\(\int (c+d x)^2 \csc (x) \sin (3 x) \, dx\) [363]

Optimal result

Integrand size = 14, antiderivative size = 73 \[ \int (c+d x)^2 \csc (x) \sin (3 x) \, dx=-\frac {d^2 x}{2}+\frac {(c+d x)^3}{3 d}+\frac {3}{2} d (c+d x) \cos ^2(x)-d^2 \cos (x) \sin (x)+2 (c+d x)^2 \cos (x) \sin (x)-\frac {1}{2} d (c+d x) \sin ^2(x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {4516, 3392, 32, 2715, 8} \[ \int (c+d x)^2 \csc (x) \sin (3 x) \, dx=\frac {(c+d x)^3}{3 d}-\frac {1}{2} d \sin ^2(x) (c+d x)+\frac {3}{2} d \cos ^2(x) (c+d x)+2 \sin (x) \cos (x) (c+d x)^2-\frac {d^2 x}{2}-d^2 \sin (x) \cos (x) \]

[In]

Int[(c + d*x)^2*Csc[x]*Sin[3*x],x]

[Out]

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.82 \[ \int (c+d x)^2 \csc (x) \sin (3 x) \, dx=c^2 x+c d x^2+\frac {d^2 x^3}{3}+d (c+d x) \cos (2 x)+\left (2 c^2+4 c d x+d^2 \left (-1+2 x^2\right )\right ) \cos (x) \sin (x) \]

[In]

Integrate[(c + d*x)^2*Csc[x]*Sin[3*x],x]

[Out]

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

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.96

method	result	size
risch	\(\frac {d^{2} x^{3}}{3}+d c \,x^{2}+c^{2} x +\frac {c^{3}}{3 d}+d \left (d x +c \right ) \cos \left (2 x \right )+\frac {\left (2 x^{2} d^{2}+4 c d x +2 c^{2}-d^{2}\right ) \sin \left (2 x \right )}{2}\)	\(70\)
default	\(4 d^{2} \left (x^{2} \left (\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}\right )+\frac {x \cos \left (x \right )^{2}}{2}-\frac {\cos \left (x \right ) \sin \left (x \right )}{4}-\frac {x}{4}-\frac {x^{3}}{3}\right )+8 c d \left (x \left (\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}\right )-\frac {x^{2}}{4}-\frac {\sin \left (x \right )^{2}}{4}\right )+4 c^{2} \left (\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}\right )-\frac {d^{2} x^{3}}{3}-d c \,x^{2}-c^{2} x\)	\(107\)

[In]

int((d*x+c)^2*csc(x)*sin(3*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.96 \[ \int (c+d x)^2 \csc (x) \sin (3 x) \, dx=\frac {1}{3} \, d^{2} x^{3} + c d x^{2} + 2 \, {\left (d^{2} x + c d\right )} \cos \left (x\right )^{2} + {\left (2 \, d^{2} x^{2} + 4 \, c d x + 2 \, c^{2} - d^{2}\right )} \cos \left (x\right ) \sin \left (x\right ) + {\left (c^{2} - d^{2}\right )} x \]

[In]

integrate((d*x+c)^2*csc(x)*sin(3*x),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 3.89 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.03 \[ \int (c+d x)^2 \csc (x) \sin (3 x) \, dx=c^{2} \left (x + \sin {\left (2 x \right )}\right ) + 2 c d \left (- \frac {x^{2}}{2} + x \left (x + \sin {\left (2 x \right )}\right ) + \frac {\cos {\left (2 x \right )}}{2}\right ) + d^{2} \left (\frac {x^{3}}{3} + x^{2} \left (x + \sin {\left (2 x \right )}\right ) - 2 x \left (\frac {x^{2}}{2} - \frac {\cos {\left (2 x \right )}}{2}\right ) - \frac {\sin {\left (2 x \right )}}{2}\right ) \]

[In]

integrate((d*x+c)**2*csc(x)*sin(3*x),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.82 \[ \int (c+d x)^2 \csc (x) \sin (3 x) \, dx={\left (x^{2} + 2 \, x \sin \left (2 \, x\right ) + \cos \left (2 \, x\right )\right )} c d + \frac {1}{6} \, {\left (2 \, x^{3} + 6 \, x \cos \left (2 \, x\right ) + 3 \, {\left (2 \, x^{2} - 1\right )} \sin \left (2 \, x\right )\right )} d^{2} + c^{2} {\left (x + \sin \left (2 \, x\right )\right )} \]

[In]

integrate((d*x+c)^2*csc(x)*sin(3*x),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.88 \[ \int (c+d x)^2 \csc (x) \sin (3 x) \, dx=\frac {1}{3} \, d^{2} x^{3} + c d x^{2} + c^{2} x + {\left (d^{2} x + c d\right )} \cos \left (2 \, x\right ) + \frac {1}{2} \, {\left (2 \, d^{2} x^{2} + 4 \, c d x + 2 \, c^{2} - d^{2}\right )} \sin \left (2 \, x\right ) \]

[In]

integrate((d*x+c)^2*csc(x)*sin(3*x),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 25.72 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00 \[ \int (c+d x)^2 \csc (x) \sin (3 x) \, dx=c^2\,\sin \left (2\,x\right )-\frac {d^2\,\sin \left (2\,x\right )}{2}+c^2\,x+\frac {d^2\,x^3}{3}+d^2\,x^2\,\sin \left (2\,x\right )+c\,d\,\cos \left (2\,x\right )+d^2\,x\,\cos \left (2\,x\right )+c\,d\,x^2+2\,c\,d\,x\,\sin \left (2\,x\right ) \]

[In]

int((sin(3*x)*(c + d*x)^2)/sin(x),x)

[Out]

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