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

Optimal. Leaf size=115 \[ -\frac{3}{2} c d^2 x-3 d^2 \sin (x) \cos (x) (c+d x)+\frac{(c+d x)^4}{4 d}-\frac{3}{4} d \sin ^2(x) (c+d x)^2+\frac{9}{4} d \cos ^2(x) (c+d x)^2+2 \sin (x) \cos (x) (c+d x)^3-\frac{3 d^3 x^2}{4}+\frac{3}{8} d^3 \sin ^2(x)-\frac{9}{8} d^3 \cos ^2(x) \]

[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

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Rubi [A]  time = 0.141352, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4431, 3311, 32, 3310} \[ -\frac{3}{2} c d^2 x-3 d^2 \sin (x) \cos (x) (c+d x)+\frac{(c+d x)^4}{4 d}-\frac{3}{4} d \sin ^2(x) (c+d x)^2+\frac{9}{4} d \cos ^2(x) (c+d x)^2+2 \sin (x) \cos (x) (c+d x)^3-\frac{3 d^3 x^2}{4}+\frac{3}{8} d^3 \sin ^2(x)-\frac{9}{8} d^3 \cos ^2(x) \]

Antiderivative was successfully verified.

[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 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 3310

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]

Rubi steps

\begin{align*} \int (c+d x)^3 \csc (x) \sin (3 x) \, dx &=\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]  time = 0.157289, size = 109, normalized size = 0.95 \[ \frac{1}{4} \left (x \left (6 c^2 d x+4 c^3+4 c d^2 x^2+d^3 x^3\right )+2 \sin (2 x) \left (6 c^2 d x+2 c^3+3 c d^2 \left (2 x^2-1\right )+d^3 x \left (2 x^2-3\right )\right )+3 d \cos (2 x) \left (2 c^2+4 c d x+d^2 \left (2 x^2-1\right )\right )\right ) \]

Antiderivative was successfully verified.

[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

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Maple [A]  time = 0.051, size = 179, normalized size = 1.6 \begin{align*} 4\,{d}^{3} \left ({x}^{3} \left ( 1/2\,\cos \left ( x \right ) \sin \left ( x \right ) +x/2 \right ) +3/4\,{x}^{2} \left ( \cos \left ( x \right ) \right ) ^{2}-3/2\,x \left ( 1/2\,\cos \left ( x \right ) \sin \left ( x \right ) +x/2 \right ) +3/8\,{x}^{2}+3/8\, \left ( \sin \left ( x \right ) \right ) ^{2}-3/8\,{x}^{4} \right ) +12\,{d}^{2}c \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 ) +12\,{c}^{2}d \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 ) -{\frac{{d}^{3}{x}^{4}}{4}}+4\,{c}^{3} \left ( 1/2\,\cos \left ( x \right ) \sin \left ( x \right ) +x/2 \right ) -c{d}^{2}{x}^{3}-{\frac{3\,{c}^{2}d{x}^{2}}{2}}-{c}^{3}x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.02836, size = 136, normalized size = 1.18 \begin{align*} \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Fricas [A]  time = 0.496439, size = 278, normalized size = 2.42 \begin{align*} \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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [B]  time = 146.651, size = 289, normalized size = 2.51 \begin{align*} c^{3} x + c^{3} \sin{\left (2 x \right )} - 3 c^{2} d x^{2} \sin ^{2}{\left (x \right )} - 3 c^{2} d x^{2} \cos ^{2}{\left (x \right )} + \frac{9 c^{2} d x^{2}}{2} + 6 c^{2} d x \sin{\left (x \right )} \cos{\left (x \right )} - 3 c^{2} d \sin ^{2}{\left (x \right )} - 2 c d^{2} x^{3} \sin ^{2}{\left (x \right )} - 2 c d^{2} x^{3} \cos ^{2}{\left (x \right )} + 3 c d^{2} x^{3} + 6 c d^{2} x^{2} \sin{\left (x \right )} \cos{\left (x \right )} - 3 c d^{2} x \sin ^{2}{\left (x \right )} + 3 c d^{2} x \cos ^{2}{\left (x \right )} - 3 c d^{2} \sin{\left (x \right )} \cos{\left (x \right )} - \frac{d^{3} x^{4} \sin ^{2}{\left (x \right )}}{2} - \frac{d^{3} x^{4} \cos ^{2}{\left (x \right )}}{2} + \frac{3 d^{3} x^{4}}{4} + 2 d^{3} x^{3} \sin{\left (x \right )} \cos{\left (x \right )} - \frac{3 d^{3} x^{2} \sin ^{2}{\left (x \right )}}{2} + \frac{3 d^{3} x^{2} \cos ^{2}{\left (x \right )}}{2} - 3 d^{3} x \sin{\left (x \right )} \cos{\left (x \right )} + \frac{3 d^{3} \sin ^{2}{\left (x \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14367, size = 151, normalized size = 1.31 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)