Integrand size = 20, antiderivative size = 28 \[ \int \cos ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=-\frac {4 \cos ^6(a+b x)}{3 b}+\frac {\cos ^8(a+b x)}{b} \] Output:
-4/3*cos(b*x+a)^6/b+cos(b*x+a)^8/b
Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71 \[ \int \cos ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {-72 \cos (2 (a+b x))-12 \cos (4 (a+b x))+8 \cos (6 (a+b x))+3 \cos (8 (a+b x))}{384 b} \] Input:
Integrate[Cos[a + b*x]^2*Sin[2*a + 2*b*x]^3,x]
Output:
(-72*Cos[2*(a + b*x)] - 12*Cos[4*(a + b*x)] + 8*Cos[6*(a + b*x)] + 3*Cos[8 *(a + b*x)])/(384*b)
Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07, 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.
\(\displaystyle \int \sin ^3(2 a+2 b x) \cos ^2(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (2 a+2 b x)^3 \cos (a+b x)^2dx\) |
\(\Big \downarrow \) 4775 |
\(\displaystyle 8 \int \cos ^5(a+b x) \sin ^3(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 8 \int \cos (a+b x)^5 \sin (a+b x)^3dx\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle -\frac {8 \int \cos ^5(a+b x) \left (1-\cos ^2(a+b x)\right )d\cos (a+b x)}{b}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -\frac {8 \int \left (\cos ^5(a+b x)-\cos ^7(a+b x)\right )d\cos (a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {8 \left (\frac {1}{6} \cos ^6(a+b x)-\frac {1}{8} \cos ^8(a+b x)\right )}{b}\) |
Input:
Int[Cos[a + b*x]^2*Sin[2*a + 2*b*x]^3,x]
Output:
(-8*(Cos[a + b*x]^6/6 - Cos[a + b*x]^8/8))/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 = 3.92 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.86
method | result | size |
parallelrisch | \(\frac {-12 \cos \left (4 b x +4 a \right )-199+3 \cos \left (8 b x +8 a \right )-72 \cos \left (2 b x +2 a \right )+8 \cos \left (6 b x +6 a \right )}{384 b}\) | \(52\) |
default | \(-\frac {3 \cos \left (2 b x +2 a \right )}{16 b}-\frac {\cos \left (4 b x +4 a \right )}{32 b}+\frac {\cos \left (6 b x +6 a \right )}{48 b}+\frac {\cos \left (8 b x +8 a \right )}{128 b}\) | \(58\) |
risch | \(-\frac {3 \cos \left (2 b x +2 a \right )}{16 b}-\frac {\cos \left (4 b x +4 a \right )}{32 b}+\frac {\cos \left (6 b x +6 a \right )}{48 b}+\frac {\cos \left (8 b x +8 a \right )}{128 b}\) | \(58\) |
orering | \(-\frac {205 \left (-2 \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{3} b \sin \left (b x +a \right )+6 \cos \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right )^{2} b \cos \left (2 b x +2 a \right )\right )}{576 b^{2}}-\frac {91 \left (80 \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{3} \sin \left (b x +a \right ) b^{3}+36 \sin \left (2 b x +2 a \right )^{2} \sin \left (b x +a \right )^{2} \cos \left (2 b x +2 a \right ) b^{3}-144 \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right ) \sin \left (b x +a \right ) \cos \left (2 b x +2 a \right )^{2} b^{3}-204 \cos \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right )^{2} \cos \left (2 b x +2 a \right ) b^{3}+48 \cos \left (b x +a \right )^{2} \cos \left (2 b x +2 a \right )^{3} b^{3}\right )}{3072 b^{4}}-\frac {5 \left (-4352 \sin \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{3} b^{5} \cos \left (b x +a \right )-3600 \sin \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right )^{2} b^{5} \cos \left (2 b x +2 a \right )+11520 \sin \left (b x +a \right ) \sin \left (2 b x +2 a \right ) b^{5} \cos \left (b x +a \right ) \cos \left (2 b x +2 a \right )^{2}+9456 b^{5} \cos \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right )^{2} \cos \left (2 b x +2 a \right )+960 b^{5} \sin \left (b x +a \right )^{2} \cos \left (2 b x +2 a \right )^{3}-2880 b^{5} \cos \left (b x +a \right )^{2} \cos \left (2 b x +2 a \right )^{3}\right )}{6144 b^{6}}-\frac {266240 b^{7} \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{3} \sin \left (b x +a \right )-504384 b^{7} \cos \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right )^{2} \cos \left (2 b x +2 a \right )-774144 \sin \left (b x +a \right ) \sin \left (2 b x +2 a \right ) b^{7} \cos \left (b x +a \right ) \cos \left (2 b x +2 a \right )^{2}+294336 \sin \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right )^{2} b^{7} \cos \left (2 b x +2 a \right )-94080 \sin \left (b x +a \right )^{2} b^{7} \cos \left (2 b x +2 a \right )^{3}+163968 b^{7} \cos \left (b x +a \right )^{2} \cos \left (2 b x +2 a \right )^{3}}{147456 b^{8}}\) | \(596\) |
Input:
int(cos(b*x+a)^2*sin(2*b*x+2*a)^3,x,method=_RETURNVERBOSE)
Output:
1/384*(-12*cos(4*b*x+4*a)-199+3*cos(8*b*x+8*a)-72*cos(2*b*x+2*a)+8*cos(6*b *x+6*a))/b
Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \cos ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {3 \, \cos \left (b x + a\right )^{8} - 4 \, \cos \left (b x + a\right )^{6}}{3 \, b} \] Input:
integrate(cos(b*x+a)^2*sin(2*b*x+2*a)^3,x, algorithm="fricas")
Output:
1/3*(3*cos(b*x + a)^8 - 4*cos(b*x + a)^6)/b
Leaf count of result is larger than twice the leaf count of optimal. 362 vs. \(2 (22) = 44\).
Time = 2.20 (sec) , antiderivative size = 362, normalized size of antiderivative = 12.93 \[ \int \cos ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=\begin {cases} - \frac {3 x \sin ^{2}{\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )}}{16} - \frac {3 x \sin ^{2}{\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{16} - \frac {3 x \sin {\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{8} - \frac {3 x \sin {\left (a + b x \right )} \cos {\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{8} + \frac {3 x \sin ^{3}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )}}{16} + \frac {3 x \sin {\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{16} - \frac {\sin ^{2}{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (2 a + 2 b x \right )}}{2 b} - \frac {49 \sin ^{2}{\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{96 b} + \frac {13 \sin {\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )}}{16 b} + \frac {7 \sin {\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{8 b} + \frac {17 \cos ^{2}{\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{96 b} & \text {for}\: b \neq 0 \\x \sin ^{3}{\left (2 a \right )} \cos ^{2}{\left (a \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(b*x+a)**2*sin(2*b*x+2*a)**3,x)
Output:
Piecewise((-3*x*sin(a + b*x)**2*sin(2*a + 2*b*x)**3/16 - 3*x*sin(a + b*x)* *2*sin(2*a + 2*b*x)*cos(2*a + 2*b*x)**2/16 - 3*x*sin(a + b*x)*sin(2*a + 2* b*x)**2*cos(a + b*x)*cos(2*a + 2*b*x)/8 - 3*x*sin(a + b*x)*cos(a + b*x)*co s(2*a + 2*b*x)**3/8 + 3*x*sin(2*a + 2*b*x)**3*cos(a + b*x)**2/16 + 3*x*sin (2*a + 2*b*x)*cos(a + b*x)**2*cos(2*a + 2*b*x)**2/16 - sin(a + b*x)**2*sin (2*a + 2*b*x)**2*cos(2*a + 2*b*x)/(2*b) - 49*sin(a + b*x)**2*cos(2*a + 2*b *x)**3/(96*b) + 13*sin(a + b*x)*sin(2*a + 2*b*x)**3*cos(a + b*x)/(16*b) + 7*sin(a + b*x)*sin(2*a + 2*b*x)*cos(a + b*x)*cos(2*a + 2*b*x)**2/(8*b) + 1 7*cos(a + b*x)**2*cos(2*a + 2*b*x)**3/(96*b), Ne(b, 0)), (x*sin(2*a)**3*co s(a)**2, True))
Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.79 \[ \int \cos ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {3 \, \cos \left (8 \, b x + 8 \, a\right ) + 8 \, \cos \left (6 \, b x + 6 \, a\right ) - 12 \, \cos \left (4 \, b x + 4 \, a\right ) - 72 \, \cos \left (2 \, b x + 2 \, a\right )}{384 \, b} \] Input:
integrate(cos(b*x+a)^2*sin(2*b*x+2*a)^3,x, algorithm="maxima")
Output:
1/384*(3*cos(8*b*x + 8*a) + 8*cos(6*b*x + 6*a) - 12*cos(4*b*x + 4*a) - 72* cos(2*b*x + 2*a))/b
Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \cos ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {3 \, \cos \left (b x + a\right )^{8} - 4 \, \cos \left (b x + a\right )^{6}}{3 \, b} \] Input:
integrate(cos(b*x+a)^2*sin(2*b*x+2*a)^3,x, algorithm="giac")
Output:
1/3*(3*cos(b*x + a)^8 - 4*cos(b*x + a)^6)/b
Time = 19.35 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \cos ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=-\frac {\frac {4\,{\cos \left (a+b\,x\right )}^6}{3}-{\cos \left (a+b\,x\right )}^8}{b} \] Input:
int(cos(a + b*x)^2*sin(2*a + 2*b*x)^3,x)
Output:
-((4*cos(a + b*x)^6)/3 - cos(a + b*x)^8)/b
Time = 0.16 (sec) , antiderivative size = 173, normalized size of antiderivative = 6.18 \[ \int \cos ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {-36 \cos \left (2 b x +2 a \right ) \cos \left (b x +a \right ) \sin \left (b x +a \right ) b x +18 \cos \left (2 b x +2 a \right ) \sin \left (2 b x +2 a \right )^{2} \sin \left (b x +a \right )^{2}-17 \cos \left (2 b x +2 a \right ) \sin \left (2 b x +2 a \right )^{2}+18 \cos \left (2 b x +2 a \right ) \sin \left (b x +a \right )^{2}-25 \cos \left (2 b x +2 a \right )-6 \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{3} \sin \left (b x +a \right )-36 \sin \left (2 b x +2 a \right ) \sin \left (b x +a \right )^{2} b x +18 \sin \left (2 b x +2 a \right ) b x +31}{96 b} \] Input:
int(cos(b*x+a)^2*sin(2*b*x+2*a)^3,x)
Output:
( - 36*cos(2*a + 2*b*x)*cos(a + b*x)*sin(a + b*x)*b*x + 18*cos(2*a + 2*b*x )*sin(2*a + 2*b*x)**2*sin(a + b*x)**2 - 17*cos(2*a + 2*b*x)*sin(2*a + 2*b* x)**2 + 18*cos(2*a + 2*b*x)*sin(a + b*x)**2 - 25*cos(2*a + 2*b*x) - 6*cos( a + b*x)*sin(2*a + 2*b*x)**3*sin(a + b*x) - 36*sin(2*a + 2*b*x)*sin(a + b* x)**2*b*x + 18*sin(2*a + 2*b*x)*b*x + 31)/(96*b)