Integrand size = 19, antiderivative size = 92 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {3 b x}{8}+\frac {a \sin (c+d x)}{d}+\frac {3 b \cos (c+d x) \sin (c+d x)}{8 d}+\frac {b \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^5(c+d x)}{5 d} \] Output:
3/8*b*x+a*sin(d*x+c)/d+3/8*b*cos(d*x+c)*sin(d*x+c)/d+1/4*b*cos(d*x+c)^3*si n(d*x+c)/d-2/3*a*sin(d*x+c)^3/d+1/5*a*sin(d*x+c)^5/d
Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.97 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {3 b (c+d x)}{8 d}+\frac {a \sin (c+d x)}{d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^5(c+d x)}{5 d}+\frac {b \sin (2 (c+d x))}{4 d}+\frac {b \sin (4 (c+d x))}{32 d} \] Input:
Integrate[Cos[c + d*x]^5*(a + b*Sec[c + d*x]),x]
Output:
(3*b*(c + d*x))/(8*d) + (a*Sin[c + d*x])/d - (2*a*Sin[c + d*x]^3)/(3*d) + (a*Sin[c + d*x]^5)/(5*d) + (b*Sin[2*(c + d*x)])/(4*d) + (b*Sin[4*(c + d*x) ])/(32*d)
Time = 0.40 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {3042, 4274, 3042, 3113, 2009, 3115, 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.
\(\displaystyle \int \cos ^5(c+d x) (a+b \sec (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle a \int \cos ^5(c+d x)dx+b \int \cos ^4(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \sin \left (c+d x+\frac {\pi }{2}\right )^5dx+b \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx\) |
\(\Big \downarrow \) 3113 |
\(\displaystyle b \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {a \int \left (\sin ^4(c+d x)-2 \sin ^2(c+d x)+1\right )d(-\sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle b \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {a \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle b \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {a \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {a \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle b \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {a \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle b \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {a \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
Input:
Int[Cos[c + d*x]^5*(a + b*Sec[c + d*x]),x]
Output:
-((a*(-Sin[c + d*x] + (2*Sin[c + d*x]^3)/3 - Sin[c + d*x]^5/5))/d) + b*((C os[c + d*x]^3*Sin[c + d*x])/(4*d) + (3*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/ (2*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]
Time = 1.47 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.75
method | result | size |
parallelrisch | \(\frac {180 b d x +300 a \sin \left (d x +c \right )+6 a \sin \left (5 d x +5 c \right )+15 b \sin \left (4 d x +4 c \right )+50 a \sin \left (3 d x +3 c \right )+120 b \sin \left (2 d x +2 c \right )}{480 d}\) | \(69\) |
derivativedivides | \(\frac {\frac {a \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(70\) |
default | \(\frac {\frac {a \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(70\) |
risch | \(\frac {3 b x}{8}+\frac {5 a \sin \left (d x +c \right )}{8 d}+\frac {a \sin \left (5 d x +5 c \right )}{80 d}+\frac {b \sin \left (4 d x +4 c \right )}{32 d}+\frac {5 a \sin \left (3 d x +3 c \right )}{48 d}+\frac {b \sin \left (2 d x +2 c \right )}{4 d}\) | \(78\) |
norman | \(\frac {\frac {3 b x}{8}+\frac {116 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {15 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8}+\frac {15 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4}+\frac {15 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{4}+\frac {15 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}+\frac {3 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8}+\frac {\left (8 a -5 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}+\frac {\left (8 a +5 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (16 a -3 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}+\frac {\left (16 a +3 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}\) | \(204\) |
Input:
int(cos(d*x+c)^5*(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/480*(180*b*d*x+300*a*sin(d*x+c)+6*a*sin(5*d*x+5*c)+15*b*sin(4*d*x+4*c)+5 0*a*sin(3*d*x+3*c)+120*b*sin(2*d*x+2*c))/d
Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.70 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {45 \, b d x + {\left (24 \, a \cos \left (d x + c\right )^{4} + 30 \, b \cos \left (d x + c\right )^{3} + 32 \, a \cos \left (d x + c\right )^{2} + 45 \, b \cos \left (d x + c\right ) + 64 \, a\right )} \sin \left (d x + c\right )}{120 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+b*sec(d*x+c)),x, algorithm="fricas")
Output:
1/120*(45*b*d*x + (24*a*cos(d*x + c)^4 + 30*b*cos(d*x + c)^3 + 32*a*cos(d* x + c)^2 + 45*b*cos(d*x + c) + 64*a)*sin(d*x + c))/d
\[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right ) \cos ^{5}{\left (c + d x \right )}\, dx \] Input:
integrate(cos(d*x+c)**5*(a+b*sec(d*x+c)),x)
Output:
Integral((a + b*sec(c + d*x))*cos(c + d*x)**5, x)
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.75 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} b}{480 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+b*sec(d*x+c)),x, algorithm="maxima")
Output:
1/480*(32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a + 15* (12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*b)/d
Time = 0.12 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.67 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {45 \, {\left (d x + c\right )} b + \frac {2 \, {\left (120 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 160 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 160 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+b*sec(d*x+c)),x, algorithm="giac")
Output:
1/120*(45*(d*x + c)*b + 2*(120*a*tan(1/2*d*x + 1/2*c)^9 - 75*b*tan(1/2*d*x + 1/2*c)^9 + 160*a*tan(1/2*d*x + 1/2*c)^7 - 30*b*tan(1/2*d*x + 1/2*c)^7 + 464*a*tan(1/2*d*x + 1/2*c)^5 + 160*a*tan(1/2*d*x + 1/2*c)^3 + 30*b*tan(1/ 2*d*x + 1/2*c)^3 + 120*a*tan(1/2*d*x + 1/2*c) + 75*b*tan(1/2*d*x + 1/2*c)) /(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d
Time = 14.40 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.23 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {3\,b\,x}{8}+\frac {\left (2\,a-\frac {5\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {8\,a}{3}-\frac {b}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {116\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{15}+\left (\frac {8\,a}{3}+\frac {b}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a+\frac {5\,b}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \] Input:
int(cos(c + d*x)^5*(a + b/cos(c + d*x)),x)
Output:
(3*b*x)/8 + (tan(c/2 + (d*x)/2)*(2*a + (5*b)/4) + tan(c/2 + (d*x)/2)^3*((8 *a)/3 + b/2) + tan(c/2 + (d*x)/2)^9*(2*a - (5*b)/4) + tan(c/2 + (d*x)/2)^7 *((8*a)/3 - b/2) + (116*a*tan(c/2 + (d*x)/2)^5)/15)/(d*(tan(c/2 + (d*x)/2) ^2 + 1)^5)
Time = 0.16 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.80 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {-30 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b +75 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b +24 \sin \left (d x +c \right )^{5} a -80 \sin \left (d x +c \right )^{3} a +120 \sin \left (d x +c \right ) a +45 b d x}{120 d} \] Input:
int(cos(d*x+c)^5*(a+b*sec(d*x+c)),x)
Output:
( - 30*cos(c + d*x)*sin(c + d*x)**3*b + 75*cos(c + d*x)*sin(c + d*x)*b + 2 4*sin(c + d*x)**5*a - 80*sin(c + d*x)**3*a + 120*sin(c + d*x)*a + 45*b*d*x )/(120*d)