\(\int \sin ^3(a+b x) \sin (2 a+2 b x) \, dx\) [27]

Optimal result

Integrand size = 18, antiderivative size = 15 \[ \int \sin ^3(a+b x) \sin (2 a+2 b x) \, dx=\frac {2 \sin ^5(a+b x)}{5 b} \]

[Out]

2/5*sin(b*x+a)^5/b

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4373, 2644, 30} \[ \int \sin ^3(a+b x) \sin (2 a+2 b x) \, dx=\frac {2 \sin ^5(a+b x)}{5 b} \]

[In]

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

[Out]

(2*Sin[a + b*x]^5)/(5*b)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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}& = 2 \int \cos (a+b x) \sin ^4(a+b x) \, dx \\ & = \frac {2 \text {Subst}\left (\int x^4 \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {2 \sin ^5(a+b x)}{5 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \sin ^3(a+b x) \sin (2 a+2 b x) \, dx=\frac {2 \sin ^5(a+b x)}{5 b} \]

[In]

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

[Out]

(2*Sin[a + b*x]^5)/(5*b)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(40\) vs. \(2(13)=26\).

Time = 0.76 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.73

method	result	size
default	\(\frac {\sin \left (x b +a \right )}{4 b}-\frac {\sin \left (3 x b +3 a \right )}{8 b}+\frac {\sin \left (5 x b +5 a \right )}{40 b}\)	\(41\)
risch	\(\frac {\sin \left (x b +a \right )}{4 b}-\frac {\sin \left (3 x b +3 a \right )}{8 b}+\frac {\sin \left (5 x b +5 a \right )}{40 b}\)	\(41\)
parallelrisch	\(\frac {-\frac {16 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{5}+\frac {8 \tan \left (x b +a \right )}{5}+\frac {16 \tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \tan \left (x b +a \right )^{2}}{5}-8 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2} \tan \left (x b +a \right )}{b \left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right )^{3} \left (1+\tan \left (x b +a \right )^{2}\right )}\)	\(89\)

[In]

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

[Out]

1/4*sin(b*x+a)/b-1/8*sin(3*b*x+3*a)/b+1/40/b*sin(5*b*x+5*a)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (13) = 26\).

Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 2.07 \[ \int \sin ^3(a+b x) \sin (2 a+2 b x) \, dx=\frac {2 \, {\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right )}{5 \, b} \]

[In]

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

[Out]

2/5*(cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1)*sin(b*x + a)/b

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (12) = 24\).

Time = 0.79 (sec) , antiderivative size = 117, normalized size of antiderivative = 7.80 \[ \int \sin ^3(a+b x) \sin (2 a+2 b x) \, dx=\begin {cases} - \frac {2 \sin ^{3}{\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{5 b} - \frac {\sin ^{2}{\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )}}{5 b} - \frac {4 \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{5 b} + \frac {2 \sin {\left (2 a + 2 b x \right )} \cos ^{3}{\left (a + b x \right )}}{5 b} & \text {for}\: b \neq 0 \\x \sin ^{3}{\left (a \right )} \sin {\left (2 a \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-2*sin(a + b*x)**3*cos(2*a + 2*b*x)/(5*b) - sin(a + b*x)**2*sin(2*a + 2*b*x)*cos(a + b*x)/(5*b) - 4
*sin(a + b*x)*cos(a + b*x)**2*cos(2*a + 2*b*x)/(5*b) + 2*sin(2*a + 2*b*x)*cos(a + b*x)**3/(5*b), Ne(b, 0)), (x
*sin(a)**3*sin(2*a), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (13) = 26\).

Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 2.27 \[ \int \sin ^3(a+b x) \sin (2 a+2 b x) \, dx=\frac {\sin \left (5 \, b x + 5 \, a\right ) - 5 \, \sin \left (3 \, b x + 3 \, a\right ) + 10 \, \sin \left (b x + a\right )}{40 \, b} \]

[In]

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

[Out]

1/40*(sin(5*b*x + 5*a) - 5*sin(3*b*x + 3*a) + 10*sin(b*x + a))/b

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \sin ^3(a+b x) \sin (2 a+2 b x) \, dx=\frac {2 \, \sin \left (b x + a\right )^{5}}{5 \, b} \]

[In]

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

[Out]

2/5*sin(b*x + a)^5/b

Mupad [B] (verification not implemented)

Time = 19.23 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \sin ^3(a+b x) \sin (2 a+2 b x) \, dx=\frac {2\,{\sin \left (a+b\,x\right )}^5}{5\,b} \]

[In]

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

[Out]

(2*sin(a + b*x)^5)/(5*b)