Integrand size = 20, antiderivative size = 46 \[ \int \sin ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=-\frac {4 \cos ^3(a+b x)}{3 b}+\frac {8 \cos ^5(a+b x)}{5 b}-\frac {4 \cos ^7(a+b x)}{7 b} \] Output:
-4/3*cos(b*x+a)^3/b+8/5*cos(b*x+a)^5/b-4/7*cos(b*x+a)^7/b
Time = 0.07 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80 \[ \int \sin ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {\cos ^3(a+b x) (-157+108 \cos (2 (a+b x))-15 \cos (4 (a+b x)))}{210 b} \] Input:
Integrate[Sin[a + b*x]^3*Sin[2*a + 2*b*x]^2,x]
Output:
(Cos[a + b*x]^3*(-157 + 108*Cos[2*(a + b*x)] - 15*Cos[4*(a + b*x)]))/(210* b)
Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.91, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, 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.
\(\displaystyle \int \sin ^3(a+b x) \sin ^2(2 a+2 b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (a+b x)^3 \sin (2 a+2 b x)^2dx\) |
\(\Big \downarrow \) 4776 |
\(\displaystyle 4 \int \cos ^2(a+b x) \sin ^5(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 4 \int \cos (a+b x)^2 \sin (a+b x)^5dx\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle -\frac {4 \int \cos ^2(a+b x) \left (1-\cos ^2(a+b x)\right )^2d\cos (a+b x)}{b}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -\frac {4 \int \left (\cos ^6(a+b x)-2 \cos ^4(a+b x)+\cos ^2(a+b x)\right )d\cos (a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4 \left (\frac {1}{7} \cos ^7(a+b x)-\frac {2}{5} \cos ^5(a+b x)+\frac {1}{3} \cos ^3(a+b x)\right )}{b}\) |
Input:
Int[Sin[a + b*x]^3*Sin[2*a + 2*b*x]^2,x]
Output:
(-4*(Cos[a + b*x]^3/3 - (2*Cos[a + b*x]^5)/5 + Cos[a + b*x]^7/7))/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 = 2.80 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.20
method | result | size |
default | \(-\frac {5 \cos \left (b x +a \right )}{16 b}-\frac {\cos \left (3 b x +3 a \right )}{48 b}+\frac {3 \cos \left (5 b x +5 a \right )}{80 b}-\frac {\cos \left (7 b x +7 a \right )}{112 b}\) | \(55\) |
risch | \(-\frac {5 \cos \left (b x +a \right )}{16 b}-\frac {\cos \left (3 b x +3 a \right )}{48 b}+\frac {3 \cos \left (5 b x +5 a \right )}{80 b}-\frac {\cos \left (7 b x +7 a \right )}{112 b}\) | \(55\) |
parallelrisch | \(\frac {24+\left (64 \tan \left (b x +a \right )^{4}+176 \tan \left (b x +a \right )^{2}+88\right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{6}-192 \tan \left (b x +a \right )^{3} \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{5}+\left (384 \tan \left (b x +a \right )^{4}-48 \tan \left (b x +a \right )^{2}+264\right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{4}+\left (768 \tan \left (b x +a \right )^{3}-768 \tan \left (b x +a \right )\right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}+\left (576 \tan \left (b x +a \right )^{2}-120\right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}+192 \tan \left (b x +a \right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )}{105 b \left (1+\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}\right )^{3} \left (\tan \left (b x +a \right )^{2}+1\right )^{2}}\) | \(196\) |
orering | \(-\frac {12916 \left (3 \sin \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right )^{2} b \cos \left (b x +a \right )+4 \sin \left (b x +a \right )^{3} \sin \left (2 b x +2 a \right ) b \cos \left (2 b x +2 a \right )\right )}{11025 b^{2}}-\frac {94 \left (6 b^{3} \cos \left (b x +a \right )^{3} \sin \left (2 b x +2 a \right )^{2}+72 \sin \left (b x +a \right ) \sin \left (2 b x +2 a \right ) b^{3} \cos \left (b x +a \right )^{2} \cos \left (2 b x +2 a \right )-93 \sin \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right )^{2} b^{3} \cos \left (b x +a \right )+72 \sin \left (b x +a \right )^{2} b^{3} \cos \left (2 b x +2 a \right )^{2} \cos \left (b x +a \right )-100 \sin \left (b x +a \right )^{3} \sin \left (2 b x +2 a \right ) b^{3} \cos \left (2 b x +2 a \right )\right )}{525 b^{4}}-\frac {4 \left (3783 b^{5} \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{2} \sin \left (b x +a \right )^{2}-5040 b^{5} \cos \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right ) \sin \left (b x +a \right ) \cos \left (2 b x +2 a \right )-540 b^{5} \cos \left (b x +a \right )^{3} \sin \left (2 b x +2 a \right )^{2}+480 b^{5} \cos \left (b x +a \right )^{3} \cos \left (2 b x +2 a \right )^{2}-3600 \sin \left (b x +a \right )^{2} b^{5} \cos \left (2 b x +2 a \right )^{2} \cos \left (b x +a \right )+3364 \sin \left (b x +a \right )^{3} \sin \left (2 b x +2 a \right ) b^{5} \cos \left (2 b x +2 a \right )\right )}{525 b^{6}}-\frac {-169473 b^{7} \sin \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right )^{2} \cos \left (b x +a \right )-133060 b^{7} \sin \left (b x +a \right )^{3} \sin \left (2 b x +2 a \right ) \cos \left (2 b x +2 a \right )+167832 b^{7} \cos \left (b x +a \right ) \cos \left (2 b x +2 a \right )^{2} \sin \left (b x +a \right )^{2}+278712 b^{7} \cos \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right ) \sin \left (b x +a \right ) \cos \left (2 b x +2 a \right )+37506 b^{7} \cos \left (b x +a \right )^{3} \sin \left (2 b x +2 a \right )^{2}-36960 b^{7} \cos \left (b x +a \right )^{3} \cos \left (2 b x +2 a \right )^{2}}{11025 b^{8}}\) | \(587\) |
Input:
int(sin(b*x+a)^3*sin(2*b*x+2*a)^2,x,method=_RETURNVERBOSE)
Output:
-5/16*cos(b*x+a)/b-1/48*cos(3*b*x+3*a)/b+3/80*cos(5*b*x+5*a)/b-1/112*cos(7 *b*x+7*a)/b
Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \sin ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=-\frac {4 \, {\left (15 \, \cos \left (b x + a\right )^{7} - 42 \, \cos \left (b x + a\right )^{5} + 35 \, \cos \left (b x + a\right )^{3}\right )}}{105 \, b} \] Input:
integrate(sin(b*x+a)^3*sin(2*b*x+2*a)^2,x, algorithm="fricas")
Output:
-4/105*(15*cos(b*x + a)^7 - 42*cos(b*x + a)^5 + 35*cos(b*x + a)^3)/b
Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (39) = 78\).
Time = 2.17 (sec) , antiderivative size = 202, normalized size of antiderivative = 4.39 \[ \int \sin ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=\begin {cases} - \frac {12 \sin ^{3}{\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos {\left (2 a + 2 b x \right )}}{35 b} - \frac {11 \sin ^{2}{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )}}{35 b} - \frac {24 \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{35 b} + \frac {8 \sin {\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{35 b} - \frac {38 \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{3}{\left (a + b x \right )}}{105 b} - \frac {32 \cos ^{3}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{105 b} & \text {for}\: b \neq 0 \\x \sin ^{3}{\left (a \right )} \sin ^{2}{\left (2 a \right )} & \text {otherwise} \end {cases} \] Input:
integrate(sin(b*x+a)**3*sin(2*b*x+2*a)**2,x)
Output:
Piecewise((-12*sin(a + b*x)**3*sin(2*a + 2*b*x)*cos(2*a + 2*b*x)/(35*b) - 11*sin(a + b*x)**2*sin(2*a + 2*b*x)**2*cos(a + b*x)/(35*b) - 24*sin(a + b* x)**2*cos(a + b*x)*cos(2*a + 2*b*x)**2/(35*b) + 8*sin(a + b*x)*sin(2*a + 2 *b*x)*cos(a + b*x)**2*cos(2*a + 2*b*x)/(35*b) - 38*sin(2*a + 2*b*x)**2*cos (a + b*x)**3/(105*b) - 32*cos(a + b*x)**3*cos(2*a + 2*b*x)**2/(105*b), Ne( b, 0)), (x*sin(a)**3*sin(2*a)**2, True))
Time = 0.10 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.02 \[ \int \sin ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=-\frac {15 \, \cos \left (7 \, b x + 7 \, a\right ) - 63 \, \cos \left (5 \, b x + 5 \, a\right ) + 35 \, \cos \left (3 \, b x + 3 \, a\right ) + 525 \, \cos \left (b x + a\right )}{1680 \, b} \] Input:
integrate(sin(b*x+a)^3*sin(2*b*x+2*a)^2,x, algorithm="maxima")
Output:
-1/1680*(15*cos(7*b*x + 7*a) - 63*cos(5*b*x + 5*a) + 35*cos(3*b*x + 3*a) + 525*cos(b*x + a))/b
Time = 0.17 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.02 \[ \int \sin ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=-\frac {15 \, \cos \left (7 \, b x + 7 \, a\right ) - 63 \, \cos \left (5 \, b x + 5 \, a\right ) + 35 \, \cos \left (3 \, b x + 3 \, a\right ) + 525 \, \cos \left (b x + a\right )}{1680 \, b} \] Input:
integrate(sin(b*x+a)^3*sin(2*b*x+2*a)^2,x, algorithm="giac")
Output:
-1/1680*(15*cos(7*b*x + 7*a) - 63*cos(5*b*x + 5*a) + 35*cos(3*b*x + 3*a) + 525*cos(b*x + a))/b
Time = 17.71 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \sin ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=-\frac {4\,\left (15\,{\cos \left (a+b\,x\right )}^7-42\,{\cos \left (a+b\,x\right )}^5+35\,{\cos \left (a+b\,x\right )}^3\right )}{105\,b} \] Input:
int(sin(a + b*x)^3*sin(2*a + 2*b*x)^2,x)
Output:
-(4*(35*cos(a + b*x)^3 - 42*cos(a + b*x)^5 + 15*cos(a + b*x)^7))/(105*b)
Time = 0.18 (sec) , antiderivative size = 184, normalized size of antiderivative = 4.00 \[ \int \sin ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {-60 \cos \left (2 b x +2 a \right ) \sin \left (2 b x +2 a \right ) \sin \left (b x +a \right )^{3}+24 \cos \left (2 b x +2 a \right ) \sin \left (2 b x +2 a \right ) \sin \left (b x +a \right )-24 \cos \left (2 b x +2 a \right ) \sin \left (b x +a \right )^{2}+12 \cos \left (2 b x +2 a \right )+45 \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{2} \sin \left (b x +a \right )^{2}-6 \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{2}+24 \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right ) \sin \left (b x +a \right )-40 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2}-32 \cos \left (b x +a \right )-52}{105 b} \] Input:
int(sin(b*x+a)^3*sin(2*b*x+2*a)^2,x)
Output:
( - 60*cos(2*a + 2*b*x)*sin(2*a + 2*b*x)*sin(a + b*x)**3 + 24*cos(2*a + 2* b*x)*sin(2*a + 2*b*x)*sin(a + b*x) - 24*cos(2*a + 2*b*x)*sin(a + b*x)**2 + 12*cos(2*a + 2*b*x) + 45*cos(a + b*x)*sin(2*a + 2*b*x)**2*sin(a + b*x)**2 - 6*cos(a + b*x)*sin(2*a + 2*b*x)**2 + 24*cos(a + b*x)*sin(2*a + 2*b*x)*s in(a + b*x) - 40*cos(a + b*x)*sin(a + b*x)**2 - 32*cos(a + b*x) - 52)/(105 *b)