Integrand size = 20, antiderivative size = 29 \[ \int \sin ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {4 \sin ^6(a+b x)}{3 b}-\frac {\sin ^8(a+b x)}{b} \] Output:
4/3*sin(b*x+a)^6/b-sin(b*x+a)^8/b
Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.66 \[ \int \sin ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {-72 \cos (2 (a+b x))+12 \cos (4 (a+b x))+8 \cos (6 (a+b x))-3 \cos (8 (a+b x))}{384 b} \] Input:
Integrate[Sin[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 = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 4776, 3042, 3044, 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.
\(\displaystyle \int \sin ^2(a+b x) \sin ^3(2 a+2 b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (a+b x)^2 \sin (2 a+2 b x)^3dx\) |
\(\Big \downarrow \) 4776 |
\(\displaystyle 8 \int \cos ^3(a+b x) \sin ^5(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 8 \int \cos (a+b x)^3 \sin (a+b x)^5dx\) |
\(\Big \downarrow \) 3044 |
\(\displaystyle \frac {8 \int \sin ^5(a+b x) \left (1-\sin ^2(a+b x)\right )d\sin (a+b x)}{b}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {8 \int \left (\sin ^5(a+b x)-\sin ^7(a+b x)\right )d\sin (a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {8 \left (\frac {1}{6} \sin ^6(a+b x)-\frac {1}{8} \sin ^8(a+b x)\right )}{b}\) |
Input:
Int[Sin[a + b*x]^2*Sin[2*a + 2*b*x]^3,x]
Output:
(8*(Sin[a + b*x]^6/6 - Sin[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_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ Symbol] :> Simp[1/(a*f) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a *Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(I ntegerQ[(m - 1)/2] && LtQ[0, m, n])
Int[((f_.)*sin[(a_.) + (b_.)*(x_)])^(n_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_ Symbol] :> Simp[2^p/f^p Int[Cos[a + b*x]^p*(f*Sin[a + b*x])^(n + p), x], x] /; FreeQ[{a, b, c, d, f, n}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && I ntegerQ[p]
Time = 3.57 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.79
method | result | size |
parallelrisch | \(\frac {12 \cos \left (4 b x +4 a \right )+71-3 \cos \left (8 b x +8 a \right )-72 \cos \left (2 b x +2 a \right )+8 \cos \left (6 b x +6 a \right )}{384 b}\) | \(52\) |
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 \sin \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{3} b \cos \left (b x +a \right )+6 \sin \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 \sin \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{3} \cos \left (b x +a \right ) b^{3}+36 \sin \left (2 b x +2 a \right )^{2} \cos \left (b x +a \right )^{2} \cos \left (2 b x +2 a \right ) b^{3}+144 \sin \left (b x +a \right ) \sin \left (2 b x +2 a \right ) \cos \left (b x +a \right ) \cos \left (2 b x +2 a \right )^{2} b^{3}-204 \sin \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 \sin \left (b x +a \right )^{2} \cos \left (2 b x +2 a \right )^{3} b^{3}\right )}{3072 b^{4}}-\frac {5 \left (-3600 b^{5} \sin \left (2 b x +2 a \right )^{2} \cos \left (b x +a \right )^{2} \cos \left (2 b x +2 a \right )+4352 b^{5} \sin \left (2 b x +2 a \right )^{3} \cos \left (b x +a \right ) \sin \left (b x +a \right )-11520 \cos \left (2 b x +2 a \right )^{2} b^{5} \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right ) \sin \left (b x +a \right )+9456 \cos \left (2 b x +2 a \right ) \sin \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right )^{2} b^{5}+960 \cos \left (2 b x +2 a \right )^{3} b^{5} \cos \left (b x +a \right )^{2}-2880 \cos \left (2 b x +2 a \right )^{3} \sin \left (b x +a \right )^{2} b^{5}\right )}{6144 b^{6}}-\frac {-94080 b^{7} \cos \left (2 b x +2 a \right )^{3} \cos \left (b x +a \right )^{2}+774144 b^{7} \sin \left (2 b x +2 a \right ) \cos \left (b x +a \right ) \cos \left (2 b x +2 a \right )^{2} \sin \left (b x +a \right )+294336 b^{7} \sin \left (2 b x +2 a \right )^{2} \cos \left (b x +a \right )^{2} \cos \left (2 b x +2 a \right )-504384 b^{7} \sin \left (2 b x +2 a \right )^{2} \sin \left (b x +a \right )^{2} \cos \left (2 b x +2 a \right )-266240 b^{7} \sin \left (2 b x +2 a \right )^{3} \cos \left (b x +a \right ) \sin \left (b x +a \right )+163968 \cos \left (2 b x +2 a \right )^{3} b^{7} \sin \left (b x +a \right )^{2}}{147456 b^{8}}\) | \(596\) |
Input:
int(sin(b*x+a)^2*sin(2*b*x+2*a)^3,x,method=_RETURNVERBOSE)
Output:
1/384*(12*cos(4*b*x+4*a)+71-3*cos(8*b*x+8*a)-72*cos(2*b*x+2*a)+8*cos(6*b*x +6*a))/b
Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \sin ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=-\frac {3 \, \cos \left (b x + a\right )^{8} - 8 \, \cos \left (b x + a\right )^{6} + 6 \, \cos \left (b x + a\right )^{4}}{3 \, b} \] Input:
integrate(sin(b*x+a)^2*sin(2*b*x+2*a)^3,x, algorithm="fricas")
Output:
-1/3*(3*cos(b*x + a)^8 - 8*cos(b*x + a)^6 + 6*cos(b*x + a)^4)/b
Leaf count of result is larger than twice the leaf count of optimal. 359 vs. \(2 (22) = 44\).
Time = 2.17 (sec) , antiderivative size = 359, normalized size of antiderivative = 12.38 \[ \int \sin ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=\begin {cases} \frac {3 x \sin ^{2}{\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )}}{16} + \frac {3 x \sin ^{2}{\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{16} + \frac {3 x \sin {\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{8} + \frac {3 x \sin {\left (a + b x \right )} \cos {\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{8} - \frac {3 x \sin ^{3}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )}}{16} - \frac {3 x \sin {\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{16} - \frac {\sin ^{2}{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (2 a + 2 b x \right )}}{2 b} - \frac {31 \sin ^{2}{\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{96 b} + \frac {3 \sin {\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )}}{16 b} + \frac {\sin {\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{8 b} - \frac {\cos ^{2}{\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{96 b} & \text {for}\: b \neq 0 \\x \sin ^{2}{\left (a \right )} \sin ^{3}{\left (2 a \right )} & \text {otherwise} \end {cases} \] Input:
integrate(sin(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)*cos (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) - 31*sin(a + b*x)**2*cos(2*a + 2*b* x)**3/(96*b) + 3*sin(a + b*x)*sin(2*a + 2*b*x)**3*cos(a + b*x)/(16*b) + si n(a + b*x)*sin(2*a + 2*b*x)*cos(a + b*x)*cos(2*a + 2*b*x)**2/(8*b) - cos(a + b*x)**2*cos(2*a + 2*b*x)**3/(96*b), Ne(b, 0)), (x*sin(a)**2*sin(2*a)**3 , True))
Time = 0.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.72 \[ \int \sin ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=-\frac {3 \, \cos \left (8 \, b x + 8 \, a\right ) - 8 \, \cos \left (6 \, b x + 6 \, a\right ) - 12 \, \cos \left (4 \, b x + 4 \, a\right ) + 72 \, \cos \left (2 \, b x + 2 \, a\right )}{384 \, b} \] Input:
integrate(sin(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 = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \sin ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=-\frac {3 \, \sin \left (b x + a\right )^{8} - 4 \, \sin \left (b x + a\right )^{6}}{3 \, b} \] Input:
integrate(sin(b*x+a)^2*sin(2*b*x+2*a)^3,x, algorithm="giac")
Output:
-1/3*(3*sin(b*x + a)^8 - 4*sin(b*x + a)^6)/b
Time = 0.05 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \sin ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=-\frac {{\cos \left (a+b\,x\right )}^4\,\left ({\cos \left (a+b\,x\right )}^4-\frac {8\,{\cos \left (a+b\,x\right )}^2}{3}+2\right )}{b} \] Input:
int(sin(a + b*x)^2*sin(2*a + 2*b*x)^3,x)
Output:
-(cos(a + b*x)^4*(cos(a + b*x)^4 - (8*cos(a + b*x)^2)/3 + 2))/b
Time = 0.18 (sec) , antiderivative size = 172, normalized size of antiderivative = 5.93 \[ \int \sin ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {36 \cos \left (2 b x +2 a \right ) \cos \left (b x +a \right ) \sin \left (b x +a \right ) b x -18 \cos \left (2 b x +2 a \right ) \sin \left (2 b x +2 a \right )^{2} \sin \left (b x +a \right )^{2}+\cos \left (2 b x +2 a \right ) \sin \left (2 b x +2 a \right )^{2}-18 \cos \left (2 b x +2 a \right ) \sin \left (b x +a \right )^{2}-7 \cos \left (2 b x +2 a \right )+6 \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{3} \sin \left (b x +a \right )+36 \sin \left (2 b x +2 a \right ) \sin \left (b x +a \right )^{2} b x -18 \sin \left (2 b x +2 a \right ) b x +1}{96 b} \] Input:
int(sin(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)*s in(2*a + 2*b*x)**2*sin(a + b*x)**2 + 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 - 7*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 + 1)/(96*b)