3.229 \(\int \cos (a+b x) \csc ^3(c+b x) \, dx\)

Optimal. Leaf size=38 \[ \frac{\sin (a-c) \cot (b x+c)}{b}-\frac{\cos (a-c) \csc ^2(b x+c)}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.0425368, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4581, 2606, 30, 3767, 8} \[ \frac{\sin (a-c) \cot (b x+c)}{b}-\frac{\cos (a-c) \csc ^2(b x+c)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]*Csc[c + b*x]^3,x]

[Out]

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

Rule 4581

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

Rule 2606

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 30

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

Rule 3767

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 8

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

Rubi steps

\begin{align*} \int \cos (a+b x) \csc ^3(c+b x) \, dx &=\cos (a-c) \int \cot (c+b x) \csc ^2(c+b x) \, dx-\sin (a-c) \int \csc ^2(c+b x) \, dx\\ &=-\frac{\cos (a-c) \operatorname{Subst}(\int x \, dx,x,\csc (c+b x))}{b}+\frac{\sin (a-c) \operatorname{Subst}(\int 1 \, dx,x,\cot (c+b x))}{b}\\ &=-\frac{\cos (a-c) \csc ^2(c+b x)}{2 b}+\frac{\cot (c+b x) \sin (a-c)}{b}\\ \end{align*}

Mathematica [A]  time = 0.196821, size = 35, normalized size = 0.92 \[ -\frac{\csc (c) \csc ^2(b x+c) (\sin (a)-\sin (a-c) \cos (2 b x+c))}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.544, size = 55, normalized size = 1.5 \begin{align*} -{\frac{1}{2\,b \left ( \cos \left ( a \right ) \cos \left ( c \right ) +\sin \left ( a \right ) \sin \left ( c \right ) \right ) \left ( \tan \left ( bx+a \right ) \cos \left ( a \right ) \cos \left ( c \right ) +\tan \left ( bx+a \right ) \sin \left ( a \right ) \sin \left ( c \right ) +\cos \left ( a \right ) \sin \left ( c \right ) -\sin \left ( a \right ) \cos \left ( c \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)/sin(b*x+c)^3,x)

[Out]

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

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Maxima [B]  time = 1.12199, size = 533, normalized size = 14.03 \begin{align*} \frac{{\left (2 \, \cos \left (2 \, b x + 2 \, a + 2 \, c\right ) - \cos \left (2 \, a\right ) + \cos \left (2 \, c\right )\right )} \cos \left (4 \, b x + a + 5 \, 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 )} \cos \left (2 \, b x + a + 3 \, c\right ) -{\left (\cos \left (2 \, a\right ) - \cos \left (2 \, c\right )\right )} \cos \left (a + c\right ) + 2 \, \cos \left (2 \, b x + 2 \, a + 2 \, c\right ) \cos \left (a + c\right ) +{\left (2 \, \sin \left (2 \, b x + 2 \, a + 2 \, c\right ) - \sin \left (2 \, a\right ) + \sin \left (2 \, c\right )\right )} \sin \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 )} \sin \left (2 \, b x + a + 3 \, c\right ) -{\left (\sin \left (2 \, a\right ) - \sin \left (2 \, c\right )\right )} \sin \left (a + c\right ) + 2 \, \sin \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.469258, size = 113, normalized size = 2.97 \begin{align*} \frac{2 \, \cos \left (b x + c\right ) \sin \left (b x + c\right ) \sin \left (-a + c\right ) + \cos \left (-a + c\right )}{2 \,{\left (b \cos \left (b x + c\right )^{2} - b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.26156, size = 441, normalized size = 11.61 \begin{align*} -\frac{\tan \left (\frac{1}{2} \, a\right )^{6} \tan \left (\frac{1}{2} \, c\right )^{6} + 3 \, \tan \left (\frac{1}{2} \, a\right )^{6} \tan \left (\frac{1}{2} \, c\right )^{4} + 3 \, \tan \left (\frac{1}{2} \, a\right )^{4} \tan \left (\frac{1}{2} \, c\right )^{6} + 3 \, \tan \left (\frac{1}{2} \, a\right )^{6} \tan \left (\frac{1}{2} \, c\right )^{2} + 9 \, \tan \left (\frac{1}{2} \, a\right )^{4} \tan \left (\frac{1}{2} \, c\right )^{4} + 3 \, \tan \left (\frac{1}{2} \, a\right )^{2} \tan \left (\frac{1}{2} \, c\right )^{6} + \tan \left (\frac{1}{2} \, a\right )^{6} + 9 \, \tan \left (\frac{1}{2} \, a\right )^{4} \tan \left (\frac{1}{2} \, c\right )^{2} + 9 \, \tan \left (\frac{1}{2} \, a\right )^{2} \tan \left (\frac{1}{2} \, c\right )^{4} + \tan \left (\frac{1}{2} \, c\right )^{6} + 3 \, \tan \left (\frac{1}{2} \, a\right )^{4} + 9 \, \tan \left (\frac{1}{2} \, a\right )^{2} \tan \left (\frac{1}{2} \, c\right )^{2} + 3 \, \tan \left (\frac{1}{2} \, c\right )^{4} + 3 \, \tan \left (\frac{1}{2} \, a\right )^{2} + 3 \, \tan \left (\frac{1}{2} \, c\right )^{2} + 1}{2 \,{\left (\tan \left (b x + a\right ) \tan \left (\frac{1}{2} \, a\right )^{2} \tan \left (\frac{1}{2} \, c\right )^{2} - \tan \left (b x + a\right ) \tan \left (\frac{1}{2} \, a\right )^{2} + 4 \, \tan \left (b x + a\right ) \tan \left (\frac{1}{2} \, a\right ) \tan \left (\frac{1}{2} \, c\right ) - 2 \, \tan \left (\frac{1}{2} \, a\right )^{2} \tan \left (\frac{1}{2} \, c\right ) - \tan \left (b x + a\right ) \tan \left (\frac{1}{2} \, c\right )^{2} + 2 \, \tan \left (\frac{1}{2} \, a\right ) \tan \left (\frac{1}{2} \, c\right )^{2} + \tan \left (b x + a\right ) - 2 \, \tan \left (\frac{1}{2} \, a\right ) + 2 \, \tan \left (\frac{1}{2} \, c\right )\right )}^{2}{\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} + 4 \, \tan \left (\frac{1}{2} \, a\right ) \tan \left (\frac{1}{2} \, c\right ) - \tan \left (\frac{1}{2} \, c\right )^{2} + 1\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(tan(1/2*a)^6*tan(1/2*c)^6 + 3*tan(1/2*a)^6*tan(1/2*c)^4 + 3*tan(1/2*a)^4*tan(1/2*c)^6 + 3*tan(1/2*a)^6*t
an(1/2*c)^2 + 9*tan(1/2*a)^4*tan(1/2*c)^4 + 3*tan(1/2*a)^2*tan(1/2*c)^6 + tan(1/2*a)^6 + 9*tan(1/2*a)^4*tan(1/
2*c)^2 + 9*tan(1/2*a)^2*tan(1/2*c)^4 + tan(1/2*c)^6 + 3*tan(1/2*a)^4 + 9*tan(1/2*a)^2*tan(1/2*c)^2 + 3*tan(1/2
*c)^4 + 3*tan(1/2*a)^2 + 3*tan(1/2*c)^2 + 1)/((tan(b*x + a)*tan(1/2*a)^2*tan(1/2*c)^2 - tan(b*x + a)*tan(1/2*a
)^2 + 4*tan(b*x + a)*tan(1/2*a)*tan(1/2*c) - 2*tan(1/2*a)^2*tan(1/2*c) - tan(b*x + a)*tan(1/2*c)^2 + 2*tan(1/2
*a)*tan(1/2*c)^2 + tan(b*x + a) - 2*tan(1/2*a) + 2*tan(1/2*c))^2*(tan(1/2*a)^2*tan(1/2*c)^2 - tan(1/2*a)^2 + 4
*tan(1/2*a)*tan(1/2*c) - tan(1/2*c)^2 + 1)*b)