Integrand size = 20, antiderivative size = 46 \[ \int \cos ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {4 \sin ^3(a+b x)}{3 b}-\frac {8 \sin ^5(a+b x)}{5 b}+\frac {4 \sin ^7(a+b x)}{7 b} \] Output:
4/3*sin(b*x+a)^3/b-8/5*sin(b*x+a)^5/b+4/7*sin(b*x+a)^7/b
Time = 0.04 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80 \[ \int \cos ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {(157+108 \cos (2 (a+b x))+15 \cos (4 (a+b x))) \sin ^3(a+b x)}{210 b} \] Input:
Integrate[Cos[a + b*x]^3*Sin[2*a + 2*b*x]^2,x]
Output:
((157 + 108*Cos[2*(a + b*x)] + 15*Cos[4*(a + b*x)])*Sin[a + b*x]^3)/(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, 4775, 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.
\(\displaystyle \int \sin ^2(2 a+2 b x) \cos ^3(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (2 a+2 b x)^2 \cos (a+b x)^3dx\) |
\(\Big \downarrow \) 4775 |
\(\displaystyle 4 \int \cos ^5(a+b x) \sin ^2(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 4 \int \cos (a+b x)^5 \sin (a+b x)^2dx\) |
\(\Big \downarrow \) 3044 |
\(\displaystyle \frac {4 \int \sin ^2(a+b x) \left (1-\sin ^2(a+b x)\right )^2d\sin (a+b x)}{b}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {4 \int \left (\sin ^6(a+b x)-2 \sin ^4(a+b x)+\sin ^2(a+b x)\right )d\sin (a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 \left (\frac {1}{7} \sin ^7(a+b x)-\frac {2}{5} \sin ^5(a+b x)+\frac {1}{3} \sin ^3(a+b x)\right )}{b}\) |
Input:
Int[Cos[a + b*x]^3*Sin[2*a + 2*b*x]^2,x]
Output:
(4*(Sin[a + b*x]^3/3 - (2*Sin[a + b*x]^5)/5 + Sin[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_)]^(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[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_ Symbol] :> Simp[2^p/e^p Int[(e*Cos[a + b*x])^(m + p)*Sin[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && I ntegerQ[p]
Time = 2.99 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.20
method | result | size |
default | \(\frac {5 \sin \left (b x +a \right )}{16 b}-\frac {\sin \left (3 b x +3 a \right )}{48 b}-\frac {3 \sin \left (5 b x +5 a \right )}{80 b}-\frac {\sin \left (7 b x +7 a \right )}{112 b}\) | \(55\) |
risch | \(\frac {5 \sin \left (b x +a \right )}{16 b}-\frac {\sin \left (3 b x +3 a \right )}{48 b}-\frac {3 \sin \left (5 b x +5 a \right )}{80 b}-\frac {\sin \left (7 b x +7 a \right )}{112 b}\) | \(55\) |
parallelrisch | \(\frac {144 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{6} \tan \left (b x +a \right )+\left (-24 \tan \left (b x +a \right )^{2}+288\right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{5}+\left (48 \tan \left (b x +a \right )^{3}-768 \tan \left (b x +a \right )\right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{4}+\left (-32 \tan \left (b x +a \right )^{4}+752 \tan \left (b x +a \right )^{2}-32\right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}+\left (-768 \tan \left (b x +a \right )^{3}+48 \tan \left (b x +a \right )\right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}+\left (288 \tan \left (b x +a \right )^{4}-24 \tan \left (b x +a \right )^{2}\right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )+144 \tan \left (b x +a \right )^{3}}{105 b \left (1+\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}\right )^{3} \left (\tan \left (b x +a \right )^{2}+1\right )^{2}}\) | \(214\) |
orering | \(-\frac {12916 \left (-3 \cos \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right )^{2} b \sin \left (b x +a \right )+4 \cos \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} \sin \left (b x +a \right )^{3} \sin \left (2 b x +2 a \right )^{2}+72 \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right ) b^{3} \sin \left (b x +a \right )^{2} \cos \left (2 b x +2 a \right )+93 \cos \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right )^{2} b^{3} \sin \left (b x +a \right )-72 \cos \left (b x +a \right )^{2} b^{3} \cos \left (2 b x +2 a \right )^{2} \sin \left (b x +a \right )-100 \cos \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} \sin \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{2} \cos \left (b x +a \right )^{2}-5040 b^{5} \sin \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right ) \cos \left (b x +a \right ) \cos \left (2 b x +2 a \right )+540 b^{5} \sin \left (b x +a \right )^{3} \sin \left (2 b x +2 a \right )^{2}-480 b^{5} \sin \left (b x +a \right )^{3} \cos \left (2 b x +2 a \right )^{2}+3600 \cos \left (b x +a \right )^{2} b^{5} \cos \left (2 b x +2 a \right )^{2} \sin \left (b x +a \right )+3364 \cos \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} \cos \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right )^{2} \sin \left (b x +a \right )-133060 b^{7} \cos \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} \sin \left (b x +a \right ) \cos \left (2 b x +2 a \right )^{2} \cos \left (b x +a \right )^{2}+278712 b^{7} \sin \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right ) \cos \left (b x +a \right ) \cos \left (2 b x +2 a \right )-37506 b^{7} \sin \left (b x +a \right )^{3} \sin \left (2 b x +2 a \right )^{2}+36960 b^{7} \sin \left (b x +a \right )^{3} \cos \left (2 b x +2 a \right )^{2}}{11025 b^{8}}\) | \(587\) |
Input:
int(cos(b*x+a)^3*sin(2*b*x+2*a)^2,x,method=_RETURNVERBOSE)
Output:
5/16*sin(b*x+a)/b-1/48*sin(3*b*x+3*a)/b-3/80/b*sin(5*b*x+5*a)-1/112/b*sin( 7*b*x+7*a)
Time = 0.07 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.93 \[ \int \cos ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=-\frac {4 \, {\left (15 \, \cos \left (b x + a\right )^{6} - 3 \, \cos \left (b x + a\right )^{4} - 4 \, \cos \left (b x + a\right )^{2} - 8\right )} \sin \left (b x + a\right )}{105 \, b} \] Input:
integrate(cos(b*x+a)^3*sin(2*b*x+2*a)^2,x, algorithm="fricas")
Output:
-4/105*(15*cos(b*x + a)^6 - 3*cos(b*x + a)^4 - 4*cos(b*x + a)^2 - 8)*sin(b *x + a)/b
Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (39) = 78\).
Time = 2.08 (sec) , antiderivative size = 202, normalized size of antiderivative = 4.39 \[ \int \cos ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=\begin {cases} \frac {38 \sin ^{3}{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )}}{105 b} + \frac {32 \sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{105 b} + \frac {8 \sin ^{2}{\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{35 b} + \frac {11 \sin {\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )}}{35 b} + \frac {24 \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{35 b} - \frac {12 \sin {\left (2 a + 2 b x \right )} \cos ^{3}{\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{35 b} & \text {for}\: b \neq 0 \\x \sin ^{2}{\left (2 a \right )} \cos ^{3}{\left (a \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(b*x+a)**3*sin(2*b*x+2*a)**2,x)
Output:
Piecewise((38*sin(a + b*x)**3*sin(2*a + 2*b*x)**2/(105*b) + 32*sin(a + b*x )**3*cos(2*a + 2*b*x)**2/(105*b) + 8*sin(a + b*x)**2*sin(2*a + 2*b*x)*cos( a + b*x)*cos(2*a + 2*b*x)/(35*b) + 11*sin(a + b*x)*sin(2*a + 2*b*x)**2*cos (a + b*x)**2/(35*b) + 24*sin(a + b*x)*cos(a + b*x)**2*cos(2*a + 2*b*x)**2/ (35*b) - 12*sin(2*a + 2*b*x)*cos(a + b*x)**3*cos(2*a + 2*b*x)/(35*b), Ne(b , 0)), (x*sin(2*a)**2*cos(a)**3, True))
Time = 0.04 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.02 \[ \int \cos ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=-\frac {15 \, \sin \left (7 \, b x + 7 \, a\right ) + 63 \, \sin \left (5 \, b x + 5 \, a\right ) + 35 \, \sin \left (3 \, b x + 3 \, a\right ) - 525 \, \sin \left (b x + a\right )}{1680 \, b} \] Input:
integrate(cos(b*x+a)^3*sin(2*b*x+2*a)^2,x, algorithm="maxima")
Output:
-1/1680*(15*sin(7*b*x + 7*a) + 63*sin(5*b*x + 5*a) + 35*sin(3*b*x + 3*a) - 525*sin(b*x + a))/b
Time = 0.19 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.02 \[ \int \cos ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=-\frac {15 \, \sin \left (7 \, b x + 7 \, a\right ) + 63 \, \sin \left (5 \, b x + 5 \, a\right ) + 35 \, \sin \left (3 \, b x + 3 \, a\right ) - 525 \, \sin \left (b x + a\right )}{1680 \, b} \] Input:
integrate(cos(b*x+a)^3*sin(2*b*x+2*a)^2,x, algorithm="giac")
Output:
-1/1680*(15*sin(7*b*x + 7*a) + 63*sin(5*b*x + 5*a) + 35*sin(3*b*x + 3*a) - 525*sin(b*x + a))/b
Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \cos ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {4\,\left (15\,{\sin \left (a+b\,x\right )}^7-42\,{\sin \left (a+b\,x\right )}^5+35\,{\sin \left (a+b\,x\right )}^3\right )}{105\,b} \] Input:
int(cos(a + b*x)^3*sin(2*a + 2*b*x)^2,x)
Output:
(4*(35*sin(a + b*x)^3 - 42*sin(a + b*x)^5 + 15*sin(a + b*x)^7))/(105*b)
Time = 0.18 (sec) , antiderivative size = 177, normalized size of antiderivative = 3.85 \[ \int \cos ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {60 \cos \left (2 b x +2 a \right ) \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right ) \sin \left (b x +a \right )^{2}-36 \cos \left (2 b x +2 a \right ) \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )-72 \cos \left (2 b x +2 a \right ) \cos \left (b x +a \right ) \sin \left (b x +a \right )+45 \sin \left (2 b x +2 a \right )^{2} \sin \left (b x +a \right )^{3}-39 \sin \left (2 b x +2 a \right )^{2} \sin \left (b x +a \right )-72 \sin \left (2 b x +2 a \right ) \sin \left (b x +a \right )^{2}+36 \sin \left (2 b x +2 a \right )-40 \sin \left (b x +a \right )^{3}+72 \sin \left (b x +a \right )}{105 b} \] Input:
int(cos(b*x+a)^3*sin(2*b*x+2*a)^2,x)
Output:
(60*cos(2*a + 2*b*x)*cos(a + b*x)*sin(2*a + 2*b*x)*sin(a + b*x)**2 - 36*co s(2*a + 2*b*x)*cos(a + b*x)*sin(2*a + 2*b*x) - 72*cos(2*a + 2*b*x)*cos(a + b*x)*sin(a + b*x) + 45*sin(2*a + 2*b*x)**2*sin(a + b*x)**3 - 39*sin(2*a + 2*b*x)**2*sin(a + b*x) - 72*sin(2*a + 2*b*x)*sin(a + b*x)**2 + 36*sin(2*a + 2*b*x) - 40*sin(a + b*x)**3 + 72*sin(a + b*x))/(105*b)