Integrand size = 18, antiderivative size = 61 \[ \int \cos (a+b x) \sin ^7(2 a+2 b x) \, dx=-\frac {128 \cos ^9(a+b x)}{9 b}+\frac {384 \cos ^{11}(a+b x)}{11 b}-\frac {384 \cos ^{13}(a+b x)}{13 b}+\frac {128 \cos ^{15}(a+b x)}{15 b} \] Output:
-128/9*cos(b*x+a)^9/b+384/11*cos(b*x+a)^11/b-384/13*cos(b*x+a)^13/b+128/15 *cos(b*x+a)^15/b
Time = 0.30 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.77 \[ \int \cos (a+b x) \sin ^7(2 a+2 b x) \, dx=\frac {4 \cos ^9(a+b x) (-8330+10755 \cos (2 (a+b x))-3366 \cos (4 (a+b x))+429 \cos (6 (a+b x)))}{6435 b} \] Input:
Integrate[Cos[a + b*x]*Sin[2*a + 2*b*x]^7,x]
Output:
(4*Cos[a + b*x]^9*(-8330 + 10755*Cos[2*(a + b*x)] - 3366*Cos[4*(a + b*x)] + 429*Cos[6*(a + b*x)]))/(6435*b)
Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4775, 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.
\(\displaystyle \int \sin ^7(2 a+2 b x) \cos (a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (2 a+2 b x)^7 \cos (a+b x)dx\) |
\(\Big \downarrow \) 4775 |
\(\displaystyle 128 \int \cos ^8(a+b x) \sin ^7(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 128 \int \cos (a+b x)^8 \sin (a+b x)^7dx\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle -\frac {128 \int \cos ^8(a+b x) \left (1-\cos ^2(a+b x)\right )^3d\cos (a+b x)}{b}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -\frac {128 \int \left (-\cos ^{14}(a+b x)+3 \cos ^{12}(a+b x)-3 \cos ^{10}(a+b x)+\cos ^8(a+b x)\right )d\cos (a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {128 \left (-\frac {1}{15} \cos ^{15}(a+b x)+\frac {3}{13} \cos ^{13}(a+b x)-\frac {3}{11} \cos ^{11}(a+b x)+\frac {1}{9} \cos ^9(a+b x)\right )}{b}\) |
Input:
Int[Cos[a + b*x]*Sin[2*a + 2*b*x]^7,x]
Output:
(-128*(Cos[a + b*x]^9/9 - (3*Cos[a + b*x]^11)/11 + (3*Cos[a + b*x]^13)/13 - Cos[a + b*x]^15/15))/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[(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 = 25.36 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.52
method | result | size |
parallelrisch | \(\frac {262144+19305 \cos \left (7 b x +7 a \right )-5005 \cos \left (9 b x +9 a \right )+429 \cos \left (15 b x +15 a \right )+495 \cos \left (13 b x +13 a \right )-4095 \cos \left (11 b x +11 a \right )-75075 \cos \left (3 b x +3 a \right )-225225 \cos \left (b x +a \right )+27027 \cos \left (5 b x +5 a \right )}{823680 b}\) | \(93\) |
default | \(-\frac {35 \cos \left (b x +a \right )}{128 b}-\frac {35 \cos \left (3 b x +3 a \right )}{384 b}+\frac {21 \cos \left (5 b x +5 a \right )}{640 b}+\frac {3 \cos \left (7 b x +7 a \right )}{128 b}-\frac {7 \cos \left (9 b x +9 a \right )}{1152 b}-\frac {7 \cos \left (11 b x +11 a \right )}{1408 b}+\frac {\cos \left (13 b x +13 a \right )}{1664 b}+\frac {\cos \left (15 b x +15 a \right )}{1920 b}\) | \(111\) |
risch | \(-\frac {35 \cos \left (b x +a \right )}{128 b}-\frac {35 \cos \left (3 b x +3 a \right )}{384 b}+\frac {21 \cos \left (5 b x +5 a \right )}{640 b}+\frac {3 \cos \left (7 b x +7 a \right )}{128 b}-\frac {7 \cos \left (9 b x +9 a \right )}{1152 b}-\frac {7 \cos \left (11 b x +11 a \right )}{1408 b}+\frac {\cos \left (13 b x +13 a \right )}{1664 b}+\frac {\cos \left (15 b x +15 a \right )}{1920 b}\) | \(111\) |
orering | \(\text {Expression too large to display}\) | \(1593\) |
Input:
int(cos(b*x+a)*sin(2*b*x+2*a)^7,x,method=_RETURNVERBOSE)
Output:
1/823680*(262144+19305*cos(7*b*x+7*a)-5005*cos(9*b*x+9*a)+429*cos(15*b*x+1 5*a)+495*cos(13*b*x+13*a)-4095*cos(11*b*x+11*a)-75075*cos(3*b*x+3*a)-22522 5*cos(b*x+a)+27027*cos(5*b*x+5*a))/b
Time = 0.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.75 \[ \int \cos (a+b x) \sin ^7(2 a+2 b x) \, dx=\frac {128 \, {\left (429 \, \cos \left (b x + a\right )^{15} - 1485 \, \cos \left (b x + a\right )^{13} + 1755 \, \cos \left (b x + a\right )^{11} - 715 \, \cos \left (b x + a\right )^{9}\right )}}{6435 \, b} \] Input:
integrate(cos(b*x+a)*sin(2*b*x+2*a)^7,x, algorithm="fricas")
Output:
128/6435*(429*cos(b*x + a)^15 - 1485*cos(b*x + a)^13 + 1755*cos(b*x + a)^1 1 - 715*cos(b*x + a)^9)/b
Leaf count of result is larger than twice the leaf count of optimal. 270 vs. \(2 (53) = 106\).
Time = 25.62 (sec) , antiderivative size = 270, normalized size of antiderivative = 4.43 \[ \int \cos (a+b x) \sin ^7(2 a+2 b x) \, dx=\begin {cases} - \frac {1241 \sin {\left (a + b x \right )} \sin ^{7}{\left (2 a + 2 b x \right )}}{6435 b} - \frac {376 \sin {\left (a + b x \right )} \sin ^{5}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{715 b} - \frac {640 \sin {\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{1287 b} - \frac {1024 \sin {\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos ^{6}{\left (2 a + 2 b x \right )}}{6435 b} - \frac {3838 \sin ^{6}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{6435 b} - \frac {1648 \sin ^{4}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{1287 b} - \frac {768 \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{5}{\left (2 a + 2 b x \right )}}{715 b} - \frac {2048 \cos {\left (a + b x \right )} \cos ^{7}{\left (2 a + 2 b x \right )}}{6435 b} & \text {for}\: b \neq 0 \\x \sin ^{7}{\left (2 a \right )} \cos {\left (a \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(b*x+a)*sin(2*b*x+2*a)**7,x)
Output:
Piecewise((-1241*sin(a + b*x)*sin(2*a + 2*b*x)**7/(6435*b) - 376*sin(a + b *x)*sin(2*a + 2*b*x)**5*cos(2*a + 2*b*x)**2/(715*b) - 640*sin(a + b*x)*sin (2*a + 2*b*x)**3*cos(2*a + 2*b*x)**4/(1287*b) - 1024*sin(a + b*x)*sin(2*a + 2*b*x)*cos(2*a + 2*b*x)**6/(6435*b) - 3838*sin(2*a + 2*b*x)**6*cos(a + b *x)*cos(2*a + 2*b*x)/(6435*b) - 1648*sin(2*a + 2*b*x)**4*cos(a + b*x)*cos( 2*a + 2*b*x)**3/(1287*b) - 768*sin(2*a + 2*b*x)**2*cos(a + b*x)*cos(2*a + 2*b*x)**5/(715*b) - 2048*cos(a + b*x)*cos(2*a + 2*b*x)**7/(6435*b), Ne(b, 0)), (x*sin(2*a)**7*cos(a), True))
Time = 0.04 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.49 \[ \int \cos (a+b x) \sin ^7(2 a+2 b x) \, dx=\frac {429 \, \cos \left (15 \, b x + 15 \, a\right ) + 495 \, \cos \left (13 \, b x + 13 \, a\right ) - 4095 \, \cos \left (11 \, b x + 11 \, a\right ) - 5005 \, \cos \left (9 \, b x + 9 \, a\right ) + 19305 \, \cos \left (7 \, b x + 7 \, a\right ) + 27027 \, \cos \left (5 \, b x + 5 \, a\right ) - 75075 \, \cos \left (3 \, b x + 3 \, a\right ) - 225225 \, \cos \left (b x + a\right )}{823680 \, b} \] Input:
integrate(cos(b*x+a)*sin(2*b*x+2*a)^7,x, algorithm="maxima")
Output:
1/823680*(429*cos(15*b*x + 15*a) + 495*cos(13*b*x + 13*a) - 4095*cos(11*b* x + 11*a) - 5005*cos(9*b*x + 9*a) + 19305*cos(7*b*x + 7*a) + 27027*cos(5*b *x + 5*a) - 75075*cos(3*b*x + 3*a) - 225225*cos(b*x + a))/b
Time = 0.14 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.75 \[ \int \cos (a+b x) \sin ^7(2 a+2 b x) \, dx=\frac {128 \, {\left (429 \, \cos \left (b x + a\right )^{15} - 1485 \, \cos \left (b x + a\right )^{13} + 1755 \, \cos \left (b x + a\right )^{11} - 715 \, \cos \left (b x + a\right )^{9}\right )}}{6435 \, b} \] Input:
integrate(cos(b*x+a)*sin(2*b*x+2*a)^7,x, algorithm="giac")
Output:
128/6435*(429*cos(b*x + a)^15 - 1485*cos(b*x + a)^13 + 1755*cos(b*x + a)^1 1 - 715*cos(b*x + a)^9)/b
Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.75 \[ \int \cos (a+b x) \sin ^7(2 a+2 b x) \, dx=-\frac {-\frac {128\,{\cos \left (a+b\,x\right )}^{15}}{15}+\frac {384\,{\cos \left (a+b\,x\right )}^{13}}{13}-\frac {384\,{\cos \left (a+b\,x\right )}^{11}}{11}+\frac {128\,{\cos \left (a+b\,x\right )}^9}{9}}{b} \] Input:
int(cos(a + b*x)*sin(2*a + 2*b*x)^7,x)
Output:
-((128*cos(a + b*x)^9)/9 - (384*cos(a + b*x)^11)/11 + (384*cos(a + b*x)^13 )/13 - (128*cos(a + b*x)^15)/15)/b
Time = 0.16 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.98 \[ \int \cos (a+b x) \sin ^7(2 a+2 b x) \, dx=\frac {-16170 \cos \left (2 b x +2 a \right ) \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{6}-19600 \cos \left (2 b x +2 a \right ) \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{4}-26880 \cos \left (2 b x +2 a \right ) \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{2}-71680 \cos \left (2 b x +2 a \right ) \cos \left (b x +a \right )-1155 \sin \left (2 b x +2 a \right )^{7} \sin \left (b x +a \right )-1960 \sin \left (2 b x +2 a \right )^{5} \sin \left (b x +a \right )-4480 \sin \left (2 b x +2 a \right )^{3} \sin \left (b x +a \right )-35840 \sin \left (2 b x +2 a \right ) \sin \left (b x +a \right )-54912}{225225 b} \] Input:
int(cos(b*x+a)*sin(2*b*x+2*a)^7,x)
Output:
( - 16170*cos(2*a + 2*b*x)*cos(a + b*x)*sin(2*a + 2*b*x)**6 - 19600*cos(2* a + 2*b*x)*cos(a + b*x)*sin(2*a + 2*b*x)**4 - 26880*cos(2*a + 2*b*x)*cos(a + b*x)*sin(2*a + 2*b*x)**2 - 71680*cos(2*a + 2*b*x)*cos(a + b*x) - 1155*s in(2*a + 2*b*x)**7*sin(a + b*x) - 1960*sin(2*a + 2*b*x)**5*sin(a + b*x) - 4480*sin(2*a + 2*b*x)**3*sin(a + b*x) - 35840*sin(2*a + 2*b*x)*sin(a + b*x ) - 54912)/(225225*b)