\(\int \csc (a+b x) \sin ^3(2 a+2 b x) \, dx\) [38]

Optimal result

Integrand size = 18, antiderivative size = 31 \[ \int \csc (a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {8 \sin ^3(a+b x)}{3 b}-\frac {8 \sin ^5(a+b x)}{5 b} \]

[Out]

8/3*sin(b*x+a)^3/b-8/5*sin(b*x+a)^5/b

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4373, 2644, 14} \[ \int \csc (a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {8 \sin ^3(a+b x)}{3 b}-\frac {8 \sin ^5(a+b x)}{5 b} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \csc (a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {8 \sin ^3(a+b x)}{3 b}-\frac {8 \sin ^5(a+b x)}{5 b} \]

[In]

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

[Out]

(8*Sin[a + b*x]^3)/(3*b) - (8*Sin[a + b*x]^5)/(5*b)

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87

method	result	size
default	\(\frac {-\frac {8 \sin \left (x b +a \right )^{5}}{5}+\frac {8 \sin \left (x b +a \right )^{3}}{3}}{b}\)	\(27\)
risch	\(\frac {\sin \left (x b +a \right )}{b}-\frac {\sin \left (5 x b +5 a \right )}{10 b}-\frac {\sin \left (3 x b +3 a \right )}{6 b}\)	\(40\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \csc (a+b x) \sin ^3(2 a+2 b x) \, dx=-\frac {8 \, {\left (3 \, \cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2} - 2\right )} \sin \left (b x + a\right )}{15 \, b} \]

[In]

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

[Out]

-8/15*(3*cos(b*x + a)^4 - cos(b*x + a)^2 - 2)*sin(b*x + a)/b

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \csc (a+b x) \sin ^3(2 a+2 b x) \, dx=-\frac {3 \, \sin \left (5 \, b x + 5 \, a\right ) + 5 \, \sin \left (3 \, b x + 3 \, a\right ) - 30 \, \sin \left (b x + a\right )}{30 \, b} \]

[In]

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

[Out]

-1/30*(3*sin(5*b*x + 5*a) + 5*sin(3*b*x + 3*a) - 30*sin(b*x + a))/b

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \csc (a+b x) \sin ^3(2 a+2 b x) \, dx=-\frac {8 \, {\left (3 \, \sin \left (b x + a\right )^{5} - 5 \, \sin \left (b x + a\right )^{3}\right )}}{15 \, b} \]

[In]

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

[Out]

-8/15*(3*sin(b*x + a)^5 - 5*sin(b*x + a)^3)/b

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \csc (a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {8\,\left (5\,{\sin \left (a+b\,x\right )}^3-3\,{\sin \left (a+b\,x\right )}^5\right )}{15\,b} \]

[In]

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

[Out]

(8*(5*sin(a + b*x)^3 - 3*sin(a + b*x)^5))/(15*b)