Integrand size = 20, antiderivative size = 31 \[ \int \cos ^3(a+b x) \sin ^3(2 a+2 b x) \, dx=-\frac {8 \cos ^7(a+b x)}{7 b}+\frac {8 \cos ^9(a+b x)}{9 b} \] Output:
-8/7*cos(b*x+a)^7/b+8/9*cos(b*x+a)^9/b
Time = 0.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \cos ^3(a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {4 \cos ^7(a+b x) (-11+7 \cos (2 (a+b x)))}{63 b} \] Input:
Integrate[Cos[a + b*x]^3*Sin[2*a + 2*b*x]^3,x]
Output:
(4*Cos[a + b*x]^7*(-11 + 7*Cos[2*(a + b*x)]))/(63*b)
Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 4775, 3042, 3045, 244, 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.
\(\displaystyle \int \sin ^3(2 a+2 b x) \cos ^3(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (2 a+2 b x)^3 \cos (a+b x)^3dx\) |
\(\Big \downarrow \) 4775 |
\(\displaystyle 8 \int \cos ^6(a+b x) \sin ^3(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 8 \int \cos (a+b x)^6 \sin (a+b x)^3dx\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle -\frac {8 \int \cos ^6(a+b x) \left (1-\cos ^2(a+b x)\right )d\cos (a+b x)}{b}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -\frac {8 \int \left (\cos ^6(a+b x)-\cos ^8(a+b x)\right )d\cos (a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {8 \left (\frac {1}{7} \cos ^7(a+b x)-\frac {1}{9} \cos ^9(a+b x)\right )}{b}\) |
Input:
Int[Cos[a + b*x]^3*Sin[2*a + 2*b*x]^3,x]
Output:
(-8*(Cos[a + b*x]^7/7 - Cos[a + b*x]^9/9))/b
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
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 = 5.96 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58
method | result | size |
parallelrisch | \(\frac {4864+135 \cos \left (7 b x +7 a \right )+35 \cos \left (9 b x +9 a \right )-840 \cos \left (3 b x +3 a \right )-1890 \cos \left (b x +a \right )}{10080 b}\) | \(49\) |
default | \(-\frac {3 \cos \left (b x +a \right )}{16 b}-\frac {\cos \left (3 b x +3 a \right )}{12 b}+\frac {3 \cos \left (7 b x +7 a \right )}{224 b}+\frac {\cos \left (9 b x +9 a \right )}{288 b}\) | \(55\) |
risch | \(-\frac {3 \cos \left (b x +a \right )}{16 b}-\frac {\cos \left (3 b x +3 a \right )}{12 b}+\frac {3 \cos \left (7 b x +7 a \right )}{224 b}+\frac {\cos \left (9 b x +9 a \right )}{288 b}\) | \(55\) |
orering | \(-\frac {4540 \left (-3 \cos \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right )^{3} b \sin \left (b x +a \right )+6 \cos \left (b x +a \right )^{3} \sin \left (2 b x +2 a \right )^{2} b \cos \left (2 b x +2 a \right )\right )}{3969 b^{2}}-\frac {754 \left (-6 b^{3} \sin \left (b x +a \right )^{3} \sin \left (2 b x +2 a \right )^{3}+108 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right )^{2} b^{3} \cos \left (2 b x +2 a \right )+129 \cos \left (b x +a \right )^{2} b^{3} \sin \left (2 b x +2 a \right )^{3} \sin \left (b x +a \right )-216 \cos \left (b x +a \right )^{2} \sin \left (b x +a \right ) \sin \left (2 b x +2 a \right ) b^{3} \cos \left (2 b x +2 a \right )^{2}-222 \cos \left (b x +a \right )^{3} b^{3} \sin \left (2 b x +2 a \right )^{2} \cos \left (2 b x +2 a \right )+48 \cos \left (b x +a \right )^{3} b^{3} \cos \left (2 b x +2 a \right )^{3}\right )}{5103 b^{4}}-\frac {20 \left (780 b^{5} \sin \left (b x +a \right )^{3} \sin \left (2 b x +2 a \right )^{3}-7743 \cos \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right )^{3} b^{5} \sin \left (b x +a \right )-11880 b^{5} \sin \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right )^{2} \cos \left (2 b x +2 a \right ) \cos \left (b x +a \right )-1440 b^{5} \sin \left (b x +a \right )^{3} \sin \left (2 b x +2 a \right ) \cos \left (2 b x +2 a \right )^{2}+11526 \cos \left (b x +a \right )^{3} \sin \left (2 b x +2 a \right )^{2} b^{5} \cos \left (2 b x +2 a \right )+19440 \cos \left (b x +a \right )^{2} \cos \left (2 b x +2 a \right )^{2} b^{5} \sin \left (b x +a \right ) \sin \left (2 b x +2 a \right )+2880 \cos \left (b x +a \right ) \cos \left (2 b x +2 a \right )^{3} b^{5} \sin \left (b x +a \right )^{2}-3360 \cos \left (b x +a \right )^{3} \cos \left (2 b x +2 a \right )^{3} b^{5}\right )}{5103 b^{6}}-\frac {540669 b^{7} \sin \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{3} \cos \left (b x +a \right )^{2}+1113588 b^{7} \sin \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right )^{2} \cos \left (b x +a \right ) \cos \left (2 b x +2 a \right )-86226 b^{7} \sin \left (b x +a \right )^{3} \sin \left (2 b x +2 a \right )^{3}+231840 b^{7} \sin \left (b x +a \right )^{3} \sin \left (2 b x +2 a \right ) \cos \left (2 b x +2 a \right )^{2}-1531656 \cos \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right ) b^{7} \sin \left (b x +a \right ) \cos \left (2 b x +2 a \right )^{2}-710142 \cos \left (b x +a \right )^{3} \sin \left (2 b x +2 a \right )^{2} b^{7} \cos \left (2 b x +2 a \right )-342720 b^{7} \sin \left (b x +a \right )^{2} \cos \left (2 b x +2 a \right )^{3} \cos \left (b x +a \right )+226128 \cos \left (b x +a \right )^{3} b^{7} \cos \left (2 b x +2 a \right )^{3}}{35721 b^{8}}\) | \(778\) |
Input:
int(cos(b*x+a)^3*sin(2*b*x+2*a)^3,x,method=_RETURNVERBOSE)
Output:
1/10080*(4864+135*cos(7*b*x+7*a)+35*cos(9*b*x+9*a)-840*cos(3*b*x+3*a)-1890 *cos(b*x+a))/b
Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \cos ^3(a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {8 \, {\left (7 \, \cos \left (b x + a\right )^{9} - 9 \, \cos \left (b x + a\right )^{7}\right )}}{63 \, b} \] Input:
integrate(cos(b*x+a)^3*sin(2*b*x+2*a)^3,x, algorithm="fricas")
Output:
8/63*(7*cos(b*x + a)^9 - 9*cos(b*x + a)^7)/b
Leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (26) = 52\).
Time = 4.85 (sec) , antiderivative size = 284, normalized size of antiderivative = 9.16 \[ \int \cos ^3(a+b x) \sin ^3(2 a+2 b x) \, dx=\begin {cases} - \frac {94 \sin ^{3}{\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )}}{315 b} - \frac {32 \sin ^{3}{\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{105 b} - \frac {4 \sin ^{2}{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{7 b} - \frac {64 \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{105 b} + \frac {13 \sin {\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )}}{105 b} + \frac {8 \sin {\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{35 b} - \frac {46 \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{3}{\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{105 b} - \frac {16 \cos ^{3}{\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{63 b} & \text {for}\: b \neq 0 \\x \sin ^{3}{\left (2 a \right )} \cos ^{3}{\left (a \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(b*x+a)**3*sin(2*b*x+2*a)**3,x)
Output:
Piecewise((-94*sin(a + b*x)**3*sin(2*a + 2*b*x)**3/(315*b) - 32*sin(a + b* x)**3*sin(2*a + 2*b*x)*cos(2*a + 2*b*x)**2/(105*b) - 4*sin(a + b*x)**2*sin (2*a + 2*b*x)**2*cos(a + b*x)*cos(2*a + 2*b*x)/(7*b) - 64*sin(a + b*x)**2* cos(a + b*x)*cos(2*a + 2*b*x)**3/(105*b) + 13*sin(a + b*x)*sin(2*a + 2*b*x )**3*cos(a + b*x)**2/(105*b) + 8*sin(a + b*x)*sin(2*a + 2*b*x)*cos(a + b*x )**2*cos(2*a + 2*b*x)**2/(35*b) - 46*sin(2*a + 2*b*x)**2*cos(a + b*x)**3*c os(2*a + 2*b*x)/(105*b) - 16*cos(a + b*x)**3*cos(2*a + 2*b*x)**3/(63*b), N e(b, 0)), (x*sin(2*a)**3*cos(a)**3, True))
Time = 0.04 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52 \[ \int \cos ^3(a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {7 \, \cos \left (9 \, b x + 9 \, a\right ) + 27 \, \cos \left (7 \, b x + 7 \, a\right ) - 168 \, \cos \left (3 \, b x + 3 \, a\right ) - 378 \, \cos \left (b x + a\right )}{2016 \, b} \] Input:
integrate(cos(b*x+a)^3*sin(2*b*x+2*a)^3,x, algorithm="maxima")
Output:
1/2016*(7*cos(9*b*x + 9*a) + 27*cos(7*b*x + 7*a) - 168*cos(3*b*x + 3*a) - 378*cos(b*x + a))/b
Time = 0.37 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \cos ^3(a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {8 \, {\left (7 \, \cos \left (b x + a\right )^{9} - 9 \, \cos \left (b x + a\right )^{7}\right )}}{63 \, b} \] Input:
integrate(cos(b*x+a)^3*sin(2*b*x+2*a)^3,x, algorithm="giac")
Output:
8/63*(7*cos(b*x + a)^9 - 9*cos(b*x + a)^7)/b
Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \cos ^3(a+b x) \sin ^3(2 a+2 b x) \, dx=-\frac {8\,\left (9\,{\cos \left (a+b\,x\right )}^7-7\,{\cos \left (a+b\,x\right )}^9\right )}{63\,b} \] Input:
int(cos(a + b*x)^3*sin(2*a + 2*b*x)^3,x)
Output:
-(8*(9*cos(a + b*x)^7 - 7*cos(a + b*x)^9))/(63*b)
Time = 0.17 (sec) , antiderivative size = 242, normalized size of antiderivative = 7.81 \[ \int \cos ^3(a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {70 \cos \left (2 b x +2 a \right ) \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{2} \sin \left (b x +a \right )^{2}-58 \cos \left (2 b x +2 a \right ) \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{2}-112 \cos \left (2 b x +2 a \right ) \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2}-80 \cos \left (2 b x +2 a \right ) \cos \left (b x +a \right )-72 \cos \left (2 b x +2 a \right ) \sin \left (b x +a \right )^{2}+36 \cos \left (2 b x +2 a \right )+72 \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right ) \sin \left (b x +a \right )+35 \sin \left (2 b x +2 a \right )^{3} \sin \left (b x +a \right )^{3}-33 \sin \left (2 b x +2 a \right )^{3} \sin \left (b x +a \right )-168 \sin \left (2 b x +2 a \right ) \sin \left (b x +a \right )^{3}+72 \sin \left (2 b x +2 a \right ) \sin \left (b x +a \right )-116}{315 b} \] Input:
int(cos(b*x+a)^3*sin(2*b*x+2*a)^3,x)
Output:
(70*cos(2*a + 2*b*x)*cos(a + b*x)*sin(2*a + 2*b*x)**2*sin(a + b*x)**2 - 58 *cos(2*a + 2*b*x)*cos(a + b*x)*sin(2*a + 2*b*x)**2 - 112*cos(2*a + 2*b*x)* cos(a + b*x)*sin(a + b*x)**2 - 80*cos(2*a + 2*b*x)*cos(a + b*x) - 72*cos(2 *a + 2*b*x)*sin(a + b*x)**2 + 36*cos(2*a + 2*b*x) + 72*cos(a + b*x)*sin(2* a + 2*b*x)*sin(a + b*x) + 35*sin(2*a + 2*b*x)**3*sin(a + b*x)**3 - 33*sin( 2*a + 2*b*x)**3*sin(a + b*x) - 168*sin(2*a + 2*b*x)*sin(a + b*x)**3 + 72*s in(2*a + 2*b*x)*sin(a + b*x) - 116)/(315*b)