\(\int \csc ^3(c+b x) \sin (a+b x) \, dx\) [197]

Optimal result

Integrand size = 15, antiderivative size = 39 \[ \int \csc ^3(c+b x) \sin (a+b x) \, dx=-\frac {\cos (a-c) \cot (c+b x)}{b}-\frac {\csc ^2(c+b x) \sin (a-c)}{2 b} \]

[Out]

-cos(a-c)*cot(b*x+c)/b-1/2*csc(b*x+c)^2*sin(a-c)/b

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4678, 2686, 30, 3852, 8} \[ \int \csc ^3(c+b x) \sin (a+b x) \, dx=-\frac {\cos (a-c) \cot (b x+c)}{b}-\frac {\sin (a-c) \csc ^2(b x+c)}{2 b} \]

[In]

Int[Csc[c + b*x]^3*Sin[a + b*x],x]

[Out]

-((Cos[a - c]*Cot[c + b*x])/b) - (Csc[c + b*x]^2*Sin[a - c])/(2*b)

Rule 8

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2686

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

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 4678

Int[Csc[w_]^(n_.)*Sin[v_], x_Symbol] :> Dist[Sin[v - w], Int[Cot[w]*Csc[w]^(n - 1), x], x] + Dist[Cos[v - w],
Int[Csc[w]^(n - 1), x], x] /; GtQ[n, 0] && FreeQ[v - w, x] && NeQ[w, v]

Rubi steps \begin{align*} \text {integral}& = \cos (a-c) \int \csc ^2(c+b x) \, dx+\sin (a-c) \int \cot (c+b x) \csc ^2(c+b x) \, dx \\ & = -\frac {\cos (a-c) \text {Subst}(\int 1 \, dx,x,\cot (c+b x))}{b}-\frac {\sin (a-c) \text {Subst}(\int x \, dx,x,\csc (c+b x))}{b} \\ & = -\frac {\cos (a-c) \cot (c+b x)}{b}-\frac {\csc ^2(c+b x) \sin (a-c)}{2 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.90 \[ \int \csc ^3(c+b x) \sin (a+b x) \, dx=\frac {(\cos (a)-\cos (a-c) \cos (c+2 b x)) \csc (c) \csc ^2(c+b x)}{2 b} \]

[In]

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

[Out]

((Cos[a] - Cos[a - c]*Cos[c + 2*b*x])*Csc[c]*Csc[c + b*x]^2)/(2*b)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.70 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.62

method	result	size
risch	\(\frac {i \left (-2 \,{\mathrm e}^{i \left (2 x b +5 a +c \right )}+{\mathrm e}^{i \left (5 a -c \right )}+{\mathrm e}^{i \left (3 a +c \right )}\right )}{\left (-{\mathrm e}^{2 i \left (x b +a +c \right )}+{\mathrm e}^{2 i a}\right )^{2} b}\)	\(63\)
parallelrisch	\(-\frac {\csc \left (\frac {x b}{2}+\frac {c}{2}\right ) \left (\sin \left (x b +a \right ) \left (-\frac {\sec \left (\frac {x b}{2}+\frac {c}{2}\right )^{2}}{2}+1\right ) \csc \left (\frac {x b}{2}+\frac {c}{2}\right )+\sec \left (\frac {x b}{2}+\frac {c}{2}\right ) \cos \left (x b +a \right )\right )}{4 b}\)	\(63\)
default	\(\frac {-\frac {\sin \left (a \right ) \cos \left (c \right )-\cos \left (a \right ) \sin \left (c \right )}{2 \left (\cos \left (a \right ) \cos \left (c \right )+\sin \left (a \right ) \sin \left (c \right )\right )^{2} \left (\tan \left (x b +a \right ) \cos \left (a \right ) \cos \left (c \right )+\tan \left (x b +a \right ) \sin \left (a \right ) \sin \left (c \right )-\sin \left (a \right ) \cos \left (c \right )+\cos \left (a \right ) \sin \left (c \right )\right )^{2}}-\frac {1}{\left (\cos \left (a \right ) \cos \left (c \right )+\sin \left (a \right ) \sin \left (c \right )\right )^{2} \left (\tan \left (x b +a \right ) \cos \left (a \right ) \cos \left (c \right )+\tan \left (x b +a \right ) \sin \left (a \right ) \sin \left (c \right )-\sin \left (a \right ) \cos \left (c \right )+\cos \left (a \right ) \sin \left (c \right )\right )}}{b}\)	\(120\)

[In]

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

[Out]

I/(-exp(2*I*(b*x+a+c))+exp(2*I*a))^2/b*(-2*exp(I*(2*b*x+5*a+c))+exp(I*(5*a-c))+exp(I*(3*a+c)))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.21 \[ \int \csc ^3(c+b x) \sin (a+b x) \, dx=\frac {2 \, \cos \left (b x + c\right ) \cos \left (-a + c\right ) \sin \left (b x + c\right ) - \sin \left (-a + c\right )}{2 \, {\left (b \cos \left (b x + c\right )^{2} - b\right )}} \]

[In]

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

[Out]

1/2*(2*cos(b*x + c)*cos(-a + c)*sin(b*x + c) - sin(-a + c))/(b*cos(b*x + c)^2 - b)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 399 vs. \(2 (37) = 74\).

Time = 0.22 (sec) , antiderivative size = 399, normalized size of antiderivative = 10.23 \[ \int \csc ^3(c+b x) \sin (a+b x) \, dx=\frac {{\left (2 \, \sin \left (2 \, b x + 2 \, a + 2 \, c\right ) - \sin \left (2 \, a\right ) - \sin \left (2 \, c\right )\right )} \cos \left (4 \, b x + a + 5 \, c\right ) - 2 \, {\left (2 \, \sin \left (2 \, b x + 2 \, a + 2 \, c\right ) - \sin \left (2 \, a\right ) - \sin \left (2 \, c\right )\right )} \cos \left (2 \, b x + a + 3 \, c\right ) - {\left (\sin \left (2 \, a\right ) + \sin \left (2 \, c\right )\right )} \cos \left (a + c\right ) - {\left (2 \, \cos \left (2 \, b x + 2 \, a + 2 \, c\right ) - \cos \left (2 \, a\right ) - \cos \left (2 \, c\right )\right )} \sin \left (4 \, b x + a + 5 \, c\right ) + 2 \, \cos \left (a + c\right ) \sin \left (2 \, b x + 2 \, a + 2 \, c\right ) + 2 \, {\left (2 \, \cos \left (2 \, b x + 2 \, a + 2 \, c\right ) - \cos \left (2 \, a\right ) - \cos \left (2 \, c\right )\right )} \sin \left (2 \, b x + a + 3 \, c\right ) + {\left (\cos \left (2 \, a\right ) + \cos \left (2 \, c\right )\right )} \sin \left (a + c\right ) - 2 \, \cos \left (2 \, b x + 2 \, a + 2 \, c\right ) \sin \left (a + c\right )}{b \cos \left (4 \, b x + a + 5 \, c\right )^{2} + 4 \, b \cos \left (2 \, b x + a + 3 \, c\right )^{2} - 4 \, b \cos \left (2 \, b x + a + 3 \, c\right ) \cos \left (a + c\right ) + b \cos \left (a + c\right )^{2} + b \sin \left (4 \, b x + a + 5 \, c\right )^{2} + 4 \, b \sin \left (2 \, b x + a + 3 \, c\right )^{2} - 4 \, b \sin \left (2 \, b x + a + 3 \, c\right ) \sin \left (a + c\right ) + b \sin \left (a + c\right )^{2} - 2 \, {\left (2 \, b \cos \left (2 \, b x + a + 3 \, c\right ) - b \cos \left (a + c\right )\right )} \cos \left (4 \, b x + a + 5 \, c\right ) - 2 \, {\left (2 \, b \sin \left (2 \, b x + a + 3 \, c\right ) - b \sin \left (a + c\right )\right )} \sin \left (4 \, b x + a + 5 \, c\right )} \]

[In]

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

[Out]

((2*sin(2*b*x + 2*a + 2*c) - sin(2*a) - sin(2*c))*cos(4*b*x + a + 5*c) - 2*(2*sin(2*b*x + 2*a + 2*c) - sin(2*a
) - sin(2*c))*cos(2*b*x + a + 3*c) - (sin(2*a) + sin(2*c))*cos(a + c) - (2*cos(2*b*x + 2*a + 2*c) - cos(2*a) -
cos(2*c))*sin(4*b*x + a + 5*c) + 2*cos(a + c)*sin(2*b*x + 2*a + 2*c) + 2*(2*cos(2*b*x + 2*a + 2*c) - cos(2*a)
- cos(2*c))*sin(2*b*x + a + 3*c) + (cos(2*a) + cos(2*c))*sin(a + c) - 2*cos(2*b*x + 2*a + 2*c)*sin(a + c))/(b
*cos(4*b*x + a + 5*c)^2 + 4*b*cos(2*b*x + a + 3*c)^2 - 4*b*cos(2*b*x + a + 3*c)*cos(a + c) + b*cos(a + c)^2 +
b*sin(4*b*x + a + 5*c)^2 + 4*b*sin(2*b*x + a + 3*c)^2 - 4*b*sin(2*b*x + a + 3*c)*sin(a + c) + b*sin(a + c)^2 -
2*(2*b*cos(2*b*x + a + 3*c) - b*cos(a + c))*cos(4*b*x + a + 5*c) - 2*(2*b*sin(2*b*x + a + 3*c) - b*sin(a + c)
)*sin(4*b*x + a + 5*c))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 145 vs. \(2 (37) = 74\).

Time = 0.31 (sec) , antiderivative size = 145, normalized size of antiderivative = 3.72 \[ \int \csc ^3(c+b x) \sin (a+b x) \, dx=-\frac {\tan \left (b x + c\right ) \tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - \tan \left (b x + c\right ) \tan \left (\frac {1}{2} \, a\right )^{2} + 4 \, \tan \left (b x + c\right ) \tan \left (\frac {1}{2} \, a\right ) \tan \left (\frac {1}{2} \, c\right ) + \tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - \tan \left (b x + c\right ) \tan \left (\frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, a\right ) \tan \left (\frac {1}{2} \, c\right )^{2} + \tan \left (b x + c\right ) + \tan \left (\frac {1}{2} \, a\right ) - \tan \left (\frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, a\right )^{2} + \tan \left (\frac {1}{2} \, c\right )^{2} + 1\right )} b \tan \left (b x + c\right )^{2}} \]

[In]

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

[Out]

-(tan(b*x + c)*tan(1/2*a)^2*tan(1/2*c)^2 - tan(b*x + c)*tan(1/2*a)^2 + 4*tan(b*x + c)*tan(1/2*a)*tan(1/2*c) +
tan(1/2*a)^2*tan(1/2*c) - tan(b*x + c)*tan(1/2*c)^2 - tan(1/2*a)*tan(1/2*c)^2 + tan(b*x + c) + tan(1/2*a) - ta
n(1/2*c))/((tan(1/2*a)^2*tan(1/2*c)^2 + tan(1/2*a)^2 + tan(1/2*c)^2 + 1)*b*tan(b*x + c)^2)

Mupad [F(-1)]

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

[In]

int(sin(a + b*x)/sin(c + b*x)^3,x)

[Out]

\text{Hanged}