3.1.48 \(\int \csc ^2(a+b x) \sin ^5(2 a+2 b x) \, dx\) [48]

Optimal. Leaf size=29 \[ -\frac {16 \cos ^6(a+b x)}{3 b}+\frac {4 \cos ^8(a+b x)}{b} \]

[Out]

-16/3*cos(b*x+a)^6/b+4*cos(b*x+a)^8/b

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Rubi [A]
time = 0.04, antiderivative size = 29, 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, 2645, 14} \begin {gather*} \frac {4 \cos ^8(a+b x)}{b}-\frac {16 \cos ^6(a+b x)}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^2*Sin[2*a + 2*b*x]^5,x]

[Out]

(-16*Cos[a + b*x]^6)/(3*b) + (4*Cos[a + b*x]^8)/b

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, 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 ^5(2 a+2 b x) \, dx &=32 \int \cos ^5(a+b x) \sin ^3(a+b x) \, dx\\ &=-\frac {32 \text {Subst}\left (\int x^5 \left (1-x^2\right ) \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac {32 \text {Subst}\left (\int \left (x^5-x^7\right ) \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac {16 \cos ^6(a+b x)}{3 b}+\frac {4 \cos ^8(a+b x)}{b}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 48, normalized size = 1.66 \begin {gather*} \frac {-72 \cos (2 (a+b x))-12 \cos (4 (a+b x))+8 \cos (6 (a+b x))+3 \cos (8 (a+b x))}{96 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^2*Sin[2*a + 2*b*x]^5,x]

[Out]

(-72*Cos[2*(a + b*x)] - 12*Cos[4*(a + b*x)] + 8*Cos[6*(a + b*x)] + 3*Cos[8*(a + b*x)])/(96*b)

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Maple [A]
time = 0.06, size = 35, normalized size = 1.21

method	result	size
default	\(\frac {-4 \left (\sin ^{2}\left (x b +a \right )\right ) \left (\cos ^{6}\left (x b +a \right )\right )-\frac {4 \left (\cos ^{6}\left (x b +a \right )\right )}{3}}{b}\)	\(35\)
risch	\(\frac {\cos \left (8 x b +8 a \right )}{32 b}+\frac {\cos \left (6 x b +6 a \right )}{12 b}-\frac {\cos \left (4 x b +4 a \right )}{8 b}-\frac {3 \cos \left (2 x b +2 a \right )}{4 b}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^2*sin(2*b*x+2*a)^5,x,method=_RETURNVERBOSE)

[Out]

32/b*(-1/8*sin(b*x+a)^2*cos(b*x+a)^6-1/24*cos(b*x+a)^6)

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Maxima [A]
time = 0.31, size = 50, normalized size = 1.72 \begin {gather*} \frac {3 \, \cos \left (8 \, b x + 8 \, a\right ) + 8 \, \cos \left (6 \, b x + 6 \, a\right ) - 12 \, \cos \left (4 \, b x + 4 \, a\right ) - 72 \, \cos \left (2 \, b x + 2 \, a\right )}{96 \, 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)^5,x, algorithm="maxima")

[Out]

1/96*(3*cos(8*b*x + 8*a) + 8*cos(6*b*x + 6*a) - 12*cos(4*b*x + 4*a) - 72*cos(2*b*x + 2*a))/b

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Fricas [A]
time = 3.71, size = 26, normalized size = 0.90 \begin {gather*} \frac {4 \, {\left (3 \, \cos \left (b x + a\right )^{8} - 4 \, \cos \left (b x + a\right )^{6}\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)^5,x, algorithm="fricas")

[Out]

4/3*(3*cos(b*x + a)^8 - 4*cos(b*x + a)^6)/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)**5,x)

[Out]

Timed out

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Giac [A]
time = 0.40, size = 36, normalized size = 1.24 \begin {gather*} \frac {4 \, {\left (3 \, \sin \left (b x + a\right )^{8} - 8 \, \sin \left (b x + a\right )^{6} + 6 \, \sin \left (b x + a\right )^{4}\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)^5,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.05, size = 25, normalized size = 0.86 \begin {gather*} \frac {4\,{\cos \left (a+b\,x\right )}^6\,\left (3\,{\cos \left (a+b\,x\right )}^2-4\right )}{3\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(2*a + 2*b*x)^5/sin(a + b*x)^2,x)

[Out]

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

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