\(\int (c+d x)^3 \csc (x) \sin (3 x) \, dx\) [362]

Optimal result

Integrand size = 14, antiderivative size = 115 \[ \int (c+d x)^3 \csc (x) \sin (3 x) \, dx=-\frac {3}{2} c d^2 x-\frac {3 d^3 x^2}{4}+\frac {(c+d x)^4}{4 d}-\frac {9}{8} d^3 \cos ^2(x)+\frac {9}{4} d (c+d x)^2 \cos ^2(x)-3 d^2 (c+d x) \cos (x) \sin (x)+2 (c+d x)^3 \cos (x) \sin (x)+\frac {3}{8} d^3 \sin ^2(x)-\frac {3}{4} d (c+d x)^2 \sin ^2(x) \]

[Out]

-3/2*c*d^2*x-3/4*d^3*x^2+1/4*(d*x+c)^4/d-9/8*d^3*cos(x)^2+9/4*d*(d*x+c)^2*cos(x)^2-3*d^2*(d*x+c)*cos(x)*sin(x)
+2*(d*x+c)^3*cos(x)*sin(x)+3/8*d^3*sin(x)^2-3/4*d*(d*x+c)^2*sin(x)^2

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4516, 3392, 32, 3391} \[ \int (c+d x)^3 \csc (x) \sin (3 x) \, dx=-\frac {3}{2} c d^2 x-3 d^2 \sin (x) \cos (x) (c+d x)+\frac {(c+d x)^4}{4 d}-\frac {3}{4} d \sin ^2(x) (c+d x)^2+\frac {9}{4} d \cos ^2(x) (c+d x)^2+2 \sin (x) \cos (x) (c+d x)^3-\frac {3 d^3 x^2}{4}+\frac {3}{8} d^3 \sin ^2(x)-\frac {9}{8} d^3 \cos ^2(x) \]

[In]

Int[(c + d*x)^3*Csc[x]*Sin[3*x],x]

[Out]

(-3*c*d^2*x)/2 - (3*d^3*x^2)/4 + (c + d*x)^4/(4*d) - (9*d^3*Cos[x]^2)/8 + (9*d*(c + d*x)^2*Cos[x]^2)/4 - 3*d^2
*(c + d*x)*Cos[x]*Sin[x] + 2*(c + d*x)^3*Cos[x]*Sin[x] + (3*d^3*Sin[x]^2)/8 - (3*d*(c + d*x)^2*Sin[x]^2)/4

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 3391

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

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)^3 \cos ^2(x)-(c+d x)^3 \sin ^2(x)\right ) \, dx \\ & = 3 \int (c+d x)^3 \cos ^2(x) \, dx-\int (c+d x)^3 \sin ^2(x) \, dx \\ & = \frac {9}{4} d (c+d x)^2 \cos ^2(x)+2 (c+d x)^3 \cos (x) \sin (x)-\frac {3}{4} d (c+d x)^2 \sin ^2(x)-\frac {1}{2} \int (c+d x)^3 \, dx+\frac {3}{2} \int (c+d x)^3 \, dx+\frac {1}{2} \left (3 d^2\right ) \int (c+d x) \sin ^2(x) \, dx-\frac {1}{2} \left (9 d^2\right ) \int (c+d x) \cos ^2(x) \, dx \\ & = \frac {(c+d x)^4}{4 d}-\frac {9}{8} d^3 \cos ^2(x)+\frac {9}{4} d (c+d x)^2 \cos ^2(x)-3 d^2 (c+d x) \cos (x) \sin (x)+2 (c+d x)^3 \cos (x) \sin (x)+\frac {3}{8} d^3 \sin ^2(x)-\frac {3}{4} d (c+d x)^2 \sin ^2(x)+\frac {1}{4} \left (3 d^2\right ) \int (c+d x) \, dx-\frac {1}{4} \left (9 d^2\right ) \int (c+d x) \, dx \\ & = -\frac {3}{2} c d^2 x-\frac {3 d^3 x^2}{4}+\frac {(c+d x)^4}{4 d}-\frac {9}{8} d^3 \cos ^2(x)+\frac {9}{4} d (c+d x)^2 \cos ^2(x)-3 d^2 (c+d x) \cos (x) \sin (x)+2 (c+d x)^3 \cos (x) \sin (x)+\frac {3}{8} d^3 \sin ^2(x)-\frac {3}{4} d (c+d x)^2 \sin ^2(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.95 \[ \int (c+d x)^3 \csc (x) \sin (3 x) \, dx=\frac {1}{4} \left (x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right )+3 d \left (2 c^2+4 c d x+d^2 \left (-1+2 x^2\right )\right ) \cos (2 x)+2 \left (2 c^3+6 c^2 d x+d^3 x \left (-3+2 x^2\right )+3 c d^2 \left (-1+2 x^2\right )\right ) \sin (2 x)\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.13 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.03

method	result	size
risch	\(\frac {d^{3} x^{4}}{4}+c \,d^{2} x^{3}+\frac {3 c^{2} d \,x^{2}}{2}+c^{3} x +\frac {c^{4}}{4 d}+\frac {3 d \left (2 x^{2} d^{2}+4 c d x +2 c^{2}-d^{2}\right ) \cos \left (2 x \right )}{4}+\frac {\left (2 d^{3} x^{3}+6 c \,d^{2} x^{2}+6 c^{2} d x -3 d^{3} x +2 c^{3}-3 c \,d^{2}\right ) \sin \left (2 x \right )}{2}\)	\(119\)
default	\(4 d^{3} \left (x^{3} \left (\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}\right )+\frac {3 x^{2} \cos \left (x \right )^{2}}{4}-\frac {3 x \left (\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}\right )}{2}+\frac {3 x^{2}}{8}+\frac {3 \sin \left (x \right )^{2}}{8}-\frac {3 x^{4}}{8}\right )+12 c \,d^{2} \left (x^{2} \left (\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}\right )+\frac {x \cos \left (x \right )^{2}}{2}-\frac {\cos \left (x \right ) \sin \left (x \right )}{4}-\frac {x}{4}-\frac {x^{3}}{3}\right )+12 c^{2} d \left (x \left (\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}\right )-\frac {x^{2}}{4}-\frac {\sin \left (x \right )^{2}}{4}\right )+4 c^{3} \left (\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}\right )-\frac {d^{3} x^{4}}{4}-c \,d^{2} x^{3}-\frac {3 c^{2} d \,x^{2}}{2}-c^{3} x\)	\(179\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.10 \[ \int (c+d x)^3 \csc (x) \sin (3 x) \, dx=\frac {1}{4} \, d^{3} x^{4} + c d^{2} x^{3} + \frac {3}{2} \, {\left (c^{2} d - d^{3}\right )} x^{2} + \frac {3}{2} \, {\left (2 \, d^{3} x^{2} + 4 \, c d^{2} x + 2 \, c^{2} d - d^{3}\right )} \cos \left (x\right )^{2} + {\left (2 \, d^{3} x^{3} + 6 \, c d^{2} x^{2} + 2 \, c^{3} - 3 \, c d^{2} + 3 \, {\left (2 \, c^{2} d - d^{3}\right )} x\right )} \cos \left (x\right ) \sin \left (x\right ) + {\left (c^{3} - 3 \, c d^{2}\right )} x \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 7.78 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.21 \[ \int (c+d x)^3 \csc (x) \sin (3 x) \, dx=c^{3} \left (x + \sin {\left (2 x \right )}\right ) + 3 c^{2} d \left (- \frac {x^{2}}{2} + x \left (x + \sin {\left (2 x \right )}\right ) + \frac {\cos {\left (2 x \right )}}{2}\right ) + 3 c d^{2} \left (\frac {x^{3}}{3} + x^{2} \left (x + \sin {\left (2 x \right )}\right ) - 2 x \left (\frac {x^{2}}{2} - \frac {\cos {\left (2 x \right )}}{2}\right ) - \frac {\sin {\left (2 x \right )}}{2}\right ) + d^{3} \left (- \frac {x^{4}}{4} + x^{3} \left (x + \sin {\left (2 x \right )}\right ) - 3 x^{2} \left (\frac {x^{2}}{2} - \frac {\cos {\left (2 x \right )}}{2}\right ) + 3 x \left (\frac {x^{3}}{3} - \frac {\sin {\left (2 x \right )}}{2}\right ) - \frac {3 \cos {\left (2 x \right )}}{4}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.88 \[ \int (c+d x)^3 \csc (x) \sin (3 x) \, dx=\frac {3}{2} \, {\left (x^{2} + 2 \, x \sin \left (2 \, x\right ) + \cos \left (2 \, x\right )\right )} c^{2} d + \frac {1}{2} \, {\left (2 \, x^{3} + 6 \, x \cos \left (2 \, x\right ) + 3 \, {\left (2 \, x^{2} - 1\right )} \sin \left (2 \, x\right )\right )} c d^{2} + \frac {1}{4} \, {\left (x^{4} + 3 \, {\left (2 \, x^{2} - 1\right )} \cos \left (2 \, x\right ) + 2 \, {\left (2 \, x^{3} - 3 \, x\right )} \sin \left (2 \, x\right )\right )} d^{3} + c^{3} {\left (x + \sin \left (2 \, x\right )\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.97 \[ \int (c+d x)^3 \csc (x) \sin (3 x) \, dx=\frac {1}{4} \, d^{3} x^{4} + c d^{2} x^{3} + \frac {3}{2} \, c^{2} d x^{2} + c^{3} x + \frac {3}{4} \, {\left (2 \, d^{3} x^{2} + 4 \, c d^{2} x + 2 \, c^{2} d - d^{3}\right )} \cos \left (2 \, x\right ) + \frac {1}{2} \, {\left (2 \, d^{3} x^{3} + 6 \, c d^{2} x^{2} + 6 \, c^{2} d x - 3 \, d^{3} x + 2 \, c^{3} - 3 \, c d^{2}\right )} \sin \left (2 \, x\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.18 \[ \int (c+d x)^3 \csc (x) \sin (3 x) \, dx=c^3\,\sin \left (2\,x\right )-\frac {3\,d^3\,\cos \left (2\,x\right )}{4}+c^3\,x+\frac {d^3\,x^4}{4}+\frac {3\,d^3\,x^2\,\cos \left (2\,x\right )}{2}+d^3\,x^3\,\sin \left (2\,x\right )+\frac {3\,c^2\,d\,x^2}{2}+c\,d^2\,x^3+\frac {3\,c^2\,d\,\cos \left (2\,x\right )}{2}-\frac {3\,c\,d^2\,\sin \left (2\,x\right )}{2}-\frac {3\,d^3\,x\,\sin \left (2\,x\right )}{2}+3\,c\,d^2\,x\,\cos \left (2\,x\right )+3\,c^2\,d\,x\,\sin \left (2\,x\right )+3\,c\,d^2\,x^2\,\sin \left (2\,x\right ) \]

[In]

int((sin(3*x)*(c + d*x)^3)/sin(x),x)

[Out]

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