\(\int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx\) [374]

Optimal result

Integrand size = 23, antiderivative size = 136 \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=-\frac {3 \cos ^2(a+b x)}{2 d (c+d x)^2}-\frac {4 b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}+\frac {4 b \cos (a+b x) \sin (a+b x)}{d^2 (c+d x)}+\frac {\sin ^2(a+b x)}{2 d (c+d x)^2}+\frac {4 b^2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4516, 3395, 31, 3393, 3384, 3380, 3383} \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=-\frac {4 b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}+\frac {4 b^2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}+\frac {4 b \sin (a+b x) \cos (a+b x)}{d^2 (c+d x)}+\frac {\sin ^2(a+b x)}{2 d (c+d x)^2}-\frac {3 \cos ^2(a+b x)}{2 d (c+d x)^2} \]

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 3380

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3383

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3384

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 3395

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(c + d*x)^(m + 1)*((b*Si
n[e + f*x])^n/(d*(m + 1))), x] + (Dist[b^2*f^2*n*((n - 1)/(d^2*(m + 1)*(m + 2))), Int[(c + d*x)^(m + 2)*(b*Sin
[e + f*x])^(n - 2), x], x] - Dist[f^2*(n^2/(d^2*(m + 1)*(m + 2))), Int[(c + d*x)^(m + 2)*(b*Sin[e + f*x])^n, x
], x] - Simp[b*f*n*(c + d*x)^(m + 2)*Cos[e + f*x]*((b*Sin[e + f*x])^(n - 1)/(d^2*(m + 1)*(m + 2))), x]) /; Fre
eQ[{b, c, d, e, f}, x] && GtQ[n, 1] && LtQ[m, -2]

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 (\frac {3 \cos ^2(a+b x)}{(c+d x)^3}-\frac {\sin ^2(a+b x)}{(c+d x)^3}\right ) \, dx \\ & = 3 \int \frac {\cos ^2(a+b x)}{(c+d x)^3} \, dx-\int \frac {\sin ^2(a+b x)}{(c+d x)^3} \, dx \\ & = -\frac {3 \cos ^2(a+b x)}{2 d (c+d x)^2}+\frac {4 b \cos (a+b x) \sin (a+b x)}{d^2 (c+d x)}+\frac {\sin ^2(a+b x)}{2 d (c+d x)^2}-\frac {b^2 \int \frac {1}{c+d x} \, dx}{d^2}+\frac {\left (2 b^2\right ) \int \frac {\sin ^2(a+b x)}{c+d x} \, dx}{d^2}+\frac {\left (3 b^2\right ) \int \frac {1}{c+d x} \, dx}{d^2}-\frac {\left (6 b^2\right ) \int \frac {\cos ^2(a+b x)}{c+d x} \, dx}{d^2} \\ & = -\frac {3 \cos ^2(a+b x)}{2 d (c+d x)^2}+\frac {2 b^2 \log (c+d x)}{d^3}+\frac {4 b \cos (a+b x) \sin (a+b x)}{d^2 (c+d x)}+\frac {\sin ^2(a+b x)}{2 d (c+d x)^2}+\frac {\left (2 b^2\right ) \int \left (\frac {1}{2 (c+d x)}-\frac {\cos (2 a+2 b x)}{2 (c+d x)}\right ) \, dx}{d^2}-\frac {\left (6 b^2\right ) \int \left (\frac {1}{2 (c+d x)}+\frac {\cos (2 a+2 b x)}{2 (c+d x)}\right ) \, dx}{d^2} \\ & = -\frac {3 \cos ^2(a+b x)}{2 d (c+d x)^2}+\frac {4 b \cos (a+b x) \sin (a+b x)}{d^2 (c+d x)}+\frac {\sin ^2(a+b x)}{2 d (c+d x)^2}-\frac {b^2 \int \frac {\cos (2 a+2 b x)}{c+d x} \, dx}{d^2}-\frac {\left (3 b^2\right ) \int \frac {\cos (2 a+2 b x)}{c+d x} \, dx}{d^2} \\ & = -\frac {3 \cos ^2(a+b x)}{2 d (c+d x)^2}+\frac {4 b \cos (a+b x) \sin (a+b x)}{d^2 (c+d x)}+\frac {\sin ^2(a+b x)}{2 d (c+d x)^2}-\frac {\left (b^2 \cos \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx}{d^2}-\frac {\left (3 b^2 \cos \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx}{d^2}+\frac {\left (b^2 \sin \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx}{d^2}+\frac {\left (3 b^2 \sin \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx}{d^2} \\ & = -\frac {3 \cos ^2(a+b x)}{2 d (c+d x)^2}-\frac {4 b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}+\frac {4 b \cos (a+b x) \sin (a+b x)}{d^2 (c+d x)}+\frac {\sin ^2(a+b x)}{2 d (c+d x)^2}+\frac {4 b^2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.74 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.76 \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=-\frac {8 b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b (c+d x)}{d}\right )+\frac {d (d+2 d \cos (2 (a+b x))-4 b (c+d x) \sin (2 (a+b x)))}{(c+d x)^2}-8 b^2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )}{2 d^3} \]

[In]

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

[Out]

-1/2*(8*b^2*Cos[2*a - (2*b*c)/d]*CosIntegral[(2*b*(c + d*x))/d] + (d*(d + 2*d*Cos[2*(a + b*x)] - 4*b*(c + d*x)
*Sin[2*(a + b*x)]))/(c + d*x)^2 - 8*b^2*Sin[2*a - (2*b*c)/d]*SinIntegral[(2*b*(c + d*x))/d])/d^3

Maple [A] (verified)

Time = 1.80 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.52

method	result	size
default	\(\frac {1}{2 d \left (d x +c \right )^{2}}+\frac {b^{3} \left (-\frac {\cos \left (2 x b +2 a \right )}{\left (-a d +c b +d \left (x b +a \right )\right )^{2} d}-\frac {-\frac {2 \sin \left (2 x b +2 a \right )}{\left (-a d +c b +d \left (x b +a \right )\right ) d}+\frac {\frac {4 \,\operatorname {Si}\left (2 x b +2 a +\frac {-2 a d +2 c b}{d}\right ) \sin \left (\frac {-2 a d +2 c b}{d}\right )}{d}+\frac {4 \,\operatorname {Ci}\left (2 x b +2 a +\frac {-2 a d +2 c b}{d}\right ) \cos \left (\frac {-2 a d +2 c b}{d}\right )}{d}}{d}}{d}\right )-\frac {b^{3}}{\left (-a d +c b +d \left (x b +a \right )\right )^{2} d}}{b}\)	\(207\)
risch	\(-\frac {1}{2 d \left (d x +c \right )^{2}}+\frac {2 b^{2} {\mathrm e}^{-\frac {2 i \left (a d -c b \right )}{d}} \operatorname {Ei}_{1}\left (2 i b x +2 i a -\frac {2 i \left (a d -c b \right )}{d}\right )}{d^{3}}+\frac {2 b^{2} {\mathrm e}^{\frac {2 i \left (a d -c b \right )}{d}} \operatorname {Ei}_{1}\left (-2 i b x -2 i a -\frac {2 \left (-i a d +i c b \right )}{d}\right )}{d^{3}}+\frac {\left (-2 b^{2} d^{3} x^{2}-4 b^{2} c \,d^{2} x -2 b^{2} c^{2} d \right ) \cos \left (2 x b +2 a \right )}{2 d^{2} \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right ) \left (d x +c \right )^{2}}-\frac {i \left (4 i b^{3} d^{3} x^{3}+12 i b^{3} c \,d^{2} x^{2}+12 i b^{3} c^{2} d x +4 i c^{3} b^{3}\right ) \sin \left (2 x b +2 a \right )}{2 d^{2} \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right ) \left (d x +c \right )^{2}}\)	\(290\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.34 \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=-\frac {4 \, d^{2} \cos \left (b x + a\right )^{2} + 8 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) - 8 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 8 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) - d^{2}}{2 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.97 \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=-\frac {2 \, {\left (E_{3}\left (\frac {2 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right ) + E_{3}\left (-\frac {2 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 2 \, {\left (i \, E_{3}\left (\frac {2 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right ) - i \, E_{3}\left (-\frac {2 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 1}{2 \, {\left (d^{3} x^{2} + 2 \, c d^{2} x + c^{2} d\right )}} \]

[In]

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

[Out]

-1/2*(2*(exp_integral_e(3, 2*(-I*b*d*x - I*b*c)/d) + exp_integral_e(3, -2*(-I*b*d*x - I*b*c)/d))*cos(-2*(b*c -
a*d)/d) + 2*(I*exp_integral_e(3, 2*(-I*b*d*x - I*b*c)/d) - I*exp_integral_e(3, -2*(-I*b*d*x - I*b*c)/d))*sin(
-2*(b*c - a*d)/d) + 1)/(d^3*x^2 + 2*c*d^2*x + c^2*d)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 704 vs. \(2 (132) = 264\).

Time = 0.34 (sec) , antiderivative size = 704, normalized size of antiderivative = 5.18 \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=-\frac {8 \, b^{5} c^{2} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Ci}\left (\frac {2 \, {\left (b c + {\left (b x + a\right )} d - a d\right )}}{d}\right ) + 16 \, {\left (b x + a\right )} b^{4} c d \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Ci}\left (\frac {2 \, {\left (b c + {\left (b x + a\right )} d - a d\right )}}{d}\right ) - 16 \, a b^{4} c d \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Ci}\left (\frac {2 \, {\left (b c + {\left (b x + a\right )} d - a d\right )}}{d}\right ) + 8 \, {\left (b x + a\right )}^{2} b^{3} d^{2} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Ci}\left (\frac {2 \, {\left (b c + {\left (b x + a\right )} d - a d\right )}}{d}\right ) - 16 \, {\left (b x + a\right )} a b^{3} d^{2} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Ci}\left (\frac {2 \, {\left (b c + {\left (b x + a\right )} d - a d\right )}}{d}\right ) + 8 \, a^{2} b^{3} d^{2} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Ci}\left (\frac {2 \, {\left (b c + {\left (b x + a\right )} d - a d\right )}}{d}\right ) + 8 \, b^{5} c^{2} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (-\frac {2 \, {\left (b c + {\left (b x + a\right )} d - a d\right )}}{d}\right ) + 16 \, {\left (b x + a\right )} b^{4} c d \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (-\frac {2 \, {\left (b c + {\left (b x + a\right )} d - a d\right )}}{d}\right ) - 16 \, a b^{4} c d \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (-\frac {2 \, {\left (b c + {\left (b x + a\right )} d - a d\right )}}{d}\right ) + 8 \, {\left (b x + a\right )}^{2} b^{3} d^{2} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (-\frac {2 \, {\left (b c + {\left (b x + a\right )} d - a d\right )}}{d}\right ) - 16 \, {\left (b x + a\right )} a b^{3} d^{2} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (-\frac {2 \, {\left (b c + {\left (b x + a\right )} d - a d\right )}}{d}\right ) + 8 \, a^{2} b^{3} d^{2} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (-\frac {2 \, {\left (b c + {\left (b x + a\right )} d - a d\right )}}{d}\right ) - 4 \, b^{4} c d \sin \left (2 \, b x + 2 \, a\right ) - 4 \, {\left (b x + a\right )} b^{3} d^{2} \sin \left (2 \, b x + 2 \, a\right ) + 4 \, a b^{3} d^{2} \sin \left (2 \, b x + 2 \, a\right ) + 2 \, b^{3} d^{2} \cos \left (2 \, b x + 2 \, a\right ) + b^{3} d^{2}}{2 \, {\left (b^{2} c^{2} d^{3} + 2 \, {\left (b x + a\right )} b c d^{4} - 2 \, a b c d^{4} + {\left (b x + a\right )}^{2} d^{5} - 2 \, {\left (b x + a\right )} a d^{5} + a^{2} d^{5}\right )} b} \]

[In]

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

[Out]

-1/2*(8*b^5*c^2*cos(-2*(b*c - a*d)/d)*cos_integral(2*(b*c + (b*x + a)*d - a*d)/d) + 16*(b*x + a)*b^4*c*d*cos(-
2*(b*c - a*d)/d)*cos_integral(2*(b*c + (b*x + a)*d - a*d)/d) - 16*a*b^4*c*d*cos(-2*(b*c - a*d)/d)*cos_integral
(2*(b*c + (b*x + a)*d - a*d)/d) + 8*(b*x + a)^2*b^3*d^2*cos(-2*(b*c - a*d)/d)*cos_integral(2*(b*c + (b*x + a)*
d - a*d)/d) - 16*(b*x + a)*a*b^3*d^2*cos(-2*(b*c - a*d)/d)*cos_integral(2*(b*c + (b*x + a)*d - a*d)/d) + 8*a^2
*b^3*d^2*cos(-2*(b*c - a*d)/d)*cos_integral(2*(b*c + (b*x + a)*d - a*d)/d) + 8*b^5*c^2*sin(-2*(b*c - a*d)/d)*s
in_integral(-2*(b*c + (b*x + a)*d - a*d)/d) + 16*(b*x + a)*b^4*c*d*sin(-2*(b*c - a*d)/d)*sin_integral(-2*(b*c
+ (b*x + a)*d - a*d)/d) - 16*a*b^4*c*d*sin(-2*(b*c - a*d)/d)*sin_integral(-2*(b*c + (b*x + a)*d - a*d)/d) + 8*
(b*x + a)^2*b^3*d^2*sin(-2*(b*c - a*d)/d)*sin_integral(-2*(b*c + (b*x + a)*d - a*d)/d) - 16*(b*x + a)*a*b^3*d^
2*sin(-2*(b*c - a*d)/d)*sin_integral(-2*(b*c + (b*x + a)*d - a*d)/d) + 8*a^2*b^3*d^2*sin(-2*(b*c - a*d)/d)*sin
_integral(-2*(b*c + (b*x + a)*d - a*d)/d) - 4*b^4*c*d*sin(2*b*x + 2*a) - 4*(b*x + a)*b^3*d^2*sin(2*b*x + 2*a)
+ 4*a*b^3*d^2*sin(2*b*x + 2*a) + 2*b^3*d^2*cos(2*b*x + 2*a) + b^3*d^2)/((b^2*c^2*d^3 + 2*(b*x + a)*b*c*d^4 - 2
*a*b*c*d^4 + (b*x + a)^2*d^5 - 2*(b*x + a)*a*d^5 + a^2*d^5)*b)

Mupad [F(-1)]

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

[In]

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

[Out]

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