3.62 \(\int \csc ^3(a+b x) \sin ^7(2 a+2 b x) \, dx\)

Optimal. Leaf size=61 \[ -\frac {128 \sin ^{11}(a+b x)}{11 b}+\frac {128 \sin ^9(a+b x)}{3 b}-\frac {384 \sin ^7(a+b x)}{7 b}+\frac {128 \sin ^5(a+b x)}{5 b} \]

[Out]

128/5*sin(b*x+a)^5/b-384/7*sin(b*x+a)^7/b+128/3*sin(b*x+a)^9/b-128/11*sin(b*x+a)^11/b

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Rubi [A]  time = 0.07, antiderivative size = 61, 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 = {4288, 2564, 270} \[ -\frac {128 \sin ^{11}(a+b x)}{11 b}+\frac {128 \sin ^9(a+b x)}{3 b}-\frac {384 \sin ^7(a+b x)}{7 b}+\frac {128 \sin ^5(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^3*Sin[2*a + 2*b*x]^7,x]

[Out]

(128*Sin[a + b*x]^5)/(5*b) - (384*Sin[a + b*x]^7)/(7*b) + (128*Sin[a + b*x]^9)/(3*b) - (128*Sin[a + b*x]^11)/(
11*b)

Rule 270

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 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

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]

Rubi steps

\begin {align*} \int \csc ^3(a+b x) \sin ^7(2 a+2 b x) \, dx &=128 \int \cos ^7(a+b x) \sin ^4(a+b x) \, dx\\ &=\frac {128 \operatorname {Subst}\left (\int x^4 \left (1-x^2\right )^3 \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac {128 \operatorname {Subst}\left (\int \left (x^4-3 x^6+3 x^8-x^{10}\right ) \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac {128 \sin ^5(a+b x)}{5 b}-\frac {384 \sin ^7(a+b x)}{7 b}+\frac {128 \sin ^9(a+b x)}{3 b}-\frac {128 \sin ^{11}(a+b x)}{11 b}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 48, normalized size = 0.79 \[ \frac {128 \left (-105 \sin ^{11}(a+b x)+385 \sin ^9(a+b x)-495 \sin ^7(a+b x)+231 \sin ^5(a+b x)\right )}{1155 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^3*Sin[2*a + 2*b*x]^7,x]

[Out]

(128*(231*Sin[a + b*x]^5 - 495*Sin[a + b*x]^7 + 385*Sin[a + b*x]^9 - 105*Sin[a + b*x]^11))/(1155*b)

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fricas [A]  time = 0.43, size = 63, normalized size = 1.03 \[ \frac {128 \, {\left (105 \, \cos \left (b x + a\right )^{10} - 140 \, \cos \left (b x + a\right )^{8} + 5 \, \cos \left (b x + a\right )^{6} + 6 \, \cos \left (b x + a\right )^{4} + 8 \, \cos \left (b x + a\right )^{2} + 16\right )} \sin \left (b x + a\right )}{1155 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^7,x, algorithm="fricas")

[Out]

128/1155*(105*cos(b*x + a)^10 - 140*cos(b*x + a)^8 + 5*cos(b*x + a)^6 + 6*cos(b*x + a)^4 + 8*cos(b*x + a)^2 +
16)*sin(b*x + a)/b

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giac [A]  time = 1.34, size = 46, normalized size = 0.75 \[ -\frac {128 \, {\left (105 \, \sin \left (b x + a\right )^{11} - 385 \, \sin \left (b x + a\right )^{9} + 495 \, \sin \left (b x + a\right )^{7} - 231 \, \sin \left (b x + a\right )^{5}\right )}}{1155 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^7,x, algorithm="giac")

[Out]

-128/1155*(105*sin(b*x + a)^11 - 385*sin(b*x + a)^9 + 495*sin(b*x + a)^7 - 231*sin(b*x + a)^5)/b

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maple [A]  time = 0.99, size = 79, normalized size = 1.30 \[ \frac {-\frac {128 \left (\sin ^{3}\left (b x +a \right )\right ) \left (\cos ^{8}\left (b x +a \right )\right )}{11}-\frac {128 \sin \left (b x +a \right ) \left (\cos ^{8}\left (b x +a \right )\right )}{33}+\frac {128 \left (\frac {16}{5}+\cos ^{6}\left (b x +a \right )+\frac {6 \left (\cos ^{4}\left (b x +a \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (b x +a \right )\right )}{5}\right ) \sin \left (b x +a \right )}{231}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^3*sin(2*b*x+2*a)^7,x)

[Out]

128/b*(-1/11*sin(b*x+a)^3*cos(b*x+a)^8-1/33*sin(b*x+a)*cos(b*x+a)^8+1/231*(16/5+cos(b*x+a)^6+6/5*cos(b*x+a)^4+
8/5*cos(b*x+a)^2)*sin(b*x+a))

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maxima [A]  time = 0.75, size = 69, normalized size = 1.13 \[ \frac {105 \, \sin \left (11 \, b x + 11 \, a\right ) + 385 \, \sin \left (9 \, b x + 9 \, a\right ) - 165 \, \sin \left (7 \, b x + 7 \, a\right ) - 2541 \, \sin \left (5 \, b x + 5 \, a\right ) - 2310 \, \sin \left (3 \, b x + 3 \, a\right ) + 16170 \, \sin \left (b x + a\right )}{9240 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^7,x, algorithm="maxima")

[Out]

1/9240*(105*sin(11*b*x + 11*a) + 385*sin(9*b*x + 9*a) - 165*sin(7*b*x + 7*a) - 2541*sin(5*b*x + 5*a) - 2310*si
n(3*b*x + 3*a) + 16170*sin(b*x + a))/b

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mupad [B]  time = 0.12, size = 45, normalized size = 0.74 \[ \frac {-\frac {128\,{\sin \left (a+b\,x\right )}^{11}}{11}+\frac {128\,{\sin \left (a+b\,x\right )}^9}{3}-\frac {384\,{\sin \left (a+b\,x\right )}^7}{7}+\frac {128\,{\sin \left (a+b\,x\right )}^5}{5}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(2*a + 2*b*x)^7/sin(a + b*x)^3,x)

[Out]

((128*sin(a + b*x)^5)/5 - (384*sin(a + b*x)^7)/7 + (128*sin(a + b*x)^9)/3 - (128*sin(a + b*x)^11)/11)/b

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)**3*sin(2*b*x+2*a)**7,x)

[Out]

Timed out

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