3.363 \(\int (c+d x)^2 \csc (x) \sin (3 x) \, dx\)

Optimal. Leaf size=73 \[ \frac{(c+d x)^3}{3 d}-\frac{1}{2} d \sin ^2(x) (c+d x)+\frac{3}{2} d \cos ^2(x) (c+d x)+2 \sin (x) \cos (x) (c+d x)^2-\frac{d^2 x}{2}-d^2 \sin (x) \cos (x) \]

[Out]

-(d^2*x)/2 + (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

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Rubi [A]  time = 0.102692, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {4431, 3311, 32, 2635, 8} \[ \frac{(c+d x)^3}{3 d}-\frac{1}{2} d \sin ^2(x) (c+d x)+\frac{3}{2} d \cos ^2(x) (c+d x)+2 \sin (x) \cos (x) (c+d x)^2-\frac{d^2 x}{2}-d^2 \sin (x) \cos (x) \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Csc[x]*Sin[3*x],x]

[Out]

-(d^2*x)/2 + (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 4431

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]

Rule 3311

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 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 2635

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] && Integer
Q[2*n]

Rule 8

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

Rubi steps

\begin{align*} \int (c+d x)^2 \csc (x) \sin (3 x) \, dx &=\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]  time = 0.110072, size = 60, normalized size = 0.82 \[ \sin (x) \cos (x) \left (2 c^2+4 c d x+d^2 \left (2 x^2-1\right )\right )+c^2 x+c d x^2+d \cos (2 x) (c+d x)+\frac{d^2 x^3}{3} \]

Antiderivative was successfully verified.

[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]

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Maple [A]  time = 0.054, size = 107, normalized size = 1.5 \begin{align*} 4\,{d}^{2} \left ({x}^{2} \left ( 1/2\,\cos \left ( x \right ) \sin \left ( x \right ) +x/2 \right ) +1/2\,x \left ( \cos \left ( x \right ) \right ) ^{2}-1/4\,\cos \left ( x \right ) \sin \left ( x \right ) -x/4-1/3\,{x}^{3} \right ) +8\,cd \left ( x \left ( 1/2\,\cos \left ( x \right ) \sin \left ( x \right ) +x/2 \right ) -1/4\,{x}^{2}-1/4\, \left ( \sin \left ( x \right ) \right ) ^{2} \right ) +4\,{c}^{2} \left ( 1/2\,\cos \left ( x \right ) \sin \left ( x \right ) +x/2 \right ) -{\frac{{d}^{2}{x}^{3}}{3}}-cd{x}^{2}-{c}^{2}x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.995402, size = 81, normalized size = 1.11 \begin{align*}{\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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Fricas [A]  time = 0.512624, size = 159, normalized size = 2.18 \begin{align*} \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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [B]  time = 21.0139, size = 155, normalized size = 2.12 \begin{align*} c^{2} x + c^{2} \sin{\left (2 x \right )} - 2 c d x^{2} \sin ^{2}{\left (x \right )} - 2 c d x^{2} \cos ^{2}{\left (x \right )} + 3 c d x^{2} + 4 c d x \sin{\left (x \right )} \cos{\left (x \right )} - 2 c d \sin ^{2}{\left (x \right )} - \frac{2 d^{2} x^{3} \sin ^{2}{\left (x \right )}}{3} - \frac{2 d^{2} x^{3} \cos ^{2}{\left (x \right )}}{3} + d^{2} x^{3} + 2 d^{2} x^{2} \sin{\left (x \right )} \cos{\left (x \right )} - d^{2} x \sin ^{2}{\left (x \right )} + d^{2} x \cos ^{2}{\left (x \right )} - d^{2} \sin{\left (x \right )} \cos{\left (x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12762, size = 86, normalized size = 1.18 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)