Integrand size = 20, antiderivative size = 44 \[ \int \cos ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=-\frac {4 \cos ^8(a+b x)}{b}+\frac {32 \cos ^{10}(a+b x)}{5 b}-\frac {8 \cos ^{12}(a+b x)}{3 b} \]
Time = 0.41 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.55 \[ \int \cos ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=-\frac {600 \cos (2 (a+b x))+75 \cos (4 (a+b x))-100 \cos (6 (a+b x))-30 \cos (8 (a+b x))+12 \cos (10 (a+b x))+5 \cos (12 (a+b x))}{3840 b} \]
-1/3840*(600*Cos[2*(a + b*x)] + 75*Cos[4*(a + b*x)] - 100*Cos[6*(a + b*x)] - 30*Cos[8*(a + b*x)] + 12*Cos[10*(a + b*x)] + 5*Cos[12*(a + b*x)])/b
Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3042, 4775, 3042, 3045, 243, 49, 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.
\(\displaystyle \int \sin ^5(2 a+2 b x) \cos ^2(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (2 a+2 b x)^5 \cos (a+b x)^2dx\) |
\(\Big \downarrow \) 4775 |
\(\displaystyle 32 \int \cos ^7(a+b x) \sin ^5(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 32 \int \cos (a+b x)^7 \sin (a+b x)^5dx\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle -\frac {32 \int \cos ^7(a+b x) \left (1-\cos ^2(a+b x)\right )^2d\cos (a+b x)}{b}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle -\frac {16 \int \cos ^6(a+b x) \left (1-\cos ^2(a+b x)\right )^2d\cos ^2(a+b x)}{b}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {16 \int \left (\cos ^{10}(a+b x)-2 \cos ^8(a+b x)+\cos ^6(a+b x)\right )d\cos ^2(a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {16 \left (\frac {1}{6} \cos ^{12}(a+b x)-\frac {2}{5} \cos ^{10}(a+b x)+\frac {1}{4} \cos ^8(a+b x)\right )}{b}\) |
3.2.41.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> Simp[-(a*f)^(-1) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_ Symbol] :> Simp[2^p/e^p Int[(e*Cos[a + b*x])^(m + p)*Sin[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && I ntegerQ[p]
Time = 2.98 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.68
method | result | size |
parallelrisch | \(\frac {-600 \cos \left (2 x b +2 a \right )-1662-5 \cos \left (12 x b +12 a \right )-12 \cos \left (10 x b +10 a \right )+30 \cos \left (8 x b +8 a \right )+100 \cos \left (6 x b +6 a \right )-75 \cos \left (4 x b +4 a \right )}{3840 b}\) | \(74\) |
default | \(-\frac {5 \cos \left (2 x b +2 a \right )}{32 b}-\frac {5 \cos \left (4 x b +4 a \right )}{256 b}+\frac {5 \cos \left (6 x b +6 a \right )}{192 b}+\frac {\cos \left (8 x b +8 a \right )}{128 b}-\frac {\cos \left (10 x b +10 a \right )}{320 b}-\frac {\cos \left (12 x b +12 a \right )}{768 b}\) | \(86\) |
risch | \(-\frac {5 \cos \left (2 x b +2 a \right )}{32 b}-\frac {5 \cos \left (4 x b +4 a \right )}{256 b}+\frac {5 \cos \left (6 x b +6 a \right )}{192 b}+\frac {\cos \left (8 x b +8 a \right )}{128 b}-\frac {\cos \left (10 x b +10 a \right )}{320 b}-\frac {\cos \left (12 x b +12 a \right )}{768 b}\) | \(86\) |
1/3840*(-600*cos(2*b*x+2*a)-1662-5*cos(12*b*x+12*a)-12*cos(10*b*x+10*a)+30 *cos(8*b*x+8*a)+100*cos(6*b*x+6*a)-75*cos(4*b*x+4*a))/b
Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.82 \[ \int \cos ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=-\frac {4 \, {\left (10 \, \cos \left (b x + a\right )^{12} - 24 \, \cos \left (b x + a\right )^{10} + 15 \, \cos \left (b x + a\right )^{8}\right )}}{15 \, b} \]
Leaf count of result is larger than twice the leaf count of optimal. 597 vs. \(2 (37) = 74\).
Time = 11.30 (sec) , antiderivative size = 597, normalized size of antiderivative = 13.57 \[ \int \cos ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=\begin {cases} - \frac {5 x \sin ^{2}{\left (a + b x \right )} \sin ^{5}{\left (2 a + 2 b x \right )}}{32} - \frac {5 x \sin ^{2}{\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{16} - \frac {5 x \sin ^{2}{\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{32} - \frac {5 x \sin {\left (a + b x \right )} \sin ^{4}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{16} - \frac {5 x \sin {\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{8} - \frac {5 x \sin {\left (a + b x \right )} \cos {\left (a + b x \right )} \cos ^{5}{\left (2 a + 2 b x \right )}}{16} + \frac {5 x \sin ^{5}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )}}{32} + \frac {5 x \sin ^{3}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{16} + \frac {5 x \sin {\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{32} - \frac {125 \sin ^{2}{\left (a + b x \right )} \sin ^{4}{\left (2 a + 2 b x \right )} \cos {\left (2 a + 2 b x \right )}}{384 b} - \frac {2 \sin ^{2}{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{3 b} - \frac {217 \sin ^{2}{\left (a + b x \right )} \cos ^{5}{\left (2 a + 2 b x \right )}}{640 b} + \frac {95 \sin {\left (a + b x \right )} \sin ^{5}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )}}{192 b} + \frac {13 \sin {\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{12 b} + \frac {109 \sin {\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{192 b} - \frac {67 \sin ^{4}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{384 b} + \frac {139 \cos ^{2}{\left (a + b x \right )} \cos ^{5}{\left (2 a + 2 b x \right )}}{1920 b} & \text {for}\: b \neq 0 \\x \sin ^{5}{\left (2 a \right )} \cos ^{2}{\left (a \right )} & \text {otherwise} \end {cases} \]
Piecewise((-5*x*sin(a + b*x)**2*sin(2*a + 2*b*x)**5/32 - 5*x*sin(a + b*x)* *2*sin(2*a + 2*b*x)**3*cos(2*a + 2*b*x)**2/16 - 5*x*sin(a + b*x)**2*sin(2* a + 2*b*x)*cos(2*a + 2*b*x)**4/32 - 5*x*sin(a + b*x)*sin(2*a + 2*b*x)**4*c os(a + b*x)*cos(2*a + 2*b*x)/16 - 5*x*sin(a + b*x)*sin(2*a + 2*b*x)**2*cos (a + b*x)*cos(2*a + 2*b*x)**3/8 - 5*x*sin(a + b*x)*cos(a + b*x)*cos(2*a + 2*b*x)**5/16 + 5*x*sin(2*a + 2*b*x)**5*cos(a + b*x)**2/32 + 5*x*sin(2*a + 2*b*x)**3*cos(a + b*x)**2*cos(2*a + 2*b*x)**2/16 + 5*x*sin(2*a + 2*b*x)*co s(a + b*x)**2*cos(2*a + 2*b*x)**4/32 - 125*sin(a + b*x)**2*sin(2*a + 2*b*x )**4*cos(2*a + 2*b*x)/(384*b) - 2*sin(a + b*x)**2*sin(2*a + 2*b*x)**2*cos( 2*a + 2*b*x)**3/(3*b) - 217*sin(a + b*x)**2*cos(2*a + 2*b*x)**5/(640*b) + 95*sin(a + b*x)*sin(2*a + 2*b*x)**5*cos(a + b*x)/(192*b) + 13*sin(a + b*x) *sin(2*a + 2*b*x)**3*cos(a + b*x)*cos(2*a + 2*b*x)**2/(12*b) + 109*sin(a + b*x)*sin(2*a + 2*b*x)*cos(a + b*x)*cos(2*a + 2*b*x)**4/(192*b) - 67*sin(2 *a + 2*b*x)**4*cos(a + b*x)**2*cos(2*a + 2*b*x)/(384*b) + 139*cos(a + b*x) **2*cos(2*a + 2*b*x)**5/(1920*b), Ne(b, 0)), (x*sin(2*a)**5*cos(a)**2, Tru e))
Time = 0.22 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.64 \[ \int \cos ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=-\frac {5 \, \cos \left (12 \, b x + 12 \, a\right ) + 12 \, \cos \left (10 \, b x + 10 \, a\right ) - 30 \, \cos \left (8 \, b x + 8 \, a\right ) - 100 \, \cos \left (6 \, b x + 6 \, a\right ) + 75 \, \cos \left (4 \, b x + 4 \, a\right ) + 600 \, \cos \left (2 \, b x + 2 \, a\right )}{3840 \, b} \]
-1/3840*(5*cos(12*b*x + 12*a) + 12*cos(10*b*x + 10*a) - 30*cos(8*b*x + 8*a ) - 100*cos(6*b*x + 6*a) + 75*cos(4*b*x + 4*a) + 600*cos(2*b*x + 2*a))/b
Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.82 \[ \int \cos ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=-\frac {4 \, {\left (10 \, \cos \left (b x + a\right )^{12} - 24 \, \cos \left (b x + a\right )^{10} + 15 \, \cos \left (b x + a\right )^{8}\right )}}{15 \, b} \]
Time = 19.37 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.80 \[ \int \cos ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=-\frac {4\,{\cos \left (a+b\,x\right )}^8\,\left (10\,{\cos \left (a+b\,x\right )}^4-24\,{\cos \left (a+b\,x\right )}^2+15\right )}{15\,b} \]