3.47 \(\int \csc ^2(a+b x) \sin ^6(2 a+2 b x) \, dx\)

Optimal. Leaf size=111 \[ -\frac {32 \sin ^3(a+b x) \cos ^7(a+b x)}{5 b}-\frac {12 \sin (a+b x) \cos ^7(a+b x)}{5 b}+\frac {2 \sin (a+b x) \cos ^5(a+b x)}{5 b}+\frac {\sin (a+b x) \cos ^3(a+b x)}{2 b}+\frac {3 \sin (a+b x) \cos (a+b x)}{4 b}+\frac {3 x}{4} \]

[Out]

3/4*x+3/4*cos(b*x+a)*sin(b*x+a)/b+1/2*cos(b*x+a)^3*sin(b*x+a)/b+2/5*cos(b*x+a)^5*sin(b*x+a)/b-12/5*cos(b*x+a)^
7*sin(b*x+a)/b-32/5*cos(b*x+a)^7*sin(b*x+a)^3/b

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Rubi [A]  time = 0.12, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4288, 2568, 2635, 8} \[ -\frac {32 \sin ^3(a+b x) \cos ^7(a+b x)}{5 b}-\frac {12 \sin (a+b x) \cos ^7(a+b x)}{5 b}+\frac {2 \sin (a+b x) \cos ^5(a+b x)}{5 b}+\frac {\sin (a+b x) \cos ^3(a+b x)}{2 b}+\frac {3 \sin (a+b x) \cos (a+b x)}{4 b}+\frac {3 x}{4} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^2*Sin[2*a + 2*b*x]^6,x]

[Out]

(3*x)/4 + (3*Cos[a + b*x]*Sin[a + b*x])/(4*b) + (Cos[a + b*x]^3*Sin[a + b*x])/(2*b) + (2*Cos[a + b*x]^5*Sin[a
+ b*x])/(5*b) - (12*Cos[a + b*x]^7*Sin[a + b*x])/(5*b) - (32*Cos[a + b*x]^7*Sin[a + b*x]^3)/(5*b)

Rule 8

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

Rule 2568

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 2635

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] && Integer
Q[2*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 ^2(a+b x) \sin ^6(2 a+2 b x) \, dx &=64 \int \cos ^6(a+b x) \sin ^4(a+b x) \, dx\\ &=-\frac {32 \cos ^7(a+b x) \sin ^3(a+b x)}{5 b}+\frac {96}{5} \int \cos ^6(a+b x) \sin ^2(a+b x) \, dx\\ &=-\frac {12 \cos ^7(a+b x) \sin (a+b x)}{5 b}-\frac {32 \cos ^7(a+b x) \sin ^3(a+b x)}{5 b}+\frac {12}{5} \int \cos ^6(a+b x) \, dx\\ &=\frac {2 \cos ^5(a+b x) \sin (a+b x)}{5 b}-\frac {12 \cos ^7(a+b x) \sin (a+b x)}{5 b}-\frac {32 \cos ^7(a+b x) \sin ^3(a+b x)}{5 b}+2 \int \cos ^4(a+b x) \, dx\\ &=\frac {\cos ^3(a+b x) \sin (a+b x)}{2 b}+\frac {2 \cos ^5(a+b x) \sin (a+b x)}{5 b}-\frac {12 \cos ^7(a+b x) \sin (a+b x)}{5 b}-\frac {32 \cos ^7(a+b x) \sin ^3(a+b x)}{5 b}+\frac {3}{2} \int \cos ^2(a+b x) \, dx\\ &=\frac {3 \cos (a+b x) \sin (a+b x)}{4 b}+\frac {\cos ^3(a+b x) \sin (a+b x)}{2 b}+\frac {2 \cos ^5(a+b x) \sin (a+b x)}{5 b}-\frac {12 \cos ^7(a+b x) \sin (a+b x)}{5 b}-\frac {32 \cos ^7(a+b x) \sin ^3(a+b x)}{5 b}+\frac {3 \int 1 \, dx}{4}\\ &=\frac {3 x}{4}+\frac {3 \cos (a+b x) \sin (a+b x)}{4 b}+\frac {\cos ^3(a+b x) \sin (a+b x)}{2 b}+\frac {2 \cos ^5(a+b x) \sin (a+b x)}{5 b}-\frac {12 \cos ^7(a+b x) \sin (a+b x)}{5 b}-\frac {32 \cos ^7(a+b x) \sin ^3(a+b x)}{5 b}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 62, normalized size = 0.56 \[ \frac {20 \sin (2 (a+b x))-40 \sin (4 (a+b x))-10 \sin (6 (a+b x))+5 \sin (8 (a+b x))+2 \sin (10 (a+b x))+120 b x}{160 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^2*Sin[2*a + 2*b*x]^6,x]

[Out]

(120*b*x + 20*Sin[2*(a + b*x)] - 40*Sin[4*(a + b*x)] - 10*Sin[6*(a + b*x)] + 5*Sin[8*(a + b*x)] + 2*Sin[10*(a
+ b*x)])/(160*b)

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fricas [A]  time = 0.46, size = 66, normalized size = 0.59 \[ \frac {15 \, b x + {\left (128 \, \cos \left (b x + a\right )^{9} - 176 \, \cos \left (b x + a\right )^{7} + 8 \, \cos \left (b x + a\right )^{5} + 10 \, \cos \left (b x + a\right )^{3} + 15 \, \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{20 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^6,x, algorithm="fricas")

[Out]

1/20*(15*b*x + (128*cos(b*x + a)^9 - 176*cos(b*x + a)^7 + 8*cos(b*x + a)^5 + 10*cos(b*x + a)^3 + 15*cos(b*x +
a))*sin(b*x + a))/b

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giac [A]  time = 1.02, size = 75, normalized size = 0.68 \[ \frac {15 \, b x + 15 \, a + \frac {15 \, \tan \left (b x + a\right )^{9} + 70 \, \tan \left (b x + a\right )^{7} + 128 \, \tan \left (b x + a\right )^{5} - 70 \, \tan \left (b x + a\right )^{3} - 15 \, \tan \left (b x + a\right )}{{\left (\tan \left (b x + a\right )^{2} + 1\right )}^{5}}}{20 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^6,x, algorithm="giac")

[Out]

1/20*(15*b*x + 15*a + (15*tan(b*x + a)^9 + 70*tan(b*x + a)^7 + 128*tan(b*x + a)^5 - 70*tan(b*x + a)^3 - 15*tan
(b*x + a))/(tan(b*x + a)^2 + 1)^5)/b

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maple [A]  time = 0.97, size = 83, normalized size = 0.75 \[ \frac {-\frac {32 \left (\sin ^{3}\left (b x +a \right )\right ) \left (\cos ^{7}\left (b x +a \right )\right )}{5}-\frac {12 \sin \left (b x +a \right ) \left (\cos ^{7}\left (b x +a \right )\right )}{5}+\frac {2 \left (\cos ^{5}\left (b x +a \right )+\frac {5 \left (\cos ^{3}\left (b x +a \right )\right )}{4}+\frac {15 \cos \left (b x +a \right )}{8}\right ) \sin \left (b x +a \right )}{5}+\frac {3 b x}{4}+\frac {3 a}{4}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^2*sin(2*b*x+2*a)^6,x)

[Out]

64/b*(-1/10*sin(b*x+a)^3*cos(b*x+a)^7-3/80*sin(b*x+a)*cos(b*x+a)^7+1/160*(cos(b*x+a)^5+5/4*cos(b*x+a)^3+15/8*c
os(b*x+a))*sin(b*x+a)+3/256*b*x+3/256*a)

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maxima [A]  time = 0.35, size = 65, normalized size = 0.59 \[ \frac {120 \, b x + 2 \, \sin \left (10 \, b x + 10 \, a\right ) + 5 \, \sin \left (8 \, b x + 8 \, a\right ) - 10 \, \sin \left (6 \, b x + 6 \, a\right ) - 40 \, \sin \left (4 \, b x + 4 \, a\right ) + 20 \, \sin \left (2 \, b x + 2 \, a\right )}{160 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^6,x, algorithm="maxima")

[Out]

1/160*(120*b*x + 2*sin(10*b*x + 10*a) + 5*sin(8*b*x + 8*a) - 10*sin(6*b*x + 6*a) - 40*sin(4*b*x + 4*a) + 20*si
n(2*b*x + 2*a))/b

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mupad [B]  time = 1.69, size = 109, normalized size = 0.98 \[ \frac {3\,x}{4}+\frac {\frac {3\,{\mathrm {tan}\left (a+b\,x\right )}^9}{4}+\frac {7\,{\mathrm {tan}\left (a+b\,x\right )}^7}{2}+\frac {32\,{\mathrm {tan}\left (a+b\,x\right )}^5}{5}-\frac {7\,{\mathrm {tan}\left (a+b\,x\right )}^3}{2}-\frac {3\,\mathrm {tan}\left (a+b\,x\right )}{4}}{b\,\left ({\mathrm {tan}\left (a+b\,x\right )}^{10}+5\,{\mathrm {tan}\left (a+b\,x\right )}^8+10\,{\mathrm {tan}\left (a+b\,x\right )}^6+10\,{\mathrm {tan}\left (a+b\,x\right )}^4+5\,{\mathrm {tan}\left (a+b\,x\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(2*a + 2*b*x)^6/sin(a + b*x)^2,x)

[Out]

(3*x)/4 + ((32*tan(a + b*x)^5)/5 - (7*tan(a + b*x)^3)/2 - (3*tan(a + b*x))/4 + (7*tan(a + b*x)^7)/2 + (3*tan(a
+ b*x)^9)/4)/(b*(5*tan(a + b*x)^2 + 10*tan(a + b*x)^4 + 10*tan(a + b*x)^6 + 5*tan(a + b*x)^8 + tan(a + b*x)^1
0 + 1))

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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)**2*sin(2*b*x+2*a)**6,x)

[Out]

Timed out

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