\(\int (c+d x)^4 \csc (x) \sin (3 x) \, dx\) [361]

Optimal result

Integrand size = 14, antiderivative size = 131 \[ \int (c+d x)^4 \csc (x) \sin (3 x) \, dx=\frac {3 d^4 x}{2}-d (c+d x)^3+\frac {(c+d x)^5}{5 d}-\frac {9}{2} d^3 (c+d x) \cos ^2(x)+3 d (c+d x)^3 \cos ^2(x)+3 d^4 \cos (x) \sin (x)-6 d^2 (c+d x)^2 \cos (x) \sin (x)+2 (c+d x)^4 \cos (x) \sin (x)+\frac {3}{2} d^3 (c+d x) \sin ^2(x)-d (c+d x)^3 \sin ^2(x) \]

[Out]

3/2*d^4*x-d*(d*x+c)^3+1/5*(d*x+c)^5/d-9/2*d^3*(d*x+c)*cos(x)^2+3*d*(d*x+c)^3*cos(x)^2+3*d^4*cos(x)*sin(x)-6*d^
2*(d*x+c)^2*cos(x)*sin(x)+2*(d*x+c)^4*cos(x)*sin(x)+3/2*d^3*(d*x+c)*sin(x)^2-d*(d*x+c)^3*sin(x)^2

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {4516, 3392, 32, 2715, 8} \[ \int (c+d x)^4 \csc (x) \sin (3 x) \, dx=\frac {3}{2} d^3 \sin ^2(x) (c+d x)-\frac {9}{2} d^3 \cos ^2(x) (c+d x)-6 d^2 \sin (x) \cos (x) (c+d x)^2+\frac {(c+d x)^5}{5 d}-d (c+d x)^3-d \sin ^2(x) (c+d x)^3+3 d \cos ^2(x) (c+d x)^3+2 \sin (x) \cos (x) (c+d x)^4+\frac {3 d^4 x}{2}+3 d^4 \sin (x) \cos (x) \]

[In]

Int[(c + d*x)^4*Csc[x]*Sin[3*x],x]

[Out]

(3*d^4*x)/2 - d*(c + d*x)^3 + (c + d*x)^5/(5*d) - (9*d^3*(c + d*x)*Cos[x]^2)/2 + 3*d*(c + d*x)^3*Cos[x]^2 + 3*
d^4*Cos[x]*Sin[x] - 6*d^2*(c + d*x)^2*Cos[x]*Sin[x] + 2*(c + d*x)^4*Cos[x]*Sin[x] + (3*d^3*(c + d*x)*Sin[x]^2)
/2 - d*(c + d*x)^3*Sin[x]^2

Rule 8

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2715

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] && Integ
erQ[2*n]

Rule 3392

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*m*(c + d*x)^(m - 1)*((
b*Sin[e + f*x])^n/(f^2*n^2)), x] + (Dist[b^2*((n - 1)/n), Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[d^2*m*((m - 1)/(f^2*n^2)), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[b*(c + d*x)^m*Cos[e + f
*x]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 4516

Int[((e_.) + (f_.)*(x_))^(m_.)*(F_)[(a_.) + (b_.)*(x_)]^(p_.)*(G_)[(c_.) + (d_.)*(x_)]^(q_.), x_Symbol] :> Int
[ExpandTrigExpand[(e + f*x)^m*G[c + d*x]^q, F, c + d*x, p, b/d, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && M
emberQ[{Sin, Cos}, F] && MemberQ[{Sec, Csc}, G] && IGtQ[p, 0] && IGtQ[q, 0] && EqQ[b*c - a*d, 0] && IGtQ[b/d,
1]

Rubi steps \begin{align*} \text {integral}& = \int \left (3 (c+d x)^4 \cos ^2(x)-(c+d x)^4 \sin ^2(x)\right ) \, dx \\ & = 3 \int (c+d x)^4 \cos ^2(x) \, dx-\int (c+d x)^4 \sin ^2(x) \, dx \\ & = 3 d (c+d x)^3 \cos ^2(x)+2 (c+d x)^4 \cos (x) \sin (x)-d (c+d x)^3 \sin ^2(x)-\frac {1}{2} \int (c+d x)^4 \, dx+\frac {3}{2} \int (c+d x)^4 \, dx+\left (3 d^2\right ) \int (c+d x)^2 \sin ^2(x) \, dx-\left (9 d^2\right ) \int (c+d x)^2 \cos ^2(x) \, dx \\ & = \frac {(c+d x)^5}{5 d}-\frac {9}{2} d^3 (c+d x) \cos ^2(x)+3 d (c+d x)^3 \cos ^2(x)-6 d^2 (c+d x)^2 \cos (x) \sin (x)+2 (c+d x)^4 \cos (x) \sin (x)+\frac {3}{2} d^3 (c+d x) \sin ^2(x)-d (c+d x)^3 \sin ^2(x)+\frac {1}{2} \left (3 d^2\right ) \int (c+d x)^2 \, dx-\frac {1}{2} \left (9 d^2\right ) \int (c+d x)^2 \, dx-\frac {1}{2} \left (3 d^4\right ) \int \sin ^2(x) \, dx+\frac {1}{2} \left (9 d^4\right ) \int \cos ^2(x) \, dx \\ & = -d (c+d x)^3+\frac {(c+d x)^5}{5 d}-\frac {9}{2} d^3 (c+d x) \cos ^2(x)+3 d (c+d x)^3 \cos ^2(x)+3 d^4 \cos (x) \sin (x)-6 d^2 (c+d x)^2 \cos (x) \sin (x)+2 (c+d x)^4 \cos (x) \sin (x)+\frac {3}{2} d^3 (c+d x) \sin ^2(x)-d (c+d x)^3 \sin ^2(x)-\frac {1}{4} \left (3 d^4\right ) \int 1 \, dx+\frac {1}{4} \left (9 d^4\right ) \int 1 \, dx \\ & = \frac {3 d^4 x}{2}-d (c+d x)^3+\frac {(c+d x)^5}{5 d}-\frac {9}{2} d^3 (c+d x) \cos ^2(x)+3 d (c+d x)^3 \cos ^2(x)+3 d^4 \cos (x) \sin (x)-6 d^2 (c+d x)^2 \cos (x) \sin (x)+2 (c+d x)^4 \cos (x) \sin (x)+\frac {3}{2} d^3 (c+d x) \sin ^2(x)-d (c+d x)^3 \sin ^2(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.18 \[ \int (c+d x)^4 \csc (x) \sin (3 x) \, dx=c^4 x+2 c^3 d x^2+2 c^2 d^2 x^3+c d^3 x^4+\frac {d^4 x^5}{5}+d \left (2 c^3+6 c^2 d x+d^3 x \left (-3+2 x^2\right )+3 c d^2 \left (-1+2 x^2\right )\right ) \cos (2 x)+\frac {1}{2} \left (2 c^4+8 c^3 d x+4 c d^3 x \left (-3+2 x^2\right )+6 c^2 d^2 \left (-1+2 x^2\right )+d^4 \left (3-6 x^2+2 x^4\right )\right ) \sin (2 x) \]

[In]

Integrate[(c + d*x)^4*Csc[x]*Sin[3*x],x]

[Out]

c^4*x + 2*c^3*d*x^2 + 2*c^2*d^2*x^3 + c*d^3*x^4 + (d^4*x^5)/5 + d*(2*c^3 + 6*c^2*d*x + d^3*x*(-3 + 2*x^2) + 3*
c*d^2*(-1 + 2*x^2))*Cos[2*x] + ((2*c^4 + 8*c^3*d*x + 4*c*d^3*x*(-3 + 2*x^2) + 6*c^2*d^2*(-1 + 2*x^2) + d^4*(3
- 6*x^2 + 2*x^4))*Sin[2*x])/2

Maple [A] (verified)

Time = 2.12 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.33

method	result	size
risch	\(\frac {d^{4} x^{5}}{5}+c \,d^{3} x^{4}+2 c^{2} d^{2} x^{3}+2 c^{3} d \,x^{2}+c^{4} x +\frac {c^{5}}{5 d}+d \left (2 d^{3} x^{3}+6 c \,d^{2} x^{2}+6 c^{2} d x -3 d^{3} x +2 c^{3}-3 c \,d^{2}\right ) \cos \left (2 x \right )+\frac {\left (2 d^{4} x^{4}+8 c \,d^{3} x^{3}+12 c^{2} d^{2} x^{2}-6 d^{4} x^{2}+8 c^{3} d x -12 c \,d^{3} x +2 c^{4}-6 c^{2} d^{2}+3 d^{4}\right ) \sin \left (2 x \right )}{2}\)	\(174\)
default	\(4 d^{4} \left (x^{4} \left (\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}\right )+x^{3} \cos \left (x \right )^{2}-3 x^{2} \left (\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}\right )-\frac {3 x \cos \left (x \right )^{2}}{2}+\frac {3 \cos \left (x \right ) \sin \left (x \right )}{4}+\frac {3 x}{4}+x^{3}-\frac {2 x^{5}}{5}\right )+16 c \,d^{3} \left (x^{3} \left (\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}\right )+\frac {3 x^{2} \cos \left (x \right )^{2}}{4}-\frac {3 x \left (\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}\right )}{2}+\frac {3 x^{2}}{8}+\frac {3 \sin \left (x \right )^{2}}{8}-\frac {3 x^{4}}{8}\right )+24 c^{2} d^{2} \left (x^{2} \left (\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}\right )+\frac {x \cos \left (x \right )^{2}}{2}-\frac {\cos \left (x \right ) \sin \left (x \right )}{4}-\frac {x}{4}-\frac {x^{3}}{3}\right )+16 c^{3} d \left (x \left (\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}\right )-\frac {x^{2}}{4}-\frac {\sin \left (x \right )^{2}}{4}\right )+4 c^{4} \left (\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}\right )-\frac {d^{4} x^{5}}{5}-c \,d^{3} x^{4}-2 c^{2} d^{2} x^{3}-2 c^{3} d \,x^{2}-c^{4} x\)	\(260\)

[In]

int((d*x+c)^4*csc(x)*sin(3*x),x,method=_RETURNVERBOSE)

[Out]

1/5*d^4*x^5+c*d^3*x^4+2*c^2*d^2*x^3+2*c^3*d*x^2+c^4*x+1/5/d*c^5+d*(2*d^3*x^3+6*c*d^2*x^2+6*c^2*d*x-3*d^3*x+2*c
^3-3*c*d^2)*cos(2*x)+1/2*(2*d^4*x^4+8*c*d^3*x^3+12*c^2*d^2*x^2-6*d^4*x^2+8*c^3*d*x-12*c*d^3*x+2*c^4-6*c^2*d^2+
3*d^4)*sin(2*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.53 \[ \int (c+d x)^4 \csc (x) \sin (3 x) \, dx=\frac {1}{5} \, d^{4} x^{5} + c d^{3} x^{4} + 2 \, {\left (c^{2} d^{2} - d^{4}\right )} x^{3} + 2 \, {\left (c^{3} d - 3 \, c d^{3}\right )} x^{2} + 2 \, {\left (2 \, d^{4} x^{3} + 6 \, c d^{3} x^{2} + 2 \, c^{3} d - 3 \, c d^{3} + 3 \, {\left (2 \, c^{2} d^{2} - d^{4}\right )} x\right )} \cos \left (x\right )^{2} + {\left (2 \, d^{4} x^{4} + 8 \, c d^{3} x^{3} + 2 \, c^{4} - 6 \, c^{2} d^{2} + 3 \, d^{4} + 6 \, {\left (2 \, c^{2} d^{2} - d^{4}\right )} x^{2} + 4 \, {\left (2 \, c^{3} d - 3 \, c d^{3}\right )} x\right )} \cos \left (x\right ) \sin \left (x\right ) + {\left (c^{4} - 6 \, c^{2} d^{2} + 3 \, d^{4}\right )} x \]

[In]

integrate((d*x+c)^4*csc(x)*sin(3*x),x, algorithm="fricas")

[Out]

1/5*d^4*x^5 + c*d^3*x^4 + 2*(c^2*d^2 - d^4)*x^3 + 2*(c^3*d - 3*c*d^3)*x^2 + 2*(2*d^4*x^3 + 6*c*d^3*x^2 + 2*c^3
*d - 3*c*d^3 + 3*(2*c^2*d^2 - d^4)*x)*cos(x)^2 + (2*d^4*x^4 + 8*c*d^3*x^3 + 2*c^4 - 6*c^2*d^2 + 3*d^4 + 6*(2*c
^2*d^2 - d^4)*x^2 + 4*(2*c^3*d - 3*c*d^3)*x)*cos(x)*sin(x) + (c^4 - 6*c^2*d^2 + 3*d^4)*x

Sympy [A] (verification not implemented)

Time = 14.92 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.70 \[ \int (c+d x)^4 \csc (x) \sin (3 x) \, dx=c^{4} \left (x + \sin {\left (2 x \right )}\right ) + 4 c^{3} d \left (- \frac {x^{2}}{2} + x \left (x + \sin {\left (2 x \right )}\right ) + \frac {\cos {\left (2 x \right )}}{2}\right ) + 6 c^{2} d^{2} \left (\frac {x^{3}}{3} + x^{2} \left (x + \sin {\left (2 x \right )}\right ) - 2 x \left (\frac {x^{2}}{2} - \frac {\cos {\left (2 x \right )}}{2}\right ) - \frac {\sin {\left (2 x \right )}}{2}\right ) + 4 c d^{3} \left (- \frac {x^{4}}{4} + x^{3} \left (x + \sin {\left (2 x \right )}\right ) - 3 x^{2} \left (\frac {x^{2}}{2} - \frac {\cos {\left (2 x \right )}}{2}\right ) + 3 x \left (\frac {x^{3}}{3} - \frac {\sin {\left (2 x \right )}}{2}\right ) - \frac {3 \cos {\left (2 x \right )}}{4}\right ) + d^{4} \left (\frac {x^{5}}{5} + x^{4} \left (x + \sin {\left (2 x \right )}\right ) - 4 x^{3} \left (\frac {x^{2}}{2} - \frac {\cos {\left (2 x \right )}}{2}\right ) + 6 x^{2} \left (\frac {x^{3}}{3} - \frac {\sin {\left (2 x \right )}}{2}\right ) - 2 x \left (\frac {x^{4}}{2} + \frac {3 \cos {\left (2 x \right )}}{2}\right ) + \frac {3 \sin {\left (2 x \right )}}{2}\right ) \]

[In]

integrate((d*x+c)**4*csc(x)*sin(3*x),x)

[Out]

c**4*(x + sin(2*x)) + 4*c**3*d*(-x**2/2 + x*(x + sin(2*x)) + cos(2*x)/2) + 6*c**2*d**2*(x**3/3 + x**2*(x + sin
(2*x)) - 2*x*(x**2/2 - cos(2*x)/2) - sin(2*x)/2) + 4*c*d**3*(-x**4/4 + x**3*(x + sin(2*x)) - 3*x**2*(x**2/2 -
cos(2*x)/2) + 3*x*(x**3/3 - sin(2*x)/2) - 3*cos(2*x)/4) + d**4*(x**5/5 + x**4*(x + sin(2*x)) - 4*x**3*(x**2/2
- cos(2*x)/2) + 6*x**2*(x**3/3 - sin(2*x)/2) - 2*x*(x**4/2 + 3*cos(2*x)/2) + 3*sin(2*x)/2)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.11 \[ \int (c+d x)^4 \csc (x) \sin (3 x) \, dx=2 \, {\left (x^{2} + 2 \, x \sin \left (2 \, x\right ) + \cos \left (2 \, x\right )\right )} c^{3} d + {\left (2 \, x^{3} + 6 \, x \cos \left (2 \, x\right ) + 3 \, {\left (2 \, x^{2} - 1\right )} \sin \left (2 \, x\right )\right )} c^{2} d^{2} + {\left (x^{4} + 3 \, {\left (2 \, x^{2} - 1\right )} \cos \left (2 \, x\right ) + 2 \, {\left (2 \, x^{3} - 3 \, x\right )} \sin \left (2 \, x\right )\right )} c d^{3} + \frac {1}{10} \, {\left (2 \, x^{5} + 10 \, {\left (2 \, x^{3} - 3 \, x\right )} \cos \left (2 \, x\right ) + 5 \, {\left (2 \, x^{4} - 6 \, x^{2} + 3\right )} \sin \left (2 \, x\right )\right )} d^{4} + c^{4} {\left (x + \sin \left (2 \, x\right )\right )} \]

[In]

integrate((d*x+c)^4*csc(x)*sin(3*x),x, algorithm="maxima")

[Out]

2*(x^2 + 2*x*sin(2*x) + cos(2*x))*c^3*d + (2*x^3 + 6*x*cos(2*x) + 3*(2*x^2 - 1)*sin(2*x))*c^2*d^2 + (x^4 + 3*(
2*x^2 - 1)*cos(2*x) + 2*(2*x^3 - 3*x)*sin(2*x))*c*d^3 + 1/10*(2*x^5 + 10*(2*x^3 - 3*x)*cos(2*x) + 5*(2*x^4 - 6
*x^2 + 3)*sin(2*x))*d^4 + c^4*(x + sin(2*x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.27 \[ \int (c+d x)^4 \csc (x) \sin (3 x) \, dx=\frac {1}{5} \, d^{4} x^{5} + c d^{3} x^{4} + 2 \, c^{2} d^{2} x^{3} + 2 \, c^{3} d x^{2} + c^{4} x + {\left (2 \, d^{4} x^{3} + 6 \, c d^{3} x^{2} + 6 \, c^{2} d^{2} x - 3 \, d^{4} x + 2 \, c^{3} d - 3 \, c d^{3}\right )} \cos \left (2 \, x\right ) + \frac {1}{2} \, {\left (2 \, d^{4} x^{4} + 8 \, c d^{3} x^{3} + 12 \, c^{2} d^{2} x^{2} - 6 \, d^{4} x^{2} + 8 \, c^{3} d x - 12 \, c d^{3} x + 2 \, c^{4} - 6 \, c^{2} d^{2} + 3 \, d^{4}\right )} \sin \left (2 \, x\right ) \]

[In]

integrate((d*x+c)^4*csc(x)*sin(3*x),x, algorithm="giac")

[Out]

1/5*d^4*x^5 + c*d^3*x^4 + 2*c^2*d^2*x^3 + 2*c^3*d*x^2 + c^4*x + (2*d^4*x^3 + 6*c*d^3*x^2 + 6*c^2*d^2*x - 3*d^4
*x + 2*c^3*d - 3*c*d^3)*cos(2*x) + 1/2*(2*d^4*x^4 + 8*c*d^3*x^3 + 12*c^2*d^2*x^2 - 6*d^4*x^2 + 8*c^3*d*x - 12*
c*d^3*x + 2*c^4 - 6*c^2*d^2 + 3*d^4)*sin(2*x)

Mupad [B] (verification not implemented)

Time = 26.26 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.62 \[ \int (c+d x)^4 \csc (x) \sin (3 x) \, dx=c^4\,\sin \left (2\,x\right )+\frac {3\,d^4\,\sin \left (2\,x\right )}{2}+c^4\,x+\frac {d^4\,x^5}{5}-3\,c^2\,d^2\,\sin \left (2\,x\right )+2\,d^4\,x^3\,\cos \left (2\,x\right )-3\,d^4\,x^2\,\sin \left (2\,x\right )+d^4\,x^4\,\sin \left (2\,x\right )+2\,c^3\,d\,x^2+c\,d^3\,x^4+2\,c^2\,d^2\,x^3-3\,c\,d^3\,\cos \left (2\,x\right )+2\,c^3\,d\,\cos \left (2\,x\right )-3\,d^4\,x\,\cos \left (2\,x\right )+6\,c^2\,d^2\,x^2\,\sin \left (2\,x\right )-6\,c\,d^3\,x\,\sin \left (2\,x\right )+4\,c^3\,d\,x\,\sin \left (2\,x\right )+6\,c^2\,d^2\,x\,\cos \left (2\,x\right )+6\,c\,d^3\,x^2\,\cos \left (2\,x\right )+4\,c\,d^3\,x^3\,\sin \left (2\,x\right ) \]

[In]

int((sin(3*x)*(c + d*x)^4)/sin(x),x)

[Out]

c^4*sin(2*x) + (3*d^4*sin(2*x))/2 + c^4*x + (d^4*x^5)/5 - 3*c^2*d^2*sin(2*x) + 2*d^4*x^3*cos(2*x) - 3*d^4*x^2*
sin(2*x) + d^4*x^4*sin(2*x) + 2*c^3*d*x^2 + c*d^3*x^4 + 2*c^2*d^2*x^3 - 3*c*d^3*cos(2*x) + 2*c^3*d*cos(2*x) -
3*d^4*x*cos(2*x) + 6*c^2*d^2*x^2*sin(2*x) - 6*c*d^3*x*sin(2*x) + 4*c^3*d*x*sin(2*x) + 6*c^2*d^2*x*cos(2*x) + 6
*c*d^3*x^2*cos(2*x) + 4*c*d^3*x^3*sin(2*x)