3.46 \(\int \csc ^2(a+b x) \sin ^7(2 a+2 b x) \, dx\)

Optimal. Leaf size=44 \[ -\frac{32 \cos ^{12}(a+b x)}{3 b}+\frac{128 \cos ^{10}(a+b x)}{5 b}-\frac{16 \cos ^8(a+b x)}{b} \]

[Out]

(-16*Cos[a + b*x]^8)/b + (128*Cos[a + b*x]^10)/(5*b) - (32*Cos[a + b*x]^12)/(3*b)

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Rubi [A]  time = 0.0653625, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4288, 2565, 266, 43} \[ -\frac{32 \cos ^{12}(a+b x)}{3 b}+\frac{128 \cos ^{10}(a+b x)}{5 b}-\frac{16 \cos ^8(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^2*Sin[2*a + 2*b*x]^7,x]

[Out]

(-16*Cos[a + b*x]^8)/b + (128*Cos[a + b*x]^10)/(5*b) - (32*Cos[a + b*x]^12)/(3*b)

Rule 4288

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]

Rule 2565

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \csc ^2(a+b x) \sin ^7(2 a+2 b x) \, dx &=128 \int \cos ^7(a+b x) \sin ^5(a+b x) \, dx\\ &=-\frac{128 \operatorname{Subst}\left (\int x^7 \left (1-x^2\right )^2 \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{64 \operatorname{Subst}\left (\int (1-x)^2 x^3 \, dx,x,\cos ^2(a+b x)\right )}{b}\\ &=-\frac{64 \operatorname{Subst}\left (\int \left (x^3-2 x^4+x^5\right ) \, dx,x,\cos ^2(a+b x)\right )}{b}\\ &=-\frac{16 \cos ^8(a+b x)}{b}+\frac{128 \cos ^{10}(a+b x)}{5 b}-\frac{32 \cos ^{12}(a+b x)}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.172471, size = 48, normalized size = 1.09 \[ \frac{16 \left (-10 \sin ^{12}(a+b x)+36 \sin ^{10}(a+b x)-45 \sin ^8(a+b x)+20 \sin ^6(a+b x)\right )}{15 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^2*Sin[2*a + 2*b*x]^7,x]

[Out]

(16*(20*Sin[a + b*x]^6 - 45*Sin[a + b*x]^8 + 36*Sin[a + b*x]^10 - 10*Sin[a + b*x]^12))/(15*b)

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Maple [A]  time = 0.027, size = 53, normalized size = 1.2 \begin{align*} 128\,{\frac{1}{b} \left ( -1/12\, \left ( \sin \left ( bx+a \right ) \right ) ^{4} \left ( \cos \left ( bx+a \right ) \right ) ^{8}-1/30\, \left ( \sin \left ( bx+a \right ) \right ) ^{2} \left ( \cos \left ( bx+a \right ) \right ) ^{8}-{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{8}}{120}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^2*sin(2*b*x+2*a)^7,x)

[Out]

128/b*(-1/12*sin(b*x+a)^4*cos(b*x+a)^8-1/30*sin(b*x+a)^2*cos(b*x+a)^8-1/120*cos(b*x+a)^8)

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Maxima [A]  time = 1.08059, size = 97, normalized size = 2.2 \begin{align*} -\frac{5 \, \cos \left (12 \, b x + 12 \, a\right ) + 12 \, \cos \left (10 \, b x + 10 \, a\right ) - 30 \, \cos \left (8 \, b x + 8 \, a\right ) - 100 \, \cos \left (6 \, b x + 6 \, a\right ) + 75 \, \cos \left (4 \, b x + 4 \, a\right ) + 600 \, \cos \left (2 \, b x + 2 \, a\right )}{960 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^7,x, algorithm="maxima")

[Out]

-1/960*(5*cos(12*b*x + 12*a) + 12*cos(10*b*x + 10*a) - 30*cos(8*b*x + 8*a) - 100*cos(6*b*x + 6*a) + 75*cos(4*b
*x + 4*a) + 600*cos(2*b*x + 2*a))/b

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Fricas [A]  time = 0.510969, size = 97, normalized size = 2.2 \begin{align*} -\frac{16 \,{\left (10 \, \cos \left (b x + a\right )^{12} - 24 \, \cos \left (b x + a\right )^{10} + 15 \, \cos \left (b x + a\right )^{8}\right )}}{15 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^7,x, algorithm="fricas")

[Out]

-16/15*(10*cos(b*x + a)^12 - 24*cos(b*x + a)^10 + 15*cos(b*x + a)^8)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)**2*sin(2*b*x+2*a)**7,x)

[Out]

Timed out

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Giac [B]  time = 1.6791, size = 247, normalized size = 5.61 \begin{align*} -\frac{4096 \,{\left (\frac{5 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} + \frac{15 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + \frac{39 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{5}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{5}} + \frac{42 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{6}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{6}} + \frac{39 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{7}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{7}} + \frac{15 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{8}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{8}} + \frac{5 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{9}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{9}}\right )}}{15 \, b{\left (\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1\right )}^{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^7,x, algorithm="giac")

[Out]

-4096/15*(5*(cos(b*x + a) - 1)^3/(cos(b*x + a) + 1)^3 + 15*(cos(b*x + a) - 1)^4/(cos(b*x + a) + 1)^4 + 39*(cos
(b*x + a) - 1)^5/(cos(b*x + a) + 1)^5 + 42*(cos(b*x + a) - 1)^6/(cos(b*x + a) + 1)^6 + 39*(cos(b*x + a) - 1)^7
/(cos(b*x + a) + 1)^7 + 15*(cos(b*x + a) - 1)^8/(cos(b*x + a) + 1)^8 + 5*(cos(b*x + a) - 1)^9/(cos(b*x + a) +
1)^9)/(b*((cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1)^12)