Integrand size = 21, antiderivative size = 102 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^4 \, dx=\frac {7 a^4 x}{2}+\frac {8 a^4 \sin (c+d x)}{d}+\frac {7 a^4 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a^4 \cos ^3(c+d x) \sin (c+d x)}{d}-\frac {8 a^4 \sin ^3(c+d x)}{3 d}+\frac {a^4 \sin ^5(c+d x)}{5 d} \]
7/2*a^4*x+8*a^4*sin(d*x+c)/d+7/2*a^4*cos(d*x+c)*sin(d*x+c)/d+a^4*cos(d*x+c )^3*sin(d*x+c)/d-8/3*a^4*sin(d*x+c)^3/d+1/5*a^4*sin(d*x+c)^5/d
Time = 0.14 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.62 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^4 \, dx=\frac {a^4 (840 d x+1470 \sin (c+d x)+480 \sin (2 (c+d x))+145 \sin (3 (c+d x))+30 \sin (4 (c+d x))+3 \sin (5 (c+d x)))}{240 d} \]
(a^4*(840*d*x + 1470*Sin[c + d*x] + 480*Sin[2*(c + d*x)] + 145*Sin[3*(c + d*x)] + 30*Sin[4*(c + d*x)] + 3*Sin[5*(c + d*x)]))/(240*d)
Time = 0.33 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 4278, 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.
\(\displaystyle \int \cos ^5(c+d x) (a \sec (c+d x)+a)^4 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
\(\Big \downarrow \) 4278 |
\(\displaystyle \int \left (a^4 \cos ^5(c+d x)+4 a^4 \cos ^4(c+d x)+6 a^4 \cos ^3(c+d x)+4 a^4 \cos ^2(c+d x)+a^4 \cos (c+d x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^4 \sin ^5(c+d x)}{5 d}-\frac {8 a^4 \sin ^3(c+d x)}{3 d}+\frac {8 a^4 \sin (c+d x)}{d}+\frac {a^4 \sin (c+d x) \cos ^3(c+d x)}{d}+\frac {7 a^4 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {7 a^4 x}{2}\) |
(7*a^4*x)/2 + (8*a^4*Sin[c + d*x])/d + (7*a^4*Cos[c + d*x]*Sin[c + d*x])/( 2*d) + (a^4*Cos[c + d*x]^3*Sin[c + d*x])/d - (8*a^4*Sin[c + d*x]^3)/(3*d) + (a^4*Sin[c + d*x]^5)/(5*d)
3.1.38.3.1 Defintions of rubi rules used
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[ExpandTrig[(a + b*csc[e + f*x])^m*(d*csc[e + f *x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && I GtQ[m, 0] && RationalQ[n]
Time = 0.73 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.65
method | result | size |
parallelrisch | \(\frac {a^{4} \left (840 d x +3 \sin \left (5 d x +5 c \right )+30 \sin \left (4 d x +4 c \right )+145 \sin \left (3 d x +3 c \right )+480 \sin \left (2 d x +2 c \right )+1470 \sin \left (d x +c \right )\right )}{240 d}\) | \(66\) |
risch | \(\frac {7 a^{4} x}{2}+\frac {49 a^{4} \sin \left (d x +c \right )}{8 d}+\frac {a^{4} \sin \left (5 d x +5 c \right )}{80 d}+\frac {a^{4} \sin \left (4 d x +4 c \right )}{8 d}+\frac {29 a^{4} \sin \left (3 d x +3 c \right )}{48 d}+\frac {2 a^{4} \sin \left (2 d x +2 c \right )}{d}\) | \(90\) |
derivativedivides | \(\frac {a^{4} \sin \left (d x +c \right )+4 a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a^{4} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+4 a^{4} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a^{4} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\) | \(133\) |
default | \(\frac {a^{4} \sin \left (d x +c \right )+4 a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a^{4} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+4 a^{4} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a^{4} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\) | \(133\) |
norman | \(\frac {-\frac {7 a^{4} x}{2}-\frac {25 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {67 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {349 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {203 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{15 d}-\frac {533 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{15 d}-\frac {259 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{15 d}+\frac {35 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{3 d}+\frac {7 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{d}-7 a^{4} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+7 a^{4} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+21 a^{4} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-21 a^{4} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-7 a^{4} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+7 a^{4} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\frac {7 a^{4} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}}{2}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\) | \(308\) |
1/240*a^4*(840*d*x+3*sin(5*d*x+5*c)+30*sin(4*d*x+4*c)+145*sin(3*d*x+3*c)+4 80*sin(2*d*x+2*c)+1470*sin(d*x+c))/d
Time = 0.26 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.75 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^4 \, dx=\frac {105 \, a^{4} d x + {\left (6 \, a^{4} \cos \left (d x + c\right )^{4} + 30 \, a^{4} \cos \left (d x + c\right )^{3} + 68 \, a^{4} \cos \left (d x + c\right )^{2} + 105 \, a^{4} \cos \left (d x + c\right ) + 166 \, a^{4}\right )} \sin \left (d x + c\right )}{30 \, d} \]
1/30*(105*a^4*d*x + (6*a^4*cos(d*x + c)^4 + 30*a^4*cos(d*x + c)^3 + 68*a^4 *cos(d*x + c)^2 + 105*a^4*cos(d*x + c) + 166*a^4)*sin(d*x + c))/d
Timed out. \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^4 \, dx=\text {Timed out} \]
Time = 0.22 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.25 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^4 \, dx=\frac {8 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{4} - 240 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{4} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} + 120 \, a^{4} \sin \left (d x + c\right )}{120 \, d} \]
1/120*(8*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^4 - 24 0*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^4 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^4 + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^4 + 120*a^4*sin(d*x + c))/d
Time = 0.32 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.10 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^4 \, dx=\frac {105 \, {\left (d x + c\right )} a^{4} + \frac {2 \, {\left (105 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 490 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 896 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 790 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 375 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{30 \, d} \]
1/30*(105*(d*x + c)*a^4 + 2*(105*a^4*tan(1/2*d*x + 1/2*c)^9 + 490*a^4*tan( 1/2*d*x + 1/2*c)^7 + 896*a^4*tan(1/2*d*x + 1/2*c)^5 + 790*a^4*tan(1/2*d*x + 1/2*c)^3 + 375*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5) /d
Time = 16.86 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.03 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^4 \, dx=\frac {7\,a^4\,x}{2}+\frac {7\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\frac {98\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}+\frac {896\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{15}+\frac {158\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+25\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]