3.1.49 \(\int \csc ^2(a+b x) \sin ^4(2 a+2 b x) \, dx\) [49]

Optimal. Leaf size=60 \[ x+\frac {\cos (a+b x) \sin (a+b x)}{b}+\frac {2 \cos ^3(a+b x) \sin (a+b x)}{3 b}-\frac {8 \cos ^5(a+b x) \sin (a+b x)}{3 b} \]

[Out]

x+cos(b*x+a)*sin(b*x+a)/b+2/3*cos(b*x+a)^3*sin(b*x+a)/b-8/3*cos(b*x+a)^5*sin(b*x+a)/b

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Rubi [A]
time = 0.06, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4373, 2648, 2715, 8} \begin {gather*} -\frac {8 \sin (a+b x) \cos ^5(a+b x)}{3 b}+\frac {2 \sin (a+b x) \cos ^3(a+b x)}{3 b}+\frac {\sin (a+b x) \cos (a+b x)}{b}+x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x + (Cos[a + b*x]*Sin[a + b*x])/b + (2*Cos[a + b*x]^3*Sin[a + b*x])/(3*b) - (8*Cos[a + b*x]^5*Sin[a + b*x])/(3
*b)

Rule 8

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

Rule 2648

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-a)*(b*Cos[e
+ f*x])^(n + 1)*((a*Sin[e + f*x])^(m - 1)/(b*f*(m + n))), x] + Dist[a^2*((m - 1)/(m + n)), Int[(b*Cos[e + f*x
])^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[
2*m, 2*n]

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 ^4(2 a+2 b x) \, dx &=16 \int \cos ^4(a+b x) \sin ^2(a+b x) \, dx\\ &=-\frac {8 \cos ^5(a+b x) \sin (a+b x)}{3 b}+\frac {8}{3} \int \cos ^4(a+b x) \, dx\\ &=\frac {2 \cos ^3(a+b x) \sin (a+b x)}{3 b}-\frac {8 \cos ^5(a+b x) \sin (a+b x)}{3 b}+2 \int \cos ^2(a+b x) \, dx\\ &=\frac {\cos (a+b x) \sin (a+b x)}{b}+\frac {2 \cos ^3(a+b x) \sin (a+b x)}{3 b}-\frac {8 \cos ^5(a+b x) \sin (a+b x)}{3 b}+\int 1 \, dx\\ &=x+\frac {\cos (a+b x) \sin (a+b x)}{b}+\frac {2 \cos ^3(a+b x) \sin (a+b x)}{3 b}-\frac {8 \cos ^5(a+b x) \sin (a+b x)}{3 b}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 40, normalized size = 0.67 \begin {gather*} -\frac {-12 b x-3 \sin (2 (a+b x))+3 \sin (4 (a+b x))+\sin (6 (a+b x))}{12 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(-12*b*x - 3*Sin[2*(a + b*x)] + 3*Sin[4*(a + b*x)] + Sin[6*(a + b*x)])/b

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Maple [A]
time = 0.11, size = 55, normalized size = 0.92

method	result	size
risch	\(x -\frac {\sin \left (6 x b +6 a \right )}{12 b}-\frac {\sin \left (4 x b +4 a \right )}{4 b}+\frac {\sin \left (2 x b +2 a \right )}{4 b}\)	\(45\)
default	\(\frac {-\frac {8 \sin \left (x b +a \right ) \left (\cos ^{5}\left (x b +a \right )\right )}{3}+\frac {2 \left (\cos ^{3}\left (x b +a \right )+\frac {3 \cos \left (x b +a \right )}{2}\right ) \sin \left (x b +a \right )}{3}+x b +a}{b}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/b*(-1/6*sin(b*x+a)*cos(b*x+a)^5+1/24*(cos(b*x+a)^3+3/2*cos(b*x+a))*sin(b*x+a)+1/16*x*b+1/16*a)

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Maxima [A]
time = 0.28, size = 43, normalized size = 0.72 \begin {gather*} \frac {12 \, b x - \sin \left (6 \, b x + 6 \, a\right ) - 3 \, \sin \left (4 \, b x + 4 \, a\right ) + 3 \, \sin \left (2 \, b x + 2 \, a\right )}{12 \, 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)^4,x, algorithm="maxima")

[Out]

1/12*(12*b*x - sin(6*b*x + 6*a) - 3*sin(4*b*x + 4*a) + 3*sin(2*b*x + 2*a))/b

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Fricas [A]
time = 1.92, size = 47, normalized size = 0.78 \begin {gather*} \frac {3 \, b x - {\left (8 \, \cos \left (b x + a\right )^{5} - 2 \, \cos \left (b x + a\right )^{3} - 3 \, \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{3 \, 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)^4,x, algorithm="fricas")

[Out]

1/3*(3*b*x - (8*cos(b*x + a)^5 - 2*cos(b*x + a)^3 - 3*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)**4,x)

[Out]

Timed out

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Giac [A]
time = 0.47, size = 55, normalized size = 0.92 \begin {gather*} \frac {3 \, b x + 3 \, a + \frac {3 \, \tan \left (b x + a\right )^{5} + 8 \, \tan \left (b x + a\right )^{3} - 3 \, \tan \left (b x + a\right )}{{\left (\tan \left (b x + a\right )^{2} + 1\right )}^{3}}}{3 \, 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)^4,x, algorithm="giac")

[Out]

1/3*(3*b*x + 3*a + (3*tan(b*x + a)^5 + 8*tan(b*x + a)^3 - 3*tan(b*x + a))/(tan(b*x + a)^2 + 1)^3)/b

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Mupad [B]
time = 0.49, size = 65, normalized size = 1.08 \begin {gather*} x+\frac {{\mathrm {tan}\left (a+b\,x\right )}^5+\frac {8\,{\mathrm {tan}\left (a+b\,x\right )}^3}{3}-\mathrm {tan}\left (a+b\,x\right )}{b\,\left ({\mathrm {tan}\left (a+b\,x\right )}^6+3\,{\mathrm {tan}\left (a+b\,x\right )}^4+3\,{\mathrm {tan}\left (a+b\,x\right )}^2+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(2*a + 2*b*x)^4/sin(a + b*x)^2,x)

[Out]

x + ((8*tan(a + b*x)^3)/3 - tan(a + b*x) + tan(a + b*x)^5)/(b*(3*tan(a + b*x)^2 + 3*tan(a + b*x)^4 + tan(a + b
*x)^6 + 1))

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