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

Optimal. Leaf size=44 \[ -\frac {8 \cos ^{12}(a+b x)}{3 b}+\frac {32 \cos ^{10}(a+b x)}{5 b}-\frac {4 \cos ^8(a+b x)}{b} \]

[Out]

-4*cos(b*x+a)^8/b+32/5*cos(b*x+a)^10/b-8/3*cos(b*x+a)^12/b

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Rubi [A]  time = 0.07, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4287, 2565, 266, 43} \[ -\frac {8 \cos ^{12}(a+b x)}{3 b}+\frac {32 \cos ^{10}(a+b x)}{5 b}-\frac {4 \cos ^8(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Cos[a + b*x]^8)/b + (32*Cos[a + b*x]^10)/(5*b) - (8*Cos[a + b*x]^12)/(3*b)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 4287

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

Rubi steps

\begin {align*} \int \cos ^2(a+b x) \sin ^5(2 a+2 b x) \, dx &=32 \int \cos ^7(a+b x) \sin ^5(a+b x) \, dx\\ &=-\frac {32 \operatorname {Subst}\left (\int x^7 \left (1-x^2\right )^2 \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac {16 \operatorname {Subst}\left (\int (1-x)^2 x^3 \, dx,x,\cos ^2(a+b x)\right )}{b}\\ &=-\frac {16 \operatorname {Subst}\left (\int \left (x^3-2 x^4+x^5\right ) \, dx,x,\cos ^2(a+b x)\right )}{b}\\ &=-\frac {4 \cos ^8(a+b x)}{b}+\frac {32 \cos ^{10}(a+b x)}{5 b}-\frac {8 \cos ^{12}(a+b x)}{3 b}\\ \end {align*}

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Mathematica [A]  time = 0.41, size = 68, normalized size = 1.55 \[ -\frac {600 \cos (2 (a+b x))+75 \cos (4 (a+b x))-100 \cos (6 (a+b x))-30 \cos (8 (a+b x))+12 \cos (10 (a+b x))+5 \cos (12 (a+b x))}{3840 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3840*(600*Cos[2*(a + b*x)] + 75*Cos[4*(a + b*x)] - 100*Cos[6*(a + b*x)] - 30*Cos[8*(a + b*x)] + 12*Cos[10*(
a + b*x)] + 5*Cos[12*(a + b*x)])/b

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fricas [A]  time = 0.46, size = 36, normalized size = 0.82 \[ -\frac {4 \, {\left (10 \, \cos \left (b x + a\right )^{12} - 24 \, \cos \left (b x + a\right )^{10} + 15 \, \cos \left (b x + a\right )^{8}\right )}}{15 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/15*(10*cos(b*x + a)^12 - 24*cos(b*x + a)^10 + 15*cos(b*x + a)^8)/b

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giac [B]  time = 0.49, size = 85, normalized size = 1.93 \[ -\frac {\cos \left (12 \, b x + 12 \, a\right )}{768 \, b} - \frac {\cos \left (10 \, b x + 10 \, a\right )}{320 \, b} + \frac {\cos \left (8 \, b x + 8 \, a\right )}{128 \, b} + \frac {5 \, \cos \left (6 \, b x + 6 \, a\right )}{192 \, b} - \frac {5 \, \cos \left (4 \, b x + 4 \, a\right )}{256 \, b} - \frac {5 \, \cos \left (2 \, b x + 2 \, a\right )}{32 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/768*cos(12*b*x + 12*a)/b - 1/320*cos(10*b*x + 10*a)/b + 1/128*cos(8*b*x + 8*a)/b + 5/192*cos(6*b*x + 6*a)/b
- 5/256*cos(4*b*x + 4*a)/b - 5/32*cos(2*b*x + 2*a)/b

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maple [B]  time = 0.34, size = 86, normalized size = 1.95 \[ -\frac {5 \cos \left (2 b x +2 a \right )}{32 b}-\frac {5 \cos \left (4 b x +4 a \right )}{256 b}+\frac {5 \cos \left (6 b x +6 a \right )}{192 b}+\frac {\cos \left (8 b x +8 a \right )}{128 b}-\frac {\cos \left (10 b x +10 a \right )}{320 b}-\frac {\cos \left (12 b x +12 a \right )}{768 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/32*cos(2*b*x+2*a)/b-5/256*cos(4*b*x+4*a)/b+5/192*cos(6*b*x+6*a)/b+1/128*cos(8*b*x+8*a)/b-1/320*cos(10*b*x+1
0*a)/b-1/768*cos(12*b*x+12*a)/b

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maxima [A]  time = 0.34, size = 72, normalized size = 1.64 \[ -\frac {5 \, \cos \left (12 \, b x + 12 \, a\right ) + 12 \, \cos \left (10 \, b x + 10 \, a\right ) - 30 \, \cos \left (8 \, b x + 8 \, a\right ) - 100 \, \cos \left (6 \, b x + 6 \, a\right ) + 75 \, \cos \left (4 \, b x + 4 \, a\right ) + 600 \, \cos \left (2 \, b x + 2 \, a\right )}{3840 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3840*(5*cos(12*b*x + 12*a) + 12*cos(10*b*x + 10*a) - 30*cos(8*b*x + 8*a) - 100*cos(6*b*x + 6*a) + 75*cos(4*
b*x + 4*a) + 600*cos(2*b*x + 2*a))/b

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mupad [B]  time = 0.18, size = 35, normalized size = 0.80 \[ -\frac {4\,{\cos \left (a+b\,x\right )}^8\,\left (10\,{\cos \left (a+b\,x\right )}^4-24\,{\cos \left (a+b\,x\right )}^2+15\right )}{15\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*cos(a + b*x)^8*(10*cos(a + b*x)^4 - 24*cos(a + b*x)^2 + 15))/(15*b)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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