Integrand size = 19, antiderivative size = 85 \[ \int \cos (c+d x) (a+a \cos (c+d x))^3 \, dx=\frac {15 a^3 x}{8}+\frac {4 a^3 \sin (c+d x)}{d}+\frac {15 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^3 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {a^3 \sin ^3(c+d x)}{d} \]
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 = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.60 \[ \int \cos (c+d x) (a+a \cos (c+d x))^3 \, dx=\frac {a^3 (60 d x+104 \sin (c+d x)+32 \sin (2 (c+d x))+8 \sin (3 (c+d x))+\sin (4 (c+d x)))}{32 d} \]
(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 = 0.33 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3042, 3230, 3042, 3124, 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.
\(\displaystyle \int \cos (c+d x) (a \cos (c+d x)+a)^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3dx\) |
\(\Big \downarrow \) 3230 |
\(\displaystyle \frac {3}{4} \int (\cos (c+d x) a+a)^3dx+\frac {\sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{4} \int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3dx+\frac {\sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
\(\Big \downarrow \) 3124 |
\(\displaystyle \frac {3}{4} \int \left (\cos ^3(c+d x) a^3+3 \cos ^2(c+d x) a^3+3 \cos (c+d x) a^3+a^3\right )dx+\frac {\sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3}{4} \left (-\frac {a^3 \sin ^3(c+d x)}{3 d}+\frac {4 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {5 a^3 x}{2}\right )+\frac {\sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
((a + a*Cos[c + d*x])^3*Sin[c + d*x])/(4*d) + (3*((5*a^3*x)/2 + (4*a^3*Sin [c + d*x])/d + (3*a^3*Cos[c + d*x]*Sin[c + d*x])/(2*d) - (a^3*Sin[c + d*x] ^3)/(3*d)))/4
3.1.25.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandTri g[(a + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1)) Int[(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Time = 2.31 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.62
method | result | size |
parallelrisch | \(\frac {a^{3} \left (60 d x +\sin \left (4 d x +4 c \right )+8 \sin \left (3 d x +3 c \right )+32 \sin \left (2 d x +2 c \right )+104 \sin \left (d x +c \right )\right )}{32 d}\) | \(53\) |
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 | \(\frac {a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{3} \sin \left (d x +c \right )}{d}\) | \(100\) |
default | \(\frac {a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{3} \sin \left (d x +c \right )}{d}\) | \(100\) |
parts | \(\frac {a^{3} \sin \left (d x +c \right )}{d}+\frac {a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {3 a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{d}\) | \(108\) |
norman | \(\frac {\frac {15 a^{3} x}{8}+\frac {49 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {73 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {55 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {15 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {15 a^{3} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {45 a^{3} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {15 a^{3} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {15 a^{3} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\) | \(166\) |
Time = 0.25 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.74 \[ \int \cos (c+d x) (a+a \cos (c+d x))^3 \, dx=\frac {15 \, a^{3} d x + {\left (2 \, a^{3} \cos \left (d x + c\right )^{3} + 8 \, a^{3} \cos \left (d x + c\right )^{2} + 15 \, a^{3} \cos \left (d x + c\right ) + 24 \, a^{3}\right )} \sin \left (d x + c\right )}{8 \, d} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (78) = 156\).
Time = 0.20 (sec) , antiderivative size = 224, normalized size of antiderivative = 2.64 \[ \int \cos (c+d x) (a+a \cos (c+d x))^3 \, dx=\begin {cases} \frac {3 a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 a^{3} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {5 a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {3 a^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {a^{3} \sin {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \cos {\left (c \right )} + a\right )^{3} \cos {\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((3*a**3*x*sin(c + d*x)**4/8 + 3*a**3*x*sin(c + d*x)**2*cos(c + d *x)**2/4 + 3*a**3*x*sin(c + d*x)**2/2 + 3*a**3*x*cos(c + d*x)**4/8 + 3*a** 3*x*cos(c + d*x)**2/2 + 3*a**3*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 2*a**3 *sin(c + d*x)**3/d + 5*a**3*sin(c + d*x)*cos(c + d*x)**3/(8*d) + 3*a**3*si n(c + d*x)*cos(c + d*x)**2/d + 3*a**3*sin(c + d*x)*cos(c + d*x)/(2*d) + a* *3*sin(c + d*x)/d, Ne(d, 0)), (x*(a*cos(c) + a)**3*cos(c), True))
Time = 0.24 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.11 \[ \int \cos (c+d x) (a+a \cos (c+d x))^3 \, dx=-\frac {32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{3} - {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 32 \, a^{3} \sin \left (d x + c\right )}{32 \, d} \]
-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 = 71, normalized size of antiderivative = 0.84 \[ \int \cos (c+d x) (a+a \cos (c+d x))^3 \, dx=\frac {15}{8} \, a^{3} x + \frac {a^{3} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {a^{3} \sin \left (3 \, d x + 3 \, c\right )}{4 \, d} + \frac {a^{3} \sin \left (2 \, d x + 2 \, c\right )}{d} + \frac {13 \, a^{3} \sin \left (d x + c\right )}{4 \, d} \]
15/8*a^3*x + 1/32*a^3*sin(4*d*x + 4*c)/d + 1/4*a^3*sin(3*d*x + 3*c)/d + a^ 3*sin(2*d*x + 2*c)/d + 13/4*a^3*sin(d*x + c)/d
Time = 17.36 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.05 \[ \int \cos (c+d x) (a+a \cos (c+d x))^3 \, dx=\frac {15\,a^3\,x}{8}+\frac {\frac {15\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {55\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {73\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {49\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]