3.4 \(\int \sin (a+b x) \sin ^4(2 a+2 b x) \, dx\)

Optimal. Leaf size=46 \[ -\frac {16 \cos ^9(a+b x)}{9 b}+\frac {32 \cos ^7(a+b x)}{7 b}-\frac {16 \cos ^5(a+b x)}{5 b} \]

[Out]

-16/5*cos(b*x+a)^5/b+32/7*cos(b*x+a)^7/b-16/9*cos(b*x+a)^9/b

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Rubi [A]  time = 0.06, 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, 2565, 270} \[ -\frac {16 \cos ^9(a+b x)}{9 b}+\frac {32 \cos ^7(a+b x)}{7 b}-\frac {16 \cos ^5(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]*Sin[2*a + 2*b*x]^4,x]

[Out]

(-16*Cos[a + b*x]^5)/(5*b) + (32*Cos[a + b*x]^7)/(7*b) - (16*Cos[a + b*x]^9)/(9*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 2565

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 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 ^4(2 a+2 b x) \, dx &=16 \int \cos ^4(a+b x) \sin ^5(a+b x) \, dx\\ &=-\frac {16 \operatorname {Subst}\left (\int x^4 \left (1-x^2\right )^2 \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac {16 \operatorname {Subst}\left (\int \left (x^4-2 x^6+x^8\right ) \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac {16 \cos ^5(a+b x)}{5 b}+\frac {32 \cos ^7(a+b x)}{7 b}-\frac {16 \cos ^9(a+b x)}{9 b}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 37, normalized size = 0.80 \[ \frac {2 \cos ^5(a+b x) (220 \cos (2 (a+b x))-35 \cos (4 (a+b x))-249)}{315 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]*Sin[2*a + 2*b*x]^4,x]

[Out]

(2*Cos[a + b*x]^5*(-249 + 220*Cos[2*(a + b*x)] - 35*Cos[4*(a + b*x)]))/(315*b)

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fricas [A]  time = 0.41, size = 36, normalized size = 0.78 \[ -\frac {16 \, {\left (35 \, \cos \left (b x + a\right )^{9} - 90 \, \cos \left (b x + a\right )^{7} + 63 \, \cos \left (b x + a\right )^{5}\right )}}{315 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)*sin(2*b*x+2*a)^4,x, algorithm="fricas")

[Out]

-16/315*(35*cos(b*x + a)^9 - 90*cos(b*x + a)^7 + 63*cos(b*x + a)^5)/b

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giac [A]  time = 0.45, size = 68, normalized size = 1.48 \[ -\frac {\cos \left (9 \, b x + 9 \, a\right )}{144 \, b} + \frac {\cos \left (7 \, b x + 7 \, a\right )}{112 \, b} + \frac {\cos \left (5 \, b x + 5 \, a\right )}{20 \, b} - \frac {\cos \left (3 \, b x + 3 \, a\right )}{12 \, b} - \frac {3 \, \cos \left (b x + a\right )}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)*sin(2*b*x+2*a)^4,x, algorithm="giac")

[Out]

-1/144*cos(9*b*x + 9*a)/b + 1/112*cos(7*b*x + 7*a)/b + 1/20*cos(5*b*x + 5*a)/b - 1/12*cos(3*b*x + 3*a)/b - 3/8
*cos(b*x + a)/b

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maple [A]  time = 0.24, size = 69, normalized size = 1.50 \[ -\frac {3 \cos \left (b x +a \right )}{8 b}-\frac {\cos \left (3 b x +3 a \right )}{12 b}+\frac {\cos \left (5 b x +5 a \right )}{20 b}+\frac {\cos \left (7 b x +7 a \right )}{112 b}-\frac {\cos \left (9 b x +9 a \right )}{144 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)*sin(2*b*x+2*a)^4,x)

[Out]

-3/8*cos(b*x+a)/b-1/12*cos(3*b*x+3*a)/b+1/20*cos(5*b*x+5*a)/b+1/112*cos(7*b*x+7*a)/b-1/144*cos(9*b*x+9*a)/b

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maxima [A]  time = 0.34, size = 58, normalized size = 1.26 \[ -\frac {35 \, \cos \left (9 \, b x + 9 \, a\right ) - 45 \, \cos \left (7 \, b x + 7 \, a\right ) - 252 \, \cos \left (5 \, b x + 5 \, a\right ) + 420 \, \cos \left (3 \, b x + 3 \, a\right ) + 1890 \, \cos \left (b x + a\right )}{5040 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)*sin(2*b*x+2*a)^4,x, algorithm="maxima")

[Out]

-1/5040*(35*cos(9*b*x + 9*a) - 45*cos(7*b*x + 7*a) - 252*cos(5*b*x + 5*a) + 420*cos(3*b*x + 3*a) + 1890*cos(b*
x + a))/b

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mupad [B]  time = 0.14, size = 36, normalized size = 0.78 \[ -\frac {16\,\left (35\,{\cos \left (a+b\,x\right )}^9-90\,{\cos \left (a+b\,x\right )}^7+63\,{\cos \left (a+b\,x\right )}^5\right )}{315\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)*sin(2*a + 2*b*x)^4,x)

[Out]

-(16*(63*cos(a + b*x)^5 - 90*cos(a + b*x)^7 + 35*cos(a + b*x)^9))/(315*b)

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sympy [A]  time = 40.28, size = 163, normalized size = 3.54 \[ \begin {cases} - \frac {104 \sin {\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )} \cos {\left (2 a + 2 b x \right )}}{315 b} - \frac {64 \sin {\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{315 b} - \frac {107 \sin ^{4}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )}}{315 b} - \frac {16 \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{21 b} - \frac {128 \cos {\left (a + b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{315 b} & \text {for}\: b \neq 0 \\x \sin {\relax (a )} \sin ^{4}{\left (2 a \right )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)*sin(2*b*x+2*a)**4,x)

[Out]

Piecewise((-104*sin(a + b*x)*sin(2*a + 2*b*x)**3*cos(2*a + 2*b*x)/(315*b) - 64*sin(a + b*x)*sin(2*a + 2*b*x)*c
os(2*a + 2*b*x)**3/(315*b) - 107*sin(2*a + 2*b*x)**4*cos(a + b*x)/(315*b) - 16*sin(2*a + 2*b*x)**2*cos(a + b*x
)*cos(2*a + 2*b*x)**2/(21*b) - 128*cos(a + b*x)*cos(2*a + 2*b*x)**4/(315*b), Ne(b, 0)), (x*sin(a)*sin(2*a)**4,
True))

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