\(\int \csc ^2(a+b x) \sin ^2(2 a+2 b x) \, dx\) [51]

Optimal result

Integrand size = 20, antiderivative size = 21 \[ \int \csc ^2(a+b x) \sin ^2(2 a+2 b x) \, dx=2 x+\frac {2 \cos (a+b x) \sin (a+b x)}{b} \]

[Out]

2*x+2*cos(b*x+a)*sin(b*x+a)/b

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 21, 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, 2715, 8} \[ \int \csc ^2(a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {2 \sin (a+b x) \cos (a+b x)}{b}+2 x \]

[In]

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

[Out]

2*x + (2*Cos[a + b*x]*Sin[a + b*x])/b

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*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}& = 4 \int \cos ^2(a+b x) \, dx \\ & = \frac {2 \cos (a+b x) \sin (a+b x)}{b}+2 \int 1 \, dx \\ & = 2 x+\frac {2 \cos (a+b x) \sin (a+b x)}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \csc ^2(a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {2 (a+b x)+\sin (2 (a+b x))}{b} \]

[In]

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

[Out]

(2*(a + b*x) + Sin[2*(a + b*x)])/b

Maple [A] (verified)

Time = 0.77 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86

method	result	size
risch	\(2 x +\frac {\sin \left (2 x b +2 a \right )}{b}\)	\(18\)
default	\(\frac {2 \cos \left (x b +a \right ) \sin \left (x b +a \right )+2 x b +2 a}{b}\)	\(28\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \csc ^2(a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {2 \, {\left (b x + \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )}}{b} \]

[In]

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

[Out]

2*(b*x + cos(b*x + a)*sin(b*x + a))/b

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \csc ^2(a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {2 \, b x + \sin \left (2 \, b x + 2 \, a\right )}{b} \]

[In]

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

[Out]

(2*b*x + sin(2*b*x + 2*a))/b

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \csc ^2(a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {2 \, {\left (b x + a + \frac {\tan \left (b x + a\right )}{\tan \left (b x + a\right )^{2} + 1}\right )}}{b} \]

[In]

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

[Out]

2*(b*x + a + tan(b*x + a)/(tan(b*x + a)^2 + 1))/b

Mupad [B] (verification not implemented)

Time = 19.76 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \csc ^2(a+b x) \sin ^2(2 a+2 b x) \, dx=2\,x+\frac {\sin \left (2\,a+2\,b\,x\right )}{b} \]

[In]

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

[Out]

2*x + sin(2*a + 2*b*x)/b