\(\int \csc ^3(a+b x) \sin ^5(2 a+2 b x) \, dx\) [64]

Optimal result

Integrand size = 20, antiderivative size = 46 \[ \int \csc ^3(a+b x) \sin ^5(2 a+2 b x) \, dx=\frac {32 \sin ^3(a+b x)}{3 b}-\frac {64 \sin ^5(a+b x)}{5 b}+\frac {32 \sin ^7(a+b x)}{7 b} \]

[Out]

32/3*sin(b*x+a)^3/b-64/5*sin(b*x+a)^5/b+32/7*sin(b*x+a)^7/b

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4373, 2644, 276} \[ \int \csc ^3(a+b x) \sin ^5(2 a+2 b x) \, dx=\frac {32 \sin ^7(a+b x)}{7 b}-\frac {64 \sin ^5(a+b x)}{5 b}+\frac {32 \sin ^3(a+b x)}{3 b} \]

[In]

Int[Csc[a + b*x]^3*Sin[2*a + 2*b*x]^5,x]

[Out]

(32*Sin[a + b*x]^3)/(3*b) - (64*Sin[a + b*x]^5)/(5*b) + (32*Sin[a + b*x]^7)/(7*b)

Rule 276

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 2644

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 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*} \text {integral}& = 32 \int \cos ^5(a+b x) \sin ^2(a+b x) \, dx \\ & = \frac {32 \text {Subst}\left (\int x^2 \left (1-x^2\right )^2 \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {32 \text {Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {32 \sin ^3(a+b x)}{3 b}-\frac {64 \sin ^5(a+b x)}{5 b}+\frac {32 \sin ^7(a+b x)}{7 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80 \[ \int \csc ^3(a+b x) \sin ^5(2 a+2 b x) \, dx=\frac {4 (157+108 \cos (2 (a+b x))+15 \cos (4 (a+b x))) \sin ^3(a+b x)}{105 b} \]

[In]

Integrate[Csc[a + b*x]^3*Sin[2*a + 2*b*x]^5,x]

[Out]

(4*(157 + 108*Cos[2*(a + b*x)] + 15*Cos[4*(a + b*x)])*Sin[a + b*x]^3)/(105*b)

Maple [A] (verified)

Time = 12.74 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80

method	result	size
default	\(\frac {\frac {32 \sin \left (x b +a \right )^{7}}{7}-\frac {64 \sin \left (x b +a \right )^{5}}{5}+\frac {32 \sin \left (x b +a \right )^{3}}{3}}{b}\)	\(37\)
risch	\(\frac {5 \sin \left (x b +a \right )}{2 b}-\frac {\sin \left (7 x b +7 a \right )}{14 b}-\frac {3 \sin \left (5 x b +5 a \right )}{10 b}-\frac {\sin \left (3 x b +3 a \right )}{6 b}\)	\(55\)

[In]

int(csc(b*x+a)^3*sin(2*b*x+2*a)^5,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.93 \[ \int \csc ^3(a+b x) \sin ^5(2 a+2 b x) \, dx=-\frac {32 \, {\left (15 \, \cos \left (b x + a\right )^{6} - 3 \, \cos \left (b x + a\right )^{4} - 4 \, \cos \left (b x + a\right )^{2} - 8\right )} \sin \left (b x + a\right )}{105 \, b} \]

[In]

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^5,x, algorithm="fricas")

[Out]

-32/105*(15*cos(b*x + a)^6 - 3*cos(b*x + a)^4 - 4*cos(b*x + a)^2 - 8)*sin(b*x + a)/b

Sympy [F(-1)]

Timed out. \[ \int \csc ^3(a+b x) \sin ^5(2 a+2 b x) \, dx=\text {Timed out} \]

[In]

integrate(csc(b*x+a)**3*sin(2*b*x+2*a)**5,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.02 \[ \int \csc ^3(a+b x) \sin ^5(2 a+2 b x) \, dx=-\frac {15 \, \sin \left (7 \, b x + 7 \, a\right ) + 63 \, \sin \left (5 \, b x + 5 \, a\right ) + 35 \, \sin \left (3 \, b x + 3 \, a\right ) - 525 \, \sin \left (b x + a\right )}{210 \, b} \]

[In]

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^5,x, algorithm="maxima")

[Out]

-1/210*(15*sin(7*b*x + 7*a) + 63*sin(5*b*x + 5*a) + 35*sin(3*b*x + 3*a) - 525*sin(b*x + a))/b

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \csc ^3(a+b x) \sin ^5(2 a+2 b x) \, dx=\frac {32 \, {\left (15 \, \sin \left (b x + a\right )^{7} - 42 \, \sin \left (b x + a\right )^{5} + 35 \, \sin \left (b x + a\right )^{3}\right )}}{105 \, b} \]

[In]

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^5,x, algorithm="giac")

[Out]

32/105*(15*sin(b*x + a)^7 - 42*sin(b*x + a)^5 + 35*sin(b*x + a)^3)/b

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \csc ^3(a+b x) \sin ^5(2 a+2 b x) \, dx=\frac {32\,\left (15\,{\sin \left (a+b\,x\right )}^7-42\,{\sin \left (a+b\,x\right )}^5+35\,{\sin \left (a+b\,x\right )}^3\right )}{105\,b} \]

[In]

int(sin(2*a + 2*b*x)^5/sin(a + b*x)^3,x)

[Out]

(32*(35*sin(a + b*x)^3 - 42*sin(a + b*x)^5 + 15*sin(a + b*x)^7))/(105*b)