3.1.46 \(\int \csc ^2(a+b x) \sin ^7(2 a+2 b x) \, dx\) [46]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 44, 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, 2645, 272, 45} \begin {gather*} -\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} \end {gather*}

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 45

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])

Rule 272

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 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 ^7(2 a+2 b x) \, dx &=128 \int \cos ^7(a+b x) \sin ^5(a+b x) \, dx\\ &=-\frac {128 \text {Subst}\left (\int x^7 \left (1-x^2\right )^2 \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac {64 \text {Subst}\left (\int (1-x)^2 x^3 \, dx,x,\cos ^2(a+b x)\right )}{b}\\ &=-\frac {64 \text {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*}

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Mathematica [A]
time = 0.19, size = 48, normalized size = 1.09 \begin {gather*} \frac {16 \left (20 \sin ^6(a+b x)-45 \sin ^8(a+b x)+36 \sin ^{10}(a+b x)-10 \sin ^{12}(a+b x)\right )}{15 b} \end {gather*}

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 = 0.07, size = 53, normalized size = 1.20

method	result	size
default	\(\frac {-\frac {32 \left (\sin ^{4}\left (x b +a \right )\right ) \left (\cos ^{8}\left (x b +a \right )\right )}{3}-\frac {64 \left (\sin ^{2}\left (x b +a \right )\right ) \left (\cos ^{8}\left (x b +a \right )\right )}{15}-\frac {16 \left (\cos ^{8}\left (x b +a \right )\right )}{15}}{b}\)	\(53\)
risch	\(-\frac {\cos \left (12 x b +12 a \right )}{192 b}-\frac {\cos \left (10 x b +10 a \right )}{80 b}+\frac {\cos \left (8 x b +8 a \right )}{32 b}+\frac {5 \cos \left (6 x b +6 a \right )}{48 b}-\frac {5 \cos \left (4 x b +4 a \right )}{64 b}-\frac {5 \cos \left (2 x b +2 a \right )}{8 b}\)	\(86\)

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,method=_RETURNVERBOSE)

[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 = 0.27, size = 72, normalized size = 1.64 \begin {gather*} -\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 {gather*}

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 = 4.33, size = 36, normalized size = 0.82 \begin {gather*} -\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 {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \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)**7,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3003 deep

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Giac [A]
time = 0.43, size = 46, normalized size = 1.05 \begin {gather*} -\frac {16 \, {\left (10 \, \sin \left (b x + a\right )^{12} - 36 \, \sin \left (b x + a\right )^{10} + 45 \, \sin \left (b x + a\right )^{8} - 20 \, \sin \left (b x + a\right )^{6}\right )}}{15 \, 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)^7,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.15, size = 35, normalized size = 0.80 \begin {gather*} -\frac {16\,{\cos \left (a+b\,x\right )}^8\,\left (10\,{\cos \left (a+b\,x\right )}^4-24\,{\cos \left (a+b\,x\right )}^2+15\right )}{15\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(16*cos(a + b*x)^8*(10*cos(a + b*x)^4 - 24*cos(a + b*x)^2 + 15))/(15*b)

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