\(\int \csc (a+b x) \sin ^4(2 a+2 b x) \, dx\) [37]

Optimal result

Integrand size = 18, antiderivative size = 31 \[ \int \csc (a+b x) \sin ^4(2 a+2 b x) \, dx=-\frac {16 \cos ^5(a+b x)}{5 b}+\frac {16 \cos ^7(a+b x)}{7 b} \]

[Out]

-16/5*cos(b*x+a)^5/b+16/7*cos(b*x+a)^7/b

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4373, 2645, 14} \[ \int \csc (a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {16 \cos ^7(a+b x)}{7 b}-\frac {16 \cos ^5(a+b x)}{5 b} \]

[In]

Int[Csc[a + b*x]*Sin[2*a + 2*b*x]^4,x]

[Out]

(-16*Cos[a + b*x]^5)/(5*b) + (16*Cos[a + b*x]^7)/(7*b)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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*} \text {integral}& = 16 \int \cos ^4(a+b x) \sin ^3(a+b x) \, dx \\ & = -\frac {16 \text {Subst}\left (\int x^4 \left (1-x^2\right ) \, dx,x,\cos (a+b x)\right )}{b} \\ & = -\frac {16 \text {Subst}\left (\int \left (x^4-x^6\right ) \, dx,x,\cos (a+b x)\right )}{b} \\ & = -\frac {16 \cos ^5(a+b x)}{5 b}+\frac {16 \cos ^7(a+b x)}{7 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.90 \[ \int \csc (a+b x) \sin ^4(2 a+2 b x) \, dx=-\frac {3 \cos (a+b x)}{4 b}-\frac {\cos (3 (a+b x))}{4 b}+\frac {\cos (5 (a+b x))}{20 b}+\frac {\cos (7 (a+b x))}{28 b} \]

[In]

Integrate[Csc[a + b*x]*Sin[2*a + 2*b*x]^4,x]

[Out]

(-3*Cos[a + b*x])/(4*b) - Cos[3*(a + b*x)]/(4*b) + Cos[5*(a + b*x)]/(20*b) + Cos[7*(a + b*x)]/(28*b)

Maple [A] (verified)

Time = 1.73 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87

method	result	size
default	\(\frac {\frac {16 \cos \left (x b +a \right )^{7}}{7}-\frac {16 \cos \left (x b +a \right )^{5}}{5}}{b}\)	\(27\)
risch	\(-\frac {3 \cos \left (x b +a \right )}{4 b}+\frac {\cos \left (7 x b +7 a \right )}{28 b}+\frac {\cos \left (5 x b +5 a \right )}{20 b}-\frac {\cos \left (3 x b +3 a \right )}{4 b}\)	\(55\)

[In]

int(csc(b*x+a)*sin(2*b*x+2*a)^4,x,method=_RETURNVERBOSE)

[Out]

16/b*(1/7*cos(b*x+a)^7-1/5*cos(b*x+a)^5)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \csc (a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {16 \, {\left (5 \, \cos \left (b x + a\right )^{7} - 7 \, \cos \left (b x + a\right )^{5}\right )}}{35 \, b} \]

[In]

integrate(csc(b*x+a)*sin(2*b*x+2*a)^4,x, algorithm="fricas")

[Out]

16/35*(5*cos(b*x + a)^7 - 7*cos(b*x + a)^5)/b

Sympy [F(-1)]

Timed out. \[ \int \csc (a+b x) \sin ^4(2 a+2 b x) \, dx=\text {Timed out} \]

[In]

integrate(csc(b*x+a)*sin(2*b*x+2*a)**4,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52 \[ \int \csc (a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {5 \, \cos \left (7 \, b x + 7 \, a\right ) + 7 \, \cos \left (5 \, b x + 5 \, a\right ) - 35 \, \cos \left (3 \, b x + 3 \, a\right ) - 105 \, \cos \left (b x + a\right )}{140 \, b} \]

[In]

integrate(csc(b*x+a)*sin(2*b*x+2*a)^4,x, algorithm="maxima")

[Out]

1/140*(5*cos(7*b*x + 7*a) + 7*cos(5*b*x + 5*a) - 35*cos(3*b*x + 3*a) - 105*cos(b*x + a))/b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (27) = 54\).

Time = 0.27 (sec) , antiderivative size = 138, normalized size of antiderivative = 4.45 \[ \int \csc (a+b x) \sin ^4(2 a+2 b x) \, dx=-\frac {64 \, {\left (\frac {7 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {14 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {70 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} + \frac {35 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + \frac {35 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{5}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{5}} - 1\right )}}{35 \, b {\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1\right )}^{7}} \]

[In]

integrate(csc(b*x+a)*sin(2*b*x+2*a)^4,x, algorithm="giac")

[Out]

-64/35*(7*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 14*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 70*(cos(b*x +
a) - 1)^3/(cos(b*x + a) + 1)^3 + 35*(cos(b*x + a) - 1)^4/(cos(b*x + a) + 1)^4 + 35*(cos(b*x + a) - 1)^5/(cos(
b*x + a) + 1)^5 - 1)/(b*((cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1)^7)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \csc (a+b x) \sin ^4(2 a+2 b x) \, dx=-\frac {16\,\left (7\,{\cos \left (a+b\,x\right )}^5-5\,{\cos \left (a+b\,x\right )}^7\right )}{35\,b} \]

[In]

int(sin(2*a + 2*b*x)^4/sin(a + b*x),x)

[Out]

-(16*(7*cos(a + b*x)^5 - 5*cos(a + b*x)^7))/(35*b)