (2 d x +2 c \right )}{2 d}\) | \(90\) |
| derivativedivides | \(\frac {\frac {a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+2 a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a^{2} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\) | \(96\) |
| default | \(\frac {\frac {a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+2 a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a^{2} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\) | \(96\) |
| norman | \(\frac {-\frac {3 a^{2} x}{4}-\frac {13 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}-\frac {5 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 d}-\frac {27 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5 d}+\frac {37 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{5 d}+\frac {11 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{2 d}+\frac {3 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{2 d}-3 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {15 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4}+\frac {15 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{4}+3 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\frac {3 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{4}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\) | \(236\) |
1/80*a^2*(60*d*x+sin(5*d*x+5*c)+5*sin(4*d*x+4*c)+15*sin(3*d*x+3*c)+40*sin( 2*d*x+2*c)+110*sin(d*x+c))/d
Time = 0.26 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.74 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {15 \, a^{2} d x + {\left (4 \, a^{2} \cos \left (d x + c\right )^{4} + 10 \, a^{2} \cos \left (d x + c\right )^{3} + 12 \, a^{2} \cos \left (d x + c\right )^{2} + 15 \, a^{2} \cos \left (d x + c\right ) + 24 \, a^{2}\right )} \sin \left (d x + c\right )}{20 \, d} \]
1/20*(15*a^2*d*x + (4*a^2*cos(d*x + c)^4 + 10*a^2*cos(d*x + c)^3 + 12*a^2* cos(d*x + c)^2 + 15*a^2*cos(d*x + c) + 24*a^2)*sin(d*x + c))/d
\[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \, dx=a^{2} \left (\int 2 \cos ^{5}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \cos ^{5}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \cos ^{5}{\left (c + d x \right )}\, dx\right ) \]
a**2*(Integral(2*cos(c + d*x)**5*sec(c + d*x), x) + Integral(cos(c + d*x)* *5*sec(c + d*x)**2, x) + Integral(cos(c + d*x)**5, x))
Time = 0.26 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.92 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {16 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{2} - 80 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{240 \, d} \]
1/240*(16*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^2 - 8 0*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^2 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^2)/d
Time = 0.32 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.09 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {15 \, {\left (d x + c\right )} a^{2} + \frac {2 \, {\left (15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 70 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 144 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 90 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 65 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{20 \, d} \]
1/20*(15*(d*x + c)*a^2 + 2*(15*a^2*tan(1/2*d*x + 1/2*c)^9 + 70*a^2*tan(1/2 *d*x + 1/2*c)^7 + 144*a^2*tan(1/2*d*x + 1/2*c)^5 + 90*a^2*tan(1/2*d*x + 1/ 2*c)^3 + 65*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d
Time = 16.65 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.02 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {3\,a^2\,x}{4}+\frac {\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}+7\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {72\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5}+9\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {13\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]