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

Optimal. Leaf size=21 \[ 2 x+\frac {2 \cos (a+b x) \sin (a+b x)}{b} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4373, 2715, 8} \begin {gather*} \frac {2 \sin (a+b x) \cos (a+b x)}{b}+2 x \end {gather*}

Antiderivative was successfully verified.

[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*} \int \csc ^2(a+b x) \sin ^2(2 a+2 b x) \, dx &=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*}

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Mathematica [A]
time = 0.03, size = 20, normalized size = 0.95 \begin {gather*} \frac {2 (a+b x)+\sin (2 (a+b x))}{b} \end {gather*}

Antiderivative was successfully verified.

[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

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Maple [A]
time = 0.06, size = 28, normalized size = 1.33

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 18, normalized size = 0.86 \begin {gather*} \frac {2 \, b x + \sin \left (2 \, b x + 2 \, a\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]
time = 2.53, size = 22, normalized size = 1.05 \begin {gather*} \frac {2 \, {\left (b x + \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 0.49, size = 29, normalized size = 1.38 \begin {gather*} \frac {2 \, {\left (b x + a + \frac {\tan \left (b x + a\right )}{\tan \left (b x + a\right )^{2} + 1}\right )}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Mupad [B]
time = 0.15, size = 17, normalized size = 0.81 \begin {gather*} 2\,x+\frac {\sin \left (2\,a+2\,b\,x\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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