3.1.21 \(\int \csc ^4(2 a+2 b x) \sin ^2(a+b x) \, dx\) [21]

Optimal. Leaf size=42 \[ -\frac {\cot (a+b x)}{16 b}+\frac {\tan (a+b x)}{8 b}+\frac {\tan ^3(a+b x)}{48 b} \]

[Out]

-1/16*cot(b*x+a)/b+1/8*tan(b*x+a)/b+1/48*tan(b*x+a)^3/b

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Rubi [A]
time = 0.04, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4373, 2700, 276} \begin {gather*} \frac {\tan ^3(a+b x)}{48 b}+\frac {\tan (a+b x)}{8 b}-\frac {\cot (a+b x)}{16 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*Cot[a + b*x]/b + Tan[a + b*x]/(8*b) + Tan[a + b*x]^3/(48*b)

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2700

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

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 ^4(2 a+2 b x) \sin ^2(a+b x) \, dx &=\frac {1}{16} \int \csc ^2(a+b x) \sec ^4(a+b x) \, dx\\ &=\frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x^2} \, dx,x,\tan (a+b x)\right )}{16 b}\\ &=\frac {\text {Subst}\left (\int \left (2+\frac {1}{x^2}+x^2\right ) \, dx,x,\tan (a+b x)\right )}{16 b}\\ &=-\frac {\cot (a+b x)}{16 b}+\frac {\tan (a+b x)}{8 b}+\frac {\tan ^3(a+b x)}{48 b}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 48, normalized size = 1.14 \begin {gather*} -\frac {\cot (a+b x)}{16 b}+\frac {5 \tan (a+b x)}{48 b}+\frac {\sec ^2(a+b x) \tan (a+b x)}{48 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*Cot[a + b*x]/b + (5*Tan[a + b*x])/(48*b) + (Sec[a + b*x]^2*Tan[a + b*x])/(48*b)

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Maple [A]
time = 0.11, size = 51, normalized size = 1.21

method	result	size
risch	\(-\frac {i \left (2 \,{\mathrm e}^{2 i \left (x b +a \right )}+1\right )}{3 b \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )^{3} \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )}\)	\(46\)
default	\(\frac {\frac {1}{3 \sin \left (x b +a \right ) \cos \left (x b +a \right )^{3}}+\frac {4}{3 \sin \left (x b +a \right ) \cos \left (x b +a \right )}-\frac {8 \cot \left (x b +a \right )}{3}}{16 b}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/b*(1/3/sin(b*x+a)/cos(b*x+a)^3+4/3/sin(b*x+a)/cos(b*x+a)-8/3*cot(b*x+a))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 308 vs. \(2 (36) = 72\).
time = 0.28, size = 308, normalized size = 7.33 \begin {gather*} -\frac {{\left (2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \sin \left (8 \, b x + 8 \, a\right ) + 2 \, {\left (2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \sin \left (6 \, b x + 6 \, a\right ) - 2 \, \cos \left (8 \, b x + 8 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) - 4 \, \cos \left (6 \, b x + 6 \, a\right ) \sin \left (2 \, b x + 2 \, a\right )}{3 \, {\left (b \cos \left (8 \, b x + 8 \, a\right )^{2} + 4 \, b \cos \left (6 \, b x + 6 \, a\right )^{2} + 4 \, b \cos \left (2 \, b x + 2 \, a\right )^{2} + b \sin \left (8 \, b x + 8 \, a\right )^{2} + 4 \, b \sin \left (6 \, b x + 6 \, a\right )^{2} - 8 \, b \sin \left (6 \, b x + 6 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) + 4 \, b \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, {\left (2 \, b \cos \left (6 \, b x + 6 \, a\right ) - 2 \, b \cos \left (2 \, b x + 2 \, a\right ) - b\right )} \cos \left (8 \, b x + 8 \, a\right ) - 4 \, {\left (2 \, b \cos \left (2 \, b x + 2 \, a\right ) + b\right )} \cos \left (6 \, b x + 6 \, a\right ) + 4 \, b \cos \left (2 \, b x + 2 \, a\right ) + 4 \, {\left (b \sin \left (6 \, b x + 6 \, a\right ) - b \sin \left (2 \, b x + 2 \, a\right )\right )} \sin \left (8 \, b x + 8 \, a\right ) + b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*((2*cos(2*b*x + 2*a) + 1)*sin(8*b*x + 8*a) + 2*(2*cos(2*b*x + 2*a) + 1)*sin(6*b*x + 6*a) - 2*cos(8*b*x +
8*a)*sin(2*b*x + 2*a) - 4*cos(6*b*x + 6*a)*sin(2*b*x + 2*a))/(b*cos(8*b*x + 8*a)^2 + 4*b*cos(6*b*x + 6*a)^2 +
4*b*cos(2*b*x + 2*a)^2 + b*sin(8*b*x + 8*a)^2 + 4*b*sin(6*b*x + 6*a)^2 - 8*b*sin(6*b*x + 6*a)*sin(2*b*x + 2*a)
+ 4*b*sin(2*b*x + 2*a)^2 + 2*(2*b*cos(6*b*x + 6*a) - 2*b*cos(2*b*x + 2*a) - b)*cos(8*b*x + 8*a) - 4*(2*b*cos(
2*b*x + 2*a) + b)*cos(6*b*x + 6*a) + 4*b*cos(2*b*x + 2*a) + 4*(b*sin(6*b*x + 6*a) - b*sin(2*b*x + 2*a))*sin(8*
b*x + 8*a) + b)

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Fricas [A]
time = 3.93, size = 43, normalized size = 1.02 \begin {gather*} -\frac {8 \, \cos \left (b x + a\right )^{4} - 4 \, \cos \left (b x + a\right )^{2} - 1}{48 \, b \cos \left (b x + a\right )^{3} \sin \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)^4*sin(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/48*(8*cos(b*x + a)^4 - 4*cos(b*x + a)^2 - 1)/(b*cos(b*x + a)^3*sin(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)**4*sin(b*x+a)**2,x)

[Out]

Timed out

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Giac [A]
time = 0.46, size = 32, normalized size = 0.76 \begin {gather*} \frac {\tan \left (b x + a\right )^{3} - \frac {3}{\tan \left (b x + a\right )} + 6 \, \tan \left (b x + a\right )}{48 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(tan(b*x + a)^3 - 3/tan(b*x + a) + 6*tan(b*x + a))/b

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Mupad [B]
time = 0.13, size = 33, normalized size = 0.79 \begin {gather*} \frac {{\mathrm {tan}\left (a+b\,x\right )}^4+6\,{\mathrm {tan}\left (a+b\,x\right )}^2-3}{48\,b\,\mathrm {tan}\left (a+b\,x\right )} \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)^4,x)

[Out]

(6*tan(a + b*x)^2 + tan(a + b*x)^4 - 3)/(48*b*tan(a + b*x))

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