\(\int \csc ^2(2 a+2 b x) \sin ^2(a+b x) \, dx\) [19]

Optimal result

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

[Out]

1/4*tan(b*x+a)/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, 3852, 8} \[ \int \csc ^2(2 a+2 b x) \sin ^2(a+b x) \, dx=\frac {\tan (a+b x)}{4 b} \]

[In]

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

[Out]

Tan[a + b*x]/(4*b)

Rule 8

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

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

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}& = \frac {1}{4} \int \sec ^2(a+b x) \, dx \\ & = -\frac {\text {Subst}(\int 1 \, dx,x,-\tan (a+b x))}{4 b} \\ & = \frac {\tan (a+b x)}{4 b} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

Tan[a + b*x]/(4*b)

Maple [A] (verified)

Time = 1.12 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92

method	result	size
default	\(\frac {\tan \left (x b +a \right )}{4 b}\)	\(12\)
risch	\(\frac {i}{2 b \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )}\)	\(20\)

[In]

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

[Out]

1/4*tan(b*x+a)/b

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.46 \[ \int \csc ^2(2 a+2 b x) \sin ^2(a+b x) \, dx=\frac {\sin \left (b x + a\right )}{4 \, b \cos \left (b x + a\right )} \]

[In]

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

[Out]

1/4*sin(b*x + a)/(b*cos(b*x + a))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (11) = 22\).

Time = 0.21 (sec) , antiderivative size = 53, normalized size of antiderivative = 4.08 \[ \int \csc ^2(2 a+2 b x) \sin ^2(a+b x) \, dx=\frac {\sin \left (2 \, b x + 2 \, a\right )}{2 \, {\left (b \cos \left (2 \, b x + 2 \, a\right )^{2} + b \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, b \cos \left (2 \, b x + 2 \, a\right ) + b\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \csc ^2(2 a+2 b x) \sin ^2(a+b x) \, dx=\frac {\tan \left (b x + a\right )}{4 \, b} \]

[In]

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

[Out]

1/4*tan(b*x + a)/b

Mupad [B] (verification not implemented)

Time = 20.46 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \csc ^2(2 a+2 b x) \sin ^2(a+b x) \, dx=\frac {\mathrm {tan}\left (a+b\,x\right )}{4\,b} \]

[In]

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

[Out]

tan(a + b*x)/(4*b)