\(\int \sin ^3(a+b x) \sin ^3(2 a+2 b x) \, dx\) [25]

Optimal result

Integrand size = 20, antiderivative size = 31 \[ \int \sin ^3(a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {8 \sin ^7(a+b x)}{7 b}-\frac {8 \sin ^9(a+b x)}{9 b} \]

[Out]

8/7*sin(b*x+a)^7/b-8/9*sin(b*x+a)^9/b

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4373, 2644, 14} \[ \int \sin ^3(a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {8 \sin ^7(a+b x)}{7 b}-\frac {8 \sin ^9(a+b x)}{9 b} \]

[In]

Int[Sin[a + b*x]^3*Sin[2*a + 2*b*x]^3,x]

[Out]

(8*Sin[a + b*x]^7)/(7*b) - (8*Sin[a + b*x]^9)/(9*b)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[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] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, 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}& = 8 \int \cos ^3(a+b x) \sin ^6(a+b x) \, dx \\ & = \frac {8 \text {Subst}\left (\int x^6 \left (1-x^2\right ) \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {8 \text {Subst}\left (\int \left (x^6-x^8\right ) \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {8 \sin ^7(a+b x)}{7 b}-\frac {8 \sin ^9(a+b x)}{9 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \sin ^3(a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {4 (11+7 \cos (2 (a+b x))) \sin ^7(a+b x)}{63 b} \]

[In]

Integrate[Sin[a + b*x]^3*Sin[2*a + 2*b*x]^3,x]

[Out]

(4*(11 + 7*Cos[2*(a + b*x)])*Sin[a + b*x]^7)/(63*b)

Maple [A] (verified)

Time = 2.89 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.77

method	result	size
default	\(\frac {3 \sin \left (x b +a \right )}{16 b}-\frac {\sin \left (3 x b +3 a \right )}{12 b}+\frac {3 \sin \left (7 x b +7 a \right )}{224 b}-\frac {\sin \left (9 x b +9 a \right )}{288 b}\)	\(55\)
risch	\(\frac {3 \sin \left (x b +a \right )}{16 b}-\frac {\sin \left (3 x b +3 a \right )}{12 b}+\frac {3 \sin \left (7 x b +7 a \right )}{224 b}-\frac {\sin \left (9 x b +9 a \right )}{288 b}\)	\(55\)
parallelrisch	\(\frac {16 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{6} \tan \left (x b +a \right )^{3}+\left (-96 \tan \left (x b +a \right )^{4}+96 \tan \left (x b +a \right )^{2}\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{5}+\left (192 \tan \left (x b +a \right )^{5}-720 \tan \left (x b +a \right )^{3}+192 \tan \left (x b +a \right )\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}+\left (-128 \tan \left (x b +a \right )^{6}+1920 \tan \left (x b +a \right )^{4}-1920 \tan \left (x b +a \right )^{2}+128\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}+\left (-2112 \tan \left (x b +a \right )^{5}-3120 \tan \left (x b +a \right )^{3}-2112 \tan \left (x b +a \right )\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}+\left (768 \tan \left (x b +a \right )^{6}+672 \tan \left (x b +a \right )^{4}-672 \tan \left (x b +a \right )^{2}-768\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+384 \tan \left (x b +a \right )^{5}+752 \tan \left (x b +a \right )^{3}+384 \tan \left (x b +a \right )}{315 b \left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right )^{3} \left (1+\tan \left (x b +a \right )^{2}\right )^{3}}\)	\(284\)

[In]

int(sin(b*x+a)^3*sin(2*b*x+2*a)^3,x,method=_RETURNVERBOSE)

[Out]

3/16*sin(b*x+a)/b-1/12*sin(3*b*x+3*a)/b+3/224/b*sin(7*b*x+7*a)-1/288/b*sin(9*b*x+9*a)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.71 \[ \int \sin ^3(a+b x) \sin ^3(2 a+2 b x) \, dx=-\frac {8 \, {\left (7 \, \cos \left (b x + a\right )^{8} - 19 \, \cos \left (b x + a\right )^{6} + 15 \, \cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2} - 2\right )} \sin \left (b x + a\right )}{63 \, b} \]

[In]

integrate(sin(b*x+a)^3*sin(2*b*x+2*a)^3,x, algorithm="fricas")

[Out]

-8/63*(7*cos(b*x + a)^8 - 19*cos(b*x + a)^6 + 15*cos(b*x + a)^4 - cos(b*x + a)^2 - 2)*sin(b*x + a)/b

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (26) = 52\).

Time = 4.83 (sec) , antiderivative size = 284, normalized size of antiderivative = 9.16 \[ \int \sin ^3(a+b x) \sin ^3(2 a+2 b x) \, dx=\begin {cases} - \frac {46 \sin ^{3}{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (2 a + 2 b x \right )}}{105 b} - \frac {16 \sin ^{3}{\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{63 b} - \frac {13 \sin ^{2}{\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )}}{105 b} - \frac {8 \sin ^{2}{\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{35 b} - \frac {4 \sin {\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{7 b} - \frac {64 \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{105 b} + \frac {94 \sin ^{3}{\left (2 a + 2 b x \right )} \cos ^{3}{\left (a + b x \right )}}{315 b} + \frac {32 \sin {\left (2 a + 2 b x \right )} \cos ^{3}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{105 b} & \text {for}\: b \neq 0 \\x \sin ^{3}{\left (a \right )} \sin ^{3}{\left (2 a \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(sin(b*x+a)**3*sin(2*b*x+2*a)**3,x)

[Out]

Piecewise((-46*sin(a + b*x)**3*sin(2*a + 2*b*x)**2*cos(2*a + 2*b*x)/(105*b) - 16*sin(a + b*x)**3*cos(2*a + 2*b
*x)**3/(63*b) - 13*sin(a + b*x)**2*sin(2*a + 2*b*x)**3*cos(a + b*x)/(105*b) - 8*sin(a + b*x)**2*sin(2*a + 2*b*
x)*cos(a + b*x)*cos(2*a + 2*b*x)**2/(35*b) - 4*sin(a + b*x)*sin(2*a + 2*b*x)**2*cos(a + b*x)**2*cos(2*a + 2*b*
x)/(7*b) - 64*sin(a + b*x)*cos(a + b*x)**2*cos(2*a + 2*b*x)**3/(105*b) + 94*sin(2*a + 2*b*x)**3*cos(a + b*x)**
3/(315*b) + 32*sin(2*a + 2*b*x)*cos(a + b*x)**3*cos(2*a + 2*b*x)**2/(105*b), Ne(b, 0)), (x*sin(a)**3*sin(2*a)*
*3, True))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52 \[ \int \sin ^3(a+b x) \sin ^3(2 a+2 b x) \, dx=-\frac {7 \, \sin \left (9 \, b x + 9 \, a\right ) - 27 \, \sin \left (7 \, b x + 7 \, a\right ) + 168 \, \sin \left (3 \, b x + 3 \, a\right ) - 378 \, \sin \left (b x + a\right )}{2016 \, b} \]

[In]

integrate(sin(b*x+a)^3*sin(2*b*x+2*a)^3,x, algorithm="maxima")

[Out]

-1/2016*(7*sin(9*b*x + 9*a) - 27*sin(7*b*x + 7*a) + 168*sin(3*b*x + 3*a) - 378*sin(b*x + a))/b

Giac [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \sin ^3(a+b x) \sin ^3(2 a+2 b x) \, dx=-\frac {8 \, {\left (7 \, \sin \left (b x + a\right )^{9} - 9 \, \sin \left (b x + a\right )^{7}\right )}}{63 \, b} \]

[In]

integrate(sin(b*x+a)^3*sin(2*b*x+2*a)^3,x, algorithm="giac")

[Out]

-8/63*(7*sin(b*x + a)^9 - 9*sin(b*x + a)^7)/b

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \sin ^3(a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {8\,\left (9\,{\sin \left (a+b\,x\right )}^7-7\,{\sin \left (a+b\,x\right )}^9\right )}{63\,b} \]

[In]

int(sin(a + b*x)^3*sin(2*a + 2*b*x)^3,x)

[Out]

(8*(9*sin(a + b*x)^7 - 7*sin(a + b*x)^9))/(63*b)