Integrand size = 21, antiderivative size = 85 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {15 a^3 x}{8}+\frac {4 a^3 \sin (c+d x)}{d}+\frac {15 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^3 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {a^3 \sin ^3(c+d x)}{d} \]
15/8*a^3*x+4*a^3*sin(d*x+c)/d+15/8*a^3*cos(d*x+c)*sin(d*x+c)/d+1/4*a^3*cos (d*x+c)^3*sin(d*x+c)/d-a^3*sin(d*x+c)^3/d
Time = 0.11 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.60 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {a^3 (60 d x+104 \sin (c+d x)+32 \sin (2 (c+d x))+8 \sin (3 (c+d x))+\sin (4 (c+d x)))}{32 d} \]
(a^3*(60*d*x + 104*Sin[c + d*x] + 32*Sin[2*(c + d*x)] + 8*Sin[3*(c + d*x)] + Sin[4*(c + d*x)]))/(32*d)
Time = 0.30 (sec) , antiderivative size = 85, 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.
\(\displaystyle \int \cos ^4(c+d x) (a \sec (c+d x)+a)^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
\(\Big \downarrow \) 4278 |
\(\displaystyle \int \left (a^3 \cos ^4(c+d x)+3 a^3 \cos ^3(c+d x)+3 a^3 \cos ^2(c+d x)+a^3 \cos (c+d x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^3 \sin ^3(c+d x)}{d}+\frac {4 a^3 \sin (c+d x)}{d}+\frac {a^3 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {15 a^3 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {15 a^3 x}{8}\) |
(15*a^3*x)/8 + (4*a^3*Sin[c + d*x])/d + (15*a^3*Cos[c + d*x]*Sin[c + d*x]) /(8*d) + (a^3*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) - (a^3*Sin[c + d*x]^3)/d
3.1.27.3.1 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 = 0.54 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.62
method | result | size |
parallelrisch | \(\frac {a^{3} \left (60 d x +104 \sin \left (d x +c \right )+\sin \left (4 d x +4 c \right )+8 \sin \left (3 d x +3 c \right )+32 \sin \left (2 d x +2 c \right )\right )}{32 d}\) | \(53\) |
risch | \(\frac {15 a^{3} x}{8}+\frac {13 a^{3} \sin \left (d x +c \right )}{4 d}+\frac {a^{3} \sin \left (4 d x +4 c \right )}{32 d}+\frac {a^{3} \sin \left (3 d x +3 c \right )}{4 d}+\frac {a^{3} \sin \left (2 d x +2 c \right )}{d}\) | \(72\) |
derivativedivides | \(\frac {a^{3} \sin \left (d x +c \right )+3 a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+a^{3} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(100\) |
default | \(\frac {a^{3} \sin \left (d x +c \right )+3 a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+a^{3} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(100\) |
norman | \(\frac {\frac {15 a^{3} x}{8}+\frac {49 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {25 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d}-\frac {21 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 d}-\frac {11 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{2 d}+\frac {25 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}+\frac {15 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}+\frac {15 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{4}-\frac {15 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8}-\frac {15 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}-\frac {15 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}+\frac {15 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{4}+\frac {15 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{8}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}\) | \(253\) |
Time = 0.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.74 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {15 \, a^{3} d x + {\left (2 \, a^{3} \cos \left (d x + c\right )^{3} + 8 \, a^{3} \cos \left (d x + c\right )^{2} + 15 \, a^{3} \cos \left (d x + c\right ) + 24 \, a^{3}\right )} \sin \left (d x + c\right )}{8 \, d} \]
1/8*(15*a^3*d*x + (2*a^3*cos(d*x + c)^3 + 8*a^3*cos(d*x + c)^2 + 15*a^3*co s(d*x + c) + 24*a^3)*sin(d*x + c))/d
\[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \, dx=a^{3} \left (\int 3 \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 \cos ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \cos ^{4}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \cos ^{4}{\left (c + d x \right )}\, dx\right ) \]
a**3*(Integral(3*cos(c + d*x)**4*sec(c + d*x), x) + Integral(3*cos(c + d*x )**4*sec(c + d*x)**2, x) + Integral(cos(c + d*x)**4*sec(c + d*x)**3, x) + Integral(cos(c + d*x)**4, x))
Time = 0.21 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.11 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \, dx=-\frac {32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{3} - {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 32 \, a^{3} \sin \left (d x + c\right )}{32 \, d} \]
-1/32*(32*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^3 - (12*d*x + 12*c + sin(4*d *x + 4*c) + 8*sin(2*d*x + 2*c))*a^3 - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))* a^3 - 32*a^3*sin(d*x + c))/d
Time = 0.33 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.13 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {15 \, {\left (d x + c\right )} a^{3} + \frac {2 \, {\left (15 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 55 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 73 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 49 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{8 \, d} \]
1/8*(15*(d*x + c)*a^3 + 2*(15*a^3*tan(1/2*d*x + 1/2*c)^7 + 55*a^3*tan(1/2* d*x + 1/2*c)^5 + 73*a^3*tan(1/2*d*x + 1/2*c)^3 + 49*a^3*tan(1/2*d*x + 1/2* c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d
Time = 16.72 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.05 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {15\,a^3\,x}{8}+\frac {\frac {15\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {55\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {73\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {49\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]