\(\int \csc ^2(a+b x) \sin ^3(2 a+2 b x) \, dx\) [50]

Optimal result

Integrand size = 20, antiderivative size = 13 \[ \int \csc ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=-\frac {2 \cos ^4(a+b x)}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4373, 2645, 30} \[ \int \csc ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=-\frac {2 \cos ^4(a+b x)}{b} \]

[In]

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

[Out]

(-2*Cos[a + b*x]^4)/b

Rule 30

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

Rule 2645

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 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}& = 8 \int \cos ^3(a+b x) \sin (a+b x) \, dx \\ & = -\frac {8 \text {Subst}\left (\int x^3 \, dx,x,\cos (a+b x)\right )}{b} \\ & = -\frac {2 \cos ^4(a+b x)}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \csc ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=-\frac {2 \cos ^4(a+b x)}{b} \]

[In]

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

[Out]

(-2*Cos[a + b*x]^4)/b

Maple [A] (verified)

Time = 1.53 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.08

method	result	size
default	\(-\frac {2 \cos \left (x b +a \right )^{4}}{b}\)	\(14\)
risch	\(-\frac {\cos \left (4 x b +4 a \right )}{4 b}-\frac {\cos \left (2 x b +2 a \right )}{b}\)	\(30\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \csc ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=-\frac {2 \, \cos \left (b x + a\right )^{4}}{b} \]

[In]

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

[Out]

-2*cos(b*x + a)^4/b

Sympy [F(-1)]

Timed out. \[ \int \csc ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 2.00 \[ \int \csc ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=-\frac {\cos \left (4 \, b x + 4 \, a\right ) + 4 \, \cos \left (2 \, b x + 2 \, a\right )}{4 \, b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \csc ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=-\frac {2 \, \cos \left (b x + a\right )^{4}}{b} \]

[In]

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

[Out]

-2*cos(b*x + a)^4/b

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \csc ^2(a+b x) \sin ^3(2 a+2 b x) \, dx=-\frac {2\,{\cos \left (a+b\,x\right )}^4}{b} \]

[In]

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

[Out]

-(2*cos(a + b*x)^4)/b