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

Optimal. Leaf size=13 \[ \frac {\tan (a+b x)}{4 b} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 13, 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, 3852, 8} \begin {gather*} \frac {\tan (a+b x)}{4 b} \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [A]
time = 0.01, size = 13, normalized size = 1.00 \begin {gather*} \frac {\tan (a+b x)}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Tan[a + b*x]/(4*b)

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Maple [A]
time = 0.09, size = 12, normalized size = 0.92

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

Verification of antiderivative is not currently implemented for this CAS.

[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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (11) = 22\).
time = 0.27, size = 53, normalized size = 4.08 \begin {gather*} \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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]
time = 3.43, size = 19, normalized size = 1.46 \begin {gather*} \frac {\sin \left (b x + a\right )}{4 \, b \cos \left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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

[Out]

Timed out

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Giac [A]
time = 0.44, size = 11, normalized size = 0.85 \begin {gather*} \frac {\tan \left (b x + a\right )}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Mupad [B]
time = 0.11, size = 11, normalized size = 0.85 \begin {gather*} \frac {\mathrm {tan}\left (a+b\,x\right )}{4\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(a + b*x)/(4*b)

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