Integrand size = 28, antiderivative size = 76 \[ \int \cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 C x}{8}+\frac {B \sin (c+d x)}{d}+\frac {3 C \cos (c+d x) \sin (c+d x)}{8 d}+\frac {C \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {B \sin ^3(c+d x)}{3 d} \]
3/8*C*x+B*sin(d*x+c)/d+3/8*C*cos(d*x+c)*sin(d*x+c)/d+1/4*C*cos(d*x+c)^3*si n(d*x+c)/d-1/3*B*sin(d*x+c)^3/d
Time = 0.10 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.96 \[ \int \cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 C (c+d x)}{8 d}+\frac {B \sin (c+d x)}{d}-\frac {B \sin ^3(c+d x)}{3 d}+\frac {C \sin (2 (c+d x))}{4 d}+\frac {C \sin (4 (c+d x))}{32 d} \]
(3*C*(c + d*x))/(8*d) + (B*Sin[c + d*x])/d - (B*Sin[c + d*x]^3)/(3*d) + (C *Sin[2*(c + d*x)])/(4*d) + (C*Sin[4*(c + d*x)])/(32*d)
Time = 0.46 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {3042, 3489, 3042, 3227, 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.
\(\displaystyle \int \cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 3489 |
\(\displaystyle \int \cos ^3(c+d x) (B+C \cos (c+d x))dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle B \int \cos ^3(c+d x)dx+C \int \cos ^4(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle B \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx+C \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx\) |
\(\Big \downarrow \) 3113 |
\(\displaystyle C \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {B \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle C \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {B \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle C \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {B \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle C \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {B \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle C \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {B \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle C \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {B \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
-((B*(-Sin[c + d*x] + Sin[c + d*x]^3/3))/d) + C*((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)
3.3.17.3.1 Defintions of rubi rules used
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[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b Int[(b*Sin[e + f* x])^(m + 1)*(B + C*Sin[e + f*x]), x], x] /; FreeQ[{b, e, f, B, C, m}, x]
Time = 3.88 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.75
method | result | size |
parallelrisch | \(\frac {36 d x C +72 B \sin \left (d x +c \right )+8 B \sin \left (3 d x +3 c \right )+3 \sin \left (4 d x +4 c \right ) C +24 \sin \left (2 d x +2 c \right ) C}{96 d}\) | \(57\) |
derivativedivides | \(\frac {C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(60\) |
default | \(\frac {C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(60\) |
parts | \(\frac {B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(62\) |
risch | \(\frac {3 C x}{8}+\frac {3 B \sin \left (d x +c \right )}{4 d}+\frac {C \sin \left (4 d x +4 c \right )}{32 d}+\frac {B \sin \left (3 d x +3 c \right )}{12 d}+\frac {C \sin \left (2 d x +2 c \right )}{4 d}\) | \(63\) |
norman | \(\frac {\frac {3 C x}{8}+\frac {3 C x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {9 C x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {3 C x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {3 C x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {\left (8 B -5 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (8 B +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (40 B -9 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {\left (40 B +9 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\) | \(172\) |
Time = 0.25 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.70 \[ \int \cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {9 \, C d x + {\left (6 \, C \cos \left (d x + c\right )^{3} + 8 \, B \cos \left (d x + c\right )^{2} + 9 \, C \cos \left (d x + c\right ) + 16 \, B\right )} \sin \left (d x + c\right )}{24 \, d} \]
1/24*(9*C*d*x + (6*C*cos(d*x + c)^3 + 8*B*cos(d*x + c)^2 + 9*C*cos(d*x + c ) + 16*B)*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 150 vs. \(2 (70) = 140\).
Time = 0.16 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.97 \[ \int \cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\begin {cases} \frac {2 B \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {B \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 C x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 C x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 C x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 C \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 C \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (B \cos {\left (c \right )} + C \cos ^{2}{\left (c \right )}\right ) \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((2*B*sin(c + d*x)**3/(3*d) + B*sin(c + d*x)*cos(c + d*x)**2/d + 3*C*x*sin(c + d*x)**4/8 + 3*C*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*C*x* cos(c + d*x)**4/8 + 3*C*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 5*C*sin(c + d *x)*cos(c + d*x)**3/(8*d), Ne(d, 0)), (x*(B*cos(c) + C*cos(c)**2)*cos(c)** 2, True))
Time = 0.21 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.75 \[ \int \cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=-\frac {32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C}{96 \, d} \]
-1/96*(32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B - 3*(12*d*x + 12*c + sin(4*d *x + 4*c) + 8*sin(2*d*x + 2*c))*C)/d
Time = 0.29 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.82 \[ \int \cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3}{8} \, C x + \frac {C \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {B \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {C \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {3 \, B \sin \left (d x + c\right )}{4 \, d} \]
3/8*C*x + 1/32*C*sin(4*d*x + 4*c)/d + 1/12*B*sin(3*d*x + 3*c)/d + 1/4*C*si n(2*d*x + 2*c)/d + 3/4*B*sin(d*x + c)/d
Time = 1.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.99 \[ \int \cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3\,C\,x}{8}+\frac {2\,B\,\sin \left (c+d\,x\right )}{3\,d}+\frac {3\,C\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{8\,d}+\frac {B\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}+\frac {C\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{4\,d} \]