\(\int (c+d x)^4 \csc (a+b x) \sin (3 a+3 b x) \, dx\) [368]

Optimal result

Integrand size = 23, antiderivative size = 198 \[ \int (c+d x)^4 \csc (a+b x) \sin (3 a+3 b x) \, dx=\frac {3 d^4 x}{2 b^4}-\frac {d (c+d x)^3}{b^2}+\frac {(c+d x)^5}{5 d}-\frac {9 d^3 (c+d x) \cos ^2(a+b x)}{2 b^4}+\frac {3 d (c+d x)^3 \cos ^2(a+b x)}{b^2}+\frac {3 d^4 \cos (a+b x) \sin (a+b x)}{b^5}-\frac {6 d^2 (c+d x)^2 \cos (a+b x) \sin (a+b x)}{b^3}+\frac {2 (c+d x)^4 \cos (a+b x) \sin (a+b x)}{b}+\frac {3 d^3 (c+d x) \sin ^2(a+b x)}{2 b^4}-\frac {d (c+d x)^3 \sin ^2(a+b x)}{b^2} \]

[Out]

3/2*d^4*x/b^4-d*(d*x+c)^3/b^2+1/5*(d*x+c)^5/d-9/2*d^3*(d*x+c)*cos(b*x+a)^2/b^4+3*d*(d*x+c)^3*cos(b*x+a)^2/b^2+
3*d^4*cos(b*x+a)*sin(b*x+a)/b^5-6*d^2*(d*x+c)^2*cos(b*x+a)*sin(b*x+a)/b^3+2*(d*x+c)^4*cos(b*x+a)*sin(b*x+a)/b+
3/2*d^3*(d*x+c)*sin(b*x+a)^2/b^4-d*(d*x+c)^3*sin(b*x+a)^2/b^2

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4516, 3392, 32, 2715, 8} \[ \int (c+d x)^4 \csc (a+b x) \sin (3 a+3 b x) \, dx=\frac {3 d^4 \sin (a+b x) \cos (a+b x)}{b^5}+\frac {3 d^3 (c+d x) \sin ^2(a+b x)}{2 b^4}-\frac {9 d^3 (c+d x) \cos ^2(a+b x)}{2 b^4}-\frac {6 d^2 (c+d x)^2 \sin (a+b x) \cos (a+b x)}{b^3}-\frac {d (c+d x)^3 \sin ^2(a+b x)}{b^2}+\frac {3 d (c+d x)^3 \cos ^2(a+b x)}{b^2}+\frac {2 (c+d x)^4 \sin (a+b x) \cos (a+b x)}{b}+\frac {3 d^4 x}{2 b^4}-\frac {d (c+d x)^3}{b^2}+\frac {(c+d x)^5}{5 d} \]

[In]

Int[(c + d*x)^4*Csc[a + b*x]*Sin[3*a + 3*b*x],x]

[Out]

(3*d^4*x)/(2*b^4) - (d*(c + d*x)^3)/b^2 + (c + d*x)^5/(5*d) - (9*d^3*(c + d*x)*Cos[a + b*x]^2)/(2*b^4) + (3*d*
(c + d*x)^3*Cos[a + b*x]^2)/b^2 + (3*d^4*Cos[a + b*x]*Sin[a + b*x])/b^5 - (6*d^2*(c + d*x)^2*Cos[a + b*x]*Sin[
a + b*x])/b^3 + (2*(c + d*x)^4*Cos[a + b*x]*Sin[a + b*x])/b + (3*d^3*(c + d*x)*Sin[a + b*x]^2)/(2*b^4) - (d*(c
+ d*x)^3*Sin[a + b*x]^2)/b^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)^4 \cos ^2(a+b x)-(c+d x)^4 \sin ^2(a+b x)\right ) \, dx \\ & = 3 \int (c+d x)^4 \cos ^2(a+b x) \, dx-\int (c+d x)^4 \sin ^2(a+b x) \, dx \\ & = \frac {3 d (c+d x)^3 \cos ^2(a+b x)}{b^2}+\frac {2 (c+d x)^4 \cos (a+b x) \sin (a+b x)}{b}-\frac {d (c+d x)^3 \sin ^2(a+b x)}{b^2}-\frac {1}{2} \int (c+d x)^4 \, dx+\frac {3}{2} \int (c+d x)^4 \, dx+\frac {\left (3 d^2\right ) \int (c+d x)^2 \sin ^2(a+b x) \, dx}{b^2}-\frac {\left (9 d^2\right ) \int (c+d x)^2 \cos ^2(a+b x) \, dx}{b^2} \\ & = \frac {(c+d x)^5}{5 d}-\frac {9 d^3 (c+d x) \cos ^2(a+b x)}{2 b^4}+\frac {3 d (c+d x)^3 \cos ^2(a+b x)}{b^2}-\frac {6 d^2 (c+d x)^2 \cos (a+b x) \sin (a+b x)}{b^3}+\frac {2 (c+d x)^4 \cos (a+b x) \sin (a+b x)}{b}+\frac {3 d^3 (c+d x) \sin ^2(a+b x)}{2 b^4}-\frac {d (c+d x)^3 \sin ^2(a+b x)}{b^2}+\frac {\left (3 d^2\right ) \int (c+d x)^2 \, dx}{2 b^2}-\frac {\left (9 d^2\right ) \int (c+d x)^2 \, dx}{2 b^2}-\frac {\left (3 d^4\right ) \int \sin ^2(a+b x) \, dx}{2 b^4}+\frac {\left (9 d^4\right ) \int \cos ^2(a+b x) \, dx}{2 b^4} \\ & = -\frac {d (c+d x)^3}{b^2}+\frac {(c+d x)^5}{5 d}-\frac {9 d^3 (c+d x) \cos ^2(a+b x)}{2 b^4}+\frac {3 d (c+d x)^3 \cos ^2(a+b x)}{b^2}+\frac {3 d^4 \cos (a+b x) \sin (a+b x)}{b^5}-\frac {6 d^2 (c+d x)^2 \cos (a+b x) \sin (a+b x)}{b^3}+\frac {2 (c+d x)^4 \cos (a+b x) \sin (a+b x)}{b}+\frac {3 d^3 (c+d x) \sin ^2(a+b x)}{2 b^4}-\frac {d (c+d x)^3 \sin ^2(a+b x)}{b^2}-\frac {\left (3 d^4\right ) \int 1 \, dx}{4 b^4}+\frac {\left (9 d^4\right ) \int 1 \, dx}{4 b^4} \\ & = \frac {3 d^4 x}{2 b^4}-\frac {d (c+d x)^3}{b^2}+\frac {(c+d x)^5}{5 d}-\frac {9 d^3 (c+d x) \cos ^2(a+b x)}{2 b^4}+\frac {3 d (c+d x)^3 \cos ^2(a+b x)}{b^2}+\frac {3 d^4 \cos (a+b x) \sin (a+b x)}{b^5}-\frac {6 d^2 (c+d x)^2 \cos (a+b x) \sin (a+b x)}{b^3}+\frac {2 (c+d x)^4 \cos (a+b x) \sin (a+b x)}{b}+\frac {3 d^3 (c+d x) \sin ^2(a+b x)}{2 b^4}-\frac {d (c+d x)^3 \sin ^2(a+b x)}{b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.65 \[ \int (c+d x)^4 \csc (a+b x) \sin (3 a+3 b x) \, dx=c^4 x+2 c^3 d x^2+2 c^2 d^2 x^3+c d^3 x^4+\frac {d^4 x^5}{5}+\frac {d (c+d x) \left (-3 d^2+2 b^2 (c+d x)^2\right ) \cos (2 (a+b x))}{b^4}+\frac {\left (3 d^4-6 b^2 d^2 (c+d x)^2+2 b^4 (c+d x)^4\right ) \sin (2 (a+b x))}{2 b^5} \]

[In]

Integrate[(c + d*x)^4*Csc[a + b*x]*Sin[3*a + 3*b*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*(c + d*x)*(-3*d^2 + 2*b^2*(c + d*x)^2)*Cos[
2*(a + b*x)])/b^4 + ((3*d^4 - 6*b^2*d^2*(c + d*x)^2 + 2*b^4*(c + d*x)^4)*Sin[2*(a + b*x)])/(2*b^5)

Maple [A] (verified)

Time = 1.91 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.14

method	result	size
risch	\(\frac {d^{4} x^{5}}{5}+c \,d^{3} x^{4}+2 c^{2} d^{2} x^{3}+2 c^{3} d \,x^{2}+c^{4} x +\frac {c^{5}}{5 d}+\frac {d \left (2 b^{2} d^{3} x^{3}+6 b^{2} c \,d^{2} x^{2}+6 b^{2} c^{2} d x +2 b^{2} c^{3}-3 d^{3} x -3 c \,d^{2}\right ) \cos \left (2 x b +2 a \right )}{b^{4}}+\frac {\left (2 d^{4} x^{4} b^{4}+8 b^{4} c \,d^{3} x^{3}+12 b^{4} c^{2} d^{2} x^{2}+8 b^{4} c^{3} d x +2 b^{4} c^{4}-6 b^{2} d^{4} x^{2}-12 b^{2} c \,d^{3} x -6 b^{2} c^{2} d^{2}+3 d^{4}\right ) \sin \left (2 x b +2 a \right )}{2 b^{5}}\)	\(226\)
default	\(\text {Expression too large to display}\)	\(1000\)

[In]

int((d*x+c)^4*csc(b*x+a)*sin(3*b*x+3*a),x,method=_RETURNVERBOSE)

[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+1/5/d*c^5+1/b^4*d*(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)*cos(2*b*x+2*a)+1/2*(2*b^4*d^4*x^4+8*b^4*c*d^3*x^3+12*b^4*c^2*d^2*x^2+8*b^4*c
^3*d*x+2*b^4*c^4-6*b^2*d^4*x^2-12*b^2*c*d^3*x-6*b^2*c^2*d^2+3*d^4)/b^5*sin(2*b*x+2*a)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.43 \[ \int (c+d x)^4 \csc (a+b x) \sin (3 a+3 b x) \, dx=\frac {b^{5} d^{4} x^{5} + 5 \, b^{5} c d^{3} x^{4} + 10 \, {\left (b^{5} c^{2} d^{2} - b^{3} d^{4}\right )} x^{3} + 10 \, {\left (b^{5} c^{3} d - 3 \, b^{3} c d^{3}\right )} x^{2} + 10 \, {\left (2 \, b^{3} d^{4} x^{3} + 6 \, b^{3} c d^{3} x^{2} + 2 \, b^{3} c^{3} d - 3 \, b c d^{3} + 3 \, {\left (2 \, b^{3} c^{2} d^{2} - b d^{4}\right )} x\right )} \cos \left (b x + a\right )^{2} + 5 \, {\left (2 \, b^{4} d^{4} x^{4} + 8 \, b^{4} c d^{3} x^{3} + 2 \, b^{4} c^{4} - 6 \, b^{2} c^{2} d^{2} + 3 \, d^{4} + 6 \, {\left (2 \, b^{4} c^{2} d^{2} - b^{2} d^{4}\right )} x^{2} + 4 \, {\left (2 \, b^{4} c^{3} d - 3 \, b^{2} c d^{3}\right )} x\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 5 \, {\left (b^{5} c^{4} - 6 \, b^{3} c^{2} d^{2} + 3 \, b d^{4}\right )} x}{5 \, b^{5}} \]

[In]

integrate((d*x+c)^4*csc(b*x+a)*sin(3*b*x+3*a),x, algorithm="fricas")

[Out]

1/5*(b^5*d^4*x^5 + 5*b^5*c*d^3*x^4 + 10*(b^5*c^2*d^2 - b^3*d^4)*x^3 + 10*(b^5*c^3*d - 3*b^3*c*d^3)*x^2 + 10*(2
*b^3*d^4*x^3 + 6*b^3*c*d^3*x^2 + 2*b^3*c^3*d - 3*b*c*d^3 + 3*(2*b^3*c^2*d^2 - b*d^4)*x)*cos(b*x + a)^2 + 5*(2*
b^4*d^4*x^4 + 8*b^4*c*d^3*x^3 + 2*b^4*c^4 - 6*b^2*c^2*d^2 + 3*d^4 + 6*(2*b^4*c^2*d^2 - b^2*d^4)*x^2 + 4*(2*b^4
*c^3*d - 3*b^2*c*d^3)*x)*cos(b*x + a)*sin(b*x + a) + 5*(b^5*c^4 - 6*b^3*c^2*d^2 + 3*b*d^4)*x)/b^5

Sympy [F(-1)]

Timed out. \[ \int (c+d x)^4 \csc (a+b x) \sin (3 a+3 b x) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.23 \[ \int (c+d x)^4 \csc (a+b x) \sin (3 a+3 b x) \, dx=\frac {{\left (b x + \sin \left (2 \, b x + 2 \, a\right )\right )} c^{4}}{b} + \frac {2 \, {\left (b^{2} x^{2} + 2 \, b x \sin \left (2 \, b x + 2 \, a\right ) + \cos \left (2 \, b x + 2 \, a\right )\right )} c^{3} d}{b^{2}} + \frac {{\left (2 \, b^{3} x^{3} + 6 \, b x \cos \left (2 \, b x + 2 \, a\right ) + 3 \, {\left (2 \, b^{2} x^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} c^{2} d^{2}}{b^{3}} + \frac {{\left (b^{4} x^{4} + 3 \, {\left (2 \, b^{2} x^{2} - 1\right )} \cos \left (2 \, b x + 2 \, a\right ) + 2 \, {\left (2 \, b^{3} x^{3} - 3 \, b x\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} c d^{3}}{b^{4}} + \frac {{\left (2 \, b^{5} x^{5} + 10 \, {\left (2 \, b^{3} x^{3} - 3 \, b x\right )} \cos \left (2 \, b x + 2 \, a\right ) + 5 \, {\left (2 \, b^{4} x^{4} - 6 \, b^{2} x^{2} + 3\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} d^{4}}{10 \, b^{5}} \]

[In]

integrate((d*x+c)^4*csc(b*x+a)*sin(3*b*x+3*a),x, algorithm="maxima")

[Out]

(b*x + sin(2*b*x + 2*a))*c^4/b + 2*(b^2*x^2 + 2*b*x*sin(2*b*x + 2*a) + cos(2*b*x + 2*a))*c^3*d/b^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^2*d^2/b^3 + (b^4*x^4 + 3*(2*b^2*x^2 - 1)*co
s(2*b*x + 2*a) + 2*(2*b^3*x^3 - 3*b*x)*sin(2*b*x + 2*a))*c*d^3/b^4 + 1/10*(2*b^5*x^5 + 10*(2*b^3*x^3 - 3*b*x)*
cos(2*b*x + 2*a) + 5*(2*b^4*x^4 - 6*b^2*x^2 + 3)*sin(2*b*x + 2*a))*d^4/b^5

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1255 vs. \(2 (190) = 380\).

Time = 0.44 (sec) , antiderivative size = 1255, normalized size of antiderivative = 6.34 \[ \int (c+d x)^4 \csc (a+b x) \sin (3 a+3 b x) \, dx=\text {Too large to display} \]

[In]

integrate((d*x+c)^4*csc(b*x+a)*sin(3*b*x+3*a),x, algorithm="giac")

[Out]

1/5*(10*b^4*c^4*cos(b*x + a)*sin(b*x + a) + 40*(b*x + a)*b^3*c^3*d*cos(b*x + a)*sin(b*x + a) - 40*a*b^3*c^3*d*
cos(b*x + a)*sin(b*x + a) + 60*(b*x + a)^2*b^2*c^2*d^2*cos(b*x + a)*sin(b*x + a) - 120*(b*x + a)*a*b^2*c^2*d^2
*cos(b*x + a)*sin(b*x + a) + 60*a^2*b^2*c^2*d^2*cos(b*x + a)*sin(b*x + a) + 40*(b*x + a)^3*b*c*d^3*cos(b*x + a
)*sin(b*x + a) - 120*(b*x + a)^2*a*b*c*d^3*cos(b*x + a)*sin(b*x + a) + 120*(b*x + a)*a^2*b*c*d^3*cos(b*x + a)*
sin(b*x + a) - 40*a^3*b*c*d^3*cos(b*x + a)*sin(b*x + a) + 10*(b*x + a)^4*d^4*cos(b*x + a)*sin(b*x + a) - 40*(b
*x + a)^3*a*d^4*cos(b*x + a)*sin(b*x + a) + 60*(b*x + a)^2*a^2*d^4*cos(b*x + a)*sin(b*x + a) - 40*(b*x + a)*a^
3*d^4*cos(b*x + a)*sin(b*x + a) + 10*a^4*d^4*cos(b*x + a)*sin(b*x + a) + 5*(b*x + a)*b^4*c^4 + 10*(b*x + a)^2*
b^3*c^3*d - 20*(b*x + a)*a*b^3*c^3*d + 10*(b*x + a)^3*b^2*c^2*d^2 - 30*(b*x + a)^2*a*b^2*c^2*d^2 + 30*(b*x + a
)*a^2*b^2*c^2*d^2 + 5*(b*x + a)^4*b*c*d^3 - 20*(b*x + a)^3*a*b*c*d^3 + 30*(b*x + a)^2*a^2*b*c*d^3 - 20*(b*x +
a)*a^3*b*c*d^3 + (b*x + a)^5*d^4 - 5*(b*x + a)^4*a*d^4 + 10*(b*x + a)^3*a^2*d^4 - 10*(b*x + a)^2*a^3*d^4 + 5*(
b*x + a)*a^4*d^4 + 10*b^3*c^3*d*cos(b*x + a)^2 + 30*(b*x + a)*b^2*c^2*d^2*cos(b*x + a)^2 - 30*a*b^2*c^2*d^2*co
s(b*x + a)^2 + 30*(b*x + a)^2*b*c*d^3*cos(b*x + a)^2 - 60*(b*x + a)*a*b*c*d^3*cos(b*x + a)^2 + 30*a^2*b*c*d^3*
cos(b*x + a)^2 + 10*(b*x + a)^3*d^4*cos(b*x + a)^2 - 30*(b*x + a)^2*a*d^4*cos(b*x + a)^2 + 30*(b*x + a)*a^2*d^
4*cos(b*x + a)^2 - 10*a^3*d^4*cos(b*x + a)^2 - 10*b^3*c^3*d*sin(b*x + a)^2 - 30*(b*x + a)*b^2*c^2*d^2*sin(b*x
+ a)^2 + 30*a*b^2*c^2*d^2*sin(b*x + a)^2 - 30*(b*x + a)^2*b*c*d^3*sin(b*x + a)^2 + 60*(b*x + a)*a*b*c*d^3*sin(
b*x + a)^2 - 30*a^2*b*c*d^3*sin(b*x + a)^2 - 10*(b*x + a)^3*d^4*sin(b*x + a)^2 + 30*(b*x + a)^2*a*d^4*sin(b*x
+ a)^2 - 30*(b*x + a)*a^2*d^4*sin(b*x + a)^2 + 10*a^3*d^4*sin(b*x + a)^2 - 30*b^2*c^2*d^2*cos(b*x + a)*sin(b*x
+ a) - 60*(b*x + a)*b*c*d^3*cos(b*x + a)*sin(b*x + a) + 60*a*b*c*d^3*cos(b*x + a)*sin(b*x + a) - 30*(b*x + a)
^2*d^4*cos(b*x + a)*sin(b*x + a) + 60*(b*x + a)*a*d^4*cos(b*x + a)*sin(b*x + a) - 30*a^2*d^4*cos(b*x + a)*sin(
b*x + a) - 15*b*c*d^3*cos(b*x + a)^2 - 15*(b*x + a)*d^4*cos(b*x + a)^2 + 15*a*d^4*cos(b*x + a)^2 + 15*b*c*d^3*
sin(b*x + a)^2 + 15*(b*x + a)*d^4*sin(b*x + a)^2 - 15*a*d^4*sin(b*x + a)^2 + 15*d^4*cos(b*x + a)*sin(b*x + a))
/b^5

Mupad [B] (verification not implemented)

Time = 26.45 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.74 \[ \int (c+d x)^4 \csc (a+b x) \sin (3 a+3 b x) \, dx=\frac {\frac {3\,d^4\,\sin \left (2\,a+2\,b\,x\right )}{2}+b^5\,c^4\,x+b^4\,c^4\,\sin \left (2\,a+2\,b\,x\right )+\frac {b^5\,d^4\,x^5}{5}+2\,b^3\,c^3\,d\,\cos \left (2\,a+2\,b\,x\right )+2\,b^5\,c^3\,d\,x^2+b^5\,c\,d^3\,x^4-3\,b^2\,c^2\,d^2\,\sin \left (2\,a+2\,b\,x\right )+2\,b^3\,d^4\,x^3\,\cos \left (2\,a+2\,b\,x\right )+2\,b^5\,c^2\,d^2\,x^3-3\,b^2\,d^4\,x^2\,\sin \left (2\,a+2\,b\,x\right )+b^4\,d^4\,x^4\,\sin \left (2\,a+2\,b\,x\right )-3\,b\,c\,d^3\,\cos \left (2\,a+2\,b\,x\right )-3\,b\,d^4\,x\,\cos \left (2\,a+2\,b\,x\right )+6\,b^4\,c^2\,d^2\,x^2\,\sin \left (2\,a+2\,b\,x\right )-6\,b^2\,c\,d^3\,x\,\sin \left (2\,a+2\,b\,x\right )+4\,b^4\,c^3\,d\,x\,\sin \left (2\,a+2\,b\,x\right )+6\,b^3\,c^2\,d^2\,x\,\cos \left (2\,a+2\,b\,x\right )+6\,b^3\,c\,d^3\,x^2\,\cos \left (2\,a+2\,b\,x\right )+4\,b^4\,c\,d^3\,x^3\,\sin \left (2\,a+2\,b\,x\right )}{b^5} \]

[In]

int((sin(3*a + 3*b*x)*(c + d*x)^4)/sin(a + b*x),x)

[Out]

((3*d^4*sin(2*a + 2*b*x))/2 + b^5*c^4*x + b^4*c^4*sin(2*a + 2*b*x) + (b^5*d^4*x^5)/5 + 2*b^3*c^3*d*cos(2*a + 2
*b*x) + 2*b^5*c^3*d*x^2 + b^5*c*d^3*x^4 - 3*b^2*c^2*d^2*sin(2*a + 2*b*x) + 2*b^3*d^4*x^3*cos(2*a + 2*b*x) + 2*
b^5*c^2*d^2*x^3 - 3*b^2*d^4*x^2*sin(2*a + 2*b*x) + b^4*d^4*x^4*sin(2*a + 2*b*x) - 3*b*c*d^3*cos(2*a + 2*b*x) -
3*b*d^4*x*cos(2*a + 2*b*x) + 6*b^4*c^2*d^2*x^2*sin(2*a + 2*b*x) - 6*b^2*c*d^3*x*sin(2*a + 2*b*x) + 4*b^4*c^3*
d*x*sin(2*a + 2*b*x) + 6*b^3*c^2*d^2*x*cos(2*a + 2*b*x) + 6*b^3*c*d^3*x^2*cos(2*a + 2*b*x) + 4*b^4*c*d^3*x^3*s
in(2*a + 2*b*x))/b^5