\(\int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^4} \, dx\) [375]

Optimal result

Integrand size = 23, antiderivative size = 205 \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^4} \, dx=-\frac {2 b^2}{3 d^3 (c+d x)}-\frac {\cos ^2(a+b x)}{d (c+d x)^3}+\frac {2 b^2 \cos ^2(a+b x)}{d^3 (c+d x)}+\frac {8 b^3 \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{3 d^4}+\frac {4 b \cos (a+b x) \sin (a+b x)}{3 d^2 (c+d x)^2}+\frac {\sin ^2(a+b x)}{3 d (c+d x)^3}-\frac {2 b^2 \sin ^2(a+b x)}{3 d^3 (c+d x)}+\frac {8 b^3 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{3 d^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {4516, 3395, 32, 3394, 12, 3384, 3380, 3383} \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^4} \, dx=\frac {8 b^3 \sin \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{3 d^4}+\frac {8 b^3 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{3 d^4}-\frac {2 b^2 \sin ^2(a+b x)}{3 d^3 (c+d x)}+\frac {2 b^2 \cos ^2(a+b x)}{d^3 (c+d x)}+\frac {4 b \sin (a+b x) \cos (a+b x)}{3 d^2 (c+d x)^2}+\frac {\sin ^2(a+b x)}{3 d (c+d x)^3}-\frac {\cos ^2(a+b x)}{d (c+d x)^3}-\frac {2 b^2}{3 d^3 (c+d x)} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 3394

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Simp[(c + d*x)^(m + 1)*(Sin[e + f*x]^
n/(d*(m + 1))), x] - Dist[f*(n/(d*(m + 1))), Int[ExpandTrigReduce[(c + d*x)^(m + 1), Cos[e + f*x]*Sin[e + f*x]
^(n - 1), x], x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && GeQ[m, -2] && 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)^4}-\frac {\sin ^2(a+b x)}{(c+d x)^4}\right ) \, dx \\ & = 3 \int \frac {\cos ^2(a+b x)}{(c+d x)^4} \, dx-\int \frac {\sin ^2(a+b x)}{(c+d x)^4} \, dx \\ & = -\frac {\cos ^2(a+b x)}{d (c+d x)^3}+\frac {4 b \cos (a+b x) \sin (a+b x)}{3 d^2 (c+d x)^2}+\frac {\sin ^2(a+b x)}{3 d (c+d x)^3}-\frac {b^2 \int \frac {1}{(c+d x)^2} \, dx}{3 d^2}+\frac {\left (2 b^2\right ) \int \frac {\sin ^2(a+b x)}{(c+d x)^2} \, dx}{3 d^2}+\frac {b^2 \int \frac {1}{(c+d x)^2} \, dx}{d^2}-\frac {\left (2 b^2\right ) \int \frac {\cos ^2(a+b x)}{(c+d x)^2} \, dx}{d^2} \\ & = -\frac {2 b^2}{3 d^3 (c+d x)}-\frac {\cos ^2(a+b x)}{d (c+d x)^3}+\frac {2 b^2 \cos ^2(a+b x)}{d^3 (c+d x)}+\frac {4 b \cos (a+b x) \sin (a+b x)}{3 d^2 (c+d x)^2}+\frac {\sin ^2(a+b x)}{3 d (c+d x)^3}-\frac {2 b^2 \sin ^2(a+b x)}{3 d^3 (c+d x)}+\frac {\left (4 b^3\right ) \int \frac {\sin (2 a+2 b x)}{2 (c+d x)} \, dx}{3 d^3}-\frac {\left (4 b^3\right ) \int -\frac {\sin (2 a+2 b x)}{2 (c+d x)} \, dx}{d^3} \\ & = -\frac {2 b^2}{3 d^3 (c+d x)}-\frac {\cos ^2(a+b x)}{d (c+d x)^3}+\frac {2 b^2 \cos ^2(a+b x)}{d^3 (c+d x)}+\frac {4 b \cos (a+b x) \sin (a+b x)}{3 d^2 (c+d x)^2}+\frac {\sin ^2(a+b x)}{3 d (c+d x)^3}-\frac {2 b^2 \sin ^2(a+b x)}{3 d^3 (c+d x)}+\frac {\left (2 b^3\right ) \int \frac {\sin (2 a+2 b x)}{c+d x} \, dx}{3 d^3}+\frac {\left (2 b^3\right ) \int \frac {\sin (2 a+2 b x)}{c+d x} \, dx}{d^3} \\ & = -\frac {2 b^2}{3 d^3 (c+d x)}-\frac {\cos ^2(a+b x)}{d (c+d x)^3}+\frac {2 b^2 \cos ^2(a+b x)}{d^3 (c+d x)}+\frac {4 b \cos (a+b x) \sin (a+b x)}{3 d^2 (c+d x)^2}+\frac {\sin ^2(a+b x)}{3 d (c+d x)^3}-\frac {2 b^2 \sin ^2(a+b x)}{3 d^3 (c+d x)}+\frac {\left (2 b^3 \cos \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}{3 d^3}+\frac {\left (2 b^3 \cos \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^3}+\frac {\left (2 b^3 \sin \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}{3 d^3}+\frac {\left (2 b^3 \sin \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^3} \\ & = -\frac {2 b^2}{3 d^3 (c+d x)}-\frac {\cos ^2(a+b x)}{d (c+d x)^3}+\frac {2 b^2 \cos ^2(a+b x)}{d^3 (c+d x)}+\frac {8 b^3 \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{3 d^4}+\frac {4 b \cos (a+b x) \sin (a+b x)}{3 d^2 (c+d x)^2}+\frac {\sin ^2(a+b x)}{3 d (c+d x)^3}-\frac {2 b^2 \sin ^2(a+b x)}{3 d^3 (c+d x)}+\frac {8 b^3 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{3 d^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.61 \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^4} \, dx=\frac {8 b^3 \operatorname {CosIntegral}\left (\frac {2 b (c+d x)}{d}\right ) \sin \left (2 a-\frac {2 b c}{d}\right )+\frac {d \left (\left (-2 d^2+4 b^2 (c+d x)^2\right ) \cos (2 (a+b x))+d (-d+2 b (c+d x) \sin (2 (a+b x)))\right )}{(c+d x)^3}+8 b^3 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )}{3 d^4} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.63 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.19

method	result	size
default	\(\frac {1}{3 d \left (d x +c \right )^{3}}+\frac {b^{4} \left (-\frac {2 \cos \left (2 x b +2 a \right )}{3 \left (-a d +c b +d \left (x b +a \right )\right )^{3} d}-\frac {2 \left (-\frac {\sin \left (2 x b +2 a \right )}{\left (-a d +c b +d \left (x b +a \right )\right )^{2} d}+\frac {-\frac {2 \cos \left (2 x b +2 a \right )}{\left (-a d +c b +d \left (x b +a \right )\right ) d}-\frac {2 \left (\frac {2 \,\operatorname {Si}\left (2 x b +2 a +\frac {-2 a d +2 c b}{d}\right ) \cos \left (\frac {-2 a d +2 c b}{d}\right )}{d}-\frac {2 \,\operatorname {Ci}\left (2 x b +2 a +\frac {-2 a d +2 c b}{d}\right ) \sin \left (\frac {-2 a d +2 c b}{d}\right )}{d}\right )}{d}}{d}\right )}{3 d}\right )-\frac {2 b^{4}}{3 \left (-a d +c b +d \left (x b +a \right )\right )^{3} d}}{b}\)	\(243\)
risch	\(-\frac {1}{3 d \left (d x +c \right )^{3}}-\frac {4 i b^{3} {\mathrm e}^{-\frac {2 i \left (a d -c b \right )}{d}} \operatorname {Ei}_{1}\left (2 i b x +2 i a -\frac {2 i \left (a d -c b \right )}{d}\right )}{3 d^{4}}+\frac {4 i b^{3} {\mathrm e}^{\frac {2 i \left (a d -c b \right )}{d}} \operatorname {Ei}_{1}\left (-2 i b x -2 i a -\frac {2 \left (-i a d +i c b \right )}{d}\right )}{3 d^{4}}-\frac {\left (-4 b^{5} d^{5} x^{5}-20 b^{5} c \,d^{4} x^{4}-40 b^{5} c^{2} d^{3} x^{3}-40 b^{5} c^{3} d^{2} x^{2}-20 b^{5} c^{4} d x +2 b^{3} d^{5} x^{3}-4 b^{5} c^{5}+6 b^{3} c \,d^{4} x^{2}+6 b^{3} c^{2} d^{3} x +2 b^{3} c^{3} d^{2}\right ) \cos \left (2 x b +2 a \right )}{3 d^{3} \left (d^{3} x^{3} b^{3}+3 b^{3} c \,d^{2} x^{2}+3 b^{3} c^{2} d x +b^{3} c^{3}\right ) \left (d x +c \right )^{3}}-\frac {i \left (2 i b^{4} d^{5} x^{4}+8 i b^{4} c \,d^{4} x^{3}+12 i b^{4} c^{2} d^{3} x^{2}+8 i b^{4} c^{3} d^{2} x +2 i b^{4} c^{4} d \right ) \sin \left (2 x b +2 a \right )}{3 d^{3} \left (d^{3} x^{3} b^{3}+3 b^{3} c \,d^{2} x^{2}+3 b^{3} c^{2} d x +b^{3} c^{3}\right ) \left (d x +c \right )^{3}}\)	\(423\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.40 \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^4} \, dx=-\frac {4 \, b^{2} d^{3} x^{2} + 8 \, b^{2} c d^{2} x + 4 \, b^{2} c^{2} d - d^{3} - 4 \, {\left (2 \, b^{2} d^{3} x^{2} + 4 \, b^{2} c d^{2} x + 2 \, b^{2} c^{2} d - d^{3}\right )} \cos \left (b x + a\right )^{2} - 4 \, {\left (b d^{3} x + b c d^{2}\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 8 \, {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 8 \, {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right )}{3 \, {\left (d^{7} x^{3} + 3 \, c d^{6} x^{2} + 3 \, c^{2} d^{5} x + c^{3} d^{4}\right )}} \]

[In]

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

[Out]

-1/3*(4*b^2*d^3*x^2 + 8*b^2*c*d^2*x + 4*b^2*c^2*d - d^3 - 4*(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*(b*d^3*x + b*c*d^2)*cos(b*x + a)*sin(b*x + a) - 8*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*
c^2*d*x + b^3*c^3)*cos_integral(2*(b*d*x + b*c)/d)*sin(-2*(b*c - a*d)/d) - 8*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 +
3*b^3*c^2*d*x + b^3*c^3)*cos(-2*(b*c - a*d)/d)*sin_integral(2*(b*d*x + b*c)/d))/(d^7*x^3 + 3*c*d^6*x^2 + 3*c^2
*d^5*x + c^3*d^4)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.31 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.70 \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^4} \, dx=-\frac {3 \, {\left (E_{4}\left (\frac {2 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right ) + E_{4}\left (-\frac {2 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 3 \, {\left (i \, E_{4}\left (\frac {2 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right ) - i \, E_{4}\left (-\frac {2 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 1}{3 \, {\left (d^{4} x^{3} + 3 \, c d^{3} x^{2} + 3 \, c^{2} d^{2} x + c^{3} d\right )}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1305 vs. \(2 (193) = 386\).

Time = 0.38 (sec) , antiderivative size = 1305, normalized size of antiderivative = 6.37 \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^4} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/3*(8*b^7*c^3*cos_integral(2*(b*c + (b*x + a)*d - a*d)/d)*sin(-2*(b*c - a*d)/d) + 24*(b*x + a)*b^6*c^2*d*cos_
integral(2*(b*c + (b*x + a)*d - a*d)/d)*sin(-2*(b*c - a*d)/d) - 24*a*b^6*c^2*d*cos_integral(2*(b*c + (b*x + a)
*d - a*d)/d)*sin(-2*(b*c - a*d)/d) + 24*(b*x + a)^2*b^5*c*d^2*cos_integral(2*(b*c + (b*x + a)*d - a*d)/d)*sin(
-2*(b*c - a*d)/d) - 48*(b*x + a)*a*b^5*c*d^2*cos_integral(2*(b*c + (b*x + a)*d - a*d)/d)*sin(-2*(b*c - a*d)/d)
+ 24*a^2*b^5*c*d^2*cos_integral(2*(b*c + (b*x + a)*d - a*d)/d)*sin(-2*(b*c - a*d)/d) + 8*(b*x + a)^3*b^4*d^3*
cos_integral(2*(b*c + (b*x + a)*d - a*d)/d)*sin(-2*(b*c - a*d)/d) - 24*(b*x + a)^2*a*b^4*d^3*cos_integral(2*(b
*c + (b*x + a)*d - a*d)/d)*sin(-2*(b*c - a*d)/d) + 24*(b*x + a)*a^2*b^4*d^3*cos_integral(2*(b*c + (b*x + a)*d
- a*d)/d)*sin(-2*(b*c - a*d)/d) - 8*a^3*b^4*d^3*cos_integral(2*(b*c + (b*x + a)*d - a*d)/d)*sin(-2*(b*c - a*d)
/d) - 8*b^7*c^3*cos(-2*(b*c - a*d)/d)*sin_integral(-2*(b*c + (b*x + a)*d - a*d)/d) - 24*(b*x + a)*b^6*c^2*d*co
s(-2*(b*c - a*d)/d)*sin_integral(-2*(b*c + (b*x + a)*d - a*d)/d) + 24*a*b^6*c^2*d*cos(-2*(b*c - a*d)/d)*sin_in
tegral(-2*(b*c + (b*x + a)*d - a*d)/d) - 24*(b*x + a)^2*b^5*c*d^2*cos(-2*(b*c - a*d)/d)*sin_integral(-2*(b*c +
(b*x + a)*d - a*d)/d) + 48*(b*x + a)*a*b^5*c*d^2*cos(-2*(b*c - a*d)/d)*sin_integral(-2*(b*c + (b*x + a)*d - a
*d)/d) - 24*a^2*b^5*c*d^2*cos(-2*(b*c - a*d)/d)*sin_integral(-2*(b*c + (b*x + a)*d - a*d)/d) - 8*(b*x + a)^3*b
^4*d^3*cos(-2*(b*c - a*d)/d)*sin_integral(-2*(b*c + (b*x + a)*d - a*d)/d) + 24*(b*x + a)^2*a*b^4*d^3*cos(-2*(b
*c - a*d)/d)*sin_integral(-2*(b*c + (b*x + a)*d - a*d)/d) - 24*(b*x + a)*a^2*b^4*d^3*cos(-2*(b*c - a*d)/d)*sin
_integral(-2*(b*c + (b*x + a)*d - a*d)/d) + 8*a^3*b^4*d^3*cos(-2*(b*c - a*d)/d)*sin_integral(-2*(b*c + (b*x +
a)*d - a*d)/d) + 4*b^6*c^2*d*cos(2*b*x + 2*a) + 8*(b*x + a)*b^5*c*d^2*cos(2*b*x + 2*a) - 8*a*b^5*c*d^2*cos(2*b
*x + 2*a) + 4*(b*x + a)^2*b^4*d^3*cos(2*b*x + 2*a) - 8*(b*x + a)*a*b^4*d^3*cos(2*b*x + 2*a) + 4*a^2*b^4*d^3*co
s(2*b*x + 2*a) + 2*b^5*c*d^2*sin(2*b*x + 2*a) + 2*(b*x + a)*b^4*d^3*sin(2*b*x + 2*a) - 2*a*b^4*d^3*sin(2*b*x +
2*a) - 2*b^4*d^3*cos(2*b*x + 2*a) - b^4*d^3)/((b^3*c^3*d^4 + 3*(b*x + a)*b^2*c^2*d^5 - 3*a*b^2*c^2*d^5 + 3*(b
*x + a)^2*b*c*d^6 - 6*(b*x + a)*a*b*c*d^6 + 3*a^2*b*c*d^6 + (b*x + a)^3*d^7 - 3*(b*x + a)^2*a*d^7 + 3*(b*x + a
)*a^2*d^7 - a^3*d^7)*b)

Mupad [F(-1)]

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

[In]

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

[Out]

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