\(\int (c+d x)^3 \csc (a+b x) \sin (3 a+3 b x) \, dx\) [369]

Optimal result

Integrand size = 23, antiderivative size = 171 \[ \int (c+d x)^3 \csc (a+b x) \sin (3 a+3 b x) \, dx=-\frac {3 c d^2 x}{2 b^2}-\frac {3 d^3 x^2}{4 b^2}+\frac {(c+d x)^4}{4 d}-\frac {9 d^3 \cos ^2(a+b x)}{8 b^4}+\frac {9 d (c+d x)^2 \cos ^2(a+b x)}{4 b^2}-\frac {3 d^2 (c+d x) \cos (a+b x) \sin (a+b x)}{b^3}+\frac {2 (c+d x)^3 \cos (a+b x) \sin (a+b x)}{b}+\frac {3 d^3 \sin ^2(a+b x)}{8 b^4}-\frac {3 d (c+d x)^2 \sin ^2(a+b x)}{4 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {4516, 3392, 32, 3391} \[ \int (c+d x)^3 \csc (a+b x) \sin (3 a+3 b x) \, dx=\frac {3 d^3 \sin ^2(a+b x)}{8 b^4}-\frac {9 d^3 \cos ^2(a+b x)}{8 b^4}-\frac {3 d^2 (c+d x) \sin (a+b x) \cos (a+b x)}{b^3}-\frac {3 d (c+d x)^2 \sin ^2(a+b x)}{4 b^2}+\frac {9 d (c+d x)^2 \cos ^2(a+b x)}{4 b^2}+\frac {2 (c+d x)^3 \sin (a+b x) \cos (a+b x)}{b}-\frac {3 c d^2 x}{2 b^2}-\frac {3 d^3 x^2}{4 b^2}+\frac {(c+d x)^4}{4 d} \]

[In]

Int[(c + d*x)^3*Csc[a + b*x]*Sin[3*a + 3*b*x],x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.61 \[ \int (c+d x)^3 \csc (a+b x) \sin (3 a+3 b x) \, dx=\frac {b^4 x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right )+3 d \left (-d^2+2 b^2 (c+d x)^2\right ) \cos (2 (a+b x))+2 b (c+d x) \left (-3 d^2+2 b^2 (c+d x)^2\right ) \sin (2 (a+b x))}{4 b^4} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.97 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.91

method	result	size
risch	\(\frac {d^{3} x^{4}}{4}+c \,d^{2} x^{3}+\frac {3 c^{2} d \,x^{2}}{2}+c^{3} x +\frac {c^{4}}{4 d}+\frac {3 d \left (2 x^{2} d^{2} b^{2}+4 b^{2} c d x +2 b^{2} c^{2}-d^{2}\right ) \cos \left (2 x b +2 a \right )}{4 b^{4}}+\frac {\left (2 b^{2} d^{3} x^{3}+6 b^{2} c \,d^{2} x^{2}+6 b^{2} c^{2} d x +2 b^{2} c^{3}-3 d^{3} x -3 c \,d^{2}\right ) \sin \left (2 x b +2 a \right )}{2 b^{3}}\)	\(156\)
default	\(-c^{3} x -\frac {d^{3} x^{4}}{4}+\frac {4 c^{3} \left (\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{2}+\frac {x b}{2}+\frac {a}{2}\right )}{b}-c \,d^{2} x^{3}-\frac {3 c^{2} d \,x^{2}}{2}+\frac {4 d^{3} \left (\left (x b +a \right )^{3} \left (\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{2}+\frac {x b}{2}+\frac {a}{2}\right )+\frac {3 \left (x b +a \right )^{2} \cos \left (x b +a \right )^{2}}{4}-\frac {3 \left (x b +a \right ) \left (\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{2}+\frac {x b}{2}+\frac {a}{2}\right )}{2}+\frac {3 \left (x b +a \right )^{2}}{8}+\frac {3 \sin \left (x b +a \right )^{2}}{8}-\frac {3 \left (x b +a \right )^{4}}{8}-3 a \left (\left (x b +a \right )^{2} \left (\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{2}+\frac {x b}{2}+\frac {a}{2}\right )+\frac {\left (x b +a \right ) \cos \left (x b +a \right )^{2}}{2}-\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{4}-\frac {x b}{4}-\frac {a}{4}-\frac {\left (x b +a \right )^{3}}{3}\right )+3 a^{2} \left (\left (x b +a \right ) \left (\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{2}+\frac {x b}{2}+\frac {a}{2}\right )-\frac {\left (x b +a \right )^{2}}{4}-\frac {\sin \left (x b +a \right )^{2}}{4}\right )-a^{3} \left (\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{2}+\frac {x b}{2}+\frac {a}{2}\right )\right )}{b^{4}}+\frac {12 c \,d^{2} \left (\left (x b +a \right )^{2} \left (\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{2}+\frac {x b}{2}+\frac {a}{2}\right )+\frac {\left (x b +a \right ) \cos \left (x b +a \right )^{2}}{2}-\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{4}-\frac {x b}{4}-\frac {a}{4}-\frac {\left (x b +a \right )^{3}}{3}-2 a \left (\left (x b +a \right ) \left (\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{2}+\frac {x b}{2}+\frac {a}{2}\right )-\frac {\left (x b +a \right )^{2}}{4}-\frac {\sin \left (x b +a \right )^{2}}{4}\right )+a^{2} \left (\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{2}+\frac {x b}{2}+\frac {a}{2}\right )\right )}{b^{3}}+\frac {12 c^{2} d \left (\left (x b +a \right ) \left (\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{2}+\frac {x b}{2}+\frac {a}{2}\right )-\frac {\left (x b +a \right )^{2}}{4}-\frac {\sin \left (x b +a \right )^{2}}{4}-a \left (\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{2}+\frac {x b}{2}+\frac {a}{2}\right )\right )}{b^{2}}\)	\(580\)

[In]

int((d*x+c)^3*csc(b*x+a)*sin(3*b*x+3*a),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*b^2*d^2*x^2+4*b^2*c*d*x+2*b^2*c^2-d^2)/b^4*cos(2*
b*x+2*a)+1/2/b^3*(2*b^2*d^3*x^3+6*b^2*c*d^2*x^2+6*b^2*c^2*d*x+2*b^2*c^3-3*d^3*x-3*c*d^2)*sin(2*b*x+2*a)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.10 \[ \int (c+d x)^3 \csc (a+b x) \sin (3 a+3 b x) \, dx=\frac {b^{4} d^{3} x^{4} + 4 \, b^{4} c d^{2} x^{3} + 6 \, {\left (b^{4} c^{2} d - b^{2} d^{3}\right )} x^{2} + 6 \, {\left (2 \, b^{2} d^{3} x^{2} + 4 \, b^{2} c d^{2} x + 2 \, b^{2} c^{2} d - d^{3}\right )} \cos \left (b x + a\right )^{2} + 4 \, {\left (2 \, b^{3} d^{3} x^{3} + 6 \, b^{3} c d^{2} x^{2} + 2 \, b^{3} c^{3} - 3 \, b c d^{2} + 3 \, {\left (2 \, b^{3} c^{2} d - b d^{3}\right )} x\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 4 \, {\left (b^{4} c^{3} - 3 \, b^{2} c d^{2}\right )} x}{4 \, b^{4}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int (c+d x)^3 \csc (a+b x) \sin (3 a+3 b x) \, dx=\int \left (c + d x\right )^{3} \sin {\left (3 a + 3 b x \right )} \csc {\left (a + b x \right )}\, dx \]

[In]

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

[Out]

Integral((c + d*x)**3*sin(3*a + 3*b*x)*csc(a + b*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.01 \[ \int (c+d x)^3 \csc (a+b x) \sin (3 a+3 b x) \, dx=\frac {{\left (b x + \sin \left (2 \, b x + 2 \, a\right )\right )} c^{3}}{b} + \frac {3 \, {\left (b^{2} x^{2} + 2 \, b x \sin \left (2 \, b x + 2 \, a\right ) + \cos \left (2 \, b x + 2 \, a\right )\right )} c^{2} d}{2 \, b^{2}} + \frac {{\left (2 \, b^{3} x^{3} + 6 \, b x \cos \left (2 \, b x + 2 \, a\right ) + 3 \, {\left (2 \, b^{2} x^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} c d^{2}}{2 \, b^{3}} + \frac {{\left (b^{4} x^{4} + 3 \, {\left (2 \, b^{2} x^{2} - 1\right )} \cos \left (2 \, b x + 2 \, a\right ) + 2 \, {\left (2 \, b^{3} x^{3} - 3 \, b x\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} d^{3}}{4 \, b^{4}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 682 vs. \(2 (157) = 314\).

Time = 0.37 (sec) , antiderivative size = 682, normalized size of antiderivative = 3.99 \[ \int (c+d x)^3 \csc (a+b x) \sin (3 a+3 b x) \, dx=\frac {8 \, b^{3} c^{3} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 24 \, {\left (b x + a\right )} b^{2} c^{2} d \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 24 \, a b^{2} c^{2} d \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 24 \, {\left (b x + a\right )}^{2} b c d^{2} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 48 \, {\left (b x + a\right )} a b c d^{2} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 24 \, a^{2} b c d^{2} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 8 \, {\left (b x + a\right )}^{3} d^{3} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 24 \, {\left (b x + a\right )}^{2} a d^{3} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 24 \, {\left (b x + a\right )} a^{2} d^{3} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 8 \, a^{3} d^{3} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 4 \, {\left (b x + a\right )} b^{3} c^{3} + 6 \, {\left (b x + a\right )}^{2} b^{2} c^{2} d - 12 \, {\left (b x + a\right )} a b^{2} c^{2} d + 4 \, {\left (b x + a\right )}^{3} b c d^{2} - 12 \, {\left (b x + a\right )}^{2} a b c d^{2} + 12 \, {\left (b x + a\right )} a^{2} b c d^{2} + {\left (b x + a\right )}^{4} d^{3} - 4 \, {\left (b x + a\right )}^{3} a d^{3} + 6 \, {\left (b x + a\right )}^{2} a^{2} d^{3} - 4 \, {\left (b x + a\right )} a^{3} d^{3} + 6 \, b^{2} c^{2} d \cos \left (b x + a\right )^{2} + 12 \, {\left (b x + a\right )} b c d^{2} \cos \left (b x + a\right )^{2} - 12 \, a b c d^{2} \cos \left (b x + a\right )^{2} + 6 \, {\left (b x + a\right )}^{2} d^{3} \cos \left (b x + a\right )^{2} - 12 \, {\left (b x + a\right )} a d^{3} \cos \left (b x + a\right )^{2} + 6 \, a^{2} d^{3} \cos \left (b x + a\right )^{2} - 6 \, b^{2} c^{2} d \sin \left (b x + a\right )^{2} - 12 \, {\left (b x + a\right )} b c d^{2} \sin \left (b x + a\right )^{2} + 12 \, a b c d^{2} \sin \left (b x + a\right )^{2} - 6 \, {\left (b x + a\right )}^{2} d^{3} \sin \left (b x + a\right )^{2} + 12 \, {\left (b x + a\right )} a d^{3} \sin \left (b x + a\right )^{2} - 6 \, a^{2} d^{3} \sin \left (b x + a\right )^{2} - 12 \, b c d^{2} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 12 \, {\left (b x + a\right )} d^{3} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 12 \, a d^{3} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 3 \, d^{3} \cos \left (b x + a\right )^{2} + 3 \, d^{3} \sin \left (b x + a\right )^{2}}{4 \, b^{4}} \]

[In]

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

[Out]

1/4*(8*b^3*c^3*cos(b*x + a)*sin(b*x + a) + 24*(b*x + a)*b^2*c^2*d*cos(b*x + a)*sin(b*x + a) - 24*a*b^2*c^2*d*c
os(b*x + a)*sin(b*x + a) + 24*(b*x + a)^2*b*c*d^2*cos(b*x + a)*sin(b*x + a) - 48*(b*x + a)*a*b*c*d^2*cos(b*x +
a)*sin(b*x + a) + 24*a^2*b*c*d^2*cos(b*x + a)*sin(b*x + a) + 8*(b*x + a)^3*d^3*cos(b*x + a)*sin(b*x + a) - 24
*(b*x + a)^2*a*d^3*cos(b*x + a)*sin(b*x + a) + 24*(b*x + a)*a^2*d^3*cos(b*x + a)*sin(b*x + a) - 8*a^3*d^3*cos(
b*x + a)*sin(b*x + a) + 4*(b*x + a)*b^3*c^3 + 6*(b*x + a)^2*b^2*c^2*d - 12*(b*x + a)*a*b^2*c^2*d + 4*(b*x + a)
^3*b*c*d^2 - 12*(b*x + a)^2*a*b*c*d^2 + 12*(b*x + a)*a^2*b*c*d^2 + (b*x + a)^4*d^3 - 4*(b*x + a)^3*a*d^3 + 6*(
b*x + a)^2*a^2*d^3 - 4*(b*x + a)*a^3*d^3 + 6*b^2*c^2*d*cos(b*x + a)^2 + 12*(b*x + a)*b*c*d^2*cos(b*x + a)^2 -
12*a*b*c*d^2*cos(b*x + a)^2 + 6*(b*x + a)^2*d^3*cos(b*x + a)^2 - 12*(b*x + a)*a*d^3*cos(b*x + a)^2 + 6*a^2*d^3
*cos(b*x + a)^2 - 6*b^2*c^2*d*sin(b*x + a)^2 - 12*(b*x + a)*b*c*d^2*sin(b*x + a)^2 + 12*a*b*c*d^2*sin(b*x + a)
^2 - 6*(b*x + a)^2*d^3*sin(b*x + a)^2 + 12*(b*x + a)*a*d^3*sin(b*x + a)^2 - 6*a^2*d^3*sin(b*x + a)^2 - 12*b*c*
d^2*cos(b*x + a)*sin(b*x + a) - 12*(b*x + a)*d^3*cos(b*x + a)*sin(b*x + a) + 12*a*d^3*cos(b*x + a)*sin(b*x + a
) - 3*d^3*cos(b*x + a)^2 + 3*d^3*sin(b*x + a)^2)/b^4

Mupad [B] (verification not implemented)

Time = 26.39 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.26 \[ \int (c+d x)^3 \csc (a+b x) \sin (3 a+3 b x) \, dx=c^3\,x+\frac {d^3\,x^4}{4}+\frac {3\,c^2\,d\,x^2}{2}+c\,d^2\,x^3-\frac {3\,d^3\,\cos \left (2\,a+2\,b\,x\right )}{4\,b^4}+\frac {c^3\,\sin \left (2\,a+2\,b\,x\right )}{b}+\frac {3\,c^2\,d\,\cos \left (2\,a+2\,b\,x\right )}{2\,b^2}-\frac {3\,c\,d^2\,\sin \left (2\,a+2\,b\,x\right )}{2\,b^3}-\frac {3\,d^3\,x\,\sin \left (2\,a+2\,b\,x\right )}{2\,b^3}+\frac {3\,d^3\,x^2\,\cos \left (2\,a+2\,b\,x\right )}{2\,b^2}+\frac {d^3\,x^3\,\sin \left (2\,a+2\,b\,x\right )}{b}+\frac {3\,c\,d^2\,x\,\cos \left (2\,a+2\,b\,x\right )}{b^2}+\frac {3\,c^2\,d\,x\,\sin \left (2\,a+2\,b\,x\right )}{b}+\frac {3\,c\,d^2\,x^2\,\sin \left (2\,a+2\,b\,x\right )}{b} \]

[In]

int((sin(3*a + 3*b*x)*(c + d*x)^3)/sin(a + b*x),x)

[Out]

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