Integrand size = 20, antiderivative size = 44 \[ \int \sin ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=\frac {4 \sin ^8(a+b x)}{b}-\frac {32 \sin ^{10}(a+b x)}{5 b}+\frac {8 \sin ^{12}(a+b x)}{3 b} \] Output:
4*sin(b*x+a)^8/b-32/5*sin(b*x+a)^10/b+8/3*sin(b*x+a)^12/b
Time = 0.23 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.55 \[ \int \sin ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=\frac {-600 \cos (2 (a+b x))+75 \cos (4 (a+b x))+100 \cos (6 (a+b x))-30 \cos (8 (a+b x))-12 \cos (10 (a+b x))+5 \cos (12 (a+b x))}{3840 b} \] Input:
Integrate[Sin[a + b*x]^2*Sin[2*a + 2*b*x]^5,x]
Output:
(-600*Cos[2*(a + b*x)] + 75*Cos[4*(a + b*x)] + 100*Cos[6*(a + b*x)] - 30*C os[8*(a + b*x)] - 12*Cos[10*(a + b*x)] + 5*Cos[12*(a + b*x)])/(3840*b)
Time = 0.30 (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, 4776, 3042, 3044, 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.
\(\displaystyle \int \sin ^2(a+b x) \sin ^5(2 a+2 b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (a+b x)^2 \sin (2 a+2 b x)^5dx\) |
\(\Big \downarrow \) 4776 |
\(\displaystyle 32 \int \cos ^5(a+b x) \sin ^7(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 32 \int \cos (a+b x)^5 \sin (a+b x)^7dx\) |
\(\Big \downarrow \) 3044 |
\(\displaystyle \frac {32 \int \sin ^7(a+b x) \left (1-\sin ^2(a+b x)\right )^2d\sin (a+b x)}{b}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {16 \int \sin ^6(a+b x) \left (1-\sin ^2(a+b x)\right )^2d\sin ^2(a+b x)}{b}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {16 \int \left (\sin ^{10}(a+b x)-2 \sin ^8(a+b x)+\sin ^6(a+b x)\right )d\sin ^2(a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {16 \left (\frac {1}{6} \sin ^{12}(a+b x)-\frac {2}{5} \sin ^{10}(a+b x)+\frac {1}{4} \sin ^8(a+b x)\right )}{b}\) |
Input:
Int[Sin[a + b*x]^2*Sin[2*a + 2*b*x]^5,x]
Output:
(16*(Sin[a + b*x]^8/4 - (2*Sin[a + b*x]^10)/5 + Sin[a + b*x]^12/6))/b
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_)]^(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 = 13.22 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.68
method | result | size |
parallelrisch | \(\frac {75 \cos \left (4 b x +4 a \right )+5 \cos \left (12 b x +12 a \right )+462-30 \cos \left (8 b x +8 a \right )-600 \cos \left (2 b x +2 a \right )+100 \cos \left (6 b x +6 a \right )-12 \cos \left (10 b x +10 a \right )}{3840 b}\) | \(74\) |
default | \(-\frac {5 \cos \left (2 b x +2 a \right )}{32 b}+\frac {5 \cos \left (4 b x +4 a \right )}{256 b}+\frac {5 \cos \left (6 b x +6 a \right )}{192 b}-\frac {\cos \left (8 b x +8 a \right )}{128 b}-\frac {\cos \left (10 b x +10 a \right )}{320 b}+\frac {\cos \left (12 b x +12 a \right )}{768 b}\) | \(86\) |
risch | \(-\frac {5 \cos \left (2 b x +2 a \right )}{32 b}+\frac {5 \cos \left (4 b x +4 a \right )}{256 b}+\frac {5 \cos \left (6 b x +6 a \right )}{192 b}-\frac {\cos \left (8 b x +8 a \right )}{128 b}-\frac {\cos \left (10 b x +10 a \right )}{320 b}+\frac {\cos \left (12 b x +12 a \right )}{768 b}\) | \(86\) |
orering | \(\text {Expression too large to display}\) | \(1391\) |
Input:
int(sin(b*x+a)^2*sin(2*b*x+2*a)^5,x,method=_RETURNVERBOSE)
Output:
1/3840*(75*cos(4*b*x+4*a)+5*cos(12*b*x+12*a)+462-30*cos(8*b*x+8*a)-600*cos (2*b*x+2*a)+100*cos(6*b*x+6*a)-12*cos(10*b*x+10*a))/b
Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.05 \[ \int \sin ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=\frac {4 \, {\left (10 \, \cos \left (b x + a\right )^{12} - 36 \, \cos \left (b x + a\right )^{10} + 45 \, \cos \left (b x + a\right )^{8} - 20 \, \cos \left (b x + a\right )^{6}\right )}}{15 \, b} \] Input:
integrate(sin(b*x+a)^2*sin(2*b*x+2*a)^5,x, algorithm="fricas")
Output:
4/15*(10*cos(b*x + a)^12 - 36*cos(b*x + a)^10 + 45*cos(b*x + a)^8 - 20*cos (b*x + a)^6)/b
Leaf count of result is larger than twice the leaf count of optimal. 593 vs. \(2 (37) = 74\).
Time = 11.58 (sec) , antiderivative size = 593, normalized size of antiderivative = 13.48 \[ \int \sin ^2(a+b x) \sin ^5(2 a+2 b x) \, dx =\text {Too large to display} \] Input:
integrate(sin(b*x+a)**2*sin(2*b*x+2*a)**5,x)
Output:
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*co s(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)*cos (a + b*x)**2*cos(2*a + 2*b*x)**4/32 - 65*sin(a + b*x)**2*sin(2*a + 2*b*x)* *4*cos(2*a + 2*b*x)/(128*b) - 2*sin(a + b*x)**2*sin(2*a + 2*b*x)**2*cos(2* a + 2*b*x)**3/(3*b) - 167*sin(a + b*x)**2*cos(2*a + 2*b*x)**5/(640*b) + 11 *sin(a + b*x)*sin(2*a + 2*b*x)**5*cos(a + b*x)/(64*b) + sin(a + b*x)*sin(2 *a + 2*b*x)**3*cos(a + b*x)*cos(2*a + 2*b*x)**2/(4*b) + 19*sin(a + b*x)*si n(2*a + 2*b*x)*cos(a + b*x)*cos(2*a + 2*b*x)**4/(192*b) + sin(2*a + 2*b*x) **4*cos(a + b*x)**2*cos(2*a + 2*b*x)/(128*b) - 11*cos(a + b*x)**2*cos(2*a + 2*b*x)**5/(1920*b), Ne(b, 0)), (x*sin(a)**2*sin(2*a)**5, True))
Time = 0.05 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.64 \[ \int \sin ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=\frac {5 \, \cos \left (12 \, b x + 12 \, a\right ) - 12 \, \cos \left (10 \, b x + 10 \, a\right ) - 30 \, \cos \left (8 \, b x + 8 \, a\right ) + 100 \, \cos \left (6 \, b x + 6 \, a\right ) + 75 \, \cos \left (4 \, b x + 4 \, a\right ) - 600 \, \cos \left (2 \, b x + 2 \, a\right )}{3840 \, b} \] Input:
integrate(sin(b*x+a)^2*sin(2*b*x+2*a)^5,x, algorithm="maxima")
Output:
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 = 0.13 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.82 \[ \int \sin ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=\frac {4 \, {\left (10 \, \sin \left (b x + a\right )^{12} - 24 \, \sin \left (b x + a\right )^{10} + 15 \, \sin \left (b x + a\right )^{8}\right )}}{15 \, b} \] Input:
integrate(sin(b*x+a)^2*sin(2*b*x+2*a)^5,x, algorithm="giac")
Output:
4/15*(10*sin(b*x + a)^12 - 24*sin(b*x + a)^10 + 15*sin(b*x + a)^8)/b
Time = 18.06 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.05 \[ \int \sin ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=-\frac {-\frac {8\,{\cos \left (a+b\,x\right )}^{12}}{3}+\frac {48\,{\cos \left (a+b\,x\right )}^{10}}{5}-12\,{\cos \left (a+b\,x\right )}^8+\frac {16\,{\cos \left (a+b\,x\right )}^6}{3}}{b} \] Input:
int(sin(a + b*x)^2*sin(2*a + 2*b*x)^5,x)
Output:
-((16*cos(a + b*x)^6)/3 - 12*cos(a + b*x)^8 + (48*cos(a + b*x)^10)/5 - (8* cos(a + b*x)^12)/3)/b
Time = 0.18 (sec) , antiderivative size = 250, normalized size of antiderivative = 5.68 \[ \int \sin ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=\frac {300 \cos \left (2 b x +2 a \right ) \cos \left (b x +a \right ) \sin \left (b x +a \right ) b x -100 \cos \left (2 b x +2 a \right ) \sin \left (2 b x +2 a \right )^{4} \sin \left (b x +a \right )^{2}+2 \cos \left (2 b x +2 a \right ) \sin \left (2 b x +2 a \right )^{4}-150 \cos \left (2 b x +2 a \right ) \sin \left (2 b x +2 a \right )^{2} \sin \left (b x +a \right )^{2}+11 \cos \left (2 b x +2 a \right ) \sin \left (2 b x +2 a \right )^{2}-150 \cos \left (2 b x +2 a \right ) \sin \left (b x +a \right )^{2}-53 \cos \left (2 b x +2 a \right )+20 \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{5} \sin \left (b x +a \right )+50 \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{3} \sin \left (b x +a \right )+300 \sin \left (2 b x +2 a \right ) \sin \left (b x +a \right )^{2} b x -150 \sin \left (2 b x +2 a \right ) b x +3}{960 b} \] Input:
int(sin(b*x+a)^2*sin(2*b*x+2*a)^5,x)
Output:
(300*cos(2*a + 2*b*x)*cos(a + b*x)*sin(a + b*x)*b*x - 100*cos(2*a + 2*b*x) *sin(2*a + 2*b*x)**4*sin(a + b*x)**2 + 2*cos(2*a + 2*b*x)*sin(2*a + 2*b*x) **4 - 150*cos(2*a + 2*b*x)*sin(2*a + 2*b*x)**2*sin(a + b*x)**2 + 11*cos(2* a + 2*b*x)*sin(2*a + 2*b*x)**2 - 150*cos(2*a + 2*b*x)*sin(a + b*x)**2 - 53 *cos(2*a + 2*b*x) + 20*cos(a + b*x)*sin(2*a + 2*b*x)**5*sin(a + b*x) + 50* cos(a + b*x)*sin(2*a + 2*b*x)**3*sin(a + b*x) + 300*sin(2*a + 2*b*x)*sin(a + b*x)**2*b*x - 150*sin(2*a + 2*b*x)*b*x + 3)/(960*b)