Integrand size = 21, antiderivative size = 77 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {\left (a^2-b^2\right ) (a+b \sin (c+d x))^4}{4 b^3 d}+\frac {2 a (a+b \sin (c+d x))^5}{5 b^3 d}-\frac {(a+b \sin (c+d x))^6}{6 b^3 d} \]
-1/4*(a^2-b^2)*(a+b*sin(d*x+c))^4/b^3/d+2/5*a*(a+b*sin(d*x+c))^5/b^3/d-1/6 *(a+b*sin(d*x+c))^6/b^3/d
Time = 0.11 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.73 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {(a+b \sin (c+d x))^4 \left (-a^2+10 b^2+5 b^2 \cos (2 (c+d x))+4 a b \sin (c+d x)\right )}{60 b^3 d} \]
((a + b*Sin[c + d*x])^4*(-a^2 + 10*b^2 + 5*b^2*Cos[2*(c + d*x)] + 4*a*b*Si n[c + d*x]))/(60*b^3*d)
Time = 0.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.86, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3147, 476, 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.
\(\displaystyle \int \cos ^3(c+d x) (a+b \sin (c+d x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^3 (a+b \sin (c+d x))^3dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {\int (a+b \sin (c+d x))^3 \left (b^2-b^2 \sin ^2(c+d x)\right )d(b \sin (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 476 |
\(\displaystyle \frac {\int \left (-(a+b \sin (c+d x))^5+2 a (a+b \sin (c+d x))^4+\left (b^2-a^2\right ) (a+b \sin (c+d x))^3\right )d(b \sin (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{4} \left (a^2-b^2\right ) (a+b \sin (c+d x))^4-\frac {1}{6} (a+b \sin (c+d x))^6+\frac {2}{5} a (a+b \sin (c+d x))^5}{b^3 d}\) |
(-1/4*((a^2 - b^2)*(a + b*Sin[c + d*x])^4) + (2*a*(a + b*Sin[c + d*x])^5)/ 5 - (a + b*Sin[c + d*x])^6/6)/(b^3*d)
3.5.1.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 0.70 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.31
method | result | size |
derivativedivides | \(-\frac {\frac {b^{3} \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {3 a \,b^{2} \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (3 a^{2} b -b^{3}\right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (a^{3}-3 a \,b^{2}\right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right ) a^{2} b}{2}-a^{3} \sin \left (d x +c \right )}{d}\) | \(101\) |
default | \(-\frac {\frac {b^{3} \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {3 a \,b^{2} \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (3 a^{2} b -b^{3}\right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (a^{3}-3 a \,b^{2}\right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right ) a^{2} b}{2}-a^{3} \sin \left (d x +c \right )}{d}\) | \(101\) |
parallelrisch | \(\frac {5 \cos \left (6 d x +6 c \right ) b^{3}-36 \sin \left (5 d x +5 c \right ) a \,b^{2}-90 \cos \left (4 d x +4 c \right ) a^{2} b +80 \sin \left (3 d x +3 c \right ) a^{3}-60 \sin \left (3 d x +3 c \right ) a \,b^{2}-360 \cos \left (2 d x +2 c \right ) a^{2} b -45 \cos \left (2 d x +2 c \right ) b^{3}+720 a^{3} \sin \left (d x +c \right )+360 a \,b^{2} \sin \left (d x +c \right )+450 a^{2} b +40 b^{3}}{960 d}\) | \(143\) |
risch | \(\frac {3 a^{3} \sin \left (d x +c \right )}{4 d}+\frac {3 a \,b^{2} \sin \left (d x +c \right )}{8 d}+\frac {b^{3} \cos \left (6 d x +6 c \right )}{192 d}-\frac {3 \sin \left (5 d x +5 c \right ) a \,b^{2}}{80 d}-\frac {3 b \cos \left (4 d x +4 c \right ) a^{2}}{32 d}+\frac {\sin \left (3 d x +3 c \right ) a^{3}}{12 d}-\frac {\sin \left (3 d x +3 c \right ) a \,b^{2}}{16 d}-\frac {3 b \cos \left (2 d x +2 c \right ) a^{2}}{8 d}-\frac {3 b^{3} \cos \left (2 d x +2 c \right )}{64 d}\) | \(154\) |
norman | \(\frac {\frac {\left (12 a^{2} b +4 b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (12 a^{2} b +4 b^{3}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a^{3} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a^{2} b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a^{2} b \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (9 a^{2} b -2 b^{3}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {12 a \left (5 a^{2}+2 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {12 a \left (5 a^{2}+2 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {2 a \left (11 a^{2}+12 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a \left (11 a^{2}+12 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\) | \(288\) |
-1/d*(1/6*b^3*sin(d*x+c)^6+3/5*a*b^2*sin(d*x+c)^5+1/4*(3*a^2*b-b^3)*sin(d* x+c)^4+1/3*(a^3-3*a*b^2)*sin(d*x+c)^3-3/2*sin(d*x+c)^2*a^2*b-a^3*sin(d*x+c ))
Time = 0.30 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.23 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {10 \, b^{3} \cos \left (d x + c\right )^{6} - 15 \, {\left (3 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{4} - 4 \, {\left (9 \, a b^{2} \cos \left (d x + c\right )^{4} - 10 \, a^{3} - 6 \, a b^{2} - {\left (5 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, d} \]
1/60*(10*b^3*cos(d*x + c)^6 - 15*(3*a^2*b + b^3)*cos(d*x + c)^4 - 4*(9*a*b ^2*cos(d*x + c)^4 - 10*a^3 - 6*a*b^2 - (5*a^3 + 3*a*b^2)*cos(d*x + c)^2)*s in(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (66) = 132\).
Time = 0.34 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.96 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=\begin {cases} \frac {2 a^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {a^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {3 a^{2} b \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {2 a b^{2} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {a b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {b^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} - \frac {b^{3} \cos ^{6}{\left (c + d x \right )}}{12 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right )^{3} \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((2*a**3*sin(c + d*x)**3/(3*d) + a**3*sin(c + d*x)*cos(c + d*x)** 2/d - 3*a**2*b*cos(c + d*x)**4/(4*d) + 2*a*b**2*sin(c + d*x)**5/(5*d) + a* b**2*sin(c + d*x)**3*cos(c + d*x)**2/d - b**3*sin(c + d*x)**2*cos(c + d*x) **4/(4*d) - b**3*cos(c + d*x)**6/(12*d), Ne(d, 0)), (x*(a + b*sin(c))**3*c os(c)**3, True))
Time = 0.18 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.30 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {10 \, b^{3} \sin \left (d x + c\right )^{6} + 36 \, a b^{2} \sin \left (d x + c\right )^{5} - 90 \, a^{2} b \sin \left (d x + c\right )^{2} + 15 \, {\left (3 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{4} - 60 \, a^{3} \sin \left (d x + c\right ) + 20 \, {\left (a^{3} - 3 \, a b^{2}\right )} \sin \left (d x + c\right )^{3}}{60 \, d} \]
-1/60*(10*b^3*sin(d*x + c)^6 + 36*a*b^2*sin(d*x + c)^5 - 90*a^2*b*sin(d*x + c)^2 + 15*(3*a^2*b - b^3)*sin(d*x + c)^4 - 60*a^3*sin(d*x + c) + 20*(a^3 - 3*a*b^2)*sin(d*x + c)^3)/d
Time = 0.36 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.45 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {10 \, b^{3} \sin \left (d x + c\right )^{6} + 36 \, a b^{2} \sin \left (d x + c\right )^{5} + 45 \, a^{2} b \sin \left (d x + c\right )^{4} - 15 \, b^{3} \sin \left (d x + c\right )^{4} + 20 \, a^{3} \sin \left (d x + c\right )^{3} - 60 \, a b^{2} \sin \left (d x + c\right )^{3} - 90 \, a^{2} b \sin \left (d x + c\right )^{2} - 60 \, a^{3} \sin \left (d x + c\right )}{60 \, d} \]
-1/60*(10*b^3*sin(d*x + c)^6 + 36*a*b^2*sin(d*x + c)^5 + 45*a^2*b*sin(d*x + c)^4 - 15*b^3*sin(d*x + c)^4 + 20*a^3*sin(d*x + c)^3 - 60*a*b^2*sin(d*x + c)^3 - 90*a^2*b*sin(d*x + c)^2 - 60*a^3*sin(d*x + c))/d
Time = 5.14 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.27 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {{\sin \left (c+d\,x\right )}^3\,\left (a\,b^2-\frac {a^3}{3}\right )-{\sin \left (c+d\,x\right )}^4\,\left (\frac {3\,a^2\,b}{4}-\frac {b^3}{4}\right )+a^3\,\sin \left (c+d\,x\right )-\frac {b^3\,{\sin \left (c+d\,x\right )}^6}{6}+\frac {3\,a^2\,b\,{\sin \left (c+d\,x\right )}^2}{2}-\frac {3\,a\,b^2\,{\sin \left (c+d\,x\right )}^5}{5}}{d} \]