\(\int \sin (a+b x) \sin ^6(2 a+2 b x) \, dx\) [2]

Optimal result

Integrand size = 18, antiderivative size = 61 \[ \int \sin (a+b x) \sin ^6(2 a+2 b x) \, dx=-\frac {64 \cos ^7(a+b x)}{7 b}+\frac {64 \cos ^9(a+b x)}{3 b}-\frac {192 \cos ^{11}(a+b x)}{11 b}+\frac {64 \cos ^{13}(a+b x)}{13 b} \]

[Out]

-64/7*cos(b*x+a)^7/b+64/3*cos(b*x+a)^9/b-192/11*cos(b*x+a)^11/b+64/13*cos(b*x+a)^13/b

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4373, 2645, 276} \[ \int \sin (a+b x) \sin ^6(2 a+2 b x) \, dx=\frac {64 \cos ^{13}(a+b x)}{13 b}-\frac {192 \cos ^{11}(a+b x)}{11 b}+\frac {64 \cos ^9(a+b x)}{3 b}-\frac {64 \cos ^7(a+b x)}{7 b} \]

[In]

Int[Sin[a + b*x]*Sin[2*a + 2*b*x]^6,x]

[Out]

(-64*Cos[a + b*x]^7)/(7*b) + (64*Cos[a + b*x]^9)/(3*b) - (192*Cos[a + b*x]^11)/(11*b) + (64*Cos[a + b*x]^13)/(
13*b)

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(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])

Rule 4373

Int[((f_.)*sin[(a_.) + (b_.)*(x_)])^(n_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_Symbol] :> Dist[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]
&& IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = 64 \int \cos ^6(a+b x) \sin ^7(a+b x) \, dx \\ & = -\frac {64 \text {Subst}\left (\int x^6 \left (1-x^2\right )^3 \, dx,x,\cos (a+b x)\right )}{b} \\ & = -\frac {64 \text {Subst}\left (\int \left (x^6-3 x^8+3 x^{10}-x^{12}\right ) \, dx,x,\cos (a+b x)\right )}{b} \\ & = -\frac {64 \cos ^7(a+b x)}{7 b}+\frac {64 \cos ^9(a+b x)}{3 b}-\frac {192 \cos ^{11}(a+b x)}{11 b}+\frac {64 \cos ^{13}(a+b x)}{13 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.77 \[ \int \sin (a+b x) \sin ^6(2 a+2 b x) \, dx=\frac {2 \cos ^7(a+b x) (-5230+6377 \cos (2 (a+b x))-1890 \cos (4 (a+b x))+231 \cos (6 (a+b x)))}{3003 b} \]

[In]

Integrate[Sin[a + b*x]*Sin[2*a + 2*b*x]^6,x]

[Out]

(2*Cos[a + b*x]^7*(-5230 + 6377*Cos[2*(a + b*x)] - 1890*Cos[4*(a + b*x)] + 231*Cos[6*(a + b*x)]))/(3003*b)

Maple [A] (verified)

Time = 5.26 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.59

method	result	size
default	\(-\frac {5 \cos \left (x b +a \right )}{16 b}-\frac {5 \cos \left (3 x b +3 a \right )}{64 b}+\frac {3 \cos \left (5 x b +5 a \right )}{64 b}+\frac {3 \cos \left (7 x b +7 a \right )}{224 b}-\frac {\cos \left (9 x b +9 a \right )}{96 b}-\frac {\cos \left (11 x b +11 a \right )}{704 b}+\frac {\cos \left (13 x b +13 a \right )}{832 b}\)	\(97\)
risch	\(-\frac {5 \cos \left (x b +a \right )}{16 b}-\frac {5 \cos \left (3 x b +3 a \right )}{64 b}+\frac {3 \cos \left (5 x b +5 a \right )}{64 b}+\frac {3 \cos \left (7 x b +7 a \right )}{224 b}-\frac {\cos \left (9 x b +9 a \right )}{96 b}-\frac {\cos \left (11 x b +11 a \right )}{704 b}+\frac {\cos \left (13 x b +13 a \right )}{832 b}\)	\(97\)
parallelrisch	\(\frac {\left (-512 \tan \left (x b +a \right )^{10}-2688 \tan \left (x b +a \right )^{8}-5696 \tan \left (x b +a \right )^{6}-2688 \tan \left (x b +a \right )^{4}-512 \tan \left (x b +a \right )^{2}\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}+\left (2048 \tan \left (x b +a \right )^{11}+11264 \tan \left (x b +a \right )^{9}+25344 \tan \left (x b +a \right )^{7}-25344 \tan \left (x b +a \right )^{5}-11264 \tan \left (x b +a \right )^{3}-2048 \tan \left (x b +a \right )\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )-2048 \tan \left (x b +a \right )^{12}-11776 \tan \left (x b +a \right )^{10}-28032 \tan \left (x b +a \right )^{8}-35264 \tan \left (x b +a \right )^{6}-28032 \tan \left (x b +a \right )^{4}-11776 \tan \left (x b +a \right )^{2}-2048}{3003 b \left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right ) \left (1+\tan \left (x b +a \right )^{2}\right )^{6}}\)	\(227\)

[In]

int(sin(b*x+a)*sin(2*b*x+2*a)^6,x,method=_RETURNVERBOSE)

[Out]

-5/16*cos(b*x+a)/b-5/64*cos(3*b*x+3*a)/b+3/64*cos(5*b*x+5*a)/b+3/224*cos(7*b*x+7*a)/b-1/96*cos(9*b*x+9*a)/b-1/
704*cos(11*b*x+11*a)/b+1/832*cos(13*b*x+13*a)/b

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.75 \[ \int \sin (a+b x) \sin ^6(2 a+2 b x) \, dx=\frac {64 \, {\left (231 \, \cos \left (b x + a\right )^{13} - 819 \, \cos \left (b x + a\right )^{11} + 1001 \, \cos \left (b x + a\right )^{9} - 429 \, \cos \left (b x + a\right )^{7}\right )}}{3003 \, b} \]

[In]

integrate(sin(b*x+a)*sin(2*b*x+2*a)^6,x, algorithm="fricas")

[Out]

64/3003*(231*cos(b*x + a)^13 - 819*cos(b*x + a)^11 + 1001*cos(b*x + a)^9 - 429*cos(b*x + a)^7)/b

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (53) = 106\).

Time = 10.96 (sec) , antiderivative size = 235, normalized size of antiderivative = 3.85 \[ \int \sin (a+b x) \sin ^6(2 a+2 b x) \, dx=\begin {cases} - \frac {1084 \sin {\left (a + b x \right )} \sin ^{5}{\left (2 a + 2 b x \right )} \cos {\left (2 a + 2 b x \right )}}{3003 b} - \frac {64 \sin {\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{143 b} - \frac {512 \sin {\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos ^{5}{\left (2 a + 2 b x \right )}}{3003 b} - \frac {835 \sin ^{6}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )}}{3003 b} - \frac {2776 \sin ^{4}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{3003 b} - \frac {2944 \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{3003 b} - \frac {1024 \cos {\left (a + b x \right )} \cos ^{6}{\left (2 a + 2 b x \right )}}{3003 b} & \text {for}\: b \neq 0 \\x \sin {\left (a \right )} \sin ^{6}{\left (2 a \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(sin(b*x+a)*sin(2*b*x+2*a)**6,x)

[Out]

Piecewise((-1084*sin(a + b*x)*sin(2*a + 2*b*x)**5*cos(2*a + 2*b*x)/(3003*b) - 64*sin(a + b*x)*sin(2*a + 2*b*x)
**3*cos(2*a + 2*b*x)**3/(143*b) - 512*sin(a + b*x)*sin(2*a + 2*b*x)*cos(2*a + 2*b*x)**5/(3003*b) - 835*sin(2*a
+ 2*b*x)**6*cos(a + b*x)/(3003*b) - 2776*sin(2*a + 2*b*x)**4*cos(a + b*x)*cos(2*a + 2*b*x)**2/(3003*b) - 2944
*sin(2*a + 2*b*x)**2*cos(a + b*x)*cos(2*a + 2*b*x)**4/(3003*b) - 1024*cos(a + b*x)*cos(2*a + 2*b*x)**6/(3003*b
), Ne(b, 0)), (x*sin(a)*sin(2*a)**6, True))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.31 \[ \int \sin (a+b x) \sin ^6(2 a+2 b x) \, dx=\frac {231 \, \cos \left (13 \, b x + 13 \, a\right ) - 273 \, \cos \left (11 \, b x + 11 \, a\right ) - 2002 \, \cos \left (9 \, b x + 9 \, a\right ) + 2574 \, \cos \left (7 \, b x + 7 \, a\right ) + 9009 \, \cos \left (5 \, b x + 5 \, a\right ) - 15015 \, \cos \left (3 \, b x + 3 \, a\right ) - 60060 \, \cos \left (b x + a\right )}{192192 \, b} \]

[In]

integrate(sin(b*x+a)*sin(2*b*x+2*a)^6,x, algorithm="maxima")

[Out]

1/192192*(231*cos(13*b*x + 13*a) - 273*cos(11*b*x + 11*a) - 2002*cos(9*b*x + 9*a) + 2574*cos(7*b*x + 7*a) + 90
09*cos(5*b*x + 5*a) - 15015*cos(3*b*x + 3*a) - 60060*cos(b*x + a))/b

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.31 \[ \int \sin (a+b x) \sin ^6(2 a+2 b x) \, dx=\frac {231 \, \cos \left (13 \, b x + 13 \, a\right ) - 273 \, \cos \left (11 \, b x + 11 \, a\right ) - 2002 \, \cos \left (9 \, b x + 9 \, a\right ) + 2574 \, \cos \left (7 \, b x + 7 \, a\right ) + 9009 \, \cos \left (5 \, b x + 5 \, a\right ) - 15015 \, \cos \left (3 \, b x + 3 \, a\right ) - 60060 \, \cos \left (b x + a\right )}{192192 \, b} \]

[In]

integrate(sin(b*x+a)*sin(2*b*x+2*a)^6,x, algorithm="giac")

[Out]

1/192192*(231*cos(13*b*x + 13*a) - 273*cos(11*b*x + 11*a) - 2002*cos(9*b*x + 9*a) + 2574*cos(7*b*x + 7*a) + 90
09*cos(5*b*x + 5*a) - 15015*cos(3*b*x + 3*a) - 60060*cos(b*x + a))/b

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.75 \[ \int \sin (a+b x) \sin ^6(2 a+2 b x) \, dx=-\frac {-\frac {64\,{\cos \left (a+b\,x\right )}^{13}}{13}+\frac {192\,{\cos \left (a+b\,x\right )}^{11}}{11}-\frac {64\,{\cos \left (a+b\,x\right )}^9}{3}+\frac {64\,{\cos \left (a+b\,x\right )}^7}{7}}{b} \]

[In]

int(sin(a + b*x)*sin(2*a + 2*b*x)^6,x)

[Out]

-((64*cos(a + b*x)^7)/7 - (64*cos(a + b*x)^9)/3 + (192*cos(a + b*x)^11)/11 - (64*cos(a + b*x)^13)/13)/b