\(\int \csc ^3(a+b x) \sin ^4(2 a+2 b x) \, dx\) [65]

Optimal result

Integrand size = 20, antiderivative size = 15 \[ \int \csc ^3(a+b x) \sin ^4(2 a+2 b x) \, dx=-\frac {16 \cos ^5(a+b x)}{5 b} \]

[Out]

-16/5*cos(b*x+a)^5/b

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4373, 2645, 30} \[ \int \csc ^3(a+b x) \sin ^4(2 a+2 b x) \, dx=-\frac {16 \cos ^5(a+b x)}{5 b} \]

[In]

Int[Csc[a + b*x]^3*Sin[2*a + 2*b*x]^4,x]

[Out]

(-16*Cos[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 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}& = 16 \int \cos ^4(a+b x) \sin (a+b x) \, dx \\ & = -\frac {16 \text {Subst}\left (\int x^4 \, dx,x,\cos (a+b x)\right )}{b} \\ & = -\frac {16 \cos ^5(a+b x)}{5 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \csc ^3(a+b x) \sin ^4(2 a+2 b x) \, dx=-\frac {16 \cos ^5(a+b x)}{5 b} \]

[In]

Integrate[Csc[a + b*x]^3*Sin[2*a + 2*b*x]^4,x]

[Out]

(-16*Cos[a + b*x]^5)/(5*b)

Maple [A] (verified)

Time = 5.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93

method	result	size
default	\(-\frac {16 \cos \left (x b +a \right )^{5}}{5 b}\)	\(14\)
risch	\(-\frac {2 \cos \left (x b +a \right )}{b}-\frac {\cos \left (5 x b +5 a \right )}{5 b}-\frac {\cos \left (3 x b +3 a \right )}{b}\)	\(41\)

[In]

int(csc(b*x+a)^3*sin(2*b*x+2*a)^4,x,method=_RETURNVERBOSE)

[Out]

-16/5*cos(b*x+a)^5/b

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \csc ^3(a+b x) \sin ^4(2 a+2 b x) \, dx=-\frac {16 \, \cos \left (b x + a\right )^{5}}{5 \, b} \]

[In]

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^4,x, algorithm="fricas")

[Out]

-16/5*cos(b*x + a)^5/b

Sympy [F(-1)]

Timed out. \[ \int \csc ^3(a+b x) \sin ^4(2 a+2 b x) \, dx=\text {Timed out} \]

[In]

integrate(csc(b*x+a)**3*sin(2*b*x+2*a)**4,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

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

Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 2.27 \[ \int \csc ^3(a+b x) \sin ^4(2 a+2 b x) \, dx=-\frac {\cos \left (5 \, b x + 5 \, a\right ) + 5 \, \cos \left (3 \, b x + 3 \, a\right ) + 10 \, \cos \left (b x + a\right )}{5 \, b} \]

[In]

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^4,x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (13) = 26\).

Time = 0.30 (sec) , antiderivative size = 74, normalized size of antiderivative = 4.93 \[ \int \csc ^3(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {32 \, {\left (\frac {10 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {5 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + 1\right )}}{5 \, b {\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1\right )}^{5}} \]

[In]

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^4,x, algorithm="giac")

[Out]

32/5*(10*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 5*(cos(b*x + a) - 1)^4/(cos(b*x + a) + 1)^4 + 1)/(b*((cos
(b*x + a) - 1)/(cos(b*x + a) + 1) - 1)^5)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \csc ^3(a+b x) \sin ^4(2 a+2 b x) \, dx=-\frac {16\,{\cos \left (a+b\,x\right )}^5}{5\,b} \]

[In]

int(sin(2*a + 2*b*x)^4/sin(a + b*x)^3,x)

[Out]

-(16*cos(a + b*x)^5)/(5*b)