Integrand size = 18, antiderivative size = 46 \[ \int \sin (a+b x) \sin ^4(2 a+2 b x) \, dx=-\frac {16 \cos ^5(a+b x)}{5 b}+\frac {32 \cos ^7(a+b x)}{7 b}-\frac {16 \cos ^9(a+b x)}{9 b} \] Output:
-16/5*cos(b*x+a)^5/b+32/7*cos(b*x+a)^7/b-16/9*cos(b*x+a)^9/b
Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80 \[ \int \sin (a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {2 \cos ^5(a+b x) (-249+220 \cos (2 (a+b x))-35 \cos (4 (a+b x)))}{315 b} \] Input:
Integrate[Sin[a + b*x]*Sin[2*a + 2*b*x]^4,x]
Output:
(2*Cos[a + b*x]^5*(-249 + 220*Cos[2*(a + b*x)] - 35*Cos[4*(a + b*x)]))/(31 5*b)
Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.91, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4776, 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.
\(\displaystyle \int \sin (a+b x) \sin ^4(2 a+2 b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (a+b x) \sin (2 a+2 b x)^4dx\) |
\(\Big \downarrow \) 4776 |
\(\displaystyle 16 \int \cos ^4(a+b x) \sin ^5(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 16 \int \cos (a+b x)^4 \sin (a+b x)^5dx\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle -\frac {16 \int \cos ^4(a+b x) \left (1-\cos ^2(a+b x)\right )^2d\cos (a+b x)}{b}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -\frac {16 \int \left (\cos ^8(a+b x)-2 \cos ^6(a+b x)+\cos ^4(a+b x)\right )d\cos (a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {16 \left (\frac {1}{9} \cos ^9(a+b x)-\frac {2}{7} \cos ^7(a+b x)+\frac {1}{5} \cos ^5(a+b x)\right )}{b}\) |
Input:
Int[Sin[a + b*x]*Sin[2*a + 2*b*x]^4,x]
Output:
(-16*(Cos[a + b*x]^5/5 - (2*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[((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 = 4.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.30
method | result | size |
parallelrisch | \(\frac {-2048+45 \cos \left (7 b x +7 a \right )-35 \cos \left (9 b x +9 a \right )-420 \cos \left (3 b x +3 a \right )-1890 \cos \left (b x +a \right )+252 \cos \left (5 b x +5 a \right )}{5040 b}\) | \(60\) |
default | \(-\frac {3 \cos \left (b x +a \right )}{8 b}-\frac {\cos \left (3 b x +3 a \right )}{12 b}+\frac {\cos \left (5 b x +5 a \right )}{20 b}+\frac {\cos \left (7 b x +7 a \right )}{112 b}-\frac {\cos \left (9 b x +9 a \right )}{144 b}\) | \(69\) |
risch | \(-\frac {3 \cos \left (b x +a \right )}{8 b}-\frac {\cos \left (3 b x +3 a \right )}{12 b}+\frac {\cos \left (5 b x +5 a \right )}{20 b}+\frac {\cos \left (7 b x +7 a \right )}{112 b}-\frac {\cos \left (9 b x +9 a \right )}{144 b}\) | \(69\) |
orering | \(-\frac {117469 \left (b \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{4}+8 \sin \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{3} b \cos \left (2 b x +2 a \right )\right )}{99225 b^{2}}-\frac {34562 \left (-49 b^{3} \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{4}-344 \sin \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{3} b^{3} \cos \left (2 b x +2 a \right )+144 b^{3} \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{2} \cos \left (2 b x +2 a \right )^{2}+192 \sin \left (b x +a \right ) \sin \left (2 b x +2 a \right ) b^{3} \cos \left (2 b x +2 a \right )^{3}\right )}{178605 b^{4}}-\frac {418 \left (3361 b^{5} \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{4}+20648 b^{5} \sin \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{3} \cos \left (2 b x +2 a \right )-15840 b^{5} \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{2} \cos \left (2 b x +2 a \right )^{2}-17280 b^{5} \sin \left (b x +a \right ) \sin \left (2 b x +2 a \right ) \cos \left (2 b x +2 a \right )^{3}+1920 b^{5} \cos \left (b x +a \right ) \cos \left (2 b x +2 a \right )^{4}\right )}{42525 b^{6}}-\frac {11 \left (-266449 b^{7} \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{4}-1441784 b^{7} \sin \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{3} \cos \left (2 b x +2 a \right )+1484784 b^{7} \cos \left (b x +a \right ) \cos \left (2 b x +2 a \right )^{2} \sin \left (2 b x +2 a \right )^{2}+1361472 b^{7} \sin \left (b x +a \right ) \sin \left (2 b x +2 a \right ) \cos \left (2 b x +2 a \right )^{3}-228480 b^{7} \cos \left (b x +a \right ) \cos \left (2 b x +2 a \right )^{4}\right )}{59535 b^{8}}-\frac {22175041 b^{9} \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{4}+107929928 b^{9} \sin \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{3} \cos \left (2 b x +2 a \right )-130150080 b^{9} \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{2} \cos \left (2 b x +2 a \right )^{2}-105957120 b^{9} \sin \left (b x +a \right ) \sin \left (2 b x +2 a \right ) \cos \left (2 b x +2 a \right )^{3}+21208320 b^{9} \cos \left (b x +a \right ) \cos \left (2 b x +2 a \right )^{4}}{893025 b^{10}}\) | \(614\) |
Input:
int(sin(b*x+a)*sin(2*b*x+2*a)^4,x,method=_RETURNVERBOSE)
Output:
1/5040*(-2048+45*cos(7*b*x+7*a)-35*cos(9*b*x+9*a)-420*cos(3*b*x+3*a)-1890* cos(b*x+a)+252*cos(5*b*x+5*a))/b
Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \sin (a+b x) \sin ^4(2 a+2 b x) \, dx=-\frac {16 \, {\left (35 \, \cos \left (b x + a\right )^{9} - 90 \, \cos \left (b x + a\right )^{7} + 63 \, \cos \left (b x + a\right )^{5}\right )}}{315 \, b} \] Input:
integrate(sin(b*x+a)*sin(2*b*x+2*a)^4,x, algorithm="fricas")
Output:
-16/315*(35*cos(b*x + a)^9 - 90*cos(b*x + a)^7 + 63*cos(b*x + a)^5)/b
Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (39) = 78\).
Time = 2.19 (sec) , antiderivative size = 163, normalized size of antiderivative = 3.54 \[ \int \sin (a+b x) \sin ^4(2 a+2 b x) \, dx=\begin {cases} - \frac {104 \sin {\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )} \cos {\left (2 a + 2 b x \right )}}{315 b} - \frac {64 \sin {\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{315 b} - \frac {107 \sin ^{4}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )}}{315 b} - \frac {16 \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{21 b} - \frac {128 \cos {\left (a + b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{315 b} & \text {for}\: b \neq 0 \\x \sin {\left (a \right )} \sin ^{4}{\left (2 a \right )} & \text {otherwise} \end {cases} \] Input:
integrate(sin(b*x+a)*sin(2*b*x+2*a)**4,x)
Output:
Piecewise((-104*sin(a + b*x)*sin(2*a + 2*b*x)**3*cos(2*a + 2*b*x)/(315*b) - 64*sin(a + b*x)*sin(2*a + 2*b*x)*cos(2*a + 2*b*x)**3/(315*b) - 107*sin(2 *a + 2*b*x)**4*cos(a + b*x)/(315*b) - 16*sin(2*a + 2*b*x)**2*cos(a + b*x)* cos(2*a + 2*b*x)**2/(21*b) - 128*cos(a + b*x)*cos(2*a + 2*b*x)**4/(315*b), Ne(b, 0)), (x*sin(a)*sin(2*a)**4, True))
Time = 0.04 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.26 \[ \int \sin (a+b x) \sin ^4(2 a+2 b x) \, dx=-\frac {35 \, \cos \left (9 \, b x + 9 \, a\right ) - 45 \, \cos \left (7 \, b x + 7 \, a\right ) - 252 \, \cos \left (5 \, b x + 5 \, a\right ) + 420 \, \cos \left (3 \, b x + 3 \, a\right ) + 1890 \, \cos \left (b x + a\right )}{5040 \, b} \] Input:
integrate(sin(b*x+a)*sin(2*b*x+2*a)^4,x, algorithm="maxima")
Output:
-1/5040*(35*cos(9*b*x + 9*a) - 45*cos(7*b*x + 7*a) - 252*cos(5*b*x + 5*a) + 420*cos(3*b*x + 3*a) + 1890*cos(b*x + a))/b
Time = 0.13 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.26 \[ \int \sin (a+b x) \sin ^4(2 a+2 b x) \, dx=-\frac {35 \, \cos \left (9 \, b x + 9 \, a\right ) - 45 \, \cos \left (7 \, b x + 7 \, a\right ) - 252 \, \cos \left (5 \, b x + 5 \, a\right ) + 420 \, \cos \left (3 \, b x + 3 \, a\right ) + 1890 \, \cos \left (b x + a\right )}{5040 \, b} \] Input:
integrate(sin(b*x+a)*sin(2*b*x+2*a)^4,x, algorithm="giac")
Output:
-1/5040*(35*cos(9*b*x + 9*a) - 45*cos(7*b*x + 7*a) - 252*cos(5*b*x + 5*a) + 420*cos(3*b*x + 3*a) + 1890*cos(b*x + a))/b
Time = 18.52 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \sin (a+b x) \sin ^4(2 a+2 b x) \, dx=-\frac {16\,\left (35\,{\cos \left (a+b\,x\right )}^9-90\,{\cos \left (a+b\,x\right )}^7+63\,{\cos \left (a+b\,x\right )}^5\right )}{315\,b} \] Input:
int(sin(a + b*x)*sin(2*a + 2*b*x)^4,x)
Output:
-(16*(63*cos(a + b*x)^5 - 90*cos(a + b*x)^7 + 35*cos(a + b*x)^9))/(315*b)
Time = 0.19 (sec) , antiderivative size = 107, normalized size of antiderivative = 2.33 \[ \int \sin (a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {-120 \cos \left (2 b x +2 a \right ) \sin \left (2 b x +2 a \right )^{3} \sin \left (b x +a \right )-192 \cos \left (2 b x +2 a \right ) \sin \left (2 b x +2 a \right ) \sin \left (b x +a \right )+15 \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{4}+48 \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{2}-384 \cos \left (b x +a \right )+280}{945 b} \] Input:
int(sin(b*x+a)*sin(2*b*x+2*a)^4,x)
Output:
( - 120*cos(2*a + 2*b*x)*sin(2*a + 2*b*x)**3*sin(a + b*x) - 192*cos(2*a + 2*b*x)*sin(2*a + 2*b*x)*sin(a + b*x) + 15*cos(a + b*x)*sin(2*a + 2*b*x)**4 + 48*cos(a + b*x)*sin(2*a + 2*b*x)**2 - 384*cos(a + b*x) + 280)/(945*b)