Integrand size = 20, antiderivative size = 76 \[ \int \sin ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {3 x}{16}-\frac {3 \cos (2 a+2 b x) \sin (2 a+2 b x)}{32 b}-\frac {\cos (2 a+2 b x) \sin ^3(2 a+2 b x)}{16 b}-\frac {\sin ^5(2 a+2 b x)}{20 b} \] Output:
3/16*x-3/32*cos(2*b*x+2*a)*sin(2*b*x+2*a)/b-1/16*cos(2*b*x+2*a)*sin(2*b*x+ 2*a)^3/b-1/20*sin(2*b*x+2*a)^5/b
Time = 0.14 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.82 \[ \int \sin ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {120 b x-20 \sin (2 (a+b x))-40 \sin (4 (a+b x))+10 \sin (6 (a+b x))+5 \sin (8 (a+b x))-2 \sin (10 (a+b x))}{640 b} \] Input:
Integrate[Sin[a + b*x]^2*Sin[2*a + 2*b*x]^4,x]
Output:
(120*b*x - 20*Sin[2*(a + b*x)] - 40*Sin[4*(a + b*x)] + 10*Sin[6*(a + b*x)] + 5*Sin[8*(a + b*x)] - 2*Sin[10*(a + b*x)])/(640*b)
Time = 0.40 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.13, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {3042, 4774, 3042, 3044, 15, 3115, 3042, 3115, 24}
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.
\(\displaystyle \int \sin ^2(a+b x) \sin ^4(2 a+2 b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (a+b x)^2 \sin (2 a+2 b x)^4dx\) |
\(\Big \downarrow \) 4774 |
\(\displaystyle \frac {1}{2} \int \sin ^4(2 a+2 b x)dx-\frac {1}{2} \int \cos (2 a+2 b x) \sin ^4(2 a+2 b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int \sin (2 a+2 b x)^4dx-\frac {1}{2} \int \cos (2 a+2 b x) \sin (2 a+2 b x)^4dx\) |
\(\Big \downarrow \) 3044 |
\(\displaystyle \frac {1}{2} \int \sin (2 a+2 b x)^4dx-\frac {\int \sin ^4(2 a+2 b x)d\sin (2 a+2 b x)}{4 b}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {1}{2} \int \sin (2 a+2 b x)^4dx-\frac {\sin ^5(2 a+2 b x)}{20 b}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{4} \int \sin ^2(2 a+2 b x)dx-\frac {\sin ^3(2 a+2 b x) \cos (2 a+2 b x)}{8 b}\right )-\frac {\sin ^5(2 a+2 b x)}{20 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{4} \int \sin (2 a+2 b x)^2dx-\frac {\sin ^3(2 a+2 b x) \cos (2 a+2 b x)}{8 b}\right )-\frac {\sin ^5(2 a+2 b x)}{20 b}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{4} \left (\frac {\int 1dx}{2}-\frac {\sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b}\right )-\frac {\sin ^3(2 a+2 b x) \cos (2 a+2 b x)}{8 b}\right )-\frac {\sin ^5(2 a+2 b x)}{20 b}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{4} \left (\frac {x}{2}-\frac {\sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b}\right )-\frac {\sin ^3(2 a+2 b x) \cos (2 a+2 b x)}{8 b}\right )-\frac {\sin ^5(2 a+2 b x)}{20 b}\) |
Input:
Int[Sin[a + b*x]^2*Sin[2*a + 2*b*x]^4,x]
Output:
-1/20*Sin[2*a + 2*b*x]^5/b + (-1/8*(Cos[2*a + 2*b*x]*Sin[2*a + 2*b*x]^3)/b + (3*(x/2 - (Cos[2*a + 2*b*x]*Sin[2*a + 2*b*x])/(4*b)))/4)/2
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
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[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[sin[(a_.) + (b_.)*(x_)]^2*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbo l] :> Simp[1/2 Int[(g*Sin[c + d*x])^p, x], x] - Simp[1/2 Int[Cos[c + d* x]*(g*Sin[c + d*x])^p, x], x] /; FreeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d , 0] && EqQ[d/b, 2] && IGtQ[p/2, 0]
Time = 6.92 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.87
method | result | size |
parallelrisch | \(\frac {10 \sin \left (6 b x +6 a \right )-40 \sin \left (4 b x +4 a \right )+120 b x -2 \sin \left (10 b x +10 a \right )+5 \sin \left (8 b x +8 a \right )-20 \sin \left (2 b x +2 a \right )}{640 b}\) | \(66\) |
default | \(\frac {3 x}{16}-\frac {\sin \left (2 b x +2 a \right )}{32 b}-\frac {\sin \left (4 b x +4 a \right )}{16 b}+\frac {\sin \left (6 b x +6 a \right )}{64 b}+\frac {\sin \left (8 b x +8 a \right )}{128 b}-\frac {\sin \left (10 b x +10 a \right )}{320 b}\) | \(75\) |
risch | \(\frac {3 x}{16}-\frac {\sin \left (2 b x +2 a \right )}{32 b}-\frac {\sin \left (4 b x +4 a \right )}{16 b}+\frac {\sin \left (6 b x +6 a \right )}{64 b}+\frac {\sin \left (8 b x +8 a \right )}{128 b}-\frac {\sin \left (10 b x +10 a \right )}{320 b}\) | \(75\) |
orering | \(\text {Expression too large to display}\) | \(2048\) |
Input:
int(sin(b*x+a)^2*sin(2*b*x+2*a)^4,x,method=_RETURNVERBOSE)
Output:
1/640*(10*sin(6*b*x+6*a)-40*sin(4*b*x+4*a)+120*b*x-2*sin(10*b*x+10*a)+5*si n(8*b*x+8*a)-20*sin(2*b*x+2*a))/b
Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.88 \[ \int \sin ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {15 \, b x - {\left (128 \, \cos \left (b x + a\right )^{9} - 336 \, \cos \left (b x + a\right )^{7} + 248 \, \cos \left (b x + a\right )^{5} - 10 \, \cos \left (b x + a\right )^{3} - 15 \, \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{80 \, b} \] Input:
integrate(sin(b*x+a)^2*sin(2*b*x+2*a)^4,x, algorithm="fricas")
Output:
1/80*(15*b*x - (128*cos(b*x + a)^9 - 336*cos(b*x + a)^7 + 248*cos(b*x + a) ^5 - 10*cos(b*x + a)^3 - 15*cos(b*x + a))*sin(b*x + a))/b
Leaf count of result is larger than twice the leaf count of optimal. 434 vs. \(2 (70) = 140\).
Time = 5.10 (sec) , antiderivative size = 434, normalized size of antiderivative = 5.71 \[ \int \sin ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=\begin {cases} \frac {3 x \sin ^{2}{\left (a + b x \right )} \sin ^{4}{\left (2 a + 2 b x \right )}}{16} + \frac {3 x \sin ^{2}{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{8} + \frac {3 x \sin ^{2}{\left (a + b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{16} + \frac {3 x \sin ^{4}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )}}{16} + \frac {3 x \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{8} + \frac {3 x \cos ^{2}{\left (a + b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{16} - \frac {57 \sin ^{2}{\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )} \cos {\left (2 a + 2 b x \right )}}{160 b} - \frac {109 \sin ^{2}{\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{480 b} - \frac {\sin {\left (a + b x \right )} \sin ^{4}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )}}{10 b} - \frac {2 \sin {\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{5 b} - \frac {4 \sin {\left (a + b x \right )} \cos {\left (a + b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{15 b} + \frac {7 \sin ^{3}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{160 b} + \frac {19 \sin {\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{480 b} & \text {for}\: b \neq 0 \\x \sin ^{2}{\left (a \right )} \sin ^{4}{\left (2 a \right )} & \text {otherwise} \end {cases} \] Input:
integrate(sin(b*x+a)**2*sin(2*b*x+2*a)**4,x)
Output:
Piecewise((3*x*sin(a + b*x)**2*sin(2*a + 2*b*x)**4/16 + 3*x*sin(a + b*x)** 2*sin(2*a + 2*b*x)**2*cos(2*a + 2*b*x)**2/8 + 3*x*sin(a + b*x)**2*cos(2*a + 2*b*x)**4/16 + 3*x*sin(2*a + 2*b*x)**4*cos(a + b*x)**2/16 + 3*x*sin(2*a + 2*b*x)**2*cos(a + b*x)**2*cos(2*a + 2*b*x)**2/8 + 3*x*cos(a + b*x)**2*co s(2*a + 2*b*x)**4/16 - 57*sin(a + b*x)**2*sin(2*a + 2*b*x)**3*cos(2*a + 2* b*x)/(160*b) - 109*sin(a + b*x)**2*sin(2*a + 2*b*x)*cos(2*a + 2*b*x)**3/(4 80*b) - sin(a + b*x)*sin(2*a + 2*b*x)**4*cos(a + b*x)/(10*b) - 2*sin(a + b *x)*sin(2*a + 2*b*x)**2*cos(a + b*x)*cos(2*a + 2*b*x)**2/(5*b) - 4*sin(a + b*x)*cos(a + b*x)*cos(2*a + 2*b*x)**4/(15*b) + 7*sin(2*a + 2*b*x)**3*cos( a + b*x)**2*cos(2*a + 2*b*x)/(160*b) + 19*sin(2*a + 2*b*x)*cos(a + b*x)**2 *cos(2*a + 2*b*x)**3/(480*b), Ne(b, 0)), (x*sin(a)**2*sin(2*a)**4, True))
Time = 0.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.86 \[ \int \sin ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {120 \, b x - 2 \, \sin \left (10 \, b x + 10 \, a\right ) + 5 \, \sin \left (8 \, b x + 8 \, a\right ) + 10 \, \sin \left (6 \, b x + 6 \, a\right ) - 40 \, \sin \left (4 \, b x + 4 \, a\right ) - 20 \, \sin \left (2 \, b x + 2 \, a\right )}{640 \, b} \] Input:
integrate(sin(b*x+a)^2*sin(2*b*x+2*a)^4,x, algorithm="maxima")
Output:
1/640*(120*b*x - 2*sin(10*b*x + 10*a) + 5*sin(8*b*x + 8*a) + 10*sin(6*b*x + 6*a) - 40*sin(4*b*x + 4*a) - 20*sin(2*b*x + 2*a))/b
Time = 0.13 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.89 \[ \int \sin ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {120 \, b x + 120 \, a - 2 \, \sin \left (10 \, b x + 10 \, a\right ) + 5 \, \sin \left (8 \, b x + 8 \, a\right ) + 10 \, \sin \left (6 \, b x + 6 \, a\right ) - 40 \, \sin \left (4 \, b x + 4 \, a\right ) - 20 \, \sin \left (2 \, b x + 2 \, a\right )}{640 \, b} \] Input:
integrate(sin(b*x+a)^2*sin(2*b*x+2*a)^4,x, algorithm="giac")
Output:
1/640*(120*b*x + 120*a - 2*sin(10*b*x + 10*a) + 5*sin(8*b*x + 8*a) + 10*si n(6*b*x + 6*a) - 40*sin(4*b*x + 4*a) - 20*sin(2*b*x + 2*a))/b
Time = 19.26 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.45 \[ \int \sin ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {3\,x}{16}-\frac {-\frac {3\,{\mathrm {tan}\left (a+b\,x\right )}^9}{16}-\frac {7\,{\mathrm {tan}\left (a+b\,x\right )}^7}{8}+\frac {8\,{\mathrm {tan}\left (a+b\,x\right )}^5}{5}+\frac {7\,{\mathrm {tan}\left (a+b\,x\right )}^3}{8}+\frac {3\,\mathrm {tan}\left (a+b\,x\right )}{16}}{b\,\left ({\mathrm {tan}\left (a+b\,x\right )}^{10}+5\,{\mathrm {tan}\left (a+b\,x\right )}^8+10\,{\mathrm {tan}\left (a+b\,x\right )}^6+10\,{\mathrm {tan}\left (a+b\,x\right )}^4+5\,{\mathrm {tan}\left (a+b\,x\right )}^2+1\right )} \] Input:
int(sin(a + b*x)^2*sin(2*a + 2*b*x)^4,x)
Output:
(3*x)/16 - ((3*tan(a + b*x))/16 + (7*tan(a + b*x)^3)/8 + (8*tan(a + b*x)^5 )/5 - (7*tan(a + b*x)^7)/8 - (3*tan(a + b*x)^9)/16)/(b*(5*tan(a + b*x)^2 + 10*tan(a + b*x)^4 + 10*tan(a + b*x)^6 + 5*tan(a + b*x)^8 + tan(a + b*x)^1 0 + 1))
Time = 0.22 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.99 \[ \int \sin ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {128 \cos \left (2 b x +2 a \right ) \cos \left (b x +a \right ) \sin \left (b x +a \right )-64 \cos \left (2 b x +2 a \right ) \sin \left (2 b x +2 a \right )^{3} \sin \left (b x +a \right )^{2}+2 \cos \left (2 b x +2 a \right ) \sin \left (2 b x +2 a \right )^{3}-128 \cos \left (2 b x +2 a \right ) \sin \left (2 b x +2 a \right ) \sin \left (b x +a \right )^{2}+19 \cos \left (2 b x +2 a \right ) \sin \left (2 b x +2 a \right )+16 \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{4} \sin \left (b x +a \right )+64 \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{2} \sin \left (b x +a \right )-128 \cos \left (b x +a \right ) \sin \left (b x +a \right )+128 \sin \left (2 b x +2 a \right ) \sin \left (b x +a \right )^{2}-64 \sin \left (2 b x +2 a \right )+90 b x}{480 b} \] Input:
int(sin(b*x+a)^2*sin(2*b*x+2*a)^4,x)
Output:
(128*cos(2*a + 2*b*x)*cos(a + b*x)*sin(a + b*x) - 64*cos(2*a + 2*b*x)*sin( 2*a + 2*b*x)**3*sin(a + b*x)**2 + 2*cos(2*a + 2*b*x)*sin(2*a + 2*b*x)**3 - 128*cos(2*a + 2*b*x)*sin(2*a + 2*b*x)*sin(a + b*x)**2 + 19*cos(2*a + 2*b* x)*sin(2*a + 2*b*x) + 16*cos(a + b*x)*sin(2*a + 2*b*x)**4*sin(a + b*x) + 6 4*cos(a + b*x)*sin(2*a + 2*b*x)**2*sin(a + b*x) - 128*cos(a + b*x)*sin(a + b*x) + 128*sin(2*a + 2*b*x)*sin(a + b*x)**2 - 64*sin(2*a + 2*b*x) + 90*b* x)/(480*b)