3.15 \(\int \sin ^2(a+b x) \sin ^3(2 a+2 b x) \, dx\)

Optimal. Leaf size=29 \[ \frac {4 \sin ^6(a+b x)}{3 b}-\frac {\sin ^8(a+b x)}{b} \]

[Out]

4/3*sin(b*x+a)^6/b-sin(b*x+a)^8/b

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4288, 2564, 14} \[ \frac {4 \sin ^6(a+b x)}{3 b}-\frac {\sin ^8(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sin[a + b*x]^6)/(3*b) - Sin[a + b*x]^8/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 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 ^2(a+b x) \sin ^3(2 a+2 b x) \, dx &=8 \int \cos ^3(a+b x) \sin ^5(a+b x) \, dx\\ &=\frac {8 \operatorname {Subst}\left (\int x^5 \left (1-x^2\right ) \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac {8 \operatorname {Subst}\left (\int \left (x^5-x^7\right ) \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac {4 \sin ^6(a+b x)}{3 b}-\frac {\sin ^8(a+b x)}{b}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 48, normalized size = 1.66 \[ \frac {-72 \cos (2 (a+b x))+12 \cos (4 (a+b x))+8 \cos (6 (a+b x))-3 \cos (8 (a+b x))}{384 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-72*Cos[2*(a + b*x)] + 12*Cos[4*(a + b*x)] + 8*Cos[6*(a + b*x)] - 3*Cos[8*(a + b*x)])/(384*b)

________________________________________________________________________________________

fricas [A]  time = 0.46, size = 36, normalized size = 1.24 \[ -\frac {3 \, \cos \left (b x + a\right )^{8} - 8 \, \cos \left (b x + a\right )^{6} + 6 \, \cos \left (b x + a\right )^{4}}{3 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*cos(b*x + a)^8 - 8*cos(b*x + a)^6 + 6*cos(b*x + a)^4)/b

________________________________________________________________________________________

giac [B]  time = 0.50, size = 57, normalized size = 1.97 \[ -\frac {\cos \left (8 \, b x + 8 \, a\right )}{128 \, b} + \frac {\cos \left (6 \, b x + 6 \, a\right )}{48 \, b} + \frac {\cos \left (4 \, b x + 4 \, a\right )}{32 \, b} - \frac {3 \, \cos \left (2 \, b x + 2 \, a\right )}{16 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/128*cos(8*b*x + 8*a)/b + 1/48*cos(6*b*x + 6*a)/b + 1/32*cos(4*b*x + 4*a)/b - 3/16*cos(2*b*x + 2*a)/b

________________________________________________________________________________________

maple [B]  time = 0.29, size = 58, normalized size = 2.00 \[ -\frac {3 \cos \left (2 b x +2 a \right )}{16 b}+\frac {\cos \left (4 b x +4 a \right )}{32 b}+\frac {\cos \left (6 b x +6 a \right )}{48 b}-\frac {\cos \left (8 b x +8 a \right )}{128 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/16*cos(2*b*x+2*a)/b+1/32*cos(4*b*x+4*a)/b+1/48*cos(6*b*x+6*a)/b-1/128*cos(8*b*x+8*a)/b

________________________________________________________________________________________

maxima [A]  time = 0.33, size = 50, normalized size = 1.72 \[ -\frac {3 \, \cos \left (8 \, b x + 8 \, a\right ) - 8 \, \cos \left (6 \, b x + 6 \, a\right ) - 12 \, \cos \left (4 \, b x + 4 \, a\right ) + 72 \, \cos \left (2 \, b x + 2 \, a\right )}{384 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384*(3*cos(8*b*x + 8*a) - 8*cos(6*b*x + 6*a) - 12*cos(4*b*x + 4*a) + 72*cos(2*b*x + 2*a))/b

________________________________________________________________________________________

mupad [B]  time = 0.11, size = 33, normalized size = 1.14 \[ -\frac {{\cos \left (a+b\,x\right )}^4\,\left ({\cos \left (a+b\,x\right )}^4-\frac {8\,{\cos \left (a+b\,x\right )}^2}{3}+2\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(cos(a + b*x)^4*(cos(a + b*x)^4 - (8*cos(a + b*x)^2)/3 + 2))/b

________________________________________________________________________________________

sympy [A]  time = 40.10, size = 362, normalized size = 12.48 \[ \begin {cases} \frac {3 x \sin ^{2}{\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )}}{16} + \frac {3 x \sin ^{2}{\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{16} + \frac {3 x \sin {\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{8} + \frac {3 x \sin {\left (a + b x \right )} \cos {\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{8} - \frac {3 x \sin ^{3}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )}}{16} - \frac {3 x \sin {\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{16} + \frac {17 \sin ^{2}{\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{96 b} - \frac {13 \sin {\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )}}{16 b} - \frac {7 \sin {\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{8 b} - \frac {\sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{2 b} - \frac {49 \cos ^{2}{\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{96 b} & \text {for}\: b \neq 0 \\x \sin ^{2}{\relax (a )} \sin ^{3}{\left (2 a \right )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((3*x*sin(a + b*x)**2*sin(2*a + 2*b*x)**3/16 + 3*x*sin(a + b*x)**2*sin(2*a + 2*b*x)*cos(2*a + 2*b*x)*
*2/16 + 3*x*sin(a + b*x)*sin(2*a + 2*b*x)**2*cos(a + b*x)*cos(2*a + 2*b*x)/8 + 3*x*sin(a + b*x)*cos(a + b*x)*c
os(2*a + 2*b*x)**3/8 - 3*x*sin(2*a + 2*b*x)**3*cos(a + b*x)**2/16 - 3*x*sin(2*a + 2*b*x)*cos(a + b*x)**2*cos(2
*a + 2*b*x)**2/16 + 17*sin(a + b*x)**2*cos(2*a + 2*b*x)**3/(96*b) - 13*sin(a + b*x)*sin(2*a + 2*b*x)**3*cos(a
+ b*x)/(16*b) - 7*sin(a + b*x)*sin(2*a + 2*b*x)*cos(a + b*x)*cos(2*a + 2*b*x)**2/(8*b) - sin(2*a + 2*b*x)**2*c
os(a + b*x)**2*cos(2*a + 2*b*x)/(2*b) - 49*cos(a + b*x)**2*cos(2*a + 2*b*x)**3/(96*b), Ne(b, 0)), (x*sin(a)**2
*sin(2*a)**3, True))

________________________________________________________________________________________