3.3 \(\int \sin (a+b x) \sin ^5(2 a+2 b x) \, dx\)

Optimal. Leaf size=46 \[ \frac {32 \sin ^{11}(a+b x)}{11 b}-\frac {64 \sin ^9(a+b x)}{9 b}+\frac {32 \sin ^7(a+b x)}{7 b} \]

[Out]

32/7*sin(b*x+a)^7/b-64/9*sin(b*x+a)^9/b+32/11*sin(b*x+a)^11/b

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Rubi [A]  time = 0.05, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4288, 2564, 270} \[ \frac {32 \sin ^{11}(a+b x)}{11 b}-\frac {64 \sin ^9(a+b x)}{9 b}+\frac {32 \sin ^7(a+b x)}{7 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]*Sin[2*a + 2*b*x]^5,x]

[Out]

(32*Sin[a + b*x]^7)/(7*b) - (64*Sin[a + b*x]^9)/(9*b) + (32*Sin[a + b*x]^11)/(11*b)

Rule 270

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 2564

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 4288

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*} \int \sin (a+b x) \sin ^5(2 a+2 b x) \, dx &=32 \int \cos ^5(a+b x) \sin ^6(a+b x) \, dx\\ &=\frac {32 \operatorname {Subst}\left (\int x^6 \left (1-x^2\right )^2 \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac {32 \operatorname {Subst}\left (\int \left (x^6-2 x^8+x^{10}\right ) \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac {32 \sin ^7(a+b x)}{7 b}-\frac {64 \sin ^9(a+b x)}{9 b}+\frac {32 \sin ^{11}(a+b x)}{11 b}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 37, normalized size = 0.80 \[ \frac {4 \sin ^7(a+b x) (364 \cos (2 (a+b x))+63 \cos (4 (a+b x))+365)}{693 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]*Sin[2*a + 2*b*x]^5,x]

[Out]

(4*(365 + 364*Cos[2*(a + b*x)] + 63*Cos[4*(a + b*x)])*Sin[a + b*x]^7)/(693*b)

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fricas [A]  time = 0.42, size = 63, normalized size = 1.37 \[ -\frac {32 \, {\left (63 \, \cos \left (b x + a\right )^{10} - 161 \, \cos \left (b x + a\right )^{8} + 113 \, \cos \left (b x + a\right )^{6} - 3 \, \cos \left (b x + a\right )^{4} - 4 \, \cos \left (b x + a\right )^{2} - 8\right )} \sin \left (b x + a\right )}{693 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)*sin(2*b*x+2*a)^5,x, algorithm="fricas")

[Out]

-32/693*(63*cos(b*x + a)^10 - 161*cos(b*x + a)^8 + 113*cos(b*x + a)^6 - 3*cos(b*x + a)^4 - 4*cos(b*x + a)^2 -
8)*sin(b*x + a)/b

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giac [B]  time = 0.71, size = 82, normalized size = 1.78 \[ -\frac {\sin \left (11 \, b x + 11 \, a\right )}{352 \, b} + \frac {\sin \left (9 \, b x + 9 \, a\right )}{288 \, b} + \frac {5 \, \sin \left (7 \, b x + 7 \, a\right )}{224 \, b} - \frac {\sin \left (5 \, b x + 5 \, a\right )}{32 \, b} - \frac {5 \, \sin \left (3 \, b x + 3 \, a\right )}{48 \, b} + \frac {5 \, \sin \left (b x + a\right )}{16 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)*sin(2*b*x+2*a)^5,x, algorithm="giac")

[Out]

-1/352*sin(11*b*x + 11*a)/b + 1/288*sin(9*b*x + 9*a)/b + 5/224*sin(7*b*x + 7*a)/b - 1/32*sin(5*b*x + 5*a)/b -
5/48*sin(3*b*x + 3*a)/b + 5/16*sin(b*x + a)/b

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maple [B]  time = 1.29, size = 83, normalized size = 1.80 \[ \frac {5 \sin \left (b x +a \right )}{16 b}-\frac {5 \sin \left (3 b x +3 a \right )}{48 b}-\frac {\sin \left (5 b x +5 a \right )}{32 b}+\frac {5 \sin \left (7 b x +7 a \right )}{224 b}+\frac {\sin \left (9 b x +9 a \right )}{288 b}-\frac {\sin \left (11 b x +11 a \right )}{352 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)*sin(2*b*x+2*a)^5,x)

[Out]

5/16*sin(b*x+a)/b-5/48*sin(3*b*x+3*a)/b-1/32/b*sin(5*b*x+5*a)+5/224/b*sin(7*b*x+7*a)+1/288/b*sin(9*b*x+9*a)-1/
352/b*sin(11*b*x+11*a)

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maxima [A]  time = 0.34, size = 69, normalized size = 1.50 \[ -\frac {63 \, \sin \left (11 \, b x + 11 \, a\right ) - 77 \, \sin \left (9 \, b x + 9 \, a\right ) - 495 \, \sin \left (7 \, b x + 7 \, a\right ) + 693 \, \sin \left (5 \, b x + 5 \, a\right ) + 2310 \, \sin \left (3 \, b x + 3 \, a\right ) - 6930 \, \sin \left (b x + a\right )}{22176 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)*sin(2*b*x+2*a)^5,x, algorithm="maxima")

[Out]

-1/22176*(63*sin(11*b*x + 11*a) - 77*sin(9*b*x + 9*a) - 495*sin(7*b*x + 7*a) + 693*sin(5*b*x + 5*a) + 2310*sin
(3*b*x + 3*a) - 6930*sin(b*x + a))/b

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mupad [B]  time = 0.09, size = 36, normalized size = 0.78 \[ \frac {32\,\left (63\,{\sin \left (a+b\,x\right )}^{11}-154\,{\sin \left (a+b\,x\right )}^9+99\,{\sin \left (a+b\,x\right )}^7\right )}{693\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)*sin(2*a + 2*b*x)^5,x)

[Out]

(32*(99*sin(a + b*x)^7 - 154*sin(a + b*x)^9 + 63*sin(a + b*x)^11))/(693*b)

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sympy [A]  time = 124.08, size = 197, normalized size = 4.28 \[ \begin {cases} - \frac {422 \sin {\left (a + b x \right )} \sin ^{4}{\left (2 a + 2 b x \right )} \cos {\left (2 a + 2 b x \right )}}{693 b} - \frac {608 \sin {\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{693 b} - \frac {256 \sin {\left (a + b x \right )} \cos ^{5}{\left (2 a + 2 b x \right )}}{693 b} + \frac {151 \sin ^{5}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )}}{693 b} + \frac {272 \sin ^{3}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{693 b} + \frac {128 \sin {\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{693 b} & \text {for}\: b \neq 0 \\x \sin {\relax (a )} \sin ^{5}{\left (2 a \right )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)*sin(2*b*x+2*a)**5,x)

[Out]

Piecewise((-422*sin(a + b*x)*sin(2*a + 2*b*x)**4*cos(2*a + 2*b*x)/(693*b) - 608*sin(a + b*x)*sin(2*a + 2*b*x)*
*2*cos(2*a + 2*b*x)**3/(693*b) - 256*sin(a + b*x)*cos(2*a + 2*b*x)**5/(693*b) + 151*sin(2*a + 2*b*x)**5*cos(a
+ b*x)/(693*b) + 272*sin(2*a + 2*b*x)**3*cos(a + b*x)*cos(2*a + 2*b*x)**2/(693*b) + 128*sin(2*a + 2*b*x)*cos(a
+ b*x)*cos(2*a + 2*b*x)**4/(693*b), Ne(b, 0)), (x*sin(a)*sin(2*a)**5, True))

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