3.361 \(\int (c+d x)^4 \csc (x) \sin (3 x) \, dx\)

Optimal. Leaf size=131 \[ \frac{3}{2} d^3 \sin ^2(x) (c+d x)-\frac{9}{2} d^3 \cos ^2(x) (c+d x)-6 d^2 \sin (x) \cos (x) (c+d x)^2+\frac{(c+d x)^5}{5 d}-d (c+d x)^3-d \sin ^2(x) (c+d x)^3+3 d \cos ^2(x) (c+d x)^3+2 \sin (x) \cos (x) (c+d x)^4+\frac{3 d^4 x}{2}+3 d^4 \sin (x) \cos (x) \]

[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

________________________________________________________________________________________

Rubi [A]  time = 0.188099, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 14, 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} \[ \frac{3}{2} d^3 \sin ^2(x) (c+d x)-\frac{9}{2} d^3 \cos ^2(x) (c+d x)-6 d^2 \sin (x) \cos (x) (c+d x)^2+\frac{(c+d x)^5}{5 d}-d (c+d x)^3-d \sin ^2(x) (c+d x)^3+3 d \cos ^2(x) (c+d x)^3+2 \sin (x) \cos (x) (c+d x)^4+\frac{3 d^4 x}{2}+3 d^4 \sin (x) \cos (x) \]

Antiderivative was successfully verified.

[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 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)^4 \csc (x) \sin (3 x) \, dx &=\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]  time = 0.218675, size = 154, normalized size = 1.18 \[ 2 c^2 d^2 x^3+\frac{1}{2} \sin (2 x) \left (6 c^2 d^2 \left (2 x^2-1\right )+8 c^3 d x+2 c^4+4 c d^3 x \left (2 x^2-3\right )+d^4 \left (2 x^4-6 x^2+3\right )\right )+d \cos (2 x) \left (6 c^2 d x+2 c^3+3 c d^2 \left (2 x^2-1\right )+d^3 x \left (2 x^2-3\right )\right )+2 c^3 d x^2+c^4 x+c d^3 x^4+\frac{d^4 x^5}{5} \]

Antiderivative was successfully verified.

[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 [B]  time = 0.066, size = 260, normalized size = 2. \begin{align*} 4\,{d}^{4} \left ({x}^{4} \left ( 1/2\,\cos \left ( x \right ) \sin \left ( x \right ) +x/2 \right ) +{x}^{3} \left ( \cos \left ( x \right ) \right ) ^{2}-3\,{x}^{2} \left ( 1/2\,\cos \left ( x \right ) \sin \left ( x \right ) +x/2 \right ) -3/2\,x \left ( \cos \left ( x \right ) \right ) ^{2}+3/4\,\cos \left ( x \right ) \sin \left ( x \right ) +3/4\,x+{x}^{3}-2/5\,{x}^{5} \right ) +16\,c{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 ) +24\,{c}^{2}{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 ) -{\frac{{d}^{4}{x}^{5}}{5}}+16\,{c}^{3}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 ) -c{d}^{3}{x}^{4}+4\,{c}^{4} \left ( 1/2\,\cos \left ( x \right ) \sin \left ( x \right ) +x/2 \right ) -2\,{c}^{2}{d}^{2}{x}^{3}-2\,{c}^{3}d{x}^{2}-{c}^{4}x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*d^4*(x^4*(1/2*cos(x)*sin(x)+1/2*x)+x^3*cos(x)^2-3*x^2*(1/2*cos(x)*sin(x)+1/2*x)-3/2*x*cos(x)^2+3/4*cos(x)*si
n(x)+3/4*x+x^3-2/5*x^5)+16*c*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)+24*c^2*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)-1/5*d^4*x^5+16*c^3*d*(x*(1/2*cos(x)*sin(x)+1/2*x)-1/4*x^2-1/4*sin(x)^2)-c*d^3*x^4+4*c^4*(1/2*cos(
x)*sin(x)+1/2*x)-2*c^2*d^2*x^3-2*c^3*d*x^2-c^4*x

________________________________________________________________________________________

Maxima [A]  time = 1.04852, size = 197, normalized size = 1.5 \begin{align*} 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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

Fricas [A]  time = 0.515347, size = 419, normalized size = 3.2 \begin{align*} \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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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 [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.13595, size = 225, normalized size = 1.72 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)